View Full Version : The Hodge dual: some definitions & examples
Greg Egan
Nov3-04, 10:03 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In another thread, I sketched a definition of the Hodge dual of p-forms,\nbut I\'m feeling a bit guilty that I told Oz to go and find a textbook\nwhen he needed a bit more detail. So as penance, here are some proper\ndefinitions (taken from Baez & Munian\'s _Gauge Fields, Knots and\nGravity_), and some examples.\n\nThe metric on 1-forms\n=====================\n\nFirst, we have to be sure that we have a metric defined, and this will\nlet us take the inner product of two 1-forms. If t and u are 1-forms, we\nwrite:\n\n<t,u> = g(t,u) = g^{ij} t_i u_j\n\nFor anyone more used to thinking of the metric as giving the dot product\nof two vectors, v and w:\n\nv.w = g(v,w) = g_{ij} v^i w^j\n\nthe two metrics are really just different versions of the same thing,\nwith the matrix of coordinates g^{ij} and the matrix of coordinates\ng_{ij} just being inverses of each other:\n\ng_{ij} g^{jk} = delta_i^k\n\nNB The <,> notation is used in a wide variety of slightly different\ncontexts, and its use here shouldn\'t be confused with "contracting" a\n1-form with a vector, a process that doesn\'t require a metric. It is,\nhowever, related to that other meaning in a very natural way; if v is a\nvector corresponding to a 1-form u, i.e. produced from u by "raising an\nindex" with the metric:\n\nv^i = g^{ij} u_j\n\nthen <t,v> = <t,u>, where the LHS use of <,> is the contraction of a\nvector with a 1-form, and the RHS use of <,> is the inner product of two\n1-forms:\n\n<t,v> = t_i v^i\n= t_i g^{ij} u_j\n= <t,u>\n\nAlso, we note that the inner product between two 1-forms is equal to the\ndot product of the vectors produced by raising indices on those 1-forms.\nIf {e_i} is a basis of vectors, and the 1-forms t and u have coordinates\nt_i and u_i relative to a dual basis of 1-forms, {e^i}, then the vectors\nproduced by raising an index on each 1-form are:\n\nv = g^{ij} t_i e_j\n\nw = g^{kl} u_k e_l\n\nwith dot product\n\ng(v,w)\n\n= g(g^{ij} t_i e_j, g^{kl} u_k e_l)\n\n= g^{ij} g_{jl} g^{kl} t_i u_k\n\n(where we\'ve used the linearity of the metric, and that\ng(e_j,e_l) = g_{jl})\n\n= delta^i_l g^{kl} t_i u_k\n\n= g^{ki} t_i u_k\n\n= <t,u>\n\nSo <t,u> = g(t,u) is exactly the dot product between the vectors v and w\northogonal to the hypersurfaces of the 1-forms t and u.\n\nThe inner product of p-forms\n============================\n\nWe can extend this definition of an inner product to include two forms of\nany degree. Consider "simple" p-forms produced just by taking the wedge\nproducts of 1-forms, say t^1 ^ t^2 ^ ... t^p and u^1 ^ u^2 ^ ... u^p,\nwhere the t^i and u^i are complete individual 1-forms, not coordinates.\nThen we define:\n\n<t^1 ^ t^2 ^ ... t^p, u^1 ^ u^2 ^ ... u^p>\n= det [<t^i,u^j>]\n\nwhere the RHS is the determinant of the p x p matrix of 1-form inner\nproducts of the individual 1-forms. This definition then extends by\nlinearity to other p-forms, which are linear combinations of the "simple"\nones produced just by wedging together 1-forms.\n\nTo give an example, suppose we have coordinate 1-forms dx, dy, dz in\nEuclidean space. Then these three 1-forms form an orthonormal basis,\nwith:\n\n<dx,dx> = <dy,dy> = <dz,dz> = 1\n<dx,dy> = <dy,dz> = <dx,dz> = 0\n\nWe then have:\n\n<dx ^ dy, dy ^ dz> = det [<dx,dy> <dx,dz>]\n[<dy,dy> <dy,dz>]\n= det [0 0]\n[1 0]\n= 0\n\n\n<dx ^ dy, dx ^ dy> = det [<dx,dx> <dx,dy>]\n[<dy,dx> <dy,dy>]\n= det [1 0]\n[0 1]\n= 1\n\nThe canonical volume form\n=========================\n\nIf you have a metric on an n-dimensional space, and coordinates x^1, ..,\nx^n, you can define an n-form:\n\nvol = sqrt(abs(det [g_{ij}])) dx^1 ^ dx^2 ^ ... dx^n\n\nwhich is called the canonical volume form associated with the metric. If\nyou change coordinates, or have different parts of your space covered by\ndifferent coordinate systems, it doesn\'t matter, because it turns out\nthat this definition is really independent of the choice of coordinates\nand only depends on the geometry of the metric itself, plus an\n"orientation" (loosely speaking, an orientation is a generalisation of\nusing either left-handed or right-handed coordinate systems).\n\nFor nice Cartesian coordinates, of course the metric is just a diagonal\nmatrix with 1\'s on the diagonal, so it has a determinant of 1, and we\nhave:\n\nvol = dx^1 ^ dx^2 ^ ... dx^n\n\nWhat about something more complicated, though? In spherical polar\ncoordinates in R^3, if we have x^1 = r, x^2 = theta (co-latitude) and x^3\n= phi (longitude), the coordinates of the metric are:\n\n[1 0 0 ]\ng_{ij}= [0 r^2 0 ]\n[0 0 r^2 cos(theta)^2]\n\ndet [g_{ij}] = r^4 cos(theta)^2\n\nvol = r^2 cos(theta) dr ^ dtheta ^ dphi\n\nYou might recognise r^2 cos(theta) as exactly the weight needed to\ncorrectly integrate a function in polar coordinates.\n\nWhy is "vol" called the "volume form"? If you counted the little cuboids\nin a region defined by contour surfaces of constant r, theta and phi at\nuniform intervals, and weighted the count just by the product of the\ncontour increments (e.g. you might increment all the coordinates by 0.01,\nand then weight the count of little cuboids by 0.01 x 0.01 x 0.01), you\nwouldn\'t get the actual volume of the region. But if you include the\nweight r^2 cos(theta) for each cuboid, this count *would* approximate the\nvolume of the region (arbitrarily well as you decreased the contour\nincrement).\n\nSo you can think of the volume form in n dimensions as being a mesh of\nhypercuboids, which you can count (with appropriate weighting) in any\nregion of the space to give you the volume of that region.\n\nThe Hodge *\n===========\n\nOK, now we have all the preliminaries we need to define the Hodge dual.\n\nIn n-dimensional space, if a is a p-form for any 0<=p<=n, then (*a) is an\n(n-p)-form, and it satisfies:\n\na ^ (*a) = <a, a> vol\n\nWhat does this mean? Geometrically, if a is thought of as a mesh of\nhypersurfaces, (*a) is the orthogonal mesh that when combined with that\nof a will give a hypercuboidal mesh that fills all the dimensions of the\nspace. What\'s more, the density of that mesh is very closely related to\nthe volume form\'s mesh. If <a,a>=1 everywhere, then a ^ (*a) is exactly\nequal to the volume form, but if <a,a> is greater then a ^ (*a) has a\ndenser mesh. The relationship is quadratic, because the Hodge * is a\nlinear operator from p-forms to (n-p)-forms. So if c is some scalar,\nthen:\n\n(ca) ^ *(ca) = (c^2) a ^ (*a) = (c^2) <a,a> vol\n\nExamples\n========\n\n1. If x, y and z are Cartesian coordinates in R^3, then the volume form\nis:\n\nvol = dx ^ dy ^ dz.\n\nand the inner products of 1-forms have already been given as:\n\n<dx,dx> = <dy,dy> = <dz,dz> = 1\n<dx,dy> = <dy,dz> = <dx,dz> = 0\n\nSo it\'s pretty easy to see that:\n\n(a) *dx = dy ^ dz\n\nbecause\n\ndx ^ dy ^ dz = <dx,dx> vol\n\nand\n\n(b) *dy = - dx ^ dz\n\nbecause\n\ndy ^ (- dx ^ dz) = - dy ^ dx ^ dz\n= dx ^ dy ^ dz\n= <dy,dy> vol\n\nFor 2-forms, it\'s much the same:\n\n(c) *(dx ^ dy) = dz\n\nbecause\n\ndx ^ dy ^ dz = <dx^dy,dx^dy> vol\n\nAnd for 3-forms:\n\n(d) *(dx ^ dy ^ dz) = 1\n\nbecause we define "wedging" with a 0-form, or scalar, to be just scalar\nmultiplication.\n\n2. What about spherical coordinates? We can guess the general form of\nthe answers, then solve for the fudge factors.\n\n(a) *dr = f dtheta ^ dphi\n\nf dr ^ dtheta ^ dphi\n= <dr,dr> vol\n= r^2 cos(theta) dr ^ dtheta ^ dphi\n= r^2 cos(theta) dr ^ dtheta ^ dphi\n\nSo *dr = r^2 cos(theta) dtheta ^ dphi.\n\n(b) *dtheta = f dr ^ dphi\n\nf dtheta ^ dr ^ dphi\n= <dtheta,dtheta> vol\n\nNote that <dtheta,dtheta> can be found by inverting the matrix for g_{ij}\nthat\'s given earlier in this post, to yield:\n\n<dtheta,dtheta>=1/r^2.\n\nSo:\n\nf dtheta ^ dr ^ dphi\n= (1/r^2) r^2 cos(theta) dr ^ dtheta ^ dphi\n\nSo *dtheta = -cos(theta) dr ^ dphi\n\n(c) *dphi = f dr ^ dtheta\n\nFrom the inverse matrix for the metric:\n\n<dphi,dphi> = 1/(r^2 cos(theta)^2)\n\nso we need:\n\nf dphi ^ dr ^ dtheta\n= <dphi,dphi> vol\n= 1/(r^2 cos(theta)^2) r^2 cos(theta) dr ^ dtheta ^ dphi\n\nSo *dphi = (1/cos(theta)) dr ^ dtheta\n\nAnother definition\n==================\n\nSuppose we have an orthonormal basis of 1-forms, {e^i}. Note that this\ndoesn\'t always mean that <e^i,e^i> = 1, because if we\'re dealing with\nspacetime, <dt,dt>=-1. But <e^i,e^i> = 1 or -1.\n\nThen we can define the Hodge * as the linear operator from p-forms to\n(n-p)-forms that satisfies:\n\n*(e^{i_1} ^ e^{i_2} ^ ... e^{i_p}) = +/- e^{i_{p+1}} ^ .. ^ e^{i_n}\n\nwhere the set {i_1, ... i_n} is some permutation of {1, .., n}. The +/-\nsign is the product of the sign of that permutation, and the product of\nthe signs <e^i,e^i> for i=1 to p.\n\nFor example, in Cartesian coordinates in spacetime, {dt,dx,dy,dz} form an\northonormal basis. Then we have:\n\n*(dt ^ dx) = - (dy ^ dz)\n\nbecause the {i_1, ... i_n} are just {1,2,3,4}, the identity permutation,\nand the only sign comes from <dt,dt>=-1.\n\nAnd we have:\n\n*(dy ^ dz) = (dt ^ dx)\n\nbecause {i_1, ... i_n} is {3,4,1,2}, an even permutation.\n\nThis leads to the interesting result:\n\n**(dt ^ dx) = - (dt ^ dx)\n\nand in fact, in spacetime, **a=-a for all 2-forms a.\n\nNow, how can we apply this formula to polar coordinates, to check some of\nour examples? The coordinate 1-forms are mutually orthogonal, but\nthey\'re not normalised, e.g. <dtheta,dtheta>=1/r^2. But we can easily\nconstruct an orthonormal basis:\n\ne^1 = dr\ne^2 = r dtheta\ne^3 = r cos(theta) dphi\n\nIf we translate the result of example 2(c) into the terms of this new\nbasis, we get:\n\n*dphi = (1/cos(theta)) dr ^ dtheta\n\n1/(r cos(theta)) *e^3\n= 1/cos(theta) 1/r e^1 ^ e^2\n\nor, multiplying through by r cos(theta):\n\n*e^3 = e^1 ^ e^2\n\nwhich is correct, because {3,1,2} is an even permutation.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In another thread, I sketched a definition of the Hodge dual of p-forms,
but I'm feeling a bit guilty that I told Oz to go and find a textbook
when he needed a bit more detail. So as penance, here are some proper
definitions (taken from Baez & Munian's _Gauge Fields, Knots and
Gravity_), and some examples.
The metric on 1-forms
=====================
First, we have to be sure that we have a metric defined, and this will
let us take the inner product of two 1-forms. If t and u are 1-forms, we
write:
<t,u> = g(t,u) = g^{ij} t_i u_j
For anyone more used to thinking of the metric as giving the dot product
of two vectors, v and w:
v.w = g(v,w) = g_{ij} v^i w^j
the two metrics are really just different versions of the same thing,
with the matrix of coordinates g^{ij} and the matrix of coordinates
g_{ij} just being inverses of each other:
g_{ij} g^{jk} = \delta_i^k
NB The <,> notation is used in a wide variety of slightly different
contexts, and its use here shouldn't be confused with "contracting" a
1-form with a vector, a process that doesn't require a metric. It is,
however, related to that other meaning in a very natural way; if v is a
vector corresponding to a 1-form u, i.e. produced from u by "raising an
index" with the metric:
v^i = g^{ij} u_j
then <t,v> = <t,u>, where the LHS use of <,> is the contraction of a
vector with a 1-form, and the RHS use of <,> is the inner product of two
1-forms:
<t,v> = t_i v^i= t_i g^{ij} u_j
= <t,u>
Also, we note that the inner product between two 1-forms is equal to the
dot product of the vectors produced by raising indices on those 1-forms.
If {e_i} is a basis of vectors, and the 1-forms t and u have coordinates
t_i and u_i relative to a dual basis of 1-forms, {e^i}, then the vectors
produced by raising an index on each 1-form are:
v = g^{ij} t_i e_jw = g^{kl} u_k e_l
with dot product
g(v,w)
= g(g^{ij} t_i e_j, g^{kl} u_k e_l)= g^{ij} g_{jl} g^{kl} t_i u_k
(where we've used the linearity of the metric, and that
g(e_j,e_l) = g_{jl})= \delta^i_l g^{kl} t_i u_k= g^{ki} t_i u_k
= <t,u>
So <t,u> = g(t,u) is exactly the dot product between the vectors v and w
orthogonal to the hypersurfaces of the 1-forms t and u.
The inner product of p-forms
============================
We can extend this definition of an inner product to include two forms of
any degree. Consider "simple" p-forms produced just by taking the wedge
products of 1-forms, say t^1 ^ t^2 ^ ... t^p and u^1 ^ u^2 ^ ... u^p,
where the t^i and u^i are complete individual 1-forms, not coordinates.
Then we define:
<t^1 ^ t^2 ^ .[/itex].. t^p, u^1 ^ u^2 ^ ... u^p>
= det [<t^i,u^j>]
where the RHS is the determinant of the p x p matrix of 1-form inner
products of the individual 1-forms. This definition then extends by
linearity to other p-forms, which are linear combinations of the "simple"
ones produced just by wedging together 1-forms.
To give an example, suppose we have coordinate 1-forms dx, dy, dz in
Euclidean space. Then these three 1-forms form an orthonormal basis,
with:
<dx,dx> = <dy,dy> = <dz,dz> = 1
<dx,dy> = <dy,dz> = <dx,dz> =
We then have:
<dx ^ dy, dy ^ dz> = det [<dx,dy> <dx,dz>]
[<dy,dy> <dy,dz>]
= det [0 0]
[1 0]
=
<dx ^ dy, dx ^ dy> = det [<dx,dx> <dx,dy>]
[<dy,dx> <dy,dy>]
= det [1 0]
[0 1]
= 1
The canonical volume form
=========================
If you have a metric on an n-dimensional space, and coordinates x^1, ..,
x^n, you can define an n-form:
vol = \sqrt(abs(det [g_{ij}])) dx^1 ^ dx^2 ^ ... dx^n
which is called the canonical volume form associated with the metric. If
you change coordinates, or have different parts of your space covered by
different coordinate systems, it doesn't matter, because it turns out
that this definition is really independent of the choice of coordinates
and only depends on the geometry of the metric itself, plus an
"orientation" (loosely speaking, an orientation is a generalisation of
using either left-handed or right-handed coordinate systems).
For nice Cartesian coordinates, of course the metric is just a diagonal
matrix with 1's on the diagonal, so it has a determinant of 1, and we
have:
vol = dx^1 ^ dx^2 ^ ... dx^n
What about something more complicated, though? In spherical polar
coordinates in R^3, if we have x^1 = r, x^2 = \theta (co-latitude) and x^3= \phi (longitude), the coordinates of the metric are:
[1 ]
g_{ij}= [0 r^2 ]
[0 r^2 cos(\theta)^2]
det [g_{ij}] = r^4 cos(\theta)^2
vol = r^2 cos(\theta) dr ^ dtheta ^ dphi
You might recognise r^2 cos(\theta) as exactly the weight needed to
correctly integrate a function in polar coordinates.
Why is "vol" called the "volume form"? If you counted the little cuboids
in a region defined by contour surfaces of constant r, \theta and \phi at
uniform intervals, and weighted the count just by the product of the
contour increments (e.g. you might increment all the coordinates by .01,
and then weight the count of little cuboids by .01 x .01 x .01), you
wouldn't get the actual volume of the region. But if you include the
weight r^2 cos(\theta) for each cuboid, this count *would* approximate the
volume of the region (arbitrarily well as you decreased the contour
increment).
So you can think of the volume form in n dimensions as being a mesh of
hypercuboids, which you can count (with appropriate weighting) in any
region of the space to give you the volume of that region.
The Hodge *
===========
OK, now we have all the preliminaries we need to define the Hodge dual.
In n-dimensional space, if a is a p-form for any 0<=p<=n, then (*a) is an
(n-p)-form, and it satisfies:
a ^ (*a) = <a, a> vol
What does this mean? Geometrically, if a is thought of as a mesh of
hypersurfaces, (*a) is the orthogonal mesh that when combined with that
of a will give a hypercuboidal mesh that fills all the dimensions of the
space. What's more, the density of that mesh is very closely related to
the volume form's mesh. If <a,a>=1 everywhere, then a ^ (*a) is exactly
equal to the volume form, but if <a,a> is greater then a ^ (*a) has a
denser mesh. The relationship is quadratic, because the Hodge * is a
linear operator from p-forms to (n-p)-forms. So if c is some scalar,
then:
(ca) ^ *(ca) = (c^2) a ^ (*a) = (c^2) <a,a> vol
Examples
========
1. If x, y and z are Cartesian coordinates in R^3, then the volume form
is:
vol = dx ^ dy ^ dz.
and the inner products of 1-forms have already been given as:
<dx,dx> = <dy,dy> = <dz,dz> = 1
<dx,dy> = <dy,dz> = <dx,dz> =
So it's pretty easy to see that:
(a) *dx = dy ^ dz
because
dx ^ dy ^ dz = <dx,dx> vol
and
(b) *dy = - dx ^ dz
because
dy ^ (- dx ^ dz) = - dy ^ dx ^ dz= dx ^ dy ^ dz
= <dy,dy> vol
For 2-forms, it's much the same:
(c) *(dx ^ dy) = dz
because
dx ^ dy ^ dz = <dx^dy,dx^dy> vol
And for 3-forms:
(d) *(dx ^ dy ^ dz) = 1
because we define "wedging" with a 0-form, or scalar, to be just scalar
multiplication.
2. What about spherical coordinates? We can guess the general form of
the answers, then solve for the fudge factors.
(a) *dr = f dtheta ^ dphi
f dr ^ dtheta ^ dphi
= <dr,dr> vol
= r^2 cos(\theta) dr ^ dtheta ^ dphi
= r^2 cos(\theta) dr ^ dtheta ^ dphi
So *dr = r^2 cos(\theta) dtheta ^ dphi.
(b) *dtheta = f dr ^ dphi
f dtheta ^ dr ^ dphi
= <dtheta,dtheta> vol
Note that <dtheta,dtheta> can be found by inverting the matrix for g_{ij}
that's given earlier in this post, to yield:
<dtheta,dtheta>=1/r^2.
So:
f dtheta ^ dr ^ dphi
= (1/r^2) r^2 cos(\theta) dr ^ dtheta ^ dphi
So *dtheta = -cos(\theta) dr ^ dphi
(c) *dphi = f dr ^ dtheta
From the inverse matrix for the metric:
<dphi,dphi> = 1/(r^2 cos(\theta)^2)
so we need:
f dphi ^ dr ^ dtheta
= <dphi,dphi> vol
= 1/(r^2 cos(\theta)^2) r^2 cos(\theta) dr ^ dtheta ^ dphi
So *dphi = (1/cos(\theta)) dr ^ dtheta
Another definition
==================
Suppose we have an orthonormal basis of 1-forms, {e^i}. Note that this
doesn't always mean that <e^i,e^i> = 1, because if we're dealing with
spacetime, <dt,dt>=-1. But <e^i,e^i> = 1 or -1.
Then we can define the Hodge * as the linear operator from p-forms to
(n-p)-forms that satisfies:
*(e^{i_1} ^ e^{i_2} ^ ... e^{i_p}) = +/- e^{i_{p+1}} ^ .. ^ e^{i_n}
where the set {i_1, ... i_n} is some permutation of {1, .., n}. The +/-
sign is the product of the sign of that permutation, and the product of
the signs <e^i,e^i> for i=1 to p.
For example, in Cartesian coordinates in spacetime, {dt,dx,dy,dz} form an
orthonormal basis. Then we have:
*(dt ^ dx) = - (dy ^ dz)
because the {i_1, ... i_n} are just {1,2,3,4}, the identity permutation,
and the only sign comes from <dt,dt>=-1.
And we have:
*(dy ^ dz) = (dt ^ dx)
because {i_1, ... i_n} is {3,4,1,2}, an even permutation.
This leads to the interesting result:
**(dt ^ dx) = - (dt ^ dx)
and in fact, in spacetime, **a=-a for all 2-forms a.
Now, how can we apply this formula to polar coordinates, to check some of
our examples? The coordinate 1-forms are mutually orthogonal, but
they're not normalised, e.g. <dtheta,dtheta>=1/r^2. But we can easily
construct an orthonormal basis:
e^1 = dre^2 = r dtheta
e^3 = r cos(\theta) dphi
If we translate the result of example 2(c) into the terms of this new
basis, we get:
*dphi = (1/cos(\theta)) dr ^ dtheta
[itex]1/(r cos(\theta)) *e^3= 1/cos(\theta) 1/r e^1 ^ e^2
or, multiplying through by r cos(\theta):*e^3 = e^1 ^ e^2
which is correct, because {3,1,2} is an even permutation.
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Greg Egan <gregegan@netspace.zebra.net.au> writes\n\n[NB Very many thanks for this long and detailed post]\n\n>The metric on 1-forms\n>=====================\n>\n>First, we have to be sure that we have a metric defined, and this will\n>let us take the inner product of two 1-forms. If t and u are 1-forms, we\n>write:\n>\n> <t,u> = g(t,u) = g^{ij} t_i u_j\n\nIs it valid to say g^ij t_i -> t^j ?\n\nThen we have <t,u> = t^j u_j which is a standard contraction.\n\n>For anyone more used to thinking of the metric as giving the dot product\n>of two vectors, v and w:\n>\n> v.w = g(v,w) = g_{ij} v^i w^j\n\nWhich we could express similarly.\n\nSo the metric is really an expression of the invertibility between\n1-forms and vectors.\n\nNow, I find I have been seduced by JB\'s \'GM course for idiots\' where he\nillustrated index gymnastics in a \'don\'t think, just do it\' manner.\nConsequently the indices were never properly explained. One\n\'appreciated\' them by osmosis, which is not at all ideal.\n\nI think it is (long past) time that I straightened this out.\n\nThe best way is probably for me to say how I see it, and for someone to\nwade in and point out where I have it wrong.\n\nI see it as follows (and it will probably do me good to try and work out\nhow I really *do* see it, if I can):\n\nv^i is a vector, w_j is in some sense an inverse vector.\n\nTypically the \'i\'s and \'j\'s do not match up because they are general\nindexes. For example in 3-space we could lay down a co-ordinate system\n(x,y,z) and would have three basis vectors v^x, v^y, v^z and three\n\'basis 1-forms\' w_x, w_y, w_z.\n\nIn this situation, where we have defined objects, it seems clear to me\nthat v^x w_x will give me a scalar, whilst v^x w_y will not because they\npoint in \'different directions\'.\n\nConsequently when dealing with generalised objects, without an imposed\nco-ordinate system, its important to keep track of the relationships\nbetween the various indices.\n\nThe existence of a metric says v^i is invertible to v_i.\n\n>the two metrics are really just different versions of the same thing,\n>with the matrix of coordinates g^{ij} and the matrix of coordinates\n>g_{ij} just being inverses of each other:\n>\n> g_{ij} g^{jk} = delta_i^k\n\nNow, what does this mean precisely?\nI need to keep things co-ordinate free in my thinking though.\nAm I OK to consider this as three different \'local\' co-ordinate\n*systems* i, j & k in some sense? So its really saying that going\n\ni -> j -> k is the same as going from i -> k\n\n(v^i g_ij) g^jk = w_j g^jk = u^k\n\nand we would like u^k = v^i since all we should have done is seen the\nobject v^i from \'different perspectives\'.\n\n>NB The <,> notation is used in a wide variety of slightly different\n>contexts, and its use here shouldn\'t be confused with "contracting" a\n>1-form with a vector, a process that doesn\'t require a metric.\n\nIt would be nice if this was \'by definition\'.\n\n>It is,\n>however, related to that other meaning in a very natural way; if v is a\n>vector corresponding to a 1-form u, i.e. produced from u by "raising an\n>index" with the metric:\n>\n> v^i = g^{ij} u_j\n>\n>then <t,v> = <t,u>, where the LHS use of <,> is the contraction of a\n>vector with a 1-form, and the RHS use of <,> is the inner product of two\n>1-forms:\n>\n> <t,v> = t_i v^i\n> = t_i g^{ij} u_j\n> = <t,u>\n\nI hate this sort of notation with a passion.\n\n<Rant>==========\nA notation should always reflect the underlying structure and should not\nbe \'interpretable\'. Its one reason why I am drawn to the index notation,\ndespite its obvious inelegance. The structure is clear and, at least at\nthe elementary level, unambiguous. In fact this is one of the reasons\nwhy I am not a lover of cross and dot products. These seem to me (and\nhave always seemed to me) to be confusing and concealing an underlying\nsymmetry that would be better to be expressed.\n\nIts the same thing that made me feel maxwell\'s equations were missing a\nsymmetry at a profound level which only vanished when I was introduced\nto the (3+1)D EM method using the faraday tensor, which is something I\nwould like to follow up, if only anyone was remotely interested.\n</rant>=========\n\nIt occurs to me that the metric is implicitly a spatial expression.\nAt a point the metric is always flat (away from a singularity).\nSo what the metric is really giving us is the relationship of one point\nwith \'adjacent\' points. It is, in some very real sense, a volume term.\nHmm, that\'s not quite right, is it? It must be, for an N-D space,\nan (N-1)-D term.\n\nThat\'s enough for now on this long and detailed post.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Greg Egan <gregegan@netspace.zebra.net.au> writes
[NB Very many thanks for this long and detailed post]
>The metric on 1-forms
>=====================
>
>First, we have to be sure that we have a metric defined, and this will
>let us take the inner product of two 1-forms. If t and u are 1-forms, we
>write:
>
> <t,u> = g(t,u) = g^{ij} t_i u_j
Is it valid to say g^{ij} t_i -> t^j ?
Then we have <t,u> = t^j u_j which is a standard contraction.
>For anyone more used to thinking of the metric as giving the dot product
>of two vectors, v and w:
>
> v.w = g(v,w) = g_{ij} v^i w^j
Which we could express similarly.
So the metric is really an expression of the invertibility between
1-forms and vectors.
Now, I find I have been seduced by JB's 'GM course for idiots' where he
illustrated index gymnastics in a 'don't think, just do it' manner.
Consequently the indices were never properly explained. One
'appreciated' them by osmosis, which is not at all ideal.
I think it is (long past) time that I straightened this out.
The best way is probably for me to say how I see it, and for someone to
wade in and point out where I have it wrong.
I see it as follows (and it will probably do me good to try and work out
how I really *do* see it, if I can):
v^i[/itex] is a vector, w_j is in some sense an inverse vector.
Typically the 'i's and 'j's do not match up because they are general
indexes. For example in 3-space we could lay down a co-ordinate system
(x,y,z) and would have three basis vectors v^x, v^y, v^z and three
'basis 1-forms' w_x, w_y, w_z.
In this situation, where we have defined objects, it seems clear to me
that v^x w_x will give me a scalar, whilst v^x w_y will not because they
point in 'different directions'.
Consequently when dealing with generalised objects, without an imposed
co-ordinate system, its important to keep track of the relationships
between the various indices.
The existence of a metric says v^i is invertible to v_i.
>the two metrics are really just different versions of the same thing,
>with the matrix of coordinates g^{ij} and the matrix of coordinates
>g_{ij} just being inverses of each other:
>
> g_{ij} g^{jk} = \delta_i^k
Now, what does this mean precisely?
I need to keep things co-ordinate free in my thinking though.
Am I OK to consider this as three different 'local' co-ordinate
*systems* i, j & k in some sense? So its really saying that going
i -> j -> k is the same as going from i -> k(v^i g_{ij}) g^{jk} = w_j g^{jk} = u^k
and we would like u^k = v^i since all we should have done is seen the
object v^i from 'different perspectives'.
>NB The <,> notation is used in a wide variety of slightly different
>contexts, and its use here shouldn't be confused with "contracting" a
>1-form with a vector, a process that doesn't require a metric.
It would be nice if this was 'by definition'.
>It is,
>however, related to that other meaning in a very natural way; if v is a
>vector corresponding to a 1-form u, i.e. produced from u by "raising an
>index" with the metric:
>
> [itex]v^i = g^{ij} u_j
>
>then <t,v> = <t,u>, where the LHS use of <,> is the contraction of a
>vector with a 1-form, and the RHS use of <,> is the inner product of two
>1-forms:
>
> <t,v> = t_i v^i
> = t_i g^{ij} u_j
> = <t,u>
I hate this sort of notation with a passion.
<Rant>==========
A notation should always reflect the underlying structure and should not
be 'interpretable'. Its one reason why I am drawn to the index notation,
despite its obvious inelegance. The structure is clear and, at least at
the elementary level, unambiguous. In fact this is one of the reasons
why I am not a lover of cross and dot products. These seem to me (and
have always seemed to me) to be confusing and concealing an underlying
symmetry that would be better to be expressed.
Its the same thing that made me feel maxwell's equations were missing a
symmetry at a profound level which only vanished when I was introduced
to the (3+1)D EM method using the faraday tensor, which is something I
would like to follow up, if only anyone was remotely interested.
</rant>=========
It occurs to me that the metric is implicitly a spatial expression.
At a point the metric is always flat (away from a singularity).
So what the metric is really giving us is the relationship of one point
with 'adjacent' points. It is, in some very real sense, a volume term.
Hmm, that's not quite right, is it? It must be, for an N-D space,
an (N-1)-D term.
That's enough for now on this long and detailed post.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
grubb@math.niu.edu
Nov24-04, 01:46 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Oz <oz@farmeroz.port995.com> wrote in message\nnews:<ynHzRaA32EoBFw6Y@farmeroz.port995.c om>...\n\n> Greg Egan <gregegan@netspace.zebra.net.au> writes\n>\n> [NB Very many thanks for this long and detailed post]\n>\n> >The metric on 1-forms\n> >=====================\n> >\n> >First, we have to be sure that we have a metric defined, and this will\n> >let us take the inner product of two 1-forms. If t and u are 1-forms, we\n> >write:\n> >\n> > <t,u> = g(t,u) = g^{ij} t_i u_j\n>\n> Is it valid to say g^ij t_i -> t^j ?\n\nNot quite. Given a vector with components t_i, you get a co-vector\nwith components t^i where t^j=g^ij t_i.\n\n>\n> Then we have <t,u> = t^j u_j which is a standard contraction.\n>\n> >For anyone more used to thinking of the metric as giving the dot product\n> >of two vectors, v and w:\n> >\n> > v.w = g(v,w) = g_{ij} v^i w^j\n>\n> Which we could express similarly.\n>\n> So the metric is really an expression of the invertibility between\n> 1-forms and vectors.\n\nNot quite sure what you mean by \'invertibility\' here. The metric gives\na correspondence between vectors of one type and vectors of another\ntype. But different metrics give different correspondences.\n\n>\n> Now, I find I have been seduced by JB\'s \'GM course for idiots\' where he\n> illustrated index gymnastics in a \'don\'t think, just do it\' manner.\n> Consequently the indices were never properly explained. One\n> \'appreciated\' them by osmosis, which is not at all ideal.\n>\n> I think it is (long past) time that I straightened this out.\n>\n> The best way is probably for me to say how I see it, and for someone to\n> wade in and point out where I have it wrong.\n>\n> I see it as follows (and it will probably do me good to try and work out\n> how I really *do* see it, if I can):\n>\n> v^i is a vector, w_j is in some sense an inverse vector.\n\n*Very nervous here* There is a confusion between vectors and their\ncomponents (i.e. coordiantes) in a particular basis.\n\n>\n> Typically the \'i\'s and \'j\'s do not match up because they are general\n> indexes. For example in 3-space we could lay down a co-ordinate system\n> (x,y,z) and would have three basis vectors v^x, v^y, v^z and three\n> \'basis 1-forms\' w_x, w_y, w_z.\n\nNo. We would have basis vectors e_x, e_y and e_z. A generic vector\nwould then have an expension of the form v=v^x e_x +v^y e_y +v^z e_z.\nBy using the metric, we get a dual basis e^x, e^y, e^z. A generic covector\ncan be written as w=w_x e^x +w_y e^y +w_z e^z. Covectors are\n\'linear functionals\', so things like w(v) make sense and are scalars\nwhen w is a covector and v is a vector.\n\nAlso, the components, v^x, v^y, v^z are *scalars*. It is the combination\nv^x e_x +v^y e_y +v^z e_z which is actually the vector, although\npeople to blur the distinction between v and the triple (v^x, v^y, v^z).\nSimil;arly for w and (w_x, w_y, w_z).\n\n> In this situation, where we have defined objects, it seems clear to me\n> that v^x w_x will give me a scalar, whilst v^x w_y will not because they\n> point in \'different directions\'.\n\nNo. We would have e^x (e_x)=1, e^x (e_y)=0, e^y(e_y)=1, etc.\nAll expressions of the form w(v) will be scalars. For a generic\npair v, w as above, w(v)=w_x v^x +w_y v^y +w_z v^z, a scalar.\n\nNow, if you are given a vector v=v^x e_x +v^y e_y +v^z e_z, then\nthere is a covector, call it v* corresponding to v with\nv*=v_x e^x +v_y e^y +v_z e^z where the components satisfy\nv_x=g_xx v^x +g_xy v^y +g_xz v^z, and similar expressions\nfor v_y and v_z. The short way of writing this is v_i=g_ij v^j.\n\nBut, if you cange bases from e_x, e_y, e_z, the components of v and w\nchange, but the expression w(v) does not. It is coordinate independent.\n\n\n> Consequently when dealing with generalised objects, without an imposed\n> co-ordinate system, its important to keep track of the relationships\n> between the various indices.\n\nDefinitely not. Without coordinates, there are no indices! So, for\nexample, v^i will be the i^th coordinate of the vector v in whatever\nbasis you\'ve chosen. The index tell you which coordinate you are\nworking with.\n\n>\n> The existence of a metric says v^i is invertible to v_i.\n\nThere\'s that word \'invertible\' again. \'Corresponding\' is better.\n\n>\n> >the two metrics are really just different versions of the same thing,\n> >with the matrix of coordinates g^{ij} and the matrix of coordinates\n> >g_{ij} just being inverses of each other:\n> >\n> > g_{ij} g^{jk} = delta_i^k\n>\n> Now, what does this mean precisely?\n\nIt means that g_ij is the inverse matrix of g^ij. So if we use\ng_ij to make the correspondence from vectors to covectors, then\ng^ij gives the correspondence from covectors to vectors.\n\n> I need to keep things co-ordinate free in my thinking though.\n> Am I OK to consider this as three different \'local\' co-ordinate\n> *systems* i, j & k in some sense? So its really saying that going\n>\n> i -> j -> k is the same as going from i -> k\n>\n> (v^i g_ij) g^jk = w_j g^jk = u^k\n>\n> and we would like u^k = v^i since all we should have done is seen the\n> object v^i from \'different perspectives\'.\n>\n> >NB The <,> notation is used in a wide variety of slightly different\n> >contexts, and its use here shouldn\'t be confused with "contracting" a\n> >1-form with a vector, a process that doesn\'t require a metric.\n>\n> It would be nice if this was \'by definition\'.\n\nIn some contexts, it is.\n\n>\n> >It is,\n> >however, related to that other meaning in a very natural way; if v is a\n> >vector corresponding to a 1-form u, i.e. produced from u by "raising an\n> >index" with the metric:\n> >\n> > v^i = g^{ij} u_j\n> >\n> >then <t,v> = <t,u>, where the LHS use of <,> is the contraction of a\n> >vector with a 1-form, and the RHS use of <,> is the inner product of two\n> >1-forms:\n> >\n> > <t,v> = t_i v^i\n> > = t_i g^{ij} u_j\n> > = <t,u>\n>\n> I hate this sort of notation with a passion.\n>\n> <Rant>==========\n> A notation should always reflect the underlying structure and should not\n> be \'interpretable\'. Its one reason why I am drawn to the index notation,\n> despite its obvious inelegance. The structure is clear and, at least at\n> the elementary level, unambiguous. In fact this is one of the reasons\n> why I am not a lover of cross and dot products. These seem to me (and\n> have always seemed to me) to be confusing and concealing an underlying\n> symmetry that would be better to be expressed.\n\nThis notation *does* lie closer to the underlying structure. By making so\nthat corresponding things have the same inner products, it clarifies\nwhich things are associated with which. It also tends to avoid side issues\nlike specific ways of writing a particular vector.\n\nIn my previous notation <t,u>=<*t,u>=(*t)(u).\n\n>\n> Its the same thing that made me feel maxwell\'s equations were missing a\n> symmetry at a profound level which only vanished when I was introduced\n> to the (3+1)D EM method using the faraday tensor, which is something I\n> would like to follow up, if only anyone was remotely interested.\n> </rant>=========\n\nActually, the coordinate free language is useful for exactly this\nreason. Coordinates tend to confuse the issues because you have to\nchoose some basis vectors to get coordinates. Symmetries tend to not\ndepend on the specific basis chosen and so are more obvious when no\nsuch choice is made (i.e. when we use coordiante free language).\n\n\n> It occurs to me that the metric is implicitly a spatial expression.\n> At a point the metric is always flat (away from a singularity).\n\nNot if you interpret this as saying the geometry is flat close\nto a point, but that is a completely different matter. Given a point,\nit is possible to choose a coordinate system so that the matrix for\ng is diagonal at that point. This is not the same as saying anything is\nflat.\n\n> So what the metric is really giving us is the relationship of one point\n> with \'adjacent\' points. It is, in some very real sense, a volume term.\n> Hmm, that\'s not quite right, is it? It must be, for an N-D space,\n> an (N-1)-D term.\n\nNo, a metric allows us to talk about the \'length\' of a vector. It\nalso allows a correspondence between vectors and covectors. For a given\norientation, there is a natural volume associated with every metric,\nbut that volume is given by a \'volume form\', which is another kettle of\nfish.\n\n--Dan Grubb\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz <oz@farmeroz.port995.com> wrote in message
news:<ynHzRaA32EoBFw6Y@farmeroz.port995.com>...
> Greg Egan <gregegan@netspace.zebra.net.au> writes
>
> [NB Very many thanks for this long and detailed post]
>
> >The metric on 1-forms
> >=====================
> >
> >First, we have to be sure that we have a metric defined, and this will
> >let us take the inner product of two 1-forms. If t and u are 1-forms, we
> >write:
> >
> > <t,u> = g(t,u) = g^{ij} t_i u_j
>
> Is it valid to say g^{ij} t_i -> t^j ?
Not quite. Given a vector with components t_i, you get a co-vector
with components t^i where t^j=g^{ij} t_i.
>
> Then we have <t,u> = t^j u_j which is a standard contraction.
>
> >For anyone more used to thinking of the metric as giving the dot product
> >of two vectors, v and w:
> >
> > v.w = g(v,w) = g_{ij} v^i w^j
>
> Which we could express similarly.
>
> So the metric is really an expression of the invertibility between
> 1-forms and vectors.
Not quite sure what you mean by 'invertibility' here. The metric gives
a correspondence between vectors of one type and vectors of another
type. But different metrics give different correspondences.
>
> Now, I find I have been seduced by JB's 'GM course for idiots' where he
> illustrated index gymnastics in a 'don't think, just do it' manner.
> Consequently the indices were never properly explained. One
> 'appreciated' them by osmosis, which is not at all ideal.
>
> I think it is (long past) time that I straightened this out.
>
> The best way is probably for me to say how I see it, and for someone to
> wade in and point out where I have it wrong.
>
> I see it as follows (and it will probably do me good to try and work out
> how I really *do* see it, if I can):
>
> v^i is a vector, w_j is in some sense an inverse vector.
*Very[/itex] nervous here* There is a confusion between vectors and their
components (i.e. coordiantes) in a particular basis.
>
> Typically the 'i's and 'j's do not match up because they are general
> indexes. For example in 3-space we could lay down a co-ordinate system
> (x,y,z) and would have three basis vectors v^x, v^y, v^z and three
> 'basis 1-forms' w_x, w_y, w_z.
No. We would have basis vectors e_x, e_y and e_z. A generic vector
would then have an expension of the form v=v^x e_x +v^y e_y +v^z e_z.
By using the metric, we get a dual basis e^x, e^y, e^z. A generic covector
can be written as w=w_x e^x +w_y e^y +w_z e^z. Covectors are
'linear functionals', so things like w(v) make sense and are scalars
when w is a covector and v is a vector.
Also, the components, v^x, v^y, v^z are *scalars*. It is the combination
v^x e_x +v^y e_y +v^z e_z which is actually the vector, although
people to blur the distinction between v and the triple (v^x, v^y, v^z).
Simil;arly for w and (w_x, w_y, w_z).
> In this situation, where we have defined objects, it seems clear to me
> that v^x w_x will give me a scalar, whilst v^x w_y will not because they
> point in 'different directions'.
No. We would have e^x (e_x)=1, e^x (e_y)=0, e^y(e_y)=1, etc.
All expressions of the form w(v) will be scalars. For a generic
pair v, w as above, w(v)=w_x v^x +w_y v^y +w_z v^z, a scalar.
Now, if you are given a vector v=v^x e_x +v^y e_y +v^z e_z, then
there is a covector, call it v* corresponding to v with
v*=v_x e^x +v_y e^y +v_z e^z where the components satisfy
v_x=g_{xx} v^x +g_{xy} v^y +g_{xz} v^z, and similar expressions
for v_y and v_z. The short way of writing this is v_i=g_{ij} v^j.
But, if you cange bases from e_x, e_y, e_z, the components of v and w
change, but the expression w(v) does not. It is coordinate independent.
> Consequently when dealing with generalised objects, without an imposed
> co-ordinate system, its important to keep track of the relationships
> between the various indices.
Definitely not. Without coordinates, there are no indices! So, for
example, v^i will be the i^{th} coordinate of the vector v in whatever
basis you've chosen. The index tell you which coordinate you are
working with.
>
> The existence of a metric says v^i is invertible to [itex]v_i.
There's that word 'invertible' again. 'Corresponding' is better.
>
> >the two metrics are really just different versions of the same thing,
> >with the matrix of coordinates g^{ij} and the matrix of coordinates
> >g_{ij} just being inverses of each other:
> >
> > g_{ij} g^{jk} = \delta_i^k
>
> Now, what does this mean precisely?
It means that g_{ij} is the inverse matrix of g^{ij}. So if we use
g_{ij} to make the correspondence from vectors to covectors, then
g^{ij} gives the correspondence from covectors to vectors.
> I need to keep things co-ordinate free in my thinking though.
> Am I OK to consider this as three different 'local' co-ordinate
> *systems* i, j & k in some sense? So its really saying that going
>
> i -> j -> k is the same as going from i -> k
>
> (v^i g_{ij}) g^{jk} = w_j g^{jk} = u^k
>
> and we would like u^k = v^i since all we should have done is seen the
> object v^i from 'different perspectives'.
>
> >NB The <,> notation is used in a wide variety of slightly different
> >contexts, and its use here shouldn't be confused with "contracting" a
> >1-form with a vector, a process that doesn't require a metric.
>
> It would be nice if this was 'by definition'.
In some contexts, it is.
>
> >It is,
> >however, related to that other meaning in a very natural way; if v is a
> >vector corresponding to a 1-form u, i.e. produced from u by "raising an
> >index" with the metric:
> >
> > v^i = g^{ij} u_j
> >
> >then <t,v> = <t,u>, where the LHS use of <,> is the contraction of a
> >vector with a 1-form, and the RHS use of <,> is the inner product of two
> >1-forms:
> >
> > <t,v> = t_i v^i
> > = t_i g^{ij} u_j
> > = <t,u>
>
> I hate this sort of notation with a passion.
>
> <Rant>==========
> A notation should always reflect the underlying structure and should not
> be 'interpretable'. Its one reason why I am drawn to the index notation,
> despite its obvious inelegance. The structure is clear and, at least at
> the elementary level, unambiguous. In fact this is one of the reasons
> why I am not a lover of cross and dot products. These seem to me (and
> have always seemed to me) to be confusing and concealing an underlying
> symmetry that would be better to be expressed.
This notation *does* lie closer to the underlying structure. By making so
that corresponding things have the same inner products, it clarifies
which things are associated with which. It also tends to avoid side issues
like specific ways of writing a particular vector.
In my previous notation <t,u>=<*t,u>=(*t)(u).
>
> Its the same thing that made me feel maxwell's equations were missing a
> symmetry at a profound level which only vanished when I was introduced
> to the (3+1)D EM method using the faraday tensor, which is something I
> would like to follow up, if only anyone was remotely interested.
> </rant>=========
Actually, the coordinate free language is useful for exactly this
reason. Coordinates tend to confuse the issues because you have to
choose some basis vectors to get coordinates. Symmetries tend to not
depend on the specific basis chosen and so are more obvious when no
such choice is made (i.e. when we use coordiante free language).
> It occurs to me that the metric is implicitly a spatial expression.
> At a point the metric is always flat (away from a singularity).
Not if you interpret this as saying the geometry is flat close
to a point, but that is a completely different matter. Given a point,
it is possible to choose a coordinate system so that the matrix for
g is diagonal at that point. This is not the same as saying anything is
flat.
> So what the metric is really giving us is the relationship of one point
> with 'adjacent' points. It is, in some very real sense, a volume term.
> Hmm, that's not quite right, is it? It must be, for an N-D space,
> an (N-1)-D term.
No, a metric allows us to talk about the 'length' of a vector. It
also allows a correspondence between vectors and covectors. For a given
orientation, there is a natural volume associated with every metric,
but that volume is given by a 'volume form', which is another kettle of
fish.
--Dan Grubb
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>grubb@math.niu.edu <grubb@math.niu.edu> writes\n>Oz <oz@farmeroz.port995.com> wrote in message\n\n>> >\n>> > <t,u> = g(t,u) = g^{ij} t_i u_j\n>>\n>> Is it valid to say g^ij t_i -> t^j ?\n>\n>Not quite. Given a vector with components t_i, you get a co-vector\n>with components t^i where t^j=g^ij t_i.\n\nOK. Now this is another area where I don\'t know what I am talking about.\n\nYou have identified three objects,\nt^i, t^j, t_i\n\nClearly they are not the same.\n\nBut how precisely do they differ?\n\nI would say that t^i t_i -> scalar\nOne presumes that the indices refer to the same basis (i).\n\nI therefore presume that t^j is in another basis so that t^j t_i is not\nnecessarily scalar. Presumably because g^ij s bent, and so the covector-\nvector transformation isn\'t in some sense orthogonal (not quite what I\nmean).\n\nObviously I need a bit of a discussion about these important details.\n\n>> Then we have <t,u> = t^j u_j which is a standard contraction.\n>>\n>> >For anyone more used to thinking of the metric as giving the dot product\n>> >of two vectors, v and w:\n>> >\n>> > v.w = g(v,w) = g_{ij} v^i w^j\n>>\n>> Which we could express similarly.\n>>\n>> So the metric is really an expression of the invertibility between\n>> 1-forms and vectors.\n>\n>Not quite sure what you mean by \'invertibility\' here. The metric gives\n>a correspondence between vectors of one type and vectors of another\n>type. But different metrics give different correspondences.\n\n1) I am assuming this all falls out as co-ordinate independent.\n2) I am assuming a space with a metric (any metric). In this space I can\n\'invert\' vectors and co-vectors using the (local) metric. However I\nclearly have some confusion as I identify above between\nt^i, t^i and t_i.\n\n>> v^i is a vector, w_j is in some sense an inverse vector.\n>\n>*Very nervous here* There is a confusion between vectors and their\n>components (i.e. coordiantes) in a particular basis.\n\nYes. I am not clear about what the notation precisely means.\n\n>> Typically the \'i\'s and \'j\'s do not match up because they are general\n>> indexes. For example in 3-space we could lay down a co-ordinate system\n>> (x,y,z) and would have three basis vectors v^x, v^y, v^z and three\n>> \'basis 1-forms\' w_x, w_y, w_z.\n>\n>No. We would have basis vectors e_x, e_y and e_z. A generic vector\n>would then have an expansion of the form v=v^x e_x +v^y e_y +v^z e_z.\n\nOh. That makes the components scalar, which doesn\'t quite seem right.\nSurely you must either have components of form\n(a e_x) + (b e_y) + (c e_z)\n\nor (a, b, c) with implied basis (e_x, e_y, e_z).\n\nor am I missing something utterly basic?\n\n>By using the metric, we get a dual basis e^x, e^y, e^z. A generic covector\n>can be written as w=w_x e^x +w_y e^y +w_z e^z.\n\nOK. That\'s how I read it. The metric allows us to switch between vector\nand covector \'notation\'.\n\n>Covectors are\n>\'linear functionals\', so things like w(v) make sense and are scalars\n>when w is a covector and v is a vector.\n\nI\'m relieved. I assume w(v) is w_i v^i.\nOf course I cannot but wonder if its possible to have an object with\nMIXED vector and covector description. I probably don\'t want to know....\n\n>Also, the components, v^x, v^y, v^z are *scalars*.\n\nEh? What\'s the point of having a notation where we mark vectors as t^i\nand covectors as t_i and then bog it up by calling a scalar v^x? Call it\n(vx e^x + vy e^y + vz e^x) if you must. A scalar ain\'t a vector.\n\n>It is the combination\n>v^x e_x +v^y e_y +v^z e_z which is actually the vector, although\n>people to blur the distinction between v and the triple (v^x, v^y, v^z).\n>Simil;arly for w and (w_x, w_y, w_z).\n\nHmmm. Is this physicist talk or mathmo talk?\n\n>> In this situation, where we have defined objects, it seems clear to me\n>> that v^x w_x will give me a scalar, whilst v^x w_y will not because they\n>> point in \'different directions\'.\n>\n>No. We would have e^x (e_x)=1, e^x (e_y)=0, e^y(e_y)=1, etc.\n\nYes, OK. But note we are using a slightly different notation.\n\n>All expressions of the form w(v) will be scalars. For a generic\n>pair v, w as above, w(v)=w_x v^x +w_y v^y +w_z v^z, a scalar.\n\n1) OK. So given a general vector v^i and a general covector w_j, each\nwith a *different basis* (although I don\'t think that is important, one\ncan switch readily) the result is *always* a scalar?\n\n2) Vectors and covectors (and presumably their higher order equivalents)\nare *objects*, and the basis used is irrelevant to how they are\nmanipulated. Up to some scaling.\n\n>But, if you cange bases from e_x, e_y, e_z, the components of v and w\n>change, but the expression w(v) does not. It is coordinate independent.\n\nOK. That\'s fine. That\'s how I read it subject to my confusion over\ndifferences between t^i, t_i and t^j etc.\n\n>> Consequently when dealing with generalised objects, without an imposed\n>> co-ordinate system, its important to keep track of the relationships\n>> between the various indices.\n>\n>Definitely not. Without coordinates, there are no indices! So, for\n>example, v^i will be the i^th coordinate of the vector v in whatever\n>basis you\'ve chosen. The index tell you which coordinate you are\n>working with.\n\nOk.\nSo t^i and t^j are indeed the same object expressed under different\nbases?\n\n>> The existence of a metric says v^i is invertible to v_i.\n>\n>There\'s that word \'invertible\' again. \'Corresponding\' is better.\n\nWhy? Although I guess \'invertible\' implies a matrix inversion.\n\n>> > = <t,u>\n>>\n>> I hate this sort of notation with a passion.\n>>\n>> <Rant>==========\n>> A notation should always reflect the underlying structure and should not\n>> be \'interpretable\'. Its one reason why I am drawn to the index notation,\n>> despite its obvious inelegance. The structure is clear and, at least at\n>> the elementary level, unambiguous. In fact this is one of the reasons\n>> why I am not a lover of cross and dot products. These seem to me (and\n>> have always seemed to me) to be confusing and concealing an underlying\n>> symmetry that would be better to be expressed.\n>\n>This notation *does* lie closer to the underlying structure. By making so\n>that corresponding things have the same inner products, it clarifies\n>which things are associated with which. It also tends to avoid side issues\n>like specific ways of writing a particular vector.\n>\n>In my previous notation <t,u>=<*t,u>=(*t)(u).\n\nShould that be:\n\n<t,u> is an implied vector-covector pair which,\nif written with t,u as vectors is (*t)(u)?\n\n>> Its the same thing that made me feel maxwell\'s equations were missing a\n>> symmetry at a profound level which only vanished when I was introduced\n>> to the (3+1)D EM method using the faraday tensor, which is something I\n>> would like to follow up, if only anyone was remotely interested.\n>> </rant>=========\n>\n>Actually, the coordinate free language is useful for exactly this\n>reason. Coordinates tend to confuse the issues because you have to\n>choose some basis vectors to get coordinates.\n\nI rather see it as expressed with an arbitrary co-ordinate system.\nThat is, co-ordinate system unspecified and undefined, just that it can\nexist (multiply in most cases).\n\n>Symmetries tend to not\n>depend on the specific basis chosen and so are more obvious when no\n>such choice is made (i.e. when we use coordiante free language).\n\nAgreed.\n\n>> It occurs to me that the metric is implicitly a spatial expression.\n>> At a point the metric is always flat (away from a singularity).\n>\n>Not if you interpret this as saying the geometry is flat close\n>to a point, but that is a completely different matter. Given a point,\n>it is possible to choose a coordinate system so that the matrix for\n>g is diagonal at that point. This is not the same as saying anything is\n>flat.\n\nOh. That\'s quite profound.\nIt rather solves a question I have had in the back of my mind for a long\ntime. Covectors are inherently spatial and have extent. Obviously you\ncan\'t have *planes* at a point, although admittedly having direction at\na point is also iffy. The statement that one can always find (except\npathological presumably) a basis where g is diagonal at a point\nexpresses my mental model of what\'s going on very nicely. That was a\nnaive view that a curved space is in some sense a mosaic of little flat\npatches stuck at relative angles to each other and then tended to\ninfinitesimal. That\'s just looking at one point in a preferred basis.\n\n>> So what the metric is really giving us is the relationship of one point\n>> with \'adjacent\' points. It is, in some very real sense, a volume term.\n>> Hmm, that\'s not quite right, is it? It must be, for an N-D space,\n>> an (N-1)-D term.\n>\n>No, a metric allows us to talk about the \'length\' of a vector.\n\nIsn\'t that the same thing?\nI\'m not even sure if a vector has a \'length\'.\nA force doesn\'t really have a \'length\', it\'s its effect that has a\nlength. A force has an \'implied length\', even a velocity has an implied\nlength. Its only when you measure a short interval (time say) and say\nhow something has *moved* that you get a length. That is highly\ndelightful and hugely clarifying. A magnitude is not a length,\nnecessarily.\n\n>It\n>also allows a correspondence between vectors and covectors.\n\nIndeed, now its becoming clearer. A vector specifies a direction and a\nmagnitude. The covector defines \'spacing between planes\', that is it in\nsome sense defines the effect generated by the vector. I will need to\nthink about that, its quite profound. In effect I could (if I chose)\nspecify that a body of mass m sees a certain spacing, and a body of mass\n2M sees a spacing twice as dense. Then an equation of motion would be\nF=a. Obviously this is entirely impractical because I would then have a\ndifferent spacetime for every body and the net result would be the same\nanyway.\n\n>For a given\n>orientation, there is a natural volume associated with every metric,\n\nOk, because its a g^ij, two sets of parameters.\nNo, that can\'t be quite right.\nThere ought to be a volume-like term u^i v_i = g_ij u_j v_i\nwhich doesn\'t quite make sense.\n\n>but that volume is given by a \'volume form\', which is another kettle of\n>fish.\n\nYou know me, I tread where sensible people back off ....\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>grubb@math.niu.edu <grubb@math.niu.edu> writes
>Oz <oz@farmeroz.port995.com> wrote in message
>> >
>> > <t,u> = g(t,u) = g^{ij} t_i u_j
>>
>> Is it valid to say g^{ij} t_i -> t^j ?
>
>Not quite. Given a vector with components t_i, you get a co-vector
>with components t^i where t^j=g^{ij} t_i.
OK. Now this is another area where I don't know what I am talking about.
You have identified three objects,
t^i, t^j, t_i
Clearly they are not the same.
But how precisely do they differ?
I would say that t^i t_i -> scalar
One presumes that the indices refer to the same basis (i).
I therefore presume that t^j is in another basis so that t^j t_i is not
necessarily scalar. Presumably because g^{ij} s bent, and so the covector-
vector transformation isn't in some sense orthogonal (not quite what I
mean).
Obviously I need a bit of a discussion about these important details.
>> Then we have <t,u> = t^j u_j which is a standard contraction.
>>
>> >For anyone more used to thinking of the metric as giving the dot product
>> >of two vectors, v and w:
>> >
>> > v.w = g(v,w) = g_{ij} v^i w^j
>>
>> Which we could express similarly.
>>
>> So the metric is really an expression of the invertibility between
>> 1-forms and vectors.
>
>Not quite sure what you mean by 'invertibility' here. The metric gives
>a correspondence between vectors of one type and vectors of another
>type. But different metrics give different correspondences.
1) I am assuming this all falls out as co-ordinate independent.
2) I am assuming a space with a metric (any metric). In this space I can
'invert' vectors and co-vectors using the (local) metric. However I
clearly have some confusion as I identify above between
t^i, t^i and t_i.
>> v^i[/itex] is a vector, w_j is in some sense an inverse vector.
>
>*Very nervous here* There is a confusion between vectors and their
>components (i.e. coordiantes) in a particular basis.
Yes. I am not clear about what the notation precisely means.
>> Typically the 'i's and 'j's do not match up because they are general
>> indexes. For example in 3-space we could lay down a co-ordinate system
>> (x,y,z) and would have three basis vectors v^x, v^y, v^z and three
>> 'basis 1-forms' w_x, w_y, w_z.
>
>No. We would have basis vectors e_x, e_y and e_z. A generic vector
>would then have an expansion of the form v=v^x e_x +v^y e_y +v^z e_z.
Oh. That makes the components scalar, which doesn't quite seem right.
Surely you must either have components of form
(a e_x) + (b e_y) + (c e_z)
or (a, b, c) with implied basis (e_x, e_y, e_z).
or am I missing something utterly basic?
>By using the metric, we get a dual basis e^x, e^y, e^z. A generic covector
>can be written as w=w_x e^x +w_y e^y +w_z e^z.
OK. That's how I read it. The metric allows us to switch between vector
and covector 'notation'.
>Covectors are
>'linear functionals', so things like w(v) make sense and are scalars
>when w is a covector and v is a vector.
I'm relieved. I assume w(v) is w_i v^i.
Of course I cannot but wonder if its possible to have an object with
MIXED vector and covector description. I probably don't want to know....
>Also, the components, v^x, v^y, v^z are *scalars*.
Eh? What's the point of having a notation where we mark vectors as t^i
and covectors as t_i and then bog it up by calling a scalar v^x? Call it
(vx e^x + vy e^y + vz e^x) if you must. A scalar ain't a vector.
>It is the combination
>v^x e_x +v^y e_y +v^z e_z which is actually the vector, although
>people to blur the distinction between v and the triple (v^x, v^y, v^z).
>Simil;arly for w and (w_x, w_y, w_z).
Hmmm. Is this physicist talk or mathmo talk?
>> In this situation, where we have defined objects, it seems clear to me
>> that v^x w_x will give me a scalar, whilst v^x w_y will not because they
>> point in 'different directions'.
>
>No. We would have e^x (e_x)=1, e^x (e_y)=0, e^y(e_y)=1, etc.
Yes, OK. But note we are using a slightly different notation.
>All expressions of the form w(v) will be scalars. For a generic
>pair v, w as above, w(v)=w_x v^x +w_y v^y +w_z v^z, a scalar.
1) OK. So given a general vector v^i and a general covector w_j, each
with a *different basis* (although I don't think that is important, one
can switch readily) the result is *always* a scalar?
2) Vectors and covectors (and presumably their higher order equivalents)
are *objects*, and the basis used is irrelevant to how they are
manipulated. Up to some scaling.
>But, if you cange bases from e_x, e_y, e_z, the components of v and w
>change, but the expression w(v) does not. It is coordinate independent.
OK. That's fine. That's how I read it subject to my confusion over
differences between t^i, t_i and t^j etc.
>> Consequently when dealing with generalised objects, without an imposed
>> co-ordinate system, its important to keep track of the relationships
>> between the various indices.
>
>Definitely not. Without coordinates, there are no indices! So, for
>example, v^i will be the i^{th} coordinate of the vector v in whatever
>basis you've chosen. The index tell you which coordinate you are
>working with.
Ok.
So t^i and t^j are indeed the same object expressed under different
bases?
>> The existence of a metric says v^i is invertible to [itex]v_i.
>
>There's that word 'invertible' again. 'Corresponding' is better.
Why? Although I guess 'invertible' implies a matrix inversion.
>> > = <t,u>
>>
>> I hate this sort of notation with a passion.
>>
>> <Rant>==========
>> A notation should always reflect the underlying structure and should not
>> be 'interpretable'. Its one reason why I am drawn to the index notation,
>> despite its obvious inelegance. The structure is clear and, at least at
>> the elementary level, unambiguous. In fact this is one of the reasons
>> why I am not a lover of cross and dot products. These seem to me (and
>> have always seemed to me) to be confusing and concealing an underlying
>> symmetry that would be better to be expressed.
>
>This notation *does* lie closer to the underlying structure. By making so
>that corresponding things have the same inner products, it clarifies
>which things are associated with which. It also tends to avoid side issues
>like specific ways of writing a particular vector.
>
>In my previous notation <t,u>=<*t,u>=(*t)(u).
Should that be:
<t,u> is an implied vector-covector pair which,
if written with t,u as vectors is (*t)(u)?
>> Its the same thing that made me feel maxwell's equations were missing a
>> symmetry at a profound level which only vanished when I was introduced
>> to the (3+1)D EM method using the faraday tensor, which is something I
>> would like to follow up, if only anyone was remotely interested.
>> </rant>=========
>
>Actually, the coordinate free language is useful for exactly this
>reason. Coordinates tend to confuse the issues because you have to
>choose some basis vectors to get coordinates.
I rather see it as expressed with an arbitrary co-ordinate system.
That is, co-ordinate system unspecified and undefined, just that it can
exist (multiply in most cases).
>Symmetries tend to not
>depend on the specific basis chosen and so are more obvious when no
>such choice is made (i.e. when we use coordiante free language).
Agreed.
>> It occurs to me that the metric is implicitly a spatial expression.
>> At a point the metric is always flat (away from a singularity).
>
>Not if you interpret this as saying the geometry is flat close
>to a point, but that is a completely different matter. Given a point,
>it is possible to choose a coordinate system so that the matrix for
>g is diagonal at that point. This is not the same as saying anything is
>flat.
Oh. That's quite profound.
It rather solves a question I have had in the back of my mind for a long
time. Covectors are inherently spatial and have extent. Obviously you
can't have *planes* at a point, although admittedly having direction at
a point is also iffy. The statement that one can always find (except
pathological presumably) a basis where g is diagonal at a point
expresses my mental model of what's going on very nicely. That was a
naive view that a curved space is in some sense a mosaic of little flat
patches stuck at relative angles to each other and then tended to
infinitesimal. That's just looking at one point in a preferred basis.
>> So what the metric is really giving us is the relationship of one point
>> with 'adjacent' points. It is, in some very real sense, a volume term.
>> Hmm, that's not quite right, is it? It must be, for an N-D space,
>> an (N-1)-D term.
>
>No, a metric allows us to talk about the 'length' of a vector.
Isn't that the same thing?
I'm not even sure if a vector has a 'length'.
A force doesn't really have a 'length', it's its effect that has a
length. A force has an 'implied length', even a velocity has an implied
length. Its only when you measure a short interval (time say) and say
how something has *moved* that you get a length. That is highly
delightful and hugely clarifying. A magnitude is not a length,
necessarily.
>It
>also allows a correspondence between vectors and covectors.
Indeed, now its becoming clearer. A vector specifies a direction and a
magnitude. The covector defines 'spacing between planes', that is it in
some sense defines the effect generated by the vector. I will need to
think about that, its quite profound. In effect I could (if I chose)
specify that a body of mass m sees a certain spacing, and a body of mass
2M sees a spacing twice as dense. Then an equation of motion would be
F=a. Obviously this is entirely impractical because I would then have a
different spacetime for every body and the net result would be the same
anyway.
>For a given
>orientation, there is a natural volume associated with every metric,
Ok, because its a g^{ij}, two sets of parameters.
No, that can't be quite right.
There ought to be a volume-like term u^i v_i = g_{ij} u_j v_i
which doesn't quite make sense.
>but that volume is given by a 'volume form', which is another kettle of
>fish.
You know me, I tread where sensible people back off ....
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
Greg Egan
Nov26-04, 01:20 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <ynHzRaA32EoBFw6Y@farmeroz.port995.com>, Oz\n<oz@farmeroz.port995.com> wrote:\n\n> Greg Egan <gregegan@netspace.zebra.net.au> writes\n>\n> [NB Very many thanks for this long and detailed post]\n>\n> >The metric on 1-forms\n> >=====================\n> >\n> >First, we have to be sure that we have a metric defined, and this will\n> >let us take the inner product of two 1-forms. If t and u are 1-forms, we\n> >write:\n> >\n> > <t,u> = g(t,u) = g^{ij} t_i u_j\n>\n> Is it valid to say g^ij t_i -> t^j ?\n>\n> Then we have <t,u> = t^j u_j which is a standard contraction.\n\nYes, this is all fine. The inner product of two 1-forms from a given\nmetric g is exactly what you get if you use g to "raise the index" on\neither 1-form to turn it into a vector, and then contract that vector\nwith the other 1-form. (Just keep in the back of your mind the\nrequirement that these coordinates *must* refer to dual bases, or we\ndon\'t have that nice formula for contractions.)\n\n> >For anyone more used to thinking of the metric as giving the dot product\n> >of two vectors, v and w:\n> >\n> > v.w = g(v,w) = g_{ij} v^i w^j\n>\n> Which we could express similarly.\n>\n> So the metric is really an expression of the invertibility between\n> 1-forms and vectors.\n\nI think the right word here is "inter-convertibility", which is an ugly\nword admittedly, but "invertible" is a different concept altogether.\n\nRemember, the metric lets you convert vectors into 1-forms and vice versa\nbecause it gives you an inner product on *pairs* of vectors and 1-forms,\nwhich you can then require to be equal to the contraction of a vector and\na 1-form. That\'s how we defined "the vector v is equivalent to the\n1-form t":\n\neither g(v,w) = <t,w> for all vectors w\nor g(t,u) = <u,v> for all 1-forms u\n\n> I see it as follows (and it will probably do me good to try and work out\n> how I really *do* see it, if I can):\n>\n> v^i is a vector, w_j is in some sense an inverse vector.\n\nDon\'t say inverse, say "dual vector" or "covector".\n\nThe *dual* of a vector space is the space of linear operators from that\nspace to the real numbers. If we have a basis, {e_i}, to our vector\nspace, we can define a basis of the dual space, {e^j}, by defining:\n\ne^j(e_i) = delta_i^j\n\nand then using linearity to give the value of e^j on any general vector:\n\nv = v^i e_i\ne^j(v) = e^j(v^i e_i) = v^j\n\nThat basis {e^j} is said to be the dual basis to the basis of vectors\n{e_i}.\n\nActually, we\'ve been a bit sloppy and talked about "vectors and 1-forms",\nwhen the correct terminology is to talk about the space of [tangent]\nvectors at a single point, and its dual space, the "cotangent space",\nwhich contains "cotangent vectors". Then, when you consider a region or\na whole manifold, you have *fields* of these two things: vector fields,\nand 1-forms. But I\'ll probably keep being sloppy and talking about\n"1-forms", even when discussing the dual space to the tangent vectors at\na single point.\n\n> Typically the \'i\'s and \'j\'s do not match up because they are general\n> indexes. For example in 3-space we could lay down a co-ordinate system\n> (x,y,z) and would have three basis vectors v^x, v^y, v^z and three\n> \'basis 1-forms\' w_x, w_y, w_z.\n>\n> In this situation, where we have defined objects, it seems clear to me\n> that v^x w_x will give me a scalar, whilst v^x w_y will not because they\n> point in \'different directions\'.\n\nUmm, if I\'ve read your mind correctly you\'re sort of getting there, but\nyou keep drifting off track on notation, which doesn\'t help.\n\nFor a start, if v^x, v^y and v^z are 3 individual vectors, the usual\nconvention is to label them with subscripts, e.g. {e_1,e_2,e_3} or\n{e_x,e_y,e_z} if you prefer. Superscripts are for vector *coordinates*,\nwhereas subscripts distinguish the different vectors in a basis. Then\nyou can write:\n\nv = v^i e_i\n\nand use the Einstein summation convention for matching superscripts and\nsubscripts.\n\nIn typesetting, the vectors {v, e_i} here would be bold face and the\nscalar v^i would be plain.\n\nNow if v^x is a single vector, what is v^x w_x supposed to mean? "x" is\na fixed label, not something you can sum over like i. Do you mean\n<w_x,v_x>? If so, <w_y,v_x> is also a perfectly well-defined scalar,\nalthough if your two bases are dual to each other it happens to be zero.\nIs that what you meant?\n\nLet\'s quickly recap how a coordinate system gives you bases for both\nvectors and 1-forms.\n\nIf we have a grid of coordinates (x,y,z), then we can define a tangent\nvector such as @_x as the derivative with respect to x of any scalar\nfunction f, while holding y and z at fixed values. In this way we can\ndefine @_x, @_y, @_z as differential operators on scalars, tangent\nvectors which point along the x, y and z coordinate lines: that is, the\nlines where the other coordinates are fixed while the value of x (or y or\nz) varies.\n\nDual to this basis at each point, we can define basis 1-forms dx, dy, dz.\nWe can *define* these linear operators on tangent vectors by simply\nrequiring:\n\ndx(@_x) = 1\ndx(@_y) = dx(@_z) = 0\n\nor if we can write "x^i" for an arbitrary coordinate from {x,y,z}:\n\ndx^i(@_{x^j}) = delta^i_j\n\nGeometrically, these basis 1-forms are the families of planes in which\none coordinate is *fixed* and the others vary, e.g. dx consists of the\nplanes of fixed values of x.\n\n> Consequently when dealing with generalised objects, without an imposed\n> co-ordinate system, its important to keep track of the relationships\n> between the various indices.\n\nI don\'t really know what this means, but I think the key idea to "index\ngymnastics" is the defining equation of a dual basis:\n\ne^j(e_i) = delta_i^j\n\nIf by "generalised objects" you mean using bases of vectors and 1-forms\nother than those that come from a coordinate system, the important thing\nis to use bases for the 1-forms that are dual to those for the vectors\n(which is automatically true for bases derived from coordinate systems,\nbut can still be true in other contexts).\n\n> The existence of a metric says v^i is invertible to v_i.\n\n"convertible"\n\n> >the two metrics are really just different versions of the same thing,\n> >with the matrix of coordinates g^{ij} and the matrix of coordinates\n> >g_{ij} just being inverses of each other:\n> >\n> > g_{ij} g^{jk} = delta_i^k\n>\n> Now, what does this mean precisely?\n\nThe succinct thing to say is that if you raise an index and then lower it\nagain (or lower an index and then raise it again), you ought to get back\nwhere you started.\n\nYou can also require that if t and u are 1-forms equivalent to vectors v\nand w, we must have g(v,w)=g(t,u). Again, that will require the metric\non 1-forms to have a coordinate matrix that is the inverse of that for\nthe metric on vectors (all coordinates relative to dual bases!).\n\n> I need to keep things co-ordinate free in my thinking though.\n> Am I OK to consider this as three different \'local\' co-ordinate\n> *systems* i, j & k in some sense? So its really saying that going\n\nThere aren\'t three different coordinate systems here; i, j and k are all\nindexing either the same basis or its dual.\n\n> i -> j -> k is the same as going from i -> k\n>\n> (v^i g_ij) g^jk = w_j g^jk = u^k\n>\n> and we would like u^k = v^i since all we should have done is seen the\n> object v^i from \'different perspectives\'.\n\nSome of your detailed comments in passing are confusing to me, but the\noverall statement is sound: a vector and the equivalent 1-form are\nreally two ways of talking about the same thing, and we require that (the\nvector equivalent to (the 1-form equivalent to a vector)) gives us back\nthe original vector, which is the coordinate-free way of talking about\nraising and lowering indices and ending up back where we started.\n\n> >NB The <,> notation is used in a wide variety of slightly different\n> >contexts, and its use here shouldn\'t be confused with "contracting" a\n> >1-form with a vector, a process that doesn\'t require a metric.\n>\n> It would be nice if this was \'by definition\'.\n\nWell yes, the definitions have been chosen in order to make these\ndifferent meanings of <,> agree. Thus (as you\'ve quoted back to me):\n\n> >It is,\n> >however, related to that other meaning in a very natural way; if v is a\n> >vector corresponding to a 1-form u, i.e. produced from u by "raising an\n> >index" with the metric:\n> >\n> > v^i = g^{ij} u_j\n> >\n> >then <t,v> = <t,u>, where the LHS use of <,> is the contraction of a\n> >vector with a 1-form, and the RHS use of <,> is the inner product of two\n> >1-forms:\n> >\n> > <t,v> = t_i v^i\n> > = t_i g^{ij} u_j\n> > = <t,u>\n>\n> I hate this sort of notation with a passion.\n\nWell, fine, you don\'t have to use it. You can be understood if you use\n<,> exclusively for metric-free contractions, and always explicitly\nmention the metric when it\'s used for an inner product on vectors or\n1-forms. However, other people will go on using <,> for metric-dependent\nexpressions, so you have to be alert for that.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <ynHzRaA32EoBFw6Y@farmeroz.port995.com>, Oz
<oz@farmeroz.port995.com> wrote:
> Greg Egan <gregegan@netspace.zebra.net.au> writes
>
> [NB Very many thanks for this long and detailed post]
>
> >The metric on 1-forms
> >=====================
> >
> >First, we have to be sure that we have a metric defined, and this will
> >let us take the inner product of two 1-forms. If t and u are 1-forms, we
> >write:
> >
> > <t,u> = g(t,u) = g^{ij} t_i u_j
>
> Is it valid to say g^{ij} t_i -> t^j ?
>
> Then we have <t,u> = t^j u_j which is a standard contraction.
Yes, this is all fine. The inner product of two 1-forms from a given
metric g is exactly what you get if you use g to "raise the index" on
either 1-form to turn it into a vector, and then contract that vector
with the other 1-form. (Just keep in the back of your mind the
requirement that these coordinates *must* refer to dual bases, or we
don't have that nice formula for contractions.)
> >For anyone more used to thinking of the metric as giving the dot product
> >of two vectors, v and w:
> >
> > v.w = g(v,w) = g_{ij} v^i w^j
>
> Which we could express similarly.
>
> So the metric is really an expression of the invertibility between
> 1-forms and vectors.
I think the right word here is "inter-convertibility", which is an ugly
word admittedly, but "invertible" is a different concept altogether.
Remember, the metric lets you convert vectors into 1-forms and vice versa
because it gives you an inner product on *pairs* of vectors and 1-forms,
which you can then require to be equal to the contraction of a vector and
a 1-form. That's how we defined "the vector v is equivalent to the
1-form t":
either g(v,w) = <t,w> for all vectors w
or g(t,u) = <u,v> for all 1-forms u
> I see it as follows (and it will probably do me good to try and work out
> how I really *do* see it, if I can):
>
> v^i is a vector, w_j is in some sense an inverse vector.
Don't say inverse, say "dual vector" or "covector".
The *dual* of a vector space is the space of linear operators from that
space to the real numbers. If we have a basis, {e_i}, to our vector
space, we can define a basis of the dual space, {e^j}, by defining:
e^j(e_i) = \delta_i^j
and then using linearity to give the value of e^j on any general vector:
v = v^i e_ie^j(v) = e^j(v^i e_i) = v^j
That basis {e^j} is said to be the dual basis to the basis of vectors
{e_i}.
Actually, we've been a bit sloppy and talked about "vectors and 1-forms",
when the correct terminology is to talk about the space of [tangent]
vectors at a single point, and its dual space, the "cotangent space",
which contains "cotangent vectors". Then, when you consider a region or
a whole manifold, you have *fields* of these two things: vector fields,
and 1-forms. But I'll probably keep being sloppy and talking about
"1-forms", even when discussing the dual space to the tangent vectors at
a single point.
> Typically the 'i's and 'j's do not match up because they are general
> indexes. For example in 3-space we could lay down a co-ordinate system
> (x,y,z) and would have three basis vectors v^x, v^y, v^z and three
> 'basis 1-forms' w_x, w_y, w_z.
>
> In this situation, where we have defined objects, it seems clear to me
> that v^x w_x will give me a scalar, whilst v^x w_y will not because they
> point in 'different directions'.
Umm, if I've read your mind correctly you're sort of getting there, but
you keep drifting off track on notation, which doesn't help.
For a start, if v^x, v^y and v^z are 3 individual vectors, the usual
convention is to label them with subscripts, e.g. {e_1,e_2,e_3} or
{e_x,e_y,e_z} if you prefer. Superscripts are for vector *coordinates*,
whereas subscripts distinguish the different vectors in a basis. Then
you can write:
v = v^i e_i
and use the Einstein summation convention for matching superscripts and
subscripts.
In typesetting, the vectors {v, e_i} here would be bold face and the
scalar v^i would be plain.
Now if v^x is a single vector, what is v^x w_x supposed to mean? "x" is
a fixed label, not something you can sum over like i. Do you mean
<w_x,v_x>? If so, <w_y,v_x> is also a perfectly well-defined scalar,
although if your two bases are dual to each other it happens to be zero.
Is that what you meant?
Let's quickly recap how a coordinate system gives you bases for both
vectors and 1-forms.
If we have a grid of coordinates (x,y,z), then we can define a tangent
vector such as @_x as the derivative with respect to x of any scalar
function f, while holding y and z at fixed values. In this way we can
define @_x, @_y, @_z as differential operators on scalars, tangent
vectors which point along the x, y and z coordinate lines: that is, the
lines where the other coordinates are fixed while the value of x (or y or
z) varies.
Dual to this basis at each point, we can define basis 1-forms dx, dy, dz.
We can *define* these linear operators on tangent vectors by simply
requiring:
dx(@_x) = 1dx(@_y) = dx(@_z) =
or if we can write "x^i" for an arbitrary coordinate from {x,y,z}:dx^i(@_{x^j}) = \delta^i_j
Geometrically, these basis 1-forms are the families of planes in which
one coordinate is *fixed* and the others vary, e.g. dx consists of the
planes of fixed values of x.
> Consequently when dealing with generalised objects, without an imposed
> co-ordinate system, its important to keep track of the relationships
> between the various indices.
I don't really know what this means, but I think the key idea to "index
gymnastics" is the defining equation of a dual basis:
e^j(e_i) = \delta_i^j
If by "generalised objects" you mean using bases of vectors and 1-forms
other than those that come from a coordinate system, the important thing
is to use bases for the 1-forms that are dual to those for the vectors
(which is automatically true for bases derived from coordinate systems,
but can still be true in other contexts).
> The existence of a metric says v^i is invertible to v_i.
"convertible"
> >the two metrics are really just different versions of the same thing,
> >with the matrix of coordinates g^{ij} and the matrix of coordinates
> >g_{ij} just being inverses of each other:
> >
> > g_{ij} g^{jk} = \delta_i^k
>
> Now, what does this mean precisely?
The succinct thing to say is that if you raise an index and then lower it
again (or lower an index and then raise it again), you ought to get back
where you started.
You can also require that if t and u are 1-forms equivalent to vectors v
and w, we must have g(v,w)=g(t,u). Again, that will require the metric
on 1-forms to have a coordinate matrix that is the inverse of that for
the metric on vectors (all coordinates relative to dual bases!).
> I need to keep things co-ordinate free in my thinking though.
> Am I OK to consider this as three different 'local' co-ordinate
> *systems* i, j & k in some sense? So its really saying that going
There aren't three different coordinate systems here; i, j and k are all
indexing either the same basis or its dual.
> i -> j -> k is the same as going from i -> k
>
> (v^i g_{ij}) g^{jk} = w_j g^{jk} = u^k
>
> and we would like u^k = v^i since all we should have done is seen the
> object v^i from 'different perspectives'.
Some of your detailed comments in passing are confusing to me, but the
overall statement is sound: a vector and the equivalent 1-form are
really two ways of talking about the same thing, and we require that (the
vector equivalent to (the 1-form equivalent to a vector)) gives us back
the original vector, which is the coordinate-free way of talking about
raising and lowering indices and ending up back where we started.
> >NB The <,> notation is used in a wide variety of slightly different
> >contexts, and its use here shouldn't be confused with "contracting" a
> >1-form with a vector, a process that doesn't require a metric.
>
> It would be nice if this was 'by definition'.
Well yes, the definitions have been chosen in order to make these
different meanings of <,> agree. Thus (as you've quoted back to me):
> >It is,
> >however, related to that other meaning in a very natural way; if v is a
> >vector corresponding to a 1-form u, i.e. produced from u by "raising an
> >index" with the metric:
> >
> > v^i = g^{ij} u_j
> >
> >then <t,v> = <t,u>, where the LHS use of <,> is the contraction of a
> >vector with a 1-form, and the RHS use of <,> is the inner product of two
> >1-forms:
> >
> > <t,v> = t_i v^i
> > = t_i g^{ij} u_j
> > = <t,u>
>
> I hate this sort of notation with a passion.
Well, fine, you don't have to use it. You can be understood if you use
<,> exclusively for metric-free contractions, and always explicitly
mention the metric when it's used for an inner product on vectors or
1-forms. However, other people will go on using <,> for metric-dependent
expressions, so you have to be alert for that.
grubb@math.niu.edu
Nov26-04, 01:22 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Oz <oz@farmeroz.port995.com> wrote in message news:<gWrndZAAkFpBFwiK@farmeroz.port995.com>...\n> grubb@math.niu.edu <grubb@math.niu.edu> writes\n> >Oz <oz@farmeroz.port995.com> wrote in message\n>\n> >> >\n> >> > <t,u> = g(t,u) = g^{ij} t_i u_j\n> >>\n> >> Is it valid to say g^ij t_i -> t^j ?\n> >\n> >Not quite. Given a vector with components t_i, you get a co-vector\n> >with components t^i where t^j=g^ij t_i.\n>\n> OK. Now this is another area where I don\'t know what I am talking about.\n>\n> You have identified three objects,\n> t^i, t^j, t_i\n>\n> Clearly they are not the same.\n\nThe first two are the same, just different dummy indices.\n\n>\n> But how precisely do they differ?\n>\n> I would say that t^i t_i -> scalar\n> One presumes that the indices refer to the same basis (i).\n>\n> I therefore presume that t^j is in another basis so that t^j t_i is not\n> necessarily scalar. Presumably because g^ij s bent, and so the covector-\n> vector transformation isn\'t in some sense orthogonal (not quite what I\n> mean).\n>\n> Obviously I need a bit of a discussion about these important details.\n\nPerhaps it\'s better to start over. I first want to talk about vectors\nand covectors when there isn\'t a minifold in the background. Also, I\'ll\nditch the xyz notation and number the components. That\'s more standard\nanyway.\n\nA vector space is simply a collection of objects that we can add and\nmultiply by scalars. The objects in the vector space are called vectors\n(not too surprising yet, I hope). We\'ll stick to finite dimensional\nvector spaces for the moment to avoid some pretty nasty stuff. If you\'re\ninterested, I\'ll go there, but we\'ll have enough on our plate as it is, OK?\n\nExamples of vector spaces:\n1) R^3 This is the collection of ordered triples (a,b,c) where a,b, and c are\nreal numbers. This can be generalized in two (main) ways, first to ordered\nn-tuples to get R^n, and also to ordered n-tuples of complex numbers to\nget C^n.\n\n2) The collection of solutions to some homogeneous linear differential\nequation, say y\'\'+y=0 or y\'\'\'+3xy\'\'-2y\'+y=0.\n\n3) The collection P^5 of polynomials of degree <=5 (or any n)\n\nA covector is simply a linear functional, that is to say, a linear\nmap from the vector space to the scalars. Example:\n\n1) The map that takes (a,b,c) to the scalar 3a-b+4c is a covector for the\nvector space R^3.\n\n2) The map that takes a polynomial p(x) and gives the integral of p(x)\nfrom 0 to 2 is a covector for the space of polynomials P^5.\n\nI will use the letters L, S, T for covectors to minimize confusion.\nI will use v,w for vectors. Hence an expression such as L(v) makes sense\nand is a scalar. It is common to write <L,v> for L(v), but\nI will refrain for now.\n\nFor example 1), v=(a,b,c) and L(v)=3a-b+4c.\n\nA metric is a bilinear map that takes pairs of vectors and gives scalars\nout. Hence, for vectors v and w, the expression g(v,w) makes sense and is\na scalar. Furthermore g(v_1 +v_2 ,w)=g(v_1 ,w)+g(v_2 ,w). We also require\nsymmetry so that g(v,w)=g(w,v). Finally, if we fix w and g(v,w)=0 for every v,\nwe require that w must be the 0 vector. OK?\n\nFor example,\n\n1) If v=(a,b,c) and w=(x,y,z), let g(v,w)=ax+by+cz. This gives a metric\non R^3. THis is called the standard metric of R^3.\n\n2) If v=(a,b,c) and w=(x,y,z), let g(v,w)=ax+2by+3cz. This gives a different\nmetric on R^3.\n\n3) If p and q are polynomials, let g(p,q)=int_0^1 p(x)q(x) dx. This is a metric\non P^5.\n\nIt is common to write <v,w> in place of g(v,w), but I will refrain for\nthe time being.\n\nNow, suppose we take a vector w and look at the function that takes\na vector v to the scalar g(v,w). This is a linear map, and so it is\na covector! We call this covector *w for convenience. The remarkable thing\nis that *every* covector can be obtained in this way! So if you are given\na covector L, there is a vector w so that L=*w. This is the correspondence\nthat metrics give between vectors and covectors.\n\nNext, I want to point out that the collection covectors has the property\nthat things can be added and multiplied by scalars, so the collection of\ncovectors is a vector space in its own right. Furthermore, if you are\ngiven a vector v, the map that takes L to L(v) is a co-covector. Another\nmiracle of the subject is that every co-covector comes in this way. Also,\nno metric is needed to set up this correspondence, although on is needed\nfor that between vectors and covectors above.\n\nNext, a basis is a collection of vectors where everything in the vector\nspace can be written in a unique way as a linear combination of those\nbasis elements. The coefficients used in the linear combination are\ncalled the *components* of the vector in that basis. Notice that the\ncomponents are scalars.\n\nExamples:\n\n1) The collection {(1,0,0), (0,1,0), (0,0,1)} is a basis of R^3. This\nis called the standard basis. For convenience, write e_1=(1,0,0)\ne_2=(0,1,0), and e_3=(0,0,1). Given a vector v=(a,b,c), we can write\nv=a e_1 +b e_2 +c e_3. Here, the components are a, b, and c. We write\nv^1=a, v^2 =b, and v^3 =c, so v=v^1 e_1 +v^2 e_2 +v^3 e_3 = v^i e_i\nwhere in the last expression I have used the summation notation. This\ncan also be written as v=v^j e_j since the i and j are dummy variables\nfor the summation.\n\n2) The collection {1, x, x^2, x^3, x^4, x^5} is a basis of P^5. Here,\nthe components are simply the coefficients of the polynomial.\n\n3) The collection {(1,0,0), (1,1,0), (1,1,1)} and another basis for R^3.\nhere, the components of a vector v=(a,b,c) will be a-b, b-c, and c in that\norder, so *in this basis*, v^1=a-b, v^2=b-c, and v^3=c. For later\nconvenience, let f_1=(1,0,0), f_2=(1,1,0) and f_3=(1,1,1).\n\nNext, for any basis for the collection of vectors, there is a dual\nbasis for the covectors. This is usually written with superscripts.\n\nThus, for the standard basis of R^3, the covector e^1 should satisfy\ne^1 (e_1)=1, e^1 (e_2)=0, e^1 (e_3)=0. This works out to give that\ne^1 ( (a,b,c) )=a. Similarly, e^2 (a,b,c)=b. In this way, the covector L\nabove can be written as L=3e^1 -e^2 +4 e^3. We say that L_1 =3 , L_2=-1,\nand L_3=4.\n\nBut, for the dual basis for the f\'s, we want f^1 (f_1)=1, f^1 (f_2)=0,\nand f^1 (f_3)=0. This works out to give f^1(a,b,c)=a-b. Similarly,\nf^2(a,b,c)=b-c and f^3 (a,b,c)=c. Now the covector L can be written as\nL=3f^1 +2 f^2 +6 f^3. Check this! Hence, for this dual basis, L_1 =3,\nL_2=2, and L_3 =6. Same covector, different components.\n\nNow, we can talk about the components of a metric g in a particular basis.\nIn particular, define *for a given basis* g_ij to be g applied the the i^th\nand j^th elements of the basis.\n\nExamples:\n\n1) For the standard metric and the standard basis, we have g_11=g(e_1, e_1)\n=1*1+0*0+0*0=1, similarly, g_12=0, g_13=0, g_22=1, etc.\n\n2) For the standard metric but the non-standard basis f_1, f_2, f_3, we\nhave g_22=g(f_2,f_2)=1*1+1*1+0*0=2. Similarly, g_33=3 and g_12=1.\n\nSo we have to be very clear which basis we are working in before we talk\nabout coordinates. Otherwise everything gets nasty.\n\nFinally, lets take the f basis and suppose that v=v^i f_i is the expansion\nof a vector in this basis. Also let g_ij be the components of a metric\nin this basis. The question is: What are the components of the associated\nvector *v in the basis consisting of f^1, f^2, f^3 of the collection of\ncovectors? Well, let\'s write *v=v_1 f^1 +v_2 f^2 +v_3 f^3=v_i f^i.\n\nIt will turn out that v_1 = g_11 * f^1 + g_12 * f^2 + g_13 *f^3 = g_1i f^i.\nFurthermore, for the other components, a similar expression holds.\n\nHence we write v_i=g_ij v^j in general.\n\nEven more, this expression works for *any* basis you start from as long\nas the vector v is written in that basis, the metric components are\nfor that basis and the components of *v are in the *dual* basis.\n\nNow, many people confuse the vector v with its components, v^i and even\nconfuse the covector *v with the original v and with the components\nv^i or v_i. But the vector and covector are the things related and will\nbe the geometrical things when a manifold is in the picture. The rest is\nmessy notation that depends on a choice of basis.\n\nI know this is long, but I hope it has helped. If you have any more questions,\nfeel free to ask. I can write more ;)\n\n--Dan Grubb\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz <oz@farmeroz.port995.com> wrote in message news:<gWrndZAAkFpBFwiK@farmeroz.port995.com>...
> grubb@math.niu.edu <grubb@math.niu.edu> writes
> >Oz <oz@farmeroz.port995.com> wrote in message
>
> >> >
> >> > <t,u> = g(t,u) = g^{ij} t_i u_j
> >>
> >> Is it valid to say g^{ij} t_i -> t^j ?
> >
> >Not quite. Given a vector with components t_i, you get a co-vector
> >with components t^i where t^j=g^{ij} t_i.
>
> OK. Now this is another area where I don't know what I am talking about.
>
> You have identified three objects,
> t^i, t^j, t_i
>
> Clearly they are not the same.
The first two are the same, just different dummy indices.
>
> But how precisely do they differ?
>
> I would say that t^i t_i -> scalar
> One presumes that the indices refer to the same basis (i).
>
> I therefore presume that t^j is in another basis so that t^j t_i is not
> necessarily scalar. Presumably because g^{ij} s bent, and so the covector-
> vector transformation isn't in some sense orthogonal (not quite what I
> mean).
>
> Obviously I need a bit of a discussion about these important details.
Perhaps it's better to start over. I first want to talk about vectors
and covectors when there isn't a minifold in the background. Also, I'll
ditch the xyz notation and number the components. That's more standard
anyway.
A vector space is simply a collection of objects that we can add and
multiply by scalars. The objects in the vector space are called vectors
(not too surprising yet, I hope). We'll stick to finite dimensional
vector spaces for the moment to avoid some pretty nasty stuff. If you're
interested, I'll go there, but we'll have enough on our plate as it is, OK?
Examples of vector spaces:
1) R^3 This is the collection of ordered triples (a,b,c) where a,b, and c are
real numbers. This can be generalized in two (main) ways, first to ordered
n-tuples to get R^n, and also to ordered n-tuples of complex numbers to
get C^n.
2) The collection of solutions to some homogeneous linear differential
equation, say y''+y=0 or y'''+3xy''-2y'+y=0.
3) The collection P^5 of polynomials of degree <=5 (or any n)
A covector is simply a linear functional, that is to say, a linear
map from the vector space to the scalars. Example:
1) The map that takes (a,b,c) to the scalar 3a-b+4c is a covector for the
vector space R^3.
2) The map that takes a polynomial p(x) and gives the integral of p(x)
from to 2 is a covector for the space of polynomials P^5.
I will use the letters L, S, T for covectors to minimize confusion.
I will use v,w for vectors. Hence an expression such as L(v) makes sense
and is a scalar. It is common to write <L,v> for L(v), but
I will refrain for now.
For example 1), v=(a,b,c) and L(v)=3a-b+4c.
A metric is a bilinear map that takes pairs of vectors and gives scalars
out. Hence, for vectors v and w, the expression g(v,w) makes sense and is
a scalar. Furthermore g(v_1 +v_2 ,w)=g(v_1 ,w)+g(v_2 ,w). We also require
symmetry so that g(v,w)=g(w,v). Finally, if we fix w and g(v,w)=0 for every v,
we require that w must be the vector. OK?
For example,
1) If v=(a,b,c) and w=(x,y,z), let g(v,w)=ax+by+cz. This gives a metric
on R^3. THis is called the standard metric of R^3.
2) If v=(a,b,c) and w=(x,y,z), let g(v,w)=ax+2by+3cz. This gives a different
metric on R^3.
3) If p and q are polynomials, let g(p,q)=\int_0^1 p(x)q(x) dx. This is a metric
on P^5.
It is common to write <v,w> in place of g(v,w), but I will refrain for
the time being.
Now, suppose we take a vector w and look at the function that takes
a vector v to the scalar g(v,w). This is a linear map, and so it is
a covector! We call this covector *w for convenience. The remarkable thing
is that *every* covector can be obtained in this way! So if you are given
a covector L, there is a vector w so that L=*w. This is the correspondence
that metrics give between vectors and covectors.
Next, I want to point out that the collection covectors has the property
that things can be added and multiplied by scalars, so the collection of
covectors is a vector space in its own right. Furthermore, if you are
given a vector v, the map that takes L to L(v) is a co-covector. Another
miracle of the subject is that every co-covector comes in this way. Also,
no metric is needed to set up this correspondence, although on is needed
for that between vectors and covectors above.
Next, a basis is a collection of vectors where everything in the vector
space can be written in a unique way as a linear combination of those
basis elements. The coefficients used in the linear combination are
called the *components* of the vector in that basis. Notice that the
components are scalars.
Examples:
1) The collection {(1,0,0), (0,1,0), (0,0,1)} is a basis of R^3. This
is called the standard basis. For convenience, write e_1=(1,0,0)e_2=(0,1,0), and e_3=(0,0,1). Given a vector v=(a,b,c), we can write
v=a e_1 +b e_2 +c e_3. Here, the components are a, b, and c. We write
v^1=a, v^2 =b, and v^3 =c, so v=v^1 e_1 +v^2 e_2 +v^3 e_3 = v^i e_i
where in the last expression I have used the summation notation. This
can also be written as v=v^j e_j since the i and j are dummy variables
for the summation.
2) The collection {1, x, x^2, x^3, x^4, x^5} is a basis of P^5. Here,
the components are simply the coefficients of the polynomial.
3) The collection {(1,0,0), (1,1,0), (1,1,1)} and another basis for R^3.
here, the components of a vector v=(a,b,c) will be a-b, b-c, and c in that
order, so *in this basis*, v^1=a-b, v^2=b-c, and v^3=c. For later
convenience, let f_1=(1,0,0), f_2=(1,1,0) and f_3=(1,1,1).
Next, for any basis for the collection of vectors, there is a dual
basis for the covectors. This is usually written with superscripts.
Thus, for the standard basis of R^3, the covector e^1 should satisfy
e^1 (e_1)=1, e^1 (e_2)=0, e^1 (e_3)=0. This works out to give that
e^1 ( (a,b,c) )=a. Similarly, e^2 (a,b,c)=b. In this way, the covector L
above can be written as L=3e^1 -e^2 +4 e^3. We say that L_1 =3 , L_2=-1,
and L_3=4.
But, for the dual basis for the f's, we want f^1 (f_1)=1, f^1 (f_2)=0,
and f^1 (f_3)=0. This works out to give f^1(a,b,c)=a-b. Similarly,
f^2(a,b,c)=b-c and f^3 (a,b,c)=c. Now the covector L can be written as
L=3f^1 +2 f^2 +6 f^3. Check this! Hence, for this dual basis, L_1 =3,L_2=2, and L_3 =6. Same covector, different components.
Now, we can talk about the components of a metric g in a particular basis.
In particular, define *for a given basis* g_{ij} to be g applied the the i^{th}
and j^{th} elements of the basis.
Examples:
1) For the standard metric and the standard basis, we have g_{11}=g(e_1, e_1)=1*1+0*0+0*0=1, similarly, g_{12}=0, g_{13}=0, g_{22}=1, etc.
2) For the standard metric but the non-standard basis f_1, f_2, f_3, we
have g_{22}=g(f_2,f_2)=1*1+1*1+0*0=2. Similarly, g_{33}=3 and g_{12}=1.
So we have to be very clear which basis we are working in before we talk
about coordinates. Otherwise everything gets nasty.
Finally, lets take the f basis and suppose that v=v^i f_i is the expansion
of a vector in this basis. Also let g_{ij} be the components of a metric
in this basis. The question is: What are the components of the associated
vector *v in the basis consisting of f^1, f^2, f^3 of the collection of
covectors? Well, let's write *v=v_1 f^1 +v_2 f^2 +v_3 f^3=v_i f^i.
It will turn out that v_1 = g_{11} * f^1 + g_{12} * f^2 + g_{13} *f^3 = g_{1i} f^i.
Furthermore, for the other components, a similar expression holds.
Hence we write v_i=g_{ij} v^j in general.
Even more, this expression works for *any* basis you start from as long
as the vector v is written in that basis, the metric components are
for that basis and the components of *v are in the *dual* basis.
Now, many people confuse the vector v with its components, v^i and even
confuse the covector *v with the original v and with the components
v^i or v_i. But the vector and covector are the things related and will
be the geometrical things when a manifold is in the picture. The rest is
messy notation that depends on a choice of basis.
I know this is long, but I hope it has helped. If you have any more questions,
feel free to ask. I can write more ;)
--Dan Grubb
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Blake Winter <blake.winter@houghton.edu> writes\n\n>Yes, I feel the same way about the horrors of matrix notation, what\n>with the row vectors and column vectors hiding what\'s actually going\n>on...\n\nIndeed, but at some point one has to (well, someone does) work out a\nreal example. This has a habit of being messy and I guess a matrix\nnotation for this final step is as good as any.\n\nOTOH perhaps you would claim that architects design buildings but\nbuilders build them, and mathematicians shouldn\'t be getting their hands\ndirty with manual labour. I guess that leaves it for the physicists....\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Blake Winter <blake.winter@houghton.edu> writes
>Yes, I feel the same way about the horrors of matrix notation, what
>with the row vectors and column vectors hiding what's actually going
>on...
Indeed, but at some point one has to (well, someone does) work out a
real example. This has a habit of being messy and I guess a matrix
notation for this final step is as good as any.
OTOH perhaps you would claim that architects design buildings but
builders build them, and mathematicians shouldn't be getting their hands
dirty with manual labour. I guess that leaves it for the physicists....
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
Van www
Nov27-04, 06:40 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Greg Egan wrote:\n> In another thread, I sketched a definition of the Hodge dual of p-forms,\n> A notation should always reflect the underlying structure and should not\n> be \'interpretable\'. Its one reason why I am drawn to the index notation,\n> despite its obvious inelegance. The structure is clear and, at least at\n> the elementary level, unambiguous. In fact this is one of the reasons\n> why I am not a lover of cross and dot products. These seem to me (and\n> have always seemed to me) to be confusing and concealing an underlying\n> symmetry that would be better to be expressed.\n\n\nI think this is a mistake. Indices are artifacts of coordinate systems\n(CS) and simply tell one the order of the tensor. (2 indices for a 2nd\nrank tensor, contra-variant etc.) They confuse and conceal the real\nstructure and make it difficult to see what is really going on or being\nsaid. They are occasionally useful for calculations in particular CS,\nbut even that is best done using coordinate free notation. The main\nthing is that tensors and forms represent geometric and physical\nquantities, independent of any CS. This is made obvious by using\ncoordinate free notation. We can always introduce the CS of some\nobserver and calculate what he sees in his coordinate system.\n\nAs for covariant vs contra-variant, the metric always gives us an\nisomorphism between the tangent and cotangent space.\n\n> Its the same thing that made me feel Maxwell\'s equations were missing a\n> symmetry at a profound level which only vanished when I was introduced\n> to the (3+1)D EM method using the faraday tensor, which is something I\n> would like to follow up, if only anyone was remotely interested.\n\nI would be glad to follow up on this subject.\n\nThe EM field is an excellent example of the importance of using 4D\nnotation. There is no special time, just as there is no special\norientation of the axes in 3D space. Lorentz invariance is just as\nimportant as rotational invariance in physics. One should always refer\nto the EM 2-form F and to 4 vectors like the 4-velocity u for particles\nor the matter current 4-vector J = nu for fluids. Continuing to use the\n3+1 form in mechanics is a big mistake, IMO.\n\nI learned by studying Misner, Thorne, and Wheeler\'s (MTW) "Gravitation",\nalong with Lightmann et. al. "Problem Book in Relativity and\nGravitation". I\'m sure there are many other texts like this available\nnow.\n\nI have found the material in these texts to be a foundation for all\nphysics, as all physics is relativistic, its long past time to let go of\nNewtonian mechanics and the 3+1 form of EM. I want to scream everytime I\nsee MHD or plasma done with "Galilean invariant" eqns., rather than the\ncorrect relativistic eqns. This leads to all kinds of errors, and its\nnot OK to say "we will only consider the case of velocity << speed of\nlight". An example I gave before was the Alfven speed v_a = B/sqrt(n) is\nthe Galilean result, while v_a = B/sqrt(nf+B^2) relativistically. Look\nat what happens as the density n --> 0, which is unrelated to the speed\nv of the flow. There are many such examples. > </rant>========= > > It\noccurs to me that the metric is implicitly a spatial expression.\n\nNo. Its a spacetime metric.\n\n> At a point the metric is always flat (away from a singularity).\n\nAll manifolds are locally like R^n, nothing to do with the metric.\n\n> So what the metric is really giving us is the relationship of one point\n> with \'adjacent\' points. It is, in some very real sense, a volume term.\n> Hmm, that\'s not quite right, is it? It must be, for an N-D space,\n> an (N-1)-D term.\n\n\nI don\'t get this last sentence.\nMTW and texts on differential geometry have something to say\nabout this. The spacetime metric tells us about the spacetime\ninterval between points.\n\nVan\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Greg Egan wrote:
> In another thread, I sketched a definition of the Hodge dual of p-forms,
> A notation should always reflect the underlying structure and should not
> be 'interpretable'. Its one reason why I am drawn to the index notation,
> despite its obvious inelegance. The structure is clear and, at least at
> the elementary level, unambiguous. In fact this is one of the reasons
> why I am not a lover of cross and dot products. These seem to me (and
> have always seemed to me) to be confusing and concealing an underlying
> symmetry that would be better to be expressed.
I think this is a mistake. Indices are artifacts of coordinate systems
(CS) and simply tell one the order of the tensor. (2 indices for a 2nd
rank tensor, contra-variant etc.) They confuse and conceal the real
structure and make it difficult to see what is really going on or being
said. They are occasionally useful for calculations in particular CS,
but even that is best done using coordinate free notation. The main
thing is that tensors and forms represent geometric and physical
quantities, independent of any CS. This is made obvious by using
coordinate free notation. We can always introduce the CS of some
observer and calculate what he sees in his coordinate system.
As for covariant vs contra-variant, the metric always gives us an
isomorphism between the tangent and cotangent space.
> Its the same thing that made me feel Maxwell's equations were missing a
> symmetry at a profound level which only vanished when I was introduced
> to the (3+1)D EM method using the faraday tensor, which is something I
> would like to follow up, if only anyone was remotely interested.
I would be glad to follow up on this subject.
The EM field is an excellent example of the importance of using 4D
notation. There is no special time, just as there is no special
orientation of the axes in 3D space. Lorentz invariance is just as
important as rotational invariance in physics. One should always refer
to the EM 2-form F and to 4 vectors like the 4-velocity u for particles
or the matter current 4-vector J = \nu for fluids. Continuing to use the
3+1 form in mechanics is a big mistake, IMO.
I learned by studying Misner, Thorne, and Wheeler's (MTW) "Gravitation",
along with Lightmann et. al. "Problem Book in Relativity and
Gravitation". I'm sure there are many other texts like this available
now.
I have found the material in these texts to be a foundation for all
physics, as all physics is relativistic, its long past time to let go of
Newtonian mechanics and the 3+1 form of EM. I want to scream everytime I
see MHD or plasma done with "Galilean invariant" eqns., rather than the
correct relativistic eqns. This leads to all kinds of errors, and its
not OK to say "we will only consider the case of velocity << speed of
light". An example I gave before was the Alfven speed v_a = B/\sqrt(n) is
the Galilean result, while v_a = B/\sqrt(nf+B^2) relativistically. Look
at what happens as the density n --> 0, which is unrelated to the speed
v of the flow. There are many such examples. > </rant>========= > > It
occurs to me that the metric is implicitly a spatial expression.
No. Its a spacetime metric.
> At a point the metric is always flat (away from a singularity).
All manifolds are locally like R^n, nothing to do with the metric.
> So what the metric is really giving us is the relationship of one point
> with 'adjacent' points. It is, in some very real sense, a volume term.
> Hmm, that's not quite right, is it? It must be, for an N-D space,
> an (N-1)-D term.
I don't get this last sentence.
MTW and texts on differential geometry have something to say
about this. The spacetime metric tells us about the spacetime
interval between points.
Van
Van www
Nov27-04, 12:52 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>building is, in fact,\n* occupied by Telecom. This is the part that contains the networking\n* junctions for the optical-fiber lines leased by the banks for their\n* "Electronic Funds Transfer System" (EFTS). ALL financial transactions\n* for the banks pass through there via subsidiary company, "Funds\n* Transfers Services Pty Ltd." (FTS)\n\n\n* The New York Times INTERNATIONAL Wednesday, May 21, 1997\n* by Clyde H. Farnsworth, Woomera, Australia.\n*\n* As Darth Vader\'s Death Star is blown to bits in the newly remastered "Star\n* Wars" at the local theater, the audience of Australian and American Air\n* Force personnel, and squadrons of their children, lets out a whoop. As the\n* lights go on, everyone is beaming.\n* [snip]\n*\n* Pine Gap employs nearly 1,000 people, mainly from the CIA and the U.S.\n* National Reconnaissance Office.\n*\n* It is the ground station for a U.S. satellite network that intercepts\n* telephone, radio, data links and other communications around the world.\n\nWorldwide telephone interception.\n\n\n******************************* ***********************************************\n\ nNew Zealand: Unhappy Campers\n--- ------- ------- -------\n\nYou\'d be\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>building is, in fact,
* occupied by Telecom. This is the part that contains the networking
* junctions for the optical-fiber lines leased by the banks for their
* "Electronic Funds Transfer System" (EFTS). ALL financial transactions
* for the banks pass through there via subsidiary company, "Funds
* Transfers Services Pty Ltd." (FTS)
* The New York Times INTERNATIONAL Wednesday, May 21, 1997
* by Clyde H. Farnsworth, Woomera, Australia.
*
* As Darth Vader's Death Star is blown to bits in the newly remastered "Star
* Wars" at the local theater, the audience of Australian and American Air
* Force personnel, and squadrons of their children, lets out a whoop. As the
* lights go on, everyone is beaming.
* [snip]
*
* Pine Gap employs nearly 1,000 people, mainly from the CIA and the U.S.
* National Reconnaissance Office.
*
* It is the ground station for a U.S. satellite network that intercepts
* telephone, radio, data links and other communications around the world.
Worldwide telephone interception.
************************************************** ****************************
New Zealand: Unhappy Campers
--- ------- ------- -------
You'd be
Van www
Nov27-04, 03:41 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>it is too late.\n\no 5 SLA people burned to death from a tear gas grenade.\n\no An armed wholed-up person burned to death in San Diego on May 2, 1997.\nThe police fired CS gas into the house. According to the Union Tribune\nnewspaper: as the CS gas filled the house it ignited, and caused the\nentire structure and surrounding trees to be engulfed in flames.\n\no A member of a violent militia group called \'The Posse\' was armed and\nwholed up: they simply *purposely* set the house on fire. Burned\nhim to death.\n\no Philadelphia police purposely drop an incindiary device on the top\nof a building housing an armed and holed-up African-American group\n(men, woman & children) called \'Move\'. They burned down the entire\nneighborhood of 62 homes. 9/28/96 NYT: 1.5 million dollars was awarded\nto survivors. Eleven men, women and children died in the fire purposely\nset ("a satchel of explosives") by Police Commissioner Gregore Sambor\nand Fire Chief William Richmond to open a hole in the building for tear\ngas delivery. But Pennsylvania state law was ruled to grant them\npersonal immunity from Federal civil rights charges because they were\nstate employees [what???]. The incident occurred in May 1985.\n\no Waco. CS tear gas attack by the FBI using Army tanks.\n\nThe government, across the decades, keeps managing to burn people to death,\nrather than bringing them to trial.\n\nOften, tear gas is involved.\n\n----\n\nRandy Weaver at Ruby Ridge.\n\nPersisting, a BATF informant persuaded Weaver, a DECORATED GREEN BERET\nVETERAN of Vietnam with NO CRIMINAL RECORD, to sell him two shotguns\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>it is too late.
o 5 SLA people burned to death from a tear gas grenade.
o An armed wholed-up person burned to death in San Diego on May 2, 1997.
The police fired CS gas into the house. According to the Union Tribune
newspaper: as the CS gas filled the house it ignited, and caused the
entire structure and surrounding trees to be engulfed in flames.
o A member of a violent militia group called 'The Posse' was armed and
wholed up: they simply *purposely* set the house on fire. Burned
him to death.
o Philadelphia police purposely drop an incindiary device on the top
of a building housing an armed and holed-up African-American group
(men, woman & children) called 'Move'. They burned down the entire
neighborhood of 62 homes. 9/28/96 NYT: 1.5 million dollars was awarded
to survivors. Eleven men, women and children died in the fire purposely
set ("a satchel of explosives") by Police Commissioner Gregore Sambor
and Fire Chief William Richmond to open a hole in the building for tear
gas delivery. But Pennsylvania state law was ruled to grant them
personal immunity from Federal civil rights charges because they were
state employees [what???]. The incident occurred in May 1985.
o Waco. CS tear gas attack by the FBI using Army tanks.
The government, across the decades, keeps managing to burn people to death,
rather than bringing them to trial.
Often, tear gas is involved.
----
Randy Weaver at Ruby Ridge.
Persisting, a BATF informant persuaded Weaver, a DECORATED GREEN BERET
VETERAN of Vietnam with NO CRIMINAL RECORD, to sell him two shotguns
Van www
Nov27-04, 03:45 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"\n*\n* That it was not, and never had been, simply a "telephone exchange" finally\n* came to light in the 1975 JPAC Approval Report, when it admitted that the\n* existing building had a comprehensive basement which housed NASA\'s micro-\n* wave communications headquarters in Australia. Part of the justification\n* of the "need" for the new, much larger building, was that by 1980, it was\n* expected that NASA would run out of room in their existing home.\n*\n* Apart from NASA, it is now admitted that Deakin houses the National\n* Computer Headquarters for, amongst others, the Australian Defense\n* Department, the Australian Taxation Office, the Department of Social\n* Security, the Commonwealth Department of Education, and the Department\n* of Transport and Communications.\n*\n* Both Tax and Social Security are, in turn, directly linked to Medicare.\n* In fact, the Department of Health used Social Security\'s computer\n* facilities there until their own were completed.\n*\n* A small, but highly significant, part of the building is, in fact,\n* occupied by Telecom.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"
*
* That it was not, and never had been, simply a "telephone exchange" finally
* came to light in the 1975 JPAC Approval Report, when it admitted that the
* existing building had a comprehensive basement which housed NASA's micro-
* wave communications headquarters in Australia. Part of the justification
* of the "need" for the new, much larger building, was that by 1980, it was
* expected that NASA would run out of room in their existing home.
*
* Apart from NASA, it is now admitted that Deakin houses the National
* Computer Headquarters for, amongst others, the Australian Defense
* Department, the Australian Taxation Office, the Department of Social
* Security, the Commonwealth Department of Education, and the Department
* of Transport and Communications.
*
* Both Tax and Social Security are, in turn, directly linked to Medicare.
* In fact, the Department of Health used Social Security's computer
* facilities there until their own were completed.
*
* A small, but highly significant, part of the building is, in fact,
* occupied by Telecom.
Van www
Nov27-04, 04:36 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>-SPAN Congressional Television\n*\n* Mark Klaas, father of 12-year-old Polly Klaas, who was murdered by a\n* repeat-offender that was paroled from prison, said in support of prevention\n* programs: "Building more prisons to fight crime is like building more\n* cemeteries to fight the spread of AIDS. It\'s a bad quick fix. Police\n* chiefs across the country support [me] this 4 to 1. Unfortunately, Congress\n* can\'t act "soft" on crime, and is about to pass a very bad bill on\n* juvenile crime."\n\nMore bizarre distortions in our social fabric due to Zero Tolerance:\n\n6/10/97 MSNBC: California: a ten-year-old girl who reported a classmate\nfor having a joint was also suspended by the principal, under the school\'s\nZero Tolerance for drugs policy. Her offense: handling the joint to see if\nthe other student was kidding her before reporting the other student. The\nprincipal said "too bad, that\'s what \'Zero Tolerance\' means". The little\ngirl and her mother are shocked. [I am not making these up!!!!!]\n\n6/18/97 NBC News Channel 4 NYC: A career tea\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>-SPAN Congressional Television
*
* Mark Klaas, father of 12-year-old Polly Klaas, who was murdered by a
* repeat-offender that was paroled from prison, said in support of prevention
* programs: "Building more prisons to fight crime is like building more
* cemeteries to fight the spread of AIDS. It's a bad quick fix. Police
* chiefs across the country support [me] this 4 to 1. Unfortunately, Congress
* can't act "soft" on crime, and is about to pass a very bad bill on
* juvenile crime."
More bizarre distortions in our social fabric due to Zero Tolerance:
6/10/97 MSNBC: California: a ten-year-old girl who reported a classmate
for having a joint was also suspended by the principal, under the school's
Zero Tolerance for drugs policy. Her offense: handling the joint to see if
the other student was kidding her before reporting the other student. The
principal said "too bad, that's what 'Zero Tolerance' means". The little
girl and her mother are shocked. [I am not making these up!!!!!]
6/18/97 NBC News Channel 4 NYC: A career tea
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Van www <vanjac12@yahoo.com> writes\n>Greg Egan wrote:\n\n****IMPORTANT: Greg would never write stuff like the below****\n**** it was all written by Oz.****\n\n>> In another thread, I sketched a definition of the Hodge dual of p-forms,\n>> A notation should always reflect the underlying structure and should not\n>> be \'interpretable\'. Its one reason why I am drawn to the index notation,\n>> despite its obvious inelegance. The structure is clear and, at least at\n>> the elementary level, unambiguous. In fact this is one of the reasons\n>> why I am not a lover of cross and dot products. These seem to me (and\n>> have always seemed to me) to be confusing and concealing an underlying\n>> symmetry that would be better to be expressed.\n>\n>\n>I think this is a mistake. Indices are artifacts of coordinate systems\n>(CS) and simply tell one the order of the tensor. (2 indices for a 2nd\n>rank tensor, contra-variant etc.)\n\nI agree. However at my level its important for me to be aware, very\naware, of the order of the tensor because I am very naive. I actually\nsee them as \'arbitrary co-ordinates\', probably more precisely as\n\'potential co-ordinates\'.\n\n\n>They confuse and conceal the real\n>structure and make it difficult to see what is really going on or being\n>said. They are occasionally useful for calculations in particular CS,\n>but even that is best done using coordinate free notation. The main\n>thing is that tensors and forms represent geometric and physical\n>quantities, independent of any CS. This is made obvious by using\n>coordinate free notation. We can always introduce the CS of some\n>observer and calculate what he sees in his coordinate system.\n\nYes. I agree, but showing the order is a crutch that I need right now.\n\n>As for covariant vs contra-variant, the metric always gives us an\n>isomorphism between the tangent and cotangent space.\n\nYes. However greg has pointed out that a metric may not exist.\nI find that including the metric at every possible opportunity clarifies\nwhat is actually going on. In a way its an assumption I don\'t want to\n(effectively) have set = 1.\n\n>> Its the same thing that made me feel Maxwell\'s equations were missing a\n>> symmetry at a profound level which only vanished when I was introduced\n>> to the (3+1)D EM method using the faraday tensor, which is something I\n>> would like to follow up, if only anyone was remotely interested.\n>\n>I would be glad to follow up on this subject.\n\nOh good.\nI will post a second reply to keep the threads tidy.\n\n>The EM field is an excellent example of the importance of using 4D\n>notation. There is no special time, just as there is no special\n>orientation of the axes in 3D space. Lorentz invariance is just as\n>important as rotational invariance in physics. One should always refer\n>to the EM 2-form F and to 4 vectors like the 4-velocity u for particles\n>or the matter current 4-vector J = nu for fluids. Continuing to use the\n>3+1 form in mechanics is a big mistake, IMO.\n\nWhen I said (3+1)D I meant just that.\nThat is, not separating out time and handling it separately.\n\n>No. Its a spacetime metric.\n>\n>> At a point the metric is always flat (away from a singularity).\n>\n>All manifolds are locally like R^n, nothing to do with the metric.\n\nYes. I have been taken to task on that already. I agree.\n\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Van www <vanjac12@yahoo.com> writes
>Greg Egan wrote:
****IMPORTANT: Greg would never write stuff like the below****
**** it was all written by Oz.****
>> In another thread, I sketched a definition of the Hodge dual of p-forms,
>> A notation should always reflect the underlying structure and should not
>> be 'interpretable'. Its one reason why I am drawn to the index notation,
>> despite its obvious inelegance. The structure is clear and, at least at
>> the elementary level, unambiguous. In fact this is one of the reasons
>> why I am not a lover of cross and dot products. These seem to me (and
>> have always seemed to me) to be confusing and concealing an underlying
>> symmetry that would be better to be expressed.
>
>
>I think this is a mistake. Indices are artifacts of coordinate systems
>(CS) and simply tell one the order of the tensor. (2 indices for a 2nd
>rank tensor, contra-variant etc.)
I agree. However at my level its important for me to be aware, very
aware, of the order of the tensor because I am very naive. I actually
see them as 'arbitrary co-ordinates', probably more precisely as
'potential co-ordinates'.
>They confuse and conceal the real
>structure and make it difficult to see what is really going on or being
>said. They are occasionally useful for calculations in particular CS,
>but even that is best done using coordinate free notation. The main
>thing is that tensors and forms represent geometric and physical
>quantities, independent of any CS. This is made obvious by using
>coordinate free notation. We can always introduce the CS of some
>observer and calculate what he sees in his coordinate system.
Yes. I agree, but showing the order is a crutch that I need right now.
>As for covariant vs contra-variant, the metric always gives us an
>isomorphism between the tangent and cotangent space.
Yes. However greg has pointed out that a metric may not exist.
I find that including the metric at every possible opportunity clarifies
what is actually going on. In a way its an assumption I don't want to
(effectively) have set = 1.
>> Its the same thing that made me feel Maxwell's equations were missing a
>> symmetry at a profound level which only vanished when I was introduced
>> to the (3+1)D EM method using the faraday tensor, which is something I
>> would like to follow up, if only anyone was remotely interested.
>
>I would be glad to follow up on this subject.
Oh good.
I will post a second reply to keep the threads tidy.
>The EM field is an excellent example of the importance of using 4D
>notation. There is no special time, just as there is no special
>orientation of the axes in 3D space. Lorentz invariance is just as
>important as rotational invariance in physics. One should always refer
>to the EM 2-form F and to 4 vectors like the 4-velocity u for particles
>or the matter current 4-vector J = \nu for fluids. Continuing to use the
>3+1 form in mechanics is a big mistake, IMO.
When I said (3+1)D I meant just that.
That is, not separating out time and handling it separately.
>No. Its a spacetime metric.
>
>> At a point the metric is always flat (away from a singularity).
>
>All manifolds are locally like R^n, nothing to do with the metric.
Yes. I have been taken to task on that already. I agree.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Van www <vanjac12@yahoo.com> writes\n\n>>Oz:\n>> Its the same thing that made me feel Maxwell\'s equations were missing a\n>> symmetry at a profound level which only vanished when I was introduced\n>> to the (3+1)D EM method using the faraday tensor, which is something I\n>> would like to follow up, if only anyone was remotely interested.\n>\n>I would be glad to follow up on this subject.\n\nGood.\n\nNow the first thing to remember is who I am.\n\nI have been on sci.physics and s.p.research for quite a long time, where\nI seem to be regarded a bit like a pet dog who is grateful for the\ncrumbs of knowledge dropped off the table by his masters. Many decades\nago I did physics at (UK) uni level, but can generally be considered to\nbe mathematically a complete moron, but with some grasp of physics.\n\nSo I\'m happy to take math explanations, but if you want me to follow,\nbest to give some explanations of what the maths is saying.\n\nOK. Now the faraday tensor.\n\nIts been a while since I tried to get some interest on this, so this is\nfrom memory. Feel free to correct my notation, I\'m used to being told\noff. Forgive my use of a matrix notation, at this stage I want to see\nthe nuts and bolts.\n\nI start with some electric field E. (indices suppressed)\n\n[0 -E -E -E] = F^ij\n[E 0 0 0]\n[E 0 0 0]\n[E 0 0 0]\n\nI can switch this to a moving electric (and magnetic) field using the\nlorentz transform L^i_j\n\nF^jk = L^i_k F^ij\n\nI think its true that the field around any charge can be converted to a\nsimple purely electric field by an appropriate choice of rest frame. For\nme, this is a good way to see a magnetic field. Everything all stitches\nneatly together since all magnetic fields derive from moving electric\ncharges.\n\nNow obviously the very first thing I want to do is to examine the force\non a charge adjacent to a current carrying wire.\n\nIt seems to me that I should start with the static electric field\nassociated with the charges that are involved. Now each of these has a\ndifferent rest frame. I will be (seriously) simplifying.\n\nI have the following different frames:\n\n1) The moving test charge\n2) The positive charges in the conductor (taken as the observer frame).\n3) The moving electrons in the conductor.\n\nWhat I want to find is the frame where the moving test charge sees no\nforce.\n\nI concluded that this frame is one half of the drift velocity in the\nconductor. I was told off by all and sundry, but NOBODY pointed out\nwhere my (undoubtedly naive) calculations were erroneous.\n\nAlthough I rashly attempted to show it by a series of matrix\ncalculations, it seems to me to be an unavoidable conclusion on symmetry\ngrounds.\n\nYour comments would be welcome.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Van www <vanjac12@yahoo.com> writes
>>Oz:
>> Its the same thing that made me feel Maxwell's equations were missing a
>> symmetry at a profound level which only vanished when I was introduced
>> to the (3+1)D EM method using the faraday tensor, which is something I
>> would like to follow up, if only anyone was remotely interested.
>
>I would be glad to follow up on this subject.
Good.
Now the first thing to remember is who I am.
I have been on sci.physics and s.p.research for quite a long time, where
I seem to be regarded a bit like a pet dog who is grateful for the
crumbs of knowledge dropped off the table by his masters. Many decades
ago I did physics at (UK) uni level, but can generally be considered to
be mathematically a complete moron, but with some grasp of physics.
So I'm happy to take math explanations, but if you want me to follow,
best to give some explanations of what the maths is saying.
OK. Now the faraday tensor.
Its been a while since I tried to get some interest on this, so this is
from memory. Feel free to correct my notation, I'm used to being told
off. Forgive my use of a matrix notation, at this stage I want to see
the nuts and bolts.
I start with some electric field E. (indices suppressed)
[0 -E -E -E] = F^{ij}[/itex]
[E 0]
[E 0]
[E 0]
I can switch this to a moving electric (and magnetic) field using the
lorentz transform [itex]L^{i_j}F^{jk} = L^{i_k} F^{ij}
I think its true that the field around any charge can be converted to a
simple purely electric field by an appropriate choice of rest frame. For
me, this is a good way to see a magnetic field. Everything all stitches
neatly together since all magnetic fields derive from moving electric
charges.
Now obviously the very first thing I want to do is to examine the force
on a charge adjacent to a current carrying wire.
It seems to me that I should start with the static electric field
associated with the charges that are involved. Now each of these has a
different rest frame. I will be (seriously) simplifying.
I have the following different frames:
1) The moving test charge
2) The positive charges in the conductor (taken as the observer frame).
3) The moving electrons in the conductor.
What I want to find is the frame where the moving test charge sees no
force.
I concluded that this frame is one half of the drift velocity in the
conductor. I was told off by all and sundry, but NOBODY pointed out
where my (undoubtedly naive) calculations were erroneous.
Although I rashly attempted to show it by a series of matrix
calculations, it seems to me to be an unavoidable conclusion on symmetry
grounds.
Your comments would be welcome.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Oz wrote:\n> Van www <vanjac12@yahoo.com> writes\n>\n> >>Oz:\n> >> Its the same thing that made me feel Maxwell\'s equations were\nmissing a\n> >> symmetry at a profound level which only vanished when I was\nintroduced\n> >> to the (3+1)D EM method using the faraday tensor, which is\nsomething I\n> >> would like to follow up, if only anyone was remotely interested.\n> >\n> >I would be glad to follow up on this subject.\n>\n> Good.\n>\n> Now the first thing to remember is who I am.\n>\n> I have been on sci.physics and s.p.research for quite a long time,\nwhere\n> I seem to be regarded a bit like a pet dog who is grateful for the\n> crumbs of knowledge dropped off the table by his masters. Many\ndecades\n> ago I did physics at (UK) uni level, but can generally be considered\nto\n> be mathematically a complete moron, but with some grasp of physics.\n>\n> So I\'m happy to take math explanations, but if you want me to follow,\n> best to give some explanations of what the maths is saying.\n>\n> OK. Now the faraday tensor.\n>\n> Its been a while since I tried to get some interest on this, so this\nis\n> from memory. Feel free to correct my notation, I\'m used to being told\n> off. Forgive my use of a matrix notation, at this stage I want to see\n> the nuts and bolts.\n>\n> I start with some electric field E. (indices suppressed)\n>\n> [0 -E -E -E] = F^ij\n> [E 0 0 0]\n> [E 0 0 0]\n> [E 0 0 0]\n>\n> I can switch this to a moving electric (and magnetic) field using the\n\n> lorentz transform L^i_j\n>\n> F^jk = L^i_k F^ij\n>\n> I think its true that the field around any charge can be converted to\na\n> simple purely electric field by an appropriate choice of rest frame.\n\nYes, the rest frame of a single point charge. But what if you are in a\nfluid\nwith an EM field, like MHD or plasma, or a region of space where there\nis some arbitrary F.\n\nI have found it useful to deal with F as follows.\nLet u = 4-velocity of some observer or his coordinate system (CS),\nand E = - F(u) and B = F(u) the electric and magnetic fields as\nseen in that CS. Then one can write\n\nF = u /\\ E + *(u /\\ B) ; note the E(u) = 0 = B(u) = *F(u,u) = 0 since\n\nF is antisymmetric, so that E and B are spatial vector fields.\n\nUsing these expressions its easy to calculate the the invariants\nfor the field are\n\n(F|F) = (u /\\ E + *(u /\\ B) | u /\\ E + *(u /\\ B))\n\n= (u /\\ E | u /\\ E) + (*(u /\\ B)|*(u /\\ B)) = u^2 E^2 - u^2 B^2\n\n= B^2 - E^2\n\nNote how straightforward and clear such a calculation is compared\nto one done with components. The other invariant is\n\n(F|*F) = (u /\\ E + *(u /\\ B)|*(u /\\ E) - u /\\ B) = 2 (E|B)\n\nIts useful to make a keep a table of relations like\n\nA /\\ *B = (A|B) *1 ; where the 4D volume element is *1\n= sqrt[-det(g)]* 4D Levi-Civita tensor\n= sqrt[-det(g)] dx^1 /\\ dx^2 /\\dx^3 \\/dx^4\n\nand ** = (-1)^(p+1) on p-forms in spacetime.\n\nI also use a generalized inner product, defined as the adjoint\nof the exterior product (with q >= p)\n\n(|) : L^q(M) x L^p(M) --> L^(q-p)(M) : (A,B) --> (A|B) == A(B)\n\ndefined by (A(B)|C) == A(B,C) == (A|B /\\ C)\n\nwhere A in L^q(M)== q-forms on M, B in L^p(M), C and A(B) in\nL^(q-p)(M).\n\nSo we see that one can eliminate E or B if (E|B) = 0,\nas if it is 0 in one frame it will be 0 and all.\n\nIf E^2 > B^2 and (E|B) = 0, then there is a frame where B = 0.\nIf B^2 > E^2 and (E|B) = 0, then there is a frame where E = 0.\n\nAn example of the first is the frame in which a charged particle\nis at rest.\nAn example of the 2nd case is MHD, where an observer moving with the\nfluid 4-velocity sees a pure B field, with E = 0, and we say that\nB is frozen into the fluid.\n\nVan\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz wrote:
> Van www <vanjac12@yahoo.com> writes
>
> >>Oz:
> >> Its the same thing that made me feel Maxwell's equations were
missing a
> >> symmetry at a profound level which only vanished when I was
introduced
> >> to the (3+1)D EM method using the faraday tensor, which is
something I
> >> would like to follow up, if only anyone was remotely interested.
> >
> >I would be glad to follow up on this subject.
>
> Good.
>
> Now the first thing to remember is who I am.
>
> I have been on sci.physics and s.p.research for quite a long time,
where
> I seem to be regarded a bit like a pet dog who is grateful for the
> crumbs of knowledge dropped off the table by his masters. Many
decades
> ago I did physics at (UK) uni level, but can generally be considered
to
> be mathematically a complete moron, but with some grasp of physics.
>
> So I'm happy to take math explanations, but if you want me to follow,
> best to give some explanations of what the maths is saying.
>
> OK. Now the faraday tensor.
>
> Its been a while since I tried to get some interest on this, so this
is
> from memory. Feel free to correct my notation, I'm used to being told
> off. Forgive my use of a matrix notation, at this stage I want to see
> the nuts and bolts.
>
> I start with some electric field E. (indices suppressed)
>
> [0 -E -E -E] = F^{ij}
> [E 0]
> [E 0]
> [E 0]
>
> I can switch this to a moving electric (and magnetic) field using the
> lorentz transform L^{i_j}
>
> F^{jk} = L^{i_k} F^{ij}
>
> I think its true that the field around any charge can be converted to
a
> simple purely electric field by an appropriate choice of rest frame.
Yes, the rest frame of a single point charge. But what if you are in a
fluid
with an EM field, like MHD or plasma, or a region of space where there
is some arbitrary F.
I have found it useful to deal with F as follows.
Let u = 4-velocity of some observer or his coordinate system (CS),
and E = - F(u) and B = F(u) the electric and magnetic fields as
seen in that CS. Then one can write
F = u /\ E + *(u /\ B) ;[/itex] note the E(u) = = B(u) = *F(u,u) = since
F is antisymmetric, so that E and B are spatial vector fields.
Using these expressions its easy to calculate the the invariants
for the field are
(F|F) = (u /\ E + *(u /\ B) | u /\ E + *(u /\ B))= (u /\ E | u /\ E) + (*(u /\ B)|*(u /\ B)) = u^2 E^2 - u^2 B^2= B^2 - E^2
Note how straightforward and clear such a calculation is compared
to one done with components. The other invariant is
(F|*F) = (u /\ E + *(u /\ B)|*(u /\ E) - u /\ B) = 2 (E|B)
Its useful to make a keep a table of relations like
A /\ *B = (A|B) *1 ; where the 4D volume element is *1
= \sqrt[-det(g)]* 4D Levi-Civita tensor
[itex]= \sqrt[-det(g)] dx^1 /\ dx^2 /\dx^3 \/dx^4
and ** = (-1)^(p+1) on p-forms in spacetime.
I also use a generalized inner product, defined as the adjoint
of the exterior product (with q >= p)(|) : L^q(M) x L^p(M) --> L^(q-p)(M) : (A,B) --> (A|B) == A(B)
defined by (A(B)|C) == A(B,C) == (A|B /\ C)
where A in L^q(M)== q-forms on M, B in L^p(M), C and A(B) in
L^(q-p)(M).
So we see that one can eliminate E or B if (E|B) = 0,
as if it is in one frame it will be and all.
If E^2 > B^2 and (E|B) = 0, then there is a frame where B = .
If B^2 > E^2 and (E|B) = 0, then there is a frame where E = .
An example of the first is the frame in which a charged particle
is at rest.
An example of the 2nd case is MHD, where an observer moving with the
fluid 4-velocity sees a pure B field, with E = 0, and we say that
B is frozen into the fluid.
Van
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Van www <vanjac12@yahoo.com> writes\n\n>Yes, the rest frame of a single point charge. But what if you are in a\n>fluid\n>with an EM field, like MHD or plasma, or a region of space where there\n>is some arbitrary F.\n\nWhat is an HMD?\n\n[Moderator\'s note: MHD (see stuff about dyslexia below) is\nmagnetohydrodynamics. -P.H.]\n\nI can see that in a plasma, with +ve heading one way and -ve heading the\nother way that you can find a frame with pure B. My only problem with\nthis (if this is the setup you had in mind) is that there should be an\nimpressed electric field to move the charges in that way.\n\nMy aim in looking at F was really twofold. One was to get some intuitive\ngrasp of how things behave in 4D, the other was to get a more\nsymmetrical (and intuitive) overview of maxwell.\n\nI should explain that I (and my kids) are quite dyslexic. This means we\nmake many arithmetic and symbolic errors. In order to get the right\nanswer we really want to know the answer (in general form) before we\nstart. That is gain a good intuitive grasp of the physics.\n\nAnyway, on being introduced to the general idea of F, and (very\nelementary and naive) 4D operations I came to the following conclusions,\nwhich you may wish (I hope) to criticise.\n\n1) We can ignore magnetic monopoles (for now at least).\n\n2) I consider only point charges with no magnetic field.\nIe idealised non-magnetic electrons.\n\n3) In its own rest frame, an electron only feels electric force.\n\n4) To determine the force on a test electron (e) from a distribution of\nother point charges (e_n with field F_n) we can transform each F_n to\nthe rest frame of e, sum the field (F_1 + ... F_n) and apply this to the\ncharge e.\n\n5) Now, I am certainly NOT saying that this is a smart way do solve\nproblems in general (it isn\'t). What it does do, though, is to show that\nmagnetism is just the electric field rotated by SR. If we wanted to be\nmindlessly pedantic we could in theory replace any magnetic field by an\nequivalent set of moving charged particles.\n\n6) I think one implication of this is that the electric (and thus\nmagnetic) force is always centre-to-centre between two charged\nparticles. I imagine this needs some modification if particles are\naccelerating, which is a bit of a bind since a test particle in such a\nfield will see acceleration (by definition, its acted on by a force).\n\n7) I thus am drawn to the image that the magnetic field is more closely\nrelated to the electric field than is normally held. In fact I now see a\nmagnetic field as being the electric field in a relativistic world. Its\nthe self-same thing.\n\nI also asked about the behaviour of a charged point particle in the\nvicinity of a current-carrying conductor. In particular the conditions\nwhere it feels no force. This is a hugely simple example of my image\nabove. Its a good example to see where I have it wrong. I think my\nvisualisation suggests that the condition is with a velocity of half the\ndrift current. However everyone ducks this question, even you. I suspect\nthis is because it illustrates something of a clash between the 4D way\nof doing it and naive maxwell. What do you think, and why?\n\n\n--\nOz\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Van www <vanjac12@yahoo.com> writes
>Yes, the rest frame of a single point charge. But what if you are in a
>fluid
>with an EM field, like MHD or plasma, or a region of space where there
>is some arbitrary F.
What is an HMD?
[Moderator's note: MHD (see stuff about dyslexia below) is
magnetohydrodynamics. -P.H.]
I can see that in a plasma, with +ve heading one way and -ve heading the
other way that you can find a frame with pure B. My only problem with
this (if this is the setup you had in mind) is that there should be an
impressed electric field to move the charges in that way.
My aim in looking at F was really twofold. One was to get some intuitive
grasp of how things behave in 4D, the other was to get a more
symmetrical (and intuitive) overview of maxwell.
I should explain that I (and my kids) are quite dyslexic. This means we
make many arithmetic and symbolic errors. In order to get the right
answer we really want to know the answer (in general form) before we
start. That is gain a good intuitive grasp of the physics.
Anyway, on being introduced to the general idea of F, and (very
elementary and naive) 4D operations I came to the following conclusions,
which you may wish (I hope) to criticise.
1) We can ignore magnetic monopoles (for now at least).
2) I consider only point charges with no magnetic field.
Ie idealised non-magnetic electrons.
3) In its own rest frame, an electron only feels electric force.
4) To determine the force on a test electron (e) from a distribution of
other point charges (e_n with field F_n) we can transform each F_n to
the rest frame of e, sum the field (F_1 + ... F_n) and apply this to the
charge e.
5) Now, I am certainly NOT saying that this is a smart way do solve
problems in general (it isn't). What it does do, though, is to show that
magnetism is just the electric field rotated by SR. If we wanted to be
mindlessly pedantic we could in theory replace any magnetic field by an
equivalent set of moving charged particles.
6) I think one implication of this is that the electric (and thus
magnetic) force is always centre-to-centre between two charged
particles. I imagine this needs some modification if particles are
accelerating, which is a bit of a bind since a test particle in such a
field will see acceleration (by definition, its acted on by a force).
7) I thus am drawn to the image that the magnetic field is more closely
related to the electric field than is normally held. In fact I now see a
magnetic field as being the electric field in a relativistic world. Its
the self-same thing.
I also asked about the behaviour of a charged point particle in the
vicinity of a current-carrying conductor. In particular the conditions
where it feels no force. This is a hugely simple example of my image
above. Its a good example to see where I have it wrong. I think my
visualisation suggests that the condition is with a velocity of half the
drift current. However everyone ducks this question, even you. I suspect
this is because it illustrates something of a clash between the 4D way
of doing it and naive maxwell. What do you think, and why?
--
Oz
Ralph Hartley
Dec11-04, 03:07 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Oz wrote:\n\n> I also asked about the behaviour of a charged point particle in the\n> vicinity of a current-carrying conductor.\n\nIs that a *charged* current-carrying conductor? If not, In what frame is it\nnuetral?\n\nIf the wire has an arbitrary charge density, then clearly there is not\nenough information to answer the question.\n\nIf you say "a current-carrying conductor", I have to assume that it is\nneutral. That is, that there is no electrostatic field arround it in the\n*lab* frame. If you showed be a wire, and stationary charged test particles\nwere atracted or repelled by it, I would rightly say that you had shown me\na charged wire. If I din\'t have a compass, I couldn\'t tell if there were\nany current, but I could still tell that it was charged.\n\n> In particular the conditions\n> where it feels no force. This is a hugely simple example of my image\n> above. Its a good example to see where I have it wrong. I think my\n> visualisation suggests that the condition is with a velocity of half the\n> drift current.\n\nNo. It will feel no force if it is stationary or moving tangentally (i.e.\nnot along the wire, and not towards or away from the wire).\n\nIf it is moving as you describe it will be subject to a radial force. That\nforce can be described in more than one way.\n\nIn the the lab frame, there is no electrostatic field (by definition of a\nneutral wire), but it is moving through a magnetic field. In the particle\'s\nrest frame, it is stationary so the magnetic field exerts no force, but the\nwire no longer appears neutral, there is a radial electrostatic force.\n\nIt doesn\'t matter if the whole wire is moving allong its length. If you\ntell me it isn\'t charged, I\'m not going to expect an electrostatic field.\nIf you don\'t specify a frame in which it is neutral, I\'m going to assume it\nis the obvious frame. If you only specify that it is nuetral in *some*\nframe, then you haven\'t told me anything about the charge; it could be\nanything.\n\nRalph Hartley\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz wrote:
> I also asked about the behaviour of a charged point particle in the
> vicinity of a current-carrying conductor.
Is that a *charged* current-carrying conductor? If not, In what frame is it
nuetral?
If the wire has an arbitrary charge density, then clearly there is not
enough information to answer the question.
If you say "a current-carrying conductor", I have to assume that it is
neutral. That is, that there is no electrostatic field arround it in the
*lab* frame. If you showed be a wire, and stationary charged test particles
were atracted or repelled by it, I would rightly say that you had shown me
a charged wire. If I din't have a compass, I couldn't tell if there were
any current, but I could still tell that it was charged.
> In particular the conditions
> where it feels no force. This is a hugely simple example of my image
> above. Its a good example to see where I have it wrong. I think my
> visualisation suggests that the condition is with a velocity of half the
> drift current.
No. It will feel no force if it is stationary or moving tangentally (i.e.
not along the wire, and not towards or away from the wire).
If it is moving as you describe it will be subject to a radial force. That
force can be described in more than one way.
In the the lab frame, there is no electrostatic field (by definition of a
neutral wire), but it is moving through a magnetic field. In the particle's
rest frame, it is stationary so the magnetic field exerts no force, but the
wire no longer appears neutral, there is a radial electrostatic force.
It doesn't matter if the whole wire is moving allong its length. If you
tell me it isn't charged, I'm not going to expect an electrostatic field.
If you don't specify a frame in which it is neutral, I'm going to assume it
is the obvious frame. If you only specify that it is nuetral in *some*
frame, then you haven't told me anything about the charge; it could be
anything.
Ralph Hartley
Van www
Dec13-04, 09:12 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Ralph Hartley wrote:\n> Oz wrote:\n[snip]\n\nOne must think of an electromagnetic field, which is why F is good\nto use. Think of\n\n1) light waves-- |E| = |B| for light in empty space. There is\nno primality of the electric field here. Also\n\n2) MHD = magnetohydrodynamics. A charged fluid where charges move\neasily--not like a solid--so that electric fields simply move charges\ntill the E field is 0 (a perfect conductor). Then what is left?\nA "frozen in" magnetic field B, which is non-zero even in the\nrest frame of the fluid, where E = 0.\n\nAlso, one can expand any tensor about a vector field u by\n\nT = T(u,u) u u - P(T\'(u)) u - u P(T(u)) + PTP\n\nwhere P = g + u u is the spatial projection operator = diag(0,1,1,1)\nand g = diag(-1,1,1,1) = Minkowski metric, T\' denotes transpose, and\nthe tensor product is implied (u u == u x u = u (tensor prod) u).\n\nSo T(u) = - T(u,u) u + P(T(u)) since P(u) = P\'(u) = g(u) - u = 0,\n\nsince we consider that we use g to raise and lower indices,\ng(u) - u = 0 = g_ij u^i - u_j = 0. (Its useful to do some calculations\nwith indices, and after a while you won\'t need them for most\ncalculations.\n\nAll this is to show that in my post I just decomposed F w.r.t. u to get\n\nF = u/\\E + *(u/\\B) ; E = - F(u), B = *F(u).\n\nThis is very useful for many things. Let u = e_0 = d/dx^0 = d/dt\n\nbe the unit vector along the time direction, then F, *F, and\ndF and d*F, which are 2-forms and 3-forms, can be integrated over\nvarious 2D and 3D subspaces of 4D spacetime to get the integral\nform of Maxwell\'s eqs.\n\nSee, e.g., Misner, Thorne, Wheeler, "Gravitation", which I believe\nanyone who wants to understand physics must study thoroughly.\n\nVan\nover\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Ralph Hartley wrote:
> Oz wrote:
[snip]
One must think of an electromagnetic field, which is why F is good
to use. Think of
1) light waves-- |E| = |B| for light in empty space. There is
no primality of the electric field here. Also
2) MHD = magnetohydrodynamics. A charged fluid where charges move
easily--not like a solid--so that electric fields simply move charges
till the E field is (a perfect conductor). Then what is left?
A "frozen in" magnetic field B, which is non-zero even in the
rest frame of the fluid, where E = .
Also, one can expand any tensor about a vector field u by
T = T(u,u) u u - P(T'(u)) u - u P(T(u)) +[/itex] PTP
where P = g + u u is the spatial projection operator = diag(0,1,1,1)
and g = diag(-1,1,1,1) = Minkowski metric, T' denotes transpose, and
the tensor product is implied (u u == u x u = u (tensor prod) u).
So T(u) = - T(u,u) u + P(T(u)) since P(u) = P'(u) = g(u) - u = 0,
since we consider that we use g to raise and lower indices,
g(u) - u == g_{ij} u^i - u_j = . (Its useful to do some calculations
with indices, and after a while you won't need them for most
calculations.
All this is to show that in my post I just decomposed F w.r.t. u to get
F = u/\E + *(u/\B) ; E = - F(u), B = *F(u).
This is very useful for many things. Let [itex]u = e_0 = d/dx^0 = d/dt
be the unit vector along the time direction, then F, *F, and
dF and d*F, which are 2-forms and 3-forms, can be integrated over
various 2D and 3D subspaces of 4D spacetime to get the integral
form of Maxwell's eqs.
See, e.g., Misner, Thorne, Wheeler, "Gravitation", which I believe
anyone who wants to understand physics must study thoroughly.
Van
over
Van www
Dec13-04, 09:32 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\'s security; but if that could be\ndone it would only bring back the problem of purposelessness. The real\nissue is not whether society provides well or poorly for people\'s\nsecurity; the trouble is that people are dependent on the system for\ntheir security rather than having it in their own hands. This, by the\nway, is part of the reason why some people get worked up about the\nright to bear arms; possession of a gun puts that aspect of their\nsecurity in their own hands.\n\n13. (Paragraph 66) Conservatives\' efforts to decrease the amount of\ngovernment regulation are of little benefit to the average man. For\none thing, only a fraction of the regulations can be eliminated\nbecause most regulations are necessary. For another thing, most of the\nderegulation affects business rather than the average individual, so\nthat its main effect is to take power from the government and give it\nto private corporations. What this means for the average man is that\ngovernment interference in his life is replaced by interference from\nbig corporations, which may be permitted, for example, to dump more\nchemicals that get into his water supply and give him cancer. The\nconservatives are just taking the average man for a sucker, exploiting\nhis resentment of Big Government to promote the power of Big Business.\n\n\n14. (Paragraph 73) When someone approves of the purpose for which\npropaganda is being used in a given case, he generally calls it\n"education" or applies to it some similar euphemism. But propaganda is\npropaganda regardless of the purpose for which it is used.\n\n15. (Paragraph 83) We are not expressing approval or disapproval of\nthe Panama invasion. We only use it to illustrate a point.\n\n16. (Paragraph 95) When the American colonies were under British rule\nthere were fewer and less effective legal guarantees of freedom than\nthere were after\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>'s security; but if that could be
done it would only bring back the problem of purposelessness. The real
issue is not whether society provides well or poorly for people's
security; the trouble is that people are dependent on the system for
their security rather than having it in their own hands. This, by the
way, is part of the reason why some people get worked up about the
right to bear arms; possession of a gun puts that aspect of their
security in their own hands.
13. (Paragraph 66) Conservatives' efforts to decrease the amount of
government regulation are of little benefit to the average man. For
one thing, only a fraction of the regulations can be eliminated
because most regulations are necessary. For another thing, most of the
deregulation affects business rather than the average individual, so
that its main effect is to take power from the government and give it
to private corporations. What this means for the average man is that
government interference in his life is replaced by interference from
big corporations, which may be permitted, for example, to dump more
chemicals that get into his water supply and give him cancer. The
conservatives are just taking the average man for a sucker, exploiting
his resentment of Big Government to promote the power of Big Business.
14. (Paragraph 73) When someone approves of the purpose for which
propaganda is being used in a given case, he generally calls it
"education" or applies to it some similar euphemism. But propaganda is
propaganda regardless of the purpose for which it is used.
15. (Paragraph 83) We are not expressing approval or disapproval of
the Panama invasion. We only use it to illustrate a point.
16. (Paragraph 95) When the American colonies were under British rule
there were fewer and less effective legal guarantees of freedom than
there were after
Ralph Hartley
Dec13-04, 11:46 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On the other hand there have\nalso been many pre-industrial cultures in which material acquisition\nhas played an important role. So we can\'t claim that today\'s\nacquisition-oriented culture is exclusively a creation of the\nadvertising and marketing industry. But it is clear that the\nadvertising and marketing industry has had an important part in\ncreating that culture. The big corporations that spend millions on\nadvertising wouldn\'t be spending that kind of money without solid\nproof that they were getting it back in increased sales. One member of\nFC met a sales manager a couple of years ago who was frank enough to\ntell him, "Our job is to make people buy things they don\'t want and\ndon\'t need." He then described how an untrained novice could present\npeople with the facts about a product, and make no sales at all, while\na trained and experienced professional salesman would make lots of\nsales to the same people. This shows that people are manipulated into\nbuying things they don\'t really want.\n\n12. (Paragraph 64) The problem of purposelessness seems to have become\nless serious during the last 15 years or so, because people now feel\nless secure physically and economically than they did earlier, and the\nneed for security provides them with a goal. But purposelessness has\nbeen replaced by frustration over the difficulty of attaining\nsecurity. We emphasize the problem of purposelessness because the\nliberals and leftists would wish to solve our social problems by\nhaving society guarantee everyone\'s security; but if that could be\ndone it would only bring back the problem of purposelessness. The real\nissue is not whether society provides well or poorly for people\'s\nsecurity; the trouble is that people are dependent on the system for\ntheir security rather than having it in their own hands. This, by the\nway, is part of the reason why some people get work\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On the other hand there have
also been many pre-industrial cultures in which material acquisition
has played an important role. So we can't claim that today's
acquisition-oriented culture is exclusively a creation of the
advertising and marketing industry. But it is clear that the
advertising and marketing industry has had an important part in
creating that culture. The big corporations that spend millions on
advertising wouldn't be spending that kind of money without solid
proof that they were getting it back in increased sales. One member of
FC met a sales manager a couple of years ago who was frank enough to
tell him, "Our job is to make people buy things they don't want and
don't need." He then described how an untrained novice could present
people with the facts about a product, and make no sales at all, while
a trained and experienced professional salesman would make lots of
sales to the same people. This shows that people are manipulated into
buying things they don't really want.
12. (Paragraph 64) The problem of purposelessness seems to have become
less serious during the last 15 years or so, because people now feel
less secure physically and economically than they did earlier, and the
need for security provides them with a goal. But purposelessness has
been replaced by frustration over the difficulty of attaining
security. We emphasize the problem of purposelessness because the
liberals and leftists would wish to solve our social problems by
having society guarantee everyone's security; but if that could be
done it would only bring back the problem of purposelessness. The real
issue is not whether society provides well or poorly for people's
security; the trouble is that people are dependent on the system for
their security rather than having it in their own hands. This, by the
way, is part of the reason why some people get work
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>approach. If experience indicates that some of the\nrecommendations made in the foregoing paragraphs are not going to give\ngood results, then those recommendations should be discarded.\n\nTWO KINDS OF TECHNOLOGY\n\n\n\n207. An argument likely to be raised against our proposed revolution\nis that it is bound to fail, because (it is claimed) throughout\nhistory technology has always progressed, never regressed, hence\ntechnological regression is impossible. But this claim is false.\n\n208. We distinguish between two kinds of technology, which we will\ncall small-scale technology and organization-dependent technology.\nSmall-scale technology is technology that can be used by small-scale\ncommunities without outside assistance. Organization-dependent\ntechnology is technology that depends on large-scale social\norganization. We are aware of no significant cases of regression in\nsmall-scale technology. But organization-dependent technology DOES\nregress when the social organization on which it depends breaks down.\nExample: When the Roman Empire fell apart the Romans\' small-scale\ntechnology survived because any clever village craftsman could build,\nfor instance, a water wheel, any skilled smith could make steel by\nRoman methods, and so forth. But the Romans\' organization-dependent\ntechnology DID regress. Their aqueducts fell into disrepair and were\nnever rebuilt. Their techniques of road construction were lost. The\nRoman system of urban sanitation was forgotten, so that until rather\nrecent times did the sanitation of European cities that of Ancient\nRome.\n\n209. The reason why technology has seemed always to progress is that,\nuntil perhaps a century or two before the Industrial Revolution, most\ntechnology was small-scale technology. But most of the technology\ndeveloped since the Industrial Revolution\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>approach. If experience indicates that some of the
recommendations made in the foregoing paragraphs are not going to give
good results, then those recommendations should be discarded.
TWO KINDS OF TECHNOLOGY
207. An argument likely to be raised against our proposed revolution
is that it is bound to fail, because (it is claimed) throughout
history technology has always progressed, never regressed, hence
technological regression is impossible. But this claim is false.
208. We distinguish between two kinds of technology, which we will
call small-scale technology and organization-dependent technology.
Small-scale technology is technology that can be used by small-scale
communities without outside assistance. Organization-dependent
technology is technology that depends on large-scale social
organization. We are aware of no significant cases of regression in
small-scale technology. But organization-dependent technology DOES
regress when the social organization on which it depends breaks down.
Example: When the Roman Empire fell apart the Romans' small-scale
technology survived because any clever village craftsman could build,
for instance, a water wheel, any skilled smith could make steel by
Roman methods, and so forth. But the Romans' organization-dependent
technology DID regress. Their aqueducts fell into disrepair and were
never rebuilt. Their techniques of road construction were lost. The
Roman system of urban sanitation was forgotten, so that until rather
recent times did the sanitation of European cities that of Ancient
Rome.
209. The reason why technology has seemed always to progress is that,
until perhaps a century or two before the Industrial Revolution, most
technology was small-scale technology. But most of the technology
developed since the Industrial Revolution
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nVan www <vanjac12@yahoo.com> writes\n>Ralph Hartley wrote:\n>> Oz wrote:\n>[snip]\n>\n>One must think of an electromagnetic field, which is why F is good\n>to use.\n\nI agree. However it helps if you look at things through different eyes.\nThe simplest being the example I gave. I prefer to start by crawling,\nand to get a feel for how things are seen to behave in the new\nviewpoint.\n\n>Think of\n>\n>1) light waves-- |E| = |B| for light in empty space. There is\n>no primality of the electric field here. Also\n\nOf course. I would have gotten onto this once I have a physical grasp of\nthe basics. However, one should remember that EM radiation has a source,\nthat source can be seen as (for example) a pair of oscillating point\ncharges. Consequently I can always decompose an EM wave into the\nbehaviour of point charges, that is its source. This probably isn\'t a\nvery useful way to see it, but it may well give some insight otherwise\noverlooked. As I say, though, that is for the future.\n\n>2) MHD = magnetohydrodynamics. A charged fluid where charges move\n>easily--not like a solid--so that electric fields simply move charges\n>till the E field is 0 (a perfect conductor). Then what is left?\n>A "frozen in" magnetic field B, which is non-zero even in the\n>rest frame of the fluid, where E = 0.\n\nInteresting setup. Something to examine in the future.\n\n>Also, one can expand any tensor about a vector field u by\n>\n>T = T(u,u) u u - P(T\'(u)) u - u P(T(u)) + PTP\n\nIs this an identity? It looks like it should be.\n\n>where P = g + u u is the spatial projection operator = diag(0,1,1,1)\n>and g = diag(-1,1,1,1) = Minkowski metric, T\' denotes transpose, and\n>the tensor product is implied (u u == u x u = u (tensor prod) u).\n>\n>So T(u) = - T(u,u) u + P(T(u)) since P(u) = P\'(u) = g(u) - u = 0,\n\nThis looks like a specific for minkowski spacetime and EM.\n\n>since we consider that we use g to raise and lower indices,\n\n>g(u) - u = 0 = g_ij u^i - u_j = 0.\n\nOK, got you.\n\n>(Its useful to do some calculations\n>with indices, and after a while you won\'t need them for most\n>calculations.\n\nI\'m dying to use these in anger and get out something illuminating and\nuseful.\n\n>All this is to show that in my post I just decomposed F w.r.t. u to get\n>\n>F = u/\\E + *(u/\\B) ; E = - F(u), B = *F(u).\n\nOh, very neat. I even pretty much follow this (be pleased, this is an\nachievement).\n\n>This is very useful for many things. Let u = e_0 = d/dx^0 = d/dt\n\nThis I don\'t see. I\'m sure its a matter of definition, almost, but I\'ve\nnever been able to get people to move to differentials, but yet this\nseems to be (is) critical to forms in general.\n\n>be the unit vector along the time direction,\n\n>then F, *F, and\n>dF and d*F, which are 2-forms and 3-forms, can be integrated over\n>various 2D and 3D subspaces of 4D spacetime\n\nIts very simple examples, and I mean simple bordering on the trivial, of\nthis I would like to see worked as examples. Generally once I see the\nexamples, I can see what (physically) is meant and how the though\nprocess is working.\n\n>to get the integral\n>form of Maxwell\'s eqs.\n\nEveryone keeps wanting to derive maxwell from this.\nActually I absolutely don\'t, quite the reverse. I want to ditch maxwell\nas such and see EM as F (and eventually as A). If we lived in a world\nwhere c was 10mph, we wouldn\'t separate E & B out, we would see them as\nthe same force. For this I need some simple examples.\n\n>See, e.g., Misner, Thorne, Wheeler, "Gravitation", which I believe\n>anyone who wants to understand physics must study thoroughly.\n\nYou are quite right. My problem is (even if I were smart enough) that I\nwould be dead before I reached that sort of level.\n\n--\nOz\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Van www <vanjac12@yahoo.com> writes
>Ralph Hartley wrote:
>> Oz wrote:
>[snip]
>
>One must think of an electromagnetic field, which is why F is good
>to use.
I agree. However it helps if you look at things through different eyes.
The simplest being the example I gave. I prefer to start by crawling,
and to get a feel for how things are seen to behave in the new
viewpoint.
>Think of
>
>1) light waves-- |E| = |B| for light in empty space. There is
>no primality of the electric field here. Also
Of course. I would have gotten onto this once I have a physical grasp of
the basics. However, one should remember that EM radiation has a source,
that source can be seen as (for example) a pair of oscillating point
charges. Consequently I can always decompose an EM wave into the
behaviour of point charges, that is its source. This probably isn't a
very useful way to see it, but it may well give some insight otherwise
overlooked. As I say, though, that is for the future.
>2) MHD = magnetohydrodynamics. A charged fluid where charges move
>easily--not like a solid--so that electric fields simply move charges
>till the E field is (a perfect conductor). Then what is left?
>A "frozen in" magnetic field B, which is non-zero even in the
>rest frame of the fluid, where E = .
Interesting setup. Something to examine in the future.
>Also, one can expand any tensor about a vector field u by
>
>T = T(u,u) u u - P(T'(u)) u - u P(T(u)) + PTP
Is this an identity? It looks like it should be.
>where P = g + u u is the spatial projection operator = diag(0,1,1,1)
>and g = diag(-1,1,1,1) = Minkowski metric, T' denotes transpose, and
>the tensor product is implied (u u == u x u = u (tensor prod) u).
>
>So T(u) = - T(u,u) u + P(T(u)) since P(u) = P'(u) = g(u) - u = 0,
This looks like a specific for minkowski spacetime and EM.
>since we consider that we use g to raise and lower indices,
>g(u) - u == g_{ij} u^i - u_j = .
OK, got you.
>(Its useful to do some calculations
>with indices, and after a while you won't need them for most
>calculations.
I'm dying to use these in anger and get out something illuminating and
useful.
>All this is to show that in my post I just decomposed F w.r.t. u to get
>
>F = u/\E + *(u/\B) ; E = - F(u), B = *F(u).
Oh, very neat. I even pretty much follow this (be pleased, this is an
achievement).
>This is very useful for many things. Let u = e_0 = d/dx^0 = d/dt
This I don't see. I'm sure its a matter of definition, almost, but I've
never been able to get people to move to differentials, but yet this
seems to be (is) critical to forms in general.
>be the unit vector along the time direction,
>then F, *F, and
>dF and d*F, which are 2-forms and 3-forms, can be integrated over
>various 2D and 3D subspaces of 4D spacetime
Its very simple examples, and I mean simple bordering on the trivial, of
this I would like to see worked as examples. Generally once I see the
examples, I can see what (physically) is meant and how the though
process is working.
>to get the integral
>form of Maxwell's eqs.
Everyone keeps wanting to derive maxwell from this.
Actually I absolutely don't, quite the reverse. I want to ditch maxwell
as such and see EM as F (and eventually as A). If we lived in a world
where c was 10mph, we wouldn't separate E & B out, we would see them as
the same force. For this I need some simple examples.
>See, e.g., Misner, Thorne, Wheeler, "Gravitation", which I believe
>anyone who wants to understand physics must study thoroughly.
You are quite right. My problem is (even if I were smart enough) that I
would be dead before I reached that sort of level.
--
Oz
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nRalph Hartley <hartley@aic.nrl.navy.mil> writes\n>Oz wrote:\n>\n>> I also asked about the behaviour of a charged point particle in the\n>> vicinity of a current-carrying conductor.\n>\n>Is that a *charged* current-carrying conductor? If not, In what frame is it\n>nuetral?\n\nThat\'s my question.\n\nIts a conductor of negligible resistance carrying a current.\nTo give an example experimental situation, consider a straight current-\ncarrying conductor coming into a large faraday cage where there are no\nexternal magnetic fields. The conductor is connected to the cage at one\nend (comes through a small hole in the other).\n\nTo be honest its somewhat unclear to me what the precise frame is where\nit is uncharged. Clearly electrons are moving at the drift velocity and\nthe electron stream will be contracted due to SR as seen by the metal\natoms.\n\nOne can of course say the same for the electrons who see a stream of\npositive atoms heading in the opposite direction, also contracted by SR.\n\nThat is (shock, horror) there will be a magnetic field within the wire.\nHmm, I guess that means the mobile conduction electrons will feel a\nforce pushing them to the outside of the wire (hall effect). I hadn\'t\npreviously bothered with this. The effect is tiny, but then so is the\ndrift velocity. I was hoping to be able to neglect this.\n\n>If the wire has an arbitrary charge density, then clearly there is not\n>enough information to answer the question.\n\nSee experimental setup above.\n\n>If you say "a current-carrying conductor", I have to assume that it is\n>neutral. That is, that there is no electrostatic field arround it in the\n>*lab* frame.\n\nMaybe. We can, perhaps, state that one end of the wire is earthed, see\nexperimental setup above. .\n\n>If you showed be a wire, and stationary charged test particles\n>were atracted or repelled by it, I would rightly say that you had shown me\n>a charged wire.\n\nI am not so sure I would any more. Clearly it depends on the word\n\'stationary\'. Indeed using the 4D model and F, one can always transform\na magnetic field into an electrostatic field by choice of frame. I thus\nno longer consider E & B as separate in any way at all. What I am trying\nto do is get some sort of handle on how it works. My example is the\nsimplest non-trivial one I can find. Astonishingly nobody will confirm\nor deny my conclusion, or point out where it is in error. I find this\nuncomfortable, its as if people really aren\'t sure. For such a simple\nexample in a group brimming with serious experts, that is worrying me.\n\n>If I din\'t have a compass, I couldn\'t tell if there were\n>any current, but I could still tell that it was charged.\n\nWhether a compass or a charged test particle give you a result will\ndepend on your frame. Ultimately the compass can be converted to charges\nand considered as entirely electrostatic (in the sum of rather a lot of\ndifferent frames). That\'s too complex for me now.\n\n>> In particular the conditions\n>> where it feels no force. This is a hugely simple example of my image\n>> above. Its a good example to see where I have it wrong. I think my\n>> visualisation suggests that the condition is with a velocity of half the\n>> drift current.\n>\n>No. It will feel no force if it is stationary or moving tangentally (i.e.\n>not along the wire, and not towards or away from the wire).\n\nThat\'s my question. Which frame is \'stationary\'? The frame of the metal\natoms or the frame of the electrons moving at the drift velocity, or\nwhat?\n\nI don\'t think its as trivial a problem as you think.\n\n--\nOz\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Ralph Hartley <hartley@aic.nrl.navy.mil> writes
>Oz wrote:
>
>> I also asked about the behaviour of a charged point particle in the
>> vicinity of a current-carrying conductor.
>
>Is that a *charged* current-carrying conductor? If not, In what frame is it
>nuetral?
That's my question.
Its a conductor of negligible resistance carrying a current.
To give an example experimental situation, consider a straight current-
carrying conductor coming into a large faraday cage where there are no
external magnetic fields. The conductor is connected to the cage at one
end (comes through a small hole in the other).
To be honest its somewhat unclear to me what the precise frame is where
it is uncharged. Clearly electrons are moving at the drift velocity and
the electron stream will be contracted due to SR as seen by the metal
atoms.
One can of course say the same for the electrons who see a stream of
positive atoms heading in the opposite direction, also contracted by SR.
That is (shock, horror) there will be a magnetic field within the wire.
Hmm, I guess that means the mobile conduction electrons will feel a
force pushing them to the outside of the wire (hall effect). I hadn't
previously bothered with this. The effect is tiny, but then so is the
drift velocity. I was hoping to be able to neglect this.
>If the wire has an arbitrary charge density, then clearly there is not
>enough information to answer the question.
See experimental setup above.
>If you say "a current-carrying conductor", I have to assume that it is
>neutral. That is, that there is no electrostatic field arround it in the
>*lab* frame.
Maybe. We can, perhaps, state that one end of the wire is earthed, see
experimental setup above. .
>If you showed be a wire, and stationary charged test particles
>were atracted or repelled by it, I would rightly say that you had shown me
>a charged wire.
I am not so sure I would any more. Clearly it depends on the word
'stationary'. Indeed using the 4D model and F, one can always transform
a magnetic field into an electrostatic field by choice of frame. I thus
no longer consider E & B as separate in any way at all. What I am trying
to do is get some sort of handle on how it works. My example is the
simplest non-trivial one I can find. Astonishingly nobody will confirm
or deny my conclusion, or point out where it is in error. I find this
uncomfortable, its as if people really aren't sure. For such a simple
example in a group brimming with serious experts, that is worrying me.
>If I din't have a compass, I couldn't tell if there were
>any current, but I could still tell that it was charged.
Whether a compass or a charged test particle give you a result will
depend on your frame. Ultimately the compass can be converted to charges
and considered as entirely electrostatic (in the sum of rather a lot of
different frames). That's too complex for me now.
>> In particular the conditions
>> where it feels no force. This is a hugely simple example of my image
>> above. Its a good example to see where I have it wrong. I think my
>> visualisation suggests that the condition is with a velocity of half the
>> drift current.
>
>No. It will feel no force if it is stationary or moving tangentally (i.e.
>not along the wire, and not towards or away from the wire).
That's my question. Which frame is 'stationary'? The frame of the metal
atoms or the frame of the electrons moving at the drift velocity, or
what?
I don't think its as trivial a problem as you think.
--
Oz
Ralph Hartley
Dec15-04, 12:39 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Oz wrote:\n> Ralph Hartley <hartley@aic.nrl.navy.mil> writes:\n>>Oz wrote:\n>>>I also asked about the behaviour of a charged point particle in the\n>>>vicinity of a current-carrying conductor.\n>>\n>>Is that a *charged* current-carrying conductor? If not, In what frame is it\n>>nuetral?\n>\n> That\'s my question.\n\nNo! It\'s mine!\n\nThe charge density of a conductor is independent of the current. Asking for\nthe charge density of a current carrying wire is just like asking for the\ndistance between a point on the prime meridian and the equator. It could be\nanything, and if you aren\'t told, there is no way to figure it out.\n\n> To give an example experimental situation, consider a straight current-\n> carrying conductor coming into a large faraday cage where there are no\n> external magnetic fields. The conductor is connected to the cage at one\n> end (comes through a small hole in the other).\n\nThen the wire is neutral in the frame in which the (grounded end of the)\n*cage* is stationary.\n\nLets simplify a little more, without changing anything important. Make the\ncage an infinite cylinder centered on the wire. To make sure the wire isn\'t\ncharged, we need to ground it, so add a grounding wire. In your set up one\nof the ends is the grounding wire, adding additional wires to form an end\n(or two ends) doesn\'t change anything, because additional wires will all\nact the same.\n\n------+------------------cage\n|\n|\n|Grounding wire\n|\n------+------------------wire\nCurrent-->\n\n\n\n-------------------------cage\n\nAt equilibrium the grounding wire isn\'t actually carrying any current.\n\nIn this set up the central wire is neutral in the rest frame of the\ngrounding wire. In any *other* frame, the wire is charged, because the\nmotion of the grounding wire through the magnetic field generates a\npotential difference between its ends.\n\nLets suppose the wire were neutral in some *other* frame, and describe what\nhappens, both in that frame and in the grounding wire\'s rest frame.\n\nIn the "neutral wire frame" the grounding wire is moving. Because the\nelectrons in the grounding wire are being forced to move through a magnetic\nfield, they feel a force perpendicular to B and to their motion. So there\nwill be a current in the grounding wire, which will continue until the\ncentral wire has enough charge that the electrostatic force cancels the\nmagnetic force.\n\nIn the "grounding wire frame" the magnetic field exerts no forces, because\nnothing is moving, but the density of positive and negative charges are\ndifferent, because they are subject to different amounts of SR length\ncontraction. So in this frame there is a radial electrostatic field. The\nelectrons in the grounding wire will feel an electrostatic force, producing\na current that continues until the charge is neutralized.\n\nBoth views agree that there is a current (they had better). The only\nexception is when the wire is neutral in the ground wire frame, so the two\nframes are the same. Then there is a single frame in which there is neither\nan electrostatic, nor a magnetic force on the electrons in the grounding\nwire, so it caries no current.\n\n> That is (shock, horror) there will be a magnetic field within the wire.\n> Hmm, I guess that means the mobile conduction electrons will feel a\n> force pushing them to the outside of the wire.\n\nThis is so, but irrelevant to this problem. In an ideal superconductor all\nthe current is at the surface, but for this problem we don\'t really care.\n\n>>If you say "a current-carrying conductor", I have to assume that it is\n>>neutral. That is, that there is no electrostatic field arround it in the\n>>*lab* frame.\n>\n> Maybe. We can, perhaps, state that one end of the wire is earthed, see\n> experimental setup above. .\n\nThat does it. As long as you tell me the frame in which the grounding\nconnection is at rest. (If there is more than one connection, in relative\nmotion, then there will be current in one and out the other, a dynamo.)\n\n>>If you showed me a wire, and stationary charged test particles\n>>were atracted or repelled by it, I would rightly say that you had shown me\n>>a charged wire.\n>\n> I am not so sure I would any more. Clearly it depends on the word\n> \'stationary\'.\n\nCertainly it does. But if you showed me a wire, I would use test particles\nstationary relative to *me*. I would rightly say that the wire is charged.\nIt doesn\'t matter that you, moving relative to me, see things differently.\n\n> Indeed using the 4D model and F, one can always transform\n> a magnetic field into an electrostatic field by choice of frame.\n\nThis is not so. There is current in the wire in any frame, so there is a\nmagnetic field. For any given test particle, there *is* a frame in which\nthe magnetic *force* is 0 (the particle\'s rest frame).\n\n> I thus no longer consider E & B as separate in any way at all.\n\nThey aren\'t separate.\n\nConsider this analogy. The relationship between torque and force depends on\nthe origin. A given force may or may not produce a torque, depending on\nwhere the "center" is. Force and torque are not really separate. But if you\nhave two equal and opposite forces acting along different lines, you will\nhave a torque regardless of where the origin is.\n\n> That\'s my question. Which frame is \'stationary\'?\n\nA frame is a *definition* of stationary. Like all definitions, there is no\n"correct" one.\n\n> The frame of the metal\n> atoms or the frame of the electrons moving at the drift velocity, or\n> what?\n\nEither of those or neither. A *grounded* wire will be neutral in the\nframe in which the grounding point is stationary.\n\nI think some of your confusion is because you are trying to imagine\ntaking a wire, and starting a current without adding or removing\nelectrons, as if that were well defined, but it isn\'t.\n\nIn an infinite wire, the number of electrons (and protons) is not a\nfinite number, so there is no "total number of electrons" that can be\nconserved. All you have is the electron *density* which is different in\ndifferent frames. If you say "start a current without changing the\ndensity of electrons," you are being ambiguous, because the density will\nbe unchanged in only one frame.\n\nPerhaps a *finite* wire is easier to understand. After you start the\ncurrent, there are electrons being added at one end and removed at the\nother. The total number of electrons in the wire at any one instant is\n*not* frame independent, because different frames have different\ndefinitions of "instant."\n\nSuppose you have a finite piece of wire that everyone agrees is neutral.\nYou start a current through it by simultaneously starting to add\nelectrons to one end, and removing them from the other. The total number\nof electrons, and hence the charge of the wire doesn\'t change.\n\nSomeone moving along the wire relative to you would disagree, because\nthey have a different definition of "simultaneous". They would say that\nyou started adding electrons to one end *before* (or after) you started\nremoving electrons form the other end. From *their* viewpoint you added\n(or subtracted) electrons to the wire, giving it a charge.\n\nRalph Hartley\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz wrote:
> Ralph Hartley <hartley@aic.nrl.navy.mil> writes:
>>Oz wrote:
>>>I also asked about the behaviour of a charged point particle in the
>>>vicinity of a current-carrying conductor.
>>
>>Is that a *charged* current-carrying conductor? If not, In what frame is it
>>nuetral?
>
> That's my question.
No! It's mine!
The charge density of a conductor is independent of the current. Asking for
the charge density of a current carrying wire is just like asking for the
distance between a point on the prime meridian and the equator. It could be
anything, and if you aren't told, there is no way to figure it out.
> To give an example experimental situation, consider a straight current-
> carrying conductor coming into a large faraday cage where there are no
> external magnetic fields. The conductor is connected to the cage at one
> end (comes through a small hole in the other).
Then the wire is neutral in the frame in which the (grounded end of the)
*cage* is stationary.
Lets simplify a little more, without changing anything important. Make the
cage an infinite cylinder centered on the wire. To make sure the wire isn't
charged, we need to ground it, so add a grounding wire. In your set up one
of the ends is the grounding wire, adding additional wires to form an end
(or two ends) doesn't change anything, because additional wires will all
act the same.
------+------------------cage
|
|
|Grounding wire
|
------+------------------wire
Current-->
-------------------------cage
At equilibrium the grounding wire isn't actually carrying any current.
In this set up the central wire is neutral in the rest frame of the
grounding wire. In any *other* frame, the wire is charged, because the
motion of the grounding wire through the magnetic field generates a
potential difference between its ends.
Lets suppose the wire were neutral in some *other* frame, and describe what
happens, both in that frame and in the grounding wire's rest frame.
In the "neutral wire frame" the grounding wire is moving. Because the
electrons in the grounding wire are being forced to move through a magnetic
field, they feel a force perpendicular to B and to their motion. So there
will be a current in the grounding wire, which will continue until the
central wire has enough charge that the electrostatic force cancels the
magnetic force.
In the "grounding wire frame" the magnetic field exerts no forces, because
nothing is moving, but the density of positive and negative charges are
different, because they are subject to different amounts of SR length
contraction. So in this frame there is a radial electrostatic field. The
electrons in the grounding wire will feel an electrostatic force, producing
a current that continues until the charge is neutralized.
Both views agree that there is a current (they had better). The only
exception is when the wire is neutral in the ground wire frame, so the two
frames are the same. Then there is a single frame in which there is neither
an electrostatic, nor a magnetic force on the electrons in the grounding
wire, so it caries no current.
> That is (shock, horror) there will be a magnetic field within the wire.
> Hmm, I guess that means the mobile conduction electrons will feel a
> force pushing them to the outside of the wire.
This is so, but irrelevant to this problem. In an ideal superconductor all
the current is at the surface, but for this problem we don't really care.
>>If you say "a current-carrying conductor", I have to assume that it is
>>neutral. That is, that there is no electrostatic field arround it in the
>>*lab* frame.
>
> Maybe. We can, perhaps, state that one end of the wire is earthed, see
> experimental setup above. .
That does it. As long as you tell me the frame in which the grounding
connection is at rest. (If there is more than one connection, in relative
motion, then there will be current in one and out the other, a dynamo.)
>>If you showed me a wire, and stationary charged test particles
>>were atracted or repelled by it, I would rightly say that you had shown me
>>a charged wire.
>
> I am not so sure I would any more. Clearly it depends on the word
> 'stationary'.
Certainly it does. But if you showed me a wire, I would use test particles
stationary relative to *me*. I would rightly say that the wire is charged.
It doesn't matter that you, moving relative to me, see things differently.
> Indeed using the 4D model and F, one can always transform
> a magnetic field into an electrostatic field by choice of frame.
This is not so. There is current in the wire in any frame, so there is a
magnetic field. For any given test particle, there *is* a frame in which
the magnetic *force* is (the particle's rest frame).
> I thus no longer consider E & B as separate in any way at all.
They aren't separate.
Consider this analogy. The relationship between torque and force depends on
the origin. A given force may or may not produce a torque, depending on
where the "center" is. Force and torque are not really separate. But if you
have two equal and opposite forces acting along different lines, you will
have a torque regardless of where the origin is.
> That's my question. Which frame is 'stationary'?
A frame is a *definition* of stationary. Like all definitions, there is no
"correct" one.
> The frame of the metal
> atoms or the frame of the electrons moving at the drift velocity, or
> what?
Either of those or neither. A *grounded* wire will be neutral in the
frame in which the grounding point is stationary.
I think some of your confusion is because you are trying to imagine
taking a wire, and starting a current without adding or removing
electrons, as if that were well defined, but it isn't.
In an infinite wire, the number of electrons (and protons) is not a
finite number, so there is no "total number of electrons" that can be
conserved. All you have is the electron *density* which is different in
different frames. If you say "start a current without changing the
density of electrons," you are being ambiguous, because the density will
be unchanged in only one frame.
Perhaps a *finite* wire is easier to understand. After you start the
current, there are electrons being added at one end and removed at the
other. The total number of electrons in the wire at any one instant is
*not* frame independent, because different frames have different
definitions of "instant."
Suppose you have a finite piece of wire that everyone agrees is neutral.
You start a current through it by simultaneously starting to add
electrons to one end, and removing them from the other. The total number
of electrons, and hence the charge of the wire doesn't change.
Someone moving along the wire relative to you would disagree, because
they have a different definition of "simultaneous". They would say that
you started adding electrons to one end *before* (or after) you started
removing electrons form the other end. From *their* viewpoint you added
(or subtracted) electrons to the wire, giving it a charge.
Ralph Hartley
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>[NB read to the end before replying]\n\nRalph Hartley <hartley@aic.nrl.navy.mil> writes\n>Oz wrote:\n>> Ralph Hartley <hartley@aic.nrl.navy.mil> writes:\n>>>Oz wrote:\n>>>>I also asked about the behaviour of a charged point particle in the\n>>>>vicinity of a current-carrying conductor.\n>>>\n>>>Is that a *charged* current-carrying conductor? If not, In what frame is it\n>>>nuetral?\n>>\n>> That\'s my question.\n>\n>No! It\'s mine!\n\nIs mine TOO!\n\n>> To give an example experimental situation, consider a straight current-\n>> carrying conductor coming into a large faraday cage where there are no\n>> external magnetic fields. The conductor is connected to the cage at one\n>> end (comes through a small hole in the other).\n>\n>Then the wire is neutral in the frame in which the (grounded end of the)\n>*cage* is stationary.\n\nYup. That\'s a very reasonable way to see it.\nI\'m not sure if that is precisely correct though.\nI will accept that the wire at the point of connection is the same PD as\n\'ground\'.\n\n>Lets simplify a little more, without changing anything important. Make the\n>cage an infinite cylinder centered on the wire. To make sure the wire isn\'t\n>charged, we need to ground it, so add a grounding wire. In your set up one\n>of the ends is the grounding wire, adding additional wires to form an end\n>(or two ends) doesn\'t change anything, because additional wires will all\n>act the same.\n>\n>------+------------------cage\n> |\n> |\n> |Grounding wire\n> |\n>------+------------------wire\n> Current-->\n>\n>\n>\n>-------------------------cage\n>\n>At equilibrium the grounding wire isn\'t actually carrying any current.\n\nAgreed.\n\n>In this set up the central wire is neutral in the rest frame of the\n>grounding wire. In any *other* frame, the wire is charged, because the\n>motion of the grounding wire through the magnetic field generates a\n>potential difference between its ends.\n\nAh. Hang on though.\nLets see if I can get my point across.\n\nThere are **two** frames involved in the current carrying wire.\nThey are NOT the same.\n\n1) The frame of the atoms.\n2) The frame of the drifting electrons.\n\nWhich are you going to choose, and why?\nWhat is the effect on the other frame?\n\n>Lets suppose the wire were neutral in some *other* frame, and describe what\n>happens, both in that frame and in the grounding wire\'s rest frame.\n>\n>In the "neutral wire frame" the grounding wire is moving. Because the\n>electrons in the grounding wire are being forced to move through a magnetic\n>field, they feel a force perpendicular to B and to their motion. So there\n>will be a current in the grounding wire, which will continue until the\n>central wire has enough charge that the electrostatic force cancels the\n>magnetic force.\n\nYes. But I would prefer it expressed (with verbal description) using F.\n\n>In the "grounding wire frame" the magnetic field exerts no forces, because\n>nothing is moving, but the density of positive and negative charges are\n>different, because they are subject to different amounts of SR length\n>contraction. So in this frame there is a radial electrostatic field. The\n>electrons in the grounding wire will feel an electrostatic force, producing\n>a current that continues until the charge is neutralized.\n>\n>Both views agree that there is a current (they had better). The only\n>exception is when the wire is neutral in the ground wire frame, so the two\n>frames are the same. Then there is a single frame in which there is neither\n>an electrostatic, nor a magnetic force on the electrons in the grounding\n>wire, so it caries no current.\n\nI don\'t perfectly follow you here.\nI *think* you just said:\n\n1) In the electron frame the wire is charged because the \'earthing wire\'\nis moving through a magnetic field, and this is balanced by the\n\'increase in electrons due to contraction\'.\n\n2) In the wire frame nothing is charged and there is no PD.\n\nThese don;t *quite* match up.\n\nWhich is what I was saying. Actually I wanted the frame where there was\nno force on a charged test particle.\n\n>> That is (shock, horror) there will be a magnetic field within the wire.\n>> Hmm, I guess that means the mobile conduction electrons will feel a\n>> force pushing them to the outside of the wire.\n>\n>This is so, but irrelevant to this problem. In an ideal superconductor all\n>the current is at the surface, but for this problem we don\'t really care.\n\nI agree.\n\n>>>If you say "a current-carrying conductor", I have to assume that it is\n>>>neutral. That is, that there is no electrostatic field arround it in the\n>>>*lab* frame.\n>>\n>> Maybe. We can, perhaps, state that one end of the wire is earthed, see\n>> experimental setup above. .\n>\n>That does it. As long as you tell me the frame in which the grounding\n>connection is at rest.\n\nOoops. Good point.\nI think I spot your point now.\nI will have to think about it.\n\n>(If there is more than one connection, in relative\n>motion, then there will be current in one and out the other, a dynamo.)\n\nYes.\n\n>> Indeed using the 4D model and F, one can always transform\n>> a magnetic field into an electrostatic field by choice of frame.\n>\n>This is not so. There is current in the wire in any frame,\n\nYes. I saw this immediately I first looked at this. Positive charges\ngoing one way ADD to negative charges going the other way. Given that\ngamma is not linear to velocity I don\'t quite see how this always adds\nup to the same thing in all frames, but given that we have both magnetic\nand electric fields I can see that combining them might give a quadratic\nneatly matching gamma. Obviously it does.\n\n>so there is a\n>magnetic field. For any given test particle, there *is* a frame in which\n>the magnetic *force* is 0 (the particle\'s rest frame).\n\nOK.\n\n>> I thus no longer consider E & B as separate in any way at all.\n>\n>They aren\'t separate.\n\nI know. Its just a matter, at the moment, of grokking how the mechanism\nbehaves. That is to see magnetism as intrinsically electrostatic and\ngetting a paradigm shift in my brain that sees it so.\n\n>Consider this analogy. The relationship between torque and force depends on\n>the origin. A given force may or may not produce a torque, depending on\n>where the "center" is. Force and torque are not really separate. But if you\n>have two equal and opposite forces acting along different lines, you will\n>have a torque regardless of where the origin is.\n\nOh, very neat analogy.\nHmm, actually very very neat.\nNicely illustrates how a choice of origin (= chosen co-ordinate system)\naffects the apparent physics. You have an apparent infinity in torque as\nthe force approaches the origin.\n\n>> That\'s my question. Which frame is \'stationary\'?\n>\n>A frame is a *definition* of stationary. Like all definitions, there is no\n>"correct" one.\n\nYes, my brain went awol about here.\n\n>> The frame of the metal\n>> atoms or the frame of the electrons moving at the drift velocity, or\n>> what?\n>\n>Either of those or neither. A *grounded* wire will be neutral in the\n>frame in which the grounding point is stationary.\n\nOk, OK, I admit a less than complete grasp of the basics.\nI didn\'t spot this, which was deplorable.\n\n>I think some of your confusion is because you are trying to imagine\n>taking a wire, and starting a current without adding or removing\n>electrons, as if that were well defined, but it isn\'t.\n\nNo, well maybe. I hadn\'t actually got to that example.\nIt was on the list though, but first to get the really simple stuff\nsorted out.\n\nI would expect, intuitively, a very slightly higher density of electrons\nin the wire after the initiating pulse, than before.\n\n>In an infinite wire, the number of electrons (and protons) is not a\n>finite number, so there is no "total number of electrons" that can be\n>conserved. All you have is the electron *density* which is different in\n>different frames. If you say "start a current without changing the\n>density of electrons," you are being ambiguous, because the density will\n>be unchanged in only one frame.\n\nYes. I am not sorted out as to the detail here.\nThe detail is important because we are talking about very small\ndeviations, but important deviations.\n\n>Perhaps a *finite* wire is easier to understand. After you start the\n>current, there are electrons being added at one end and removed at the\n>other. The total number of electrons in the wire at any one instant is\n>*not* frame independent, because different frames have different\n>definitions of "instant."\n\nNow you are being difficult!\n\n>Suppose you have a finite piece of wire that everyone agrees is neutral.\n>You start a current through it by simultaneously starting to add\n>electrons to one end, and removing them from the other. The total number\n>of electrons, and hence the charge of the wire doesn\'t change.\n\nNgggth!\nTotal number in which frame?\nSee, now you are getting me doing it as well.....\n\n>Someone moving along the wire relative to you would disagree, because\n>they have a different definition of "simultaneous". They would say that\n>you started adding electrons to one end *before* (or after) you started\n>removing electrons form the other end. From *their* viewpoint you added\n>(or subtracted) electrons to the wire, giving it a charge.\n\nA brilliant response. Thank you VERY much indeed.\n\nThat\'s what professors should be for. To spot where students have nearly\nthe right idea, but are missing a key insight. I wish I had had some of\nthose in my youth.\n\nA tidied up version of this should be on baez\'s website.\n\n=====================\n\nNow I can proceed without that little niggle.\n\nWhere were we?\n\nAh, yes, the hodge dual.\n\nNow, the question is why do we need this?\nWe need it to turn a 2-form magnetic field into a vector force?\n\nWas that it?\n\nI forget .....\n\n--\nOz\n-----End of original message from Oz-----\n\n--\nOz\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>[NB read to the end before replying]
Ralph Hartley <hartley@aic.nrl.navy.mil> writes
>Oz wrote:
>> Ralph Hartley <hartley@aic.nrl.navy.mil> writes:
>>>Oz wrote:
>>>>I also asked about the behaviour of a charged point particle in the
>>>>vicinity of a current-carrying conductor.
>>>
>>>Is that a *charged* current-carrying conductor? If not, In what frame is it
>>>nuetral?
>>
>> That's my question.
>
>No! It's mine!
Is mine TOO!
>> To give an example experimental situation, consider a straight current-
>> carrying conductor coming into a large faraday cage where there are no
>> external magnetic fields. The conductor is connected to the cage at one
>> end (comes through a small hole in the other).
>
>Then the wire is neutral in the frame in which the (grounded end of the)
>*cage* is stationary.
Yup. That's a very reasonable way to see it.
I'm not sure if that is precisely correct though.
I will accept that the wire at the point of connection is the same PD as
'ground'.
>Lets simplify a little more, without changing anything important. Make the
>cage an infinite cylinder centered on the wire. To make sure the wire isn't
>charged, we need to ground it, so add a grounding wire. In your set up one
>of the ends is the grounding wire, adding additional wires to form an end
>(or two ends) doesn't change anything, because additional wires will all
>act the same.
>
>------+------------------cage
> |
> |
> |Grounding wire
> |
>------+------------------wire
> Current-->
>
>
>
>-------------------------cage
>
>At equilibrium the grounding wire isn't actually carrying any current.
Agreed.
>In this set up the central wire is neutral in the rest frame of the
>grounding wire. In any *other* frame, the wire is charged, because the
>motion of the grounding wire through the magnetic field generates a
>potential difference between its ends.
Ah. Hang on though.
Lets see if I can get my point across.
There are **two** frames involved in the current carrying wire.
They are NOT the same.
1) The frame of the atoms.
2) The frame of the drifting electrons.
Which are you going to choose, and why?
What is the effect on the other frame?
>Lets suppose the wire were neutral in some *other* frame, and describe what
>happens, both in that frame and in the grounding wire's rest frame.
>
>In the "neutral wire frame" the grounding wire is moving. Because the
>electrons in the grounding wire are being forced to move through a magnetic
>field, they feel a force perpendicular to B and to their motion. So there
>will be a current in the grounding wire, which will continue until the
>central wire has enough charge that the electrostatic force cancels the
>magnetic force.
Yes. But I would prefer it expressed (with verbal description) using F.
>In the "grounding wire frame" the magnetic field exerts no forces, because
>nothing is moving, but the density of positive and negative charges are
>different, because they are subject to different amounts of SR length
>contraction. So in this frame there is a radial electrostatic field. The
>electrons in the grounding wire will feel an electrostatic force, producing
>a current that continues until the charge is neutralized.
>
>Both views agree that there is a current (they had better). The only
>exception is when the wire is neutral in the ground wire frame, so the two
>frames are the same. Then there is a single frame in which there is neither
>an electrostatic, nor a magnetic force on the electrons in the grounding
>wire, so it caries no current.
I don't perfectly follow you here.
I *think* you just said:
1) In the electron frame the wire is charged because the 'earthing wire'
is moving through a magnetic field, and this is balanced by the
'increase in electrons due to contraction'.
2) In the wire frame nothing is charged and there is no PD.
These don;t *quite* match up.
Which is what I was saying. Actually I wanted the frame where there was
no force on a charged test particle.
>> That is (shock, horror) there will be a magnetic field within the wire.
>> Hmm, I guess that means the mobile conduction electrons will feel a
>> force pushing them to the outside of the wire.
>
>This is so, but irrelevant to this problem. In an ideal superconductor all
>the current is at the surface, but for this problem we don't really care.
I agree.
>>>If you say "a current-carrying conductor", I have to assume that it is
>>>neutral. That is, that there is no electrostatic field arround it in the
>>>*lab* frame.
>>
>> Maybe. We can, perhaps, state that one end of the wire is earthed, see
>> experimental setup above. .
>
>That does it. As long as you tell me the frame in which the grounding
>connection is at rest.
Ooops. Good point.
I think I spot your point now.
I will have to think about it.
>(If there is more than one connection, in relative
>motion, then there will be current in one and out the other, a dynamo.)
Yes.
>> Indeed using the 4D model and F, one can always transform
>> a magnetic field into an electrostatic field by choice of frame.
>
>This is not so. There is current in the wire in any frame,
Yes. I saw this immediately I first looked at this. Positive charges
going one way ADD to negative charges going the other way. Given that
\gamma is not linear to velocity I don't quite see how this always adds
up to the same thing in all frames, but given that we have both magnetic
and electric fields I can see that combining them might give a quadratic
neatly matching \gamma. Obviously it does.
>so there is a
>magnetic field. For any given test particle, there *is* a frame in which
>the magnetic *force* is (the particle's rest frame).
OK.
>> I thus no longer consider E & B as separate in any way at all.
>
>They aren't separate.
I know. Its just a matter, at the moment, of grokking how the mechanism
behaves. That is to see magnetism as intrinsically electrostatic and
getting a paradigm shift in my brain that sees it so.
>Consider this analogy. The relationship between torque and force depends on
>the origin. A given force may or may not produce a torque, depending on
>where the "center" is. Force and torque are not really separate. But if you
>have two equal and opposite forces acting along different lines, you will
>have a torque regardless of where the origin is.
Oh, very neat analogy.
Hmm, actually very very neat.
Nicely illustrates how a choice of origin (= chosen co-ordinate system)
affects the apparent physics. You have an apparent infinity in torque as
the force approaches the origin.
>> That's my question. Which frame is 'stationary'?
>
>A frame is a *definition* of stationary. Like all definitions, there is no
>"correct" one.
Yes, my brain went awol about here.
>> The frame of the metal
>> atoms or the frame of the electrons moving at the drift velocity, or
>> what?
>
>Either of those or neither. A *grounded* wire will be neutral in the
>frame in which the grounding point is stationary.
Ok, OK, I admit a less than complete grasp of the basics.
I didn't spot this, which was deplorable.
>I think some of your confusion is because you are trying to imagine
>taking a wire, and starting a current without adding or removing
>electrons, as if that were well defined, but it isn't.
No, well maybe. I hadn't actually got to that example.
It was on the list though, but first to get the really simple stuff
sorted out.
I would expect, intuitively, a very slightly higher density of electrons
in the wire after the initiating pulse, than before.
>In an infinite wire, the number of electrons (and protons) is not a
>finite number, so there is no "total number of electrons" that can be
>conserved. All you have is the electron *density* which is different in
>different frames. If you say "start a current without changing the
>density of electrons," you are being ambiguous, because the density will
>be unchanged in only one frame.
Yes. I am not sorted out as to the detail here.
The detail is important because we are talking about very small
deviations, but important deviations.
>Perhaps a *finite* wire is easier to understand. After you start the
>current, there are electrons being added at one end and removed at the
>other. The total number of electrons in the wire at any one instant is
>*not* frame independent, because different frames have different
>definitions of "instant."
Now you are being difficult!
>Suppose you have a finite piece of wire that everyone agrees is neutral.
>You start a current through it by simultaneously starting to add
>electrons to one end, and removing them from the other. The total number
>of electrons, and hence the charge of the wire doesn't change.
Ngggth!
Total number in which frame?
See, now you are getting me doing it as well.....
>Someone moving along the wire relative to you would disagree, because
>they have a different definition of "simultaneous". They would say that
>you started adding electrons to one end *before* (or after) you started
>removing electrons form the other end. From *their* viewpoint you added
>(or subtracted) electrons to the wire, giving it a charge.
A brilliant response. Thank you VERY much indeed.
That's what professors should be for. To spot where students have nearly
the right idea, but are missing a key insight. I wish I had had some of
those in my youth.
A tidied up version of this should be on baez's website.
=====================
Now I can proceed without that little niggle.
Where were we?
Ah, yes, the hodge dual.
Now, the question is why do we need this?
We need it to turn a 2-form magnetic field into a vector force?
Was that it?
I forget .....
--
Oz
-----End of original message from Oz-----
--
Oz
Ralph Hartley
Dec19-04, 06:47 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Oz wrote:\n\n> There are **two** frames involved in the current carrying wire.\n> They are NOT the same.\n>\n> 1) The frame of the atoms.\n> 2) The frame of the drifting electrons.\n\nThere are at least *four* frames involved.\n\n1) The rest frame of the atoms.\n2) The rest frame of the drifting electrons.\n3) The rest frame of the grounding point = the frame in\nwhich the wire is neutral\n4) The rest frame of the test instruments.\n\nFrame 3 may, or may not, be the same as frame 1. Example: the wire is\nunrolling from a spool at one end, and being wound onto another spool at\nthe other end.\n\n> Which are you going to choose, and why?\n\nThe whole point of reference frames is that you can choose any one you want!\n\nHowever, note that frame 3 is the only one you can determine by making\nmeasurements on the electromagnetic field around the wire (presumably the\nexperimenter will also know frame 4).\n\nIn order to determine that the atoms are moving (to determine what frame 1\nis) you would have to look at the wire *itself*. If the surface of the wire\nis very smooth, you might need a microscope.\n\nTo determine what frame 2 is you would need to know how *many* electrons\nare moving, which you could only determine by knowing the properties of the\nmaterial the wire is made of, or by some sort of measurement on a sample of it.\n\n> Which is what I was saying. Actually I wanted the frame where there was\n> no force on a charged test particle.\n\nOn a *stationary* test particle? That would be the grounding frame. for a\nmoving test particle, it will depend on its velocity.\n\n>>That does it. As long as you tell me the frame in which the grounding\n>>connection is at rest.\n\nI should add that I am assuming that no part of the grounding wire is\nmoving relative to any other part. Otherwise, in *any* frame, part of it is\nmoving through a magnetic field, and it isn\'t a very good "grounding" wire.\n\n>>In an infinite wire\n....\n> Yes. I am not sorted out as to the detail here.\n> The detail is important because we are talking about very small\n> deviations, but important deviations.\n\nPerhaps I should have skipped the infinite wire, or done it *after* the\nlong but finite wire (or perhaps a marked, finite, piece of an infinite\nwire), which *is* easier to understand.\n\n> A brilliant response. Thank you VERY much indeed.\n\nI\'m glad you liked it. When I first learned about this stuff (God was that\n27 years ago?), I thought it was very cool.\n\nRalph Hartley\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz wrote:
> There are **two** frames involved in the current carrying wire.
> They are NOT the same.
>
> 1) The frame of the atoms.
> 2) The frame of the drifting electrons.
There are at least *four* frames involved.
1) The rest frame of the atoms.
2) The rest frame of the drifting electrons.
3) The rest frame of the grounding point = the frame in
which the wire is neutral
4) The rest frame of the test instruments.
Frame 3 may, or may not, be the same as frame 1. Example: the wire is
unrolling from a spool at one end, and being wound onto another spool at
the other end.
> Which are you going to choose, and why?
The whole point of reference frames is that you can choose any one you want!
However, note that frame 3 is the only one you can determine by making
measurements on the electromagnetic field around the wire (presumably the
experimenter will also know frame 4).
In order to determine that the atoms are moving (to determine what frame 1
is) you would have to look at the wire *itself*. If the surface of the wire
is very smooth, you might need a microscope.
To determine what frame 2 is you would need to know how *many* electrons
are moving, which you could only determine by knowing the properties of the
material the wire is made of, or by some sort of measurement on a sample of it.
> Which is what I was saying. Actually I wanted the frame where there was
> no force on a charged test particle.
On a *stationary* test particle? That would be the grounding frame. for a
moving test particle, it will depend on its velocity.
>>That does it. As long as you tell me the frame in which the grounding
>>connection is at rest.
I should add that I am assuming that no part of the grounding wire is
moving relative to any other part. Otherwise, in *any* frame, part of it is
moving through a magnetic field, and it isn't a very good "grounding" wire.
>>In an infinite wire
....
> Yes. I am not sorted out as to the detail here.
> The detail is important because we are talking about very small
> deviations, but important deviations.
Perhaps I should have skipped the infinite wire, or done it *after* the
long but finite wire (or perhaps a marked, finite, piece of an infinite
wire), which *is* easier to understand.
> A brilliant response. Thank you VERY much indeed.
I'm glad you liked it. When I first learned about this stuff (God was that
27 years ago?), I thought it was very cool.
Ralph Hartley
Dushan Mitrovich
Dec22-04, 05:58 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:\n>\n> There are at least *four* frames involved.\n>\n> 1) The rest frame of the atoms.\n> 2) The rest frame of the drifting electrons.\n> 3) The rest frame of the grounding point = the frame in\n> which the wire is neutral\n> 4) The rest frame of the test instruments.\n\nYour equivalence in frame #3 has me a bit confused. I think charge den-\nsity is independent of the velocity of the frame. Or are you refering\nto global rather than local characteristics?\n\n- Dushan Mitrovich\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:
>
> There are at least *four* frames involved.
>
> 1) The rest frame of the atoms.
> 2) The rest frame of the drifting electrons.
> 3) The rest frame of the grounding point = the frame in
> which the wire is neutral
> 4) The rest frame of the test instruments.
Your equivalence in frame #3 has me a bit confused. I think charge den-
sity is independent of the velocity of the frame. Or are you refering
to global rather than local characteristics?
- Dushan Mitrovich
Ralph Hartley
Dec23-04, 05:44 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Dushan Mitrovich wrote:\n> Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:\n\n>>3) The rest frame of the grounding point = the frame in\n>> which the wire is neutral\n\n> Your equivalence in frame #3 has me a bit confused. I think charge den-\n> sity is independent of the velocity of the frame.\n\nNo! What makes you think that?\n\nThis whole subthread is about how charge density depends on the velocity\n(along the wire) of the frame, and how a magnetic force in one frame may be\nan electrostatic force in another.\n\nIt isn\'t a coincidence that things work out that way. Special Relativity\nwas *designed* (or discovered, whatever) to make it so.\n\nAn even *simpler* situation will illustrate that charge density is *not*\nindependent of the velocity of the frame.\n\nConsider 11 stationary electrons evenly spaced along a line segment 1 meter\nlong.\n\n.. . . . . . . . . . .\n\nThe density is 10 electrons/meter.\n\nIn another frame, due to Lorentz contraction, the length will be different,\nsay 0.5 meters (v=c/sqrt(2)).\n\n............\n\nThe density is 20 electrons/meter.\n\nFor a wire containing both positive and negative charges, and carrying\ncurrent, even the *sign* of the charge density is frame dependent (because\nyou have two lines of charge, that are contracted by different amounts).\nThere is a unique frame (only considering motion along the wire) in which\nthe charge density is zero.\n\nAnother way to see this is to note that the total charge in a segment of\nwire can be constant in one frame, but change in another.\n\nRalph Hartley wrote (at the end of my Dec 15 post):\n> Suppose you have a finite piece of wire that everyone agrees is neutral.\n> You start a current through it by simultaneously starting to add\n> electrons to one end, and removing them from the other. The total number\n> of electrons, and hence the charge of the wire doesn\'t change.\n>\n> Someone moving along the wire relative to you would disagree, because\n> they have a different definition of "simultaneous". They would say that\n> you started adding electrons to one end *before* (or after) you started\n> removing electrons form the other end. From *their* viewpoint you added\n> (or subtracted) electrons to the wire, giving it a charge.\n\nRalph Hartley\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Dushan Mitrovich wrote:
> Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:
>>3) The rest frame of the grounding point = the frame in
>> which the wire is neutral
> Your equivalence in frame #3 has me a bit confused. I think charge den-
> sity is independent of the velocity of the frame.
No! What makes you think that?
This whole subthread is about how charge density depends on the velocity
(along the wire) of the frame, and how a magnetic force in one frame may be
an electrostatic force in another.
It isn't a coincidence that things work out that way. Special Relativity
was *designed* (or discovered, whatever) to make it so.
An even *simpler* situation will illustrate that charge density is *not*
independent of the velocity of the frame.
Consider 11 stationary electrons evenly spaced along a line segment 1 meter
long.
.. . . . . . . . . . .
The density is 10 electrons/meter.
In another frame, due to Lorentz contraction, the length will be different,
say .5 meters (v=c/\sqrt(2)).
............
The density is 20 electrons/meter.
For a wire containing both positive and negative charges, and carrying
current, even the *sign* of the charge density is frame dependent (because
you have two lines of charge, that are contracted by different amounts).
There is a unique frame (only considering motion along the wire) in which
the charge density is zero.
Another way to see this is to note that the total charge in a segment of
wire can be constant in one frame, but change in another.
Ralph Hartley wrote (at the end of my Dec 15 post):
> Suppose you have a finite piece of wire that everyone agrees is neutral.
> You start a current through it by simultaneously starting to add
> electrons to one end, and removing them from the other. The total number
> of electrons, and hence the charge of the wire doesn't change.
>
> Someone moving along the wire relative to you would disagree, because
> they have a different definition of "simultaneous". They would say that
> you started adding electrons to one end *before* (or after) you started
> removing electrons form the other end. From *their* viewpoint you added
> (or subtracted) electrons to the wire, giving it a charge.
Ralph Hartley
Dushan Mitrovich
Dec24-04, 06:50 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:\n>Dushan Mitrovich wrote:\n>> Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:\n>\n>>>3) The rest frame of the grounding point = the frame in\n>>> which the wire is neutral\n>\n>> Your equivalence in frame #3 has me a bit confused. I think charge den-\n>> sity is independent of the velocity of the frame.\n>\n> No! What makes you think that?\n\nSorry, I was a bit sloppy. What I really had in mind when I wrote the\nabove is that _zero_ charge density stays zero independent of frame velo-\ncity. But according to your argument below, even that\'s not true.\n\n> This whole subthread is about how charge density depends on the velocity\n> (along the wire) of the frame, and how a magnetic force in one frame may\n> be an electrostatic force in another.\n>\n> It isn\'t a coincidence that things work out that way. Special Relativity\n> was *designed* (or discovered, whatever) to make it so.\n>\n> An even *simpler* situation will illustrate that charge density is *not*\n> independent of the velocity of the frame.\n>\n> Consider 11 stationary electrons evenly spaced along a line segment 1 meter\n> long.\n>\n> . . . . . . . . . . .\n>\n> The density is 10 electrons/meter.\n>\n> In another frame, due to Lorentz contraction, the length will be different,\n> say 0.5 meters (v=c/sqrt(2)).\n>\n> ...........\n>\n> The density is 20 electrons/meter.\n\nYes, I knew all this, and the manifestation of the field in different\nframes.\n\n> For a wire containing both positive and negative charges, and carrying\n> current, even the *sign* of the charge density is frame dependent (because\n> you have two lines of charge, that are contracted by different amounts).\n> There is a unique frame (only considering motion along the wire) in which\n> the charge density is zero.\n\nI don\'t understand this argument. Start in one frame with two lines of op-\nposite charges, moving relative to each other but cancelling each other to\ngive a zero charge density. In another frame moving relative to the first,\nwhy should the two lines be measured as \'contracted by different amounts\'?\nThe contraction is determined by the relative frame motion, not by the\nmotions of the charges.\n\n> Another way to see this is to note that the total charge in a segment of\n> wire can be constant in one frame, but change in another.\n>\n>Ralph Hartley wrote (at the end of my Dec 15 post):\n>> Suppose you have a finite piece of wire that everyone agrees is neutral.\n>> You start a current through it by simultaneously starting to add\n>> electrons to one end, and removing them from the other. The total number\n>> of electrons, and hence the charge of the wire doesn\'t change.\n>>\n>> Someone moving along the wire relative to you would disagree, because\n>> they have a different definition of "simultaneous". They would say that\n>> you started adding electrons to one end *before* (or after) you started\n>> removing electrons form the other end. From *their* viewpoint you added\n>> (or subtracted) electrons to the wire, giving it a charge.\n\nI\'m having trouble translating this finite-size wire argument into one\nre- lating to *density*, a ratio of infinitesimals. In your example\nabove of a stationary finite length of wire, the charge of the whole\nwire, integra- ted at any instant (in that frame), stays zero. But\nbecause the charge insertion/removal signal travels at finite speed in\nthe wire, the charge density at any particular location will not remain\nzero. So in a moving frame, measuring the total charge in a section of\nwire at any instant (in the moving frame) is analogous, in the\nstationary frame, to integrating the charge density in that section\nusing different times for each slice of wire. This obviously doesn\'t\nhave to stay zero in a transient situation.\n\nSo far I understand that non-zero charge *density* is frame dependent,\nbut don\'t yet quite see that the same holds for a *zero* charge density.\n\n- Dushan Mitrovich\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:
>Dushan Mitrovich wrote:
>> Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:
>
>>>3) The rest frame of the grounding point = the frame in
>>> which the wire is neutral
>
>> Your equivalence in frame #3 has me a bit confused. I think charge den-
>> sity is independent of the velocity of the frame.
>
> No! What makes you think that?
Sorry, I was a bit sloppy. What I really had in mind when I wrote the
above is that _zero_ charge density stays zero independent of frame velo-
city. But according to your argument below, even that's not true.
> This whole subthread is about how charge density depends on the velocity
> (along the wire) of the frame, and how a magnetic force in one frame may
> be an electrostatic force in another.
>
> It isn't a coincidence that things work out that way. Special Relativity
> was *designed* (or discovered, whatever) to make it so.
>
> An even *simpler* situation will illustrate that charge density is *not*
> independent of the velocity of the frame.
>
> Consider 11 stationary electrons evenly spaced along a line segment 1 meter
> long.
>
> . . . . . . . . . . .
>
> The density is 10 electrons/meter.
>
> In another frame, due to Lorentz contraction, the length will be different,
> say .5 meters (v=c/\sqrt(2)).
>
> ...........
>
> The density is 20 electrons/meter.
Yes, I knew all this, and the manifestation of the field in different
frames.
> For a wire containing both positive and negative charges, and carrying
> current, even the *sign* of the charge density is frame dependent (because
> you have two lines of charge, that are contracted by different amounts).
> There is a unique frame (only considering motion along the wire) in which
> the charge density is zero.
I don't understand this argument. Start in one frame with two lines of op-
posite charges, moving relative to each other but cancelling each other to
give a zero charge density. In another frame moving relative to the first,
why should the two lines be measured as 'contracted by different amounts'?
The contraction is determined by the relative frame motion, not by the
motions of the charges.
> Another way to see this is to note that the total charge in a segment of
> wire can be constant in one frame, but change in another.
>
>Ralph Hartley wrote (at the end of my Dec 15 post):
>> Suppose you have a finite piece of wire that everyone agrees is neutral.
>> You start a current through it by simultaneously starting to add
>> electrons to one end, and removing them from the other. The total number
>> of electrons, and hence the charge of the wire doesn't change.
>>
>> Someone moving along the wire relative to you would disagree, because
>> they have a different definition of "simultaneous". They would say that
>> you started adding electrons to one end *before* (or after) you started
>> removing electrons form the other end. From *their* viewpoint you added
>> (or subtracted) electrons to the wire, giving it a charge.
I'm having trouble translating this finite-size wire argument into one
re- lating to *density*, a ratio of infinitesimals. In your example
above of a stationary finite length of wire, the charge of the whole
wire, integra- ted at any instant (in that frame), stays zero. But
because the charge insertion/removal signal travels at finite speed in
the wire, the charge density at any particular location will not remain
zero. So in a moving frame, measuring the total charge in a section of
wire at any instant (in the moving frame) is analogous, in the
stationary frame, to integrating the charge density in that section
using different times for each slice of wire. This obviously doesn't
have to stay zero in a transient situation.
So far I understand that non-zero charge *density* is frame dependent,
but don't yet quite see that the same holds for a *zero* charge density.
- Dushan Mitrovich
Dushan Mitrovich
Dec25-04, 02:39 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>towels.\nMake a roux with butter, oil and flour,\nbrown vegetables in the roux, then add chicken stock and\nallow to simmer for 20 minutes.\nAdd the patties or stuffed heads, and some loose crawfish,\nlobster, long piglet, or what have you.\nCook on low for 15 minutes, then allow it to set for at least\n15 minutes more.\nServe over steamed rice; this dish is very impressive!\n\n\n\nStuffed Cabbage Rolls\n\nBabies really can be found under a cabbage leaf -\nor one can arrange for ground beef to be found there instead.\n\n8 large cabbage leaves\n1 lb. lean ground newborn human filets, or ground chuck\nOnions\npeppers\ncelery\ngarlic\nsoy sauce\nsalt pepper, etc\nOlive oil\nbreadcrumbs\nTomato Gravy (see index)\n\nBoil the cabbage leaves for 2 minutes to soften.\nIn skillet, brown the meat in a little olive oil,\nthen add onions, peppers, and celery (all chopped finely)\nand season well.\nPlace in a large bowl and cool.\nAdd seasoned breadcrumbs and a little of the tomato gravy,\nenough to make the mixture pliable.\nDivide the stuffing among the cabbage leaves then roll.\nPlace seam down in a baking pan.\nLadle tomato gravy on top,\nand bake at 325° for 30 - 45 minutes.\n\n\n\nUmbilical Cordon Bleu\n\nNothing is so beautiful as the bond between mother and child,\nso why not consume it?\nChildren or chicken breasts will work wonderfully also.\n\n4 whole umbilical chords (or baby breasts, or chicken breasts)\n4 thin slices of smoked ham, and Gruyere cheese\nFlour\neggwash (milk and eggs)\nseasoned bread crumbs\n1 onion\nminced\nsalt\npepper\nbutter\nolive oil\n\nPound the breasts flat (parboil first if using umbilical\ncords so they won?t be tough).\nPlace a slice of ham and cheese on each, along with some minced onion\nthen fold in half, trimming neatly.\nDredge in flour, eggwash, then seasoned breadcrumbs;\nallow to sit for a few minutes.\nSauté in butter and olive oil until golden brown,\nabout\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>towels.
Make a roux with butter, oil and flour,
brown vegetables in the roux, then add chicken stock and
allow to simmer for 20 minutes.
Add the patties or stuffed heads, and some loose crawfish,
lobster, long piglet, or what have you.
Cook on low for 15 minutes, then allow it to set for at least
15 minutes more.
Serve over steamed rice; this dish is very impressive!
Stuffed Cabbage Rolls
Babies really can be found under a cabbage leaf -
or one can arrange for ground beef to be found there instead.
8 large cabbage leaves
1 lb. lean ground newborn human filets, or ground chuck
Onions
peppers
celery
garlic
soy sauce
salt pepper, etc
Olive oil
breadcrumbs
Tomato Gravy (see index)
Boil the cabbage leaves for 2 minutes to soften.
In skillet, brown the meat in a little olive oil,
then add onions, peppers, and celery (all chopped finely)
and season well.
Place in a large bowl and cool.
Add seasoned breadcrumbs and a little of the tomato gravy,
enough to make the mixture pliable.
Divide the stuffing among the cabbage leaves then roll.
Place seam down in a baking pan.
Ladle tomato gravy on top,
and bake at 325° for 30 - 45 minutes.
Umbilical Cordon Bleu
Nothing is so beautiful as the bond between mother and child,
so why not consume it?
Children or chicken breasts will work wonderfully also.
4 whole umbilical chords (or baby breasts, or chicken breasts)
4 thin slices of smoked ham, and Gruyere cheese
Flour
eggwash (milk and eggs)
seasoned bread crumbs
1 onion
minced
salt
pepper
butter
olive oil
Pound the breasts flat (parboil first if using umbilical
cords so they won?t be tough).
Place a slice of ham and cheese on each, along with some minced onion
then fold in half, trimming neatly.
Dredge in flour, eggwash, then seasoned breadcrumbs;
allow to sit for a few minutes.
Sauté in butter and olive oil until golden brown,
about
Dushan Mitrovich
Dec25-04, 03:05 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Ye shall eat the flesh of your sons, and the flesh of your daughters ye shall eat.\nLeviticus 26:29\n\n\nRoast Child with Cornbread Stuffing\n\nTurkey may be substituted for this classic holiday feast.\nAlthough time consuming, this dish seems to take longer than it actually does;\nas the entire house is filled with such a heavenly aroma,\nthe waiting becomes almost unbearable.\n\n1 whole child, cleaned and de-headed\n1 batch cornbread stuffing (see index)\n˝ cup melted butter\n\nRemove the giblets from the infant and set aside.\nStuff the cavity where the child?s genitals and anus were located\nusing ˝ cup per pound of meat.\nTie the arms flat to the body, then pull the skin flaps up to close the cavity.\nNow tie the thighs up tight to hold it all together.\nPlace breast side up in a large metal roasting pan.\nBake in 325° oven covered for 2 hours.\nRemove cover, stick a cooking thermometer deep into one of the\nbaby?s buttocks and cook uncovered till thermometer reads 190°,\nabout another hour.\n\n\n\nPro-Choice Po-Boy\n\nSoft-shelled crabs serve just as well in this classic so\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Ye shall eat the flesh of your sons, and the flesh of your daughters ye shall eat.
Leviticus 26:29
Roast Child with Cornbread Stuffing
Turkey may be substituted for this classic holiday feast.
Although time consuming, this dish seems to take longer than it actually does;
as the entire house is filled with such a heavenly aroma,
the waiting becomes almost unbearable.
1 whole child, cleaned and de-headed
1 batch cornbread stuffing (see index)
˝ cup melted butter
Remove the giblets from the infant and set aside.
Stuff the cavity where the child?s genitals and anus were located
using ˝ cup per pound of meat.
Tie the arms flat to the body, then pull the skin flaps up to close the cavity.
Now tie the thighs up tight to hold it all together.
Place breast side up in a large metal roasting pan.
Bake in 325° oven covered for 2 hours.
Remove cover, stick a cooking thermometer deep into one of the
baby?s buttocks and cook uncovered till thermometer reads 190°,
about another hour.
Pro-Choice Po-Boy
Soft-shelled crabs serve just as well in this classic so
Ralph Hartley
Dec28-04, 01:53 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Dushan Mitrovich wrote:\n> Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:\n>>\n> What I really had in mind when I wrote the\n> above is that _zero_ charge density stays zero independent of frame velo-\n> city. But according to your argument below, even that\'s not true.\n\nEven that\'s not true.\n\n> Yes, I knew all this, and the manifestation of the field in different\n> frames.\n\nYou don\'t know *all* that.\n\nAssume you have a neutral wire carying a current (in some frame, since it\nwill turn out to matter).\n\nA charged test particle moving along the wire feels a purely magnetic\nforce, since there is no eclectic field. In the particle\'s rest frame,\nthere cannot be a magnetic force, but the total force is not zero (because\nthe total force being zero *is* the same in all frames). In its rest frame\nthere must be an electric field. So the wire must be charged.\n\n> I don\'t understand this argument. Start in one frame with two lines of op-\n> posite charges, moving relative to each other but cancelling each other to\n> give a zero charge density. In another frame moving relative to the first,\n> why should the two lines be measured as \'contracted by different amounts\'?\n> The contraction is determined by the relative frame motion, not by the\n> motions of the charges.\n\nConsider two lines of charge, positive charges moving to the right at v\nand negative charges to the left, each with density 1 unit/m in its own\nrest frame.\n\nIn our frame they both have density slightly more (c/sqrt(c^2-v^2)), but\nstill the same, so the total charge density is zero.\n\nNow consider a new frame moving to the right with velocity v. The\ndensity of the positive charges (which are at rest) is 1, no contraction\napplies. But the negative charges are moving to the left with velocity\n-2v/(1+v^2/c^2) (remember the formula for adding velocities) the\ndensity in the new frame is (c^2+v^2)/(c^2-v*2) which is >1 if v!=0.\n\nWe didn\'t have to choose the rest frame of one of the lines of charge,\nbut it is simpler.\n\n>>Suppose you have a finite piece of wire that everyone agrees is neutral.\n>>You start a current through it by simultaneously starting to add\n>>electrons to one end, and removing them from the other. The total number\n>>of electrons, and hence the charge of the wire doesn\'t change.\n>>\n>>Someone moving along the wire relative to you would disagree, because\n>>they have a different definition of "simultaneous". They would say that\n>>you started adding electrons to one end *before* (or after) you started\n>>removing electrons form the other end. From *their* viewpoint you added\n>>(or subtracted) electrons to the wire, giving it a charge.\n\n> I\'m having trouble translating this finite-size wire argument into one\n> re- lating to *density*, a ratio of infinitesimals.\n\nJust look at the *average* density, which is the ratio of an integer and\nthe total length.\n\n> In your example\n> above of a stationary finite length of wire, the charge of the whole\n> wire, integrated at any instant (in that frame), stays zero.\n\nIn one frame it stays zero, in any other frame the total number of\nelectrons changes when the current starts, because it does not start at\nthe same time at both ends.\n\n> But\n> because the charge insertion/removal signal travels at finite speed in\n> the wire, the charge density at any particular location will not remain\n> zero.\n\nI don\'t quite understand what you mean.\n\nExcept when the current is changing, which we can assume happens very\nslowly, everyone agrees that electrons enter and leave at the same rate.\n\nIf I say the current took 100 seconds to increase from 0 to 1Amp\nstarting at t=0, any you agree about the left end of the wire, but say\nthat the current didn\'t start at the right end until t=0.0001, then we\nwill disagree about the total number of electrons in the wire at t=1000\nby 0.0001 Coulombs.\n\n> So in a moving frame, measuring the total charge in a section of\n> wire at any instant (in the moving frame) is analogous, in the\n> stationary frame, to integrating the charge density in that section\n> using different times for each slice of wire. This obviously doesn\'t\n> have to stay zero in a transient situation.\n\nSo you agree that the total number of electrons is frame dependent? The\ntotal number of protons in our finite section of wire is constant,\nbecause we never add or remove any. So the net charge changes, even if\nit was zero before.\n\nEventually, a net charge gets spread into a nonzero density.\n\n> So far I understand that non-zero charge *density* is frame dependent,\n> but don\'t yet quite see that the same holds for a *zero* charge density.\n\nBut if it starts out zero, and changes in one frame but not another,\nthen afterwards it must be zero in one frame, but not the other.\n\nRalph Hartley\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Dushan Mitrovich wrote:
> Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:
>>
> What I really had in mind when I wrote the
> above is that _zero_ charge density stays zero independent of frame velo-
> city. But according to your argument below, even that's not true.
Even that's not true.
> Yes, I knew all this, and the manifestation of the field in different
> frames.
You don't know *all* that.
Assume you have a neutral wire carying a current (in some frame, since it
will turn out to matter).
A charged test particle moving along the wire feels a purely magnetic
force, since there is no eclectic field. In the particle's rest frame,
there cannot be a magnetic force, but the total force is not zero (because
the total force being zero *is* the same in all frames). In its rest frame
there must be an electric field. So the wire must be charged.
> I don't understand this argument. Start in one frame with two lines of op-
> posite charges, moving relative to each other but cancelling each other to
> give a zero charge density. In another frame moving relative to the first,
> why should the two lines be measured as 'contracted by different amounts'?
> The contraction is determined by the relative frame motion, not by the
> motions of the charges.
Consider two lines of charge, positive charges moving to the right at v
and negative charges to the left, each with density 1 unit/m in its own
rest frame.
In our frame they both have density slightly more (c/\sqrt(c^2-v^2)), but
still the same, so the total charge density is zero.
Now consider a new frame moving to the right with velocity v. The
density of the positive charges (which are at rest) is 1, no contraction
applies. But the negative charges are moving to the left with velocity
-2v/(1+v^2/c^2) (remember the formula for adding velocities) the
density in the new frame is (c^2+v^2)/(c^2-v*2) which is >1 if v!=0.
We didn't have to choose the rest frame of one of the lines of charge,
but it is simpler.
>>Suppose you have a finite piece of wire that everyone agrees is neutral.
>>You start a current through it by simultaneously starting to add
>>electrons to one end, and removing them from the other. The total number
>>of electrons, and hence the charge of the wire doesn't change.
>>
>>Someone moving along the wire relative to you would disagree, because
>>they have a different definition of "simultaneous". They would say that
>>you started adding electrons to one end *before* (or after) you started
>>removing electrons form the other end. From *their* viewpoint you added
>>(or subtracted) electrons to the wire, giving it a charge.
> I'm having trouble translating this finite-size wire argument into one
> re- lating to *density*, a ratio of infinitesimals.
Just look at the *average* density, which is the ratio of an integer and
the total length.
> In your example
> above of a stationary finite length of wire, the charge of the whole
> wire, integrated at any instant (in that frame), stays zero.
In one frame it stays zero, in any other frame the total number of
electrons changes when the current starts, because it does not start at
the same time at both ends.
> But
> because the charge insertion/removal signal travels at finite speed in
> the wire, the charge density at any particular location will not remain
> zero.
I don't quite understand what you mean.
Except when the current is changing, which we can assume happens very
slowly, everyone agrees that electrons enter and leave at the same rate.
If I say the current took 100 seconds to increase from to 1Amp
starting at t=0, any you agree about the left end of the wire, but say
that the current didn't start at the right end until t=0.0001, then we
will disagree about the total number of electrons in the wire at t=1000
by .0001 Coulombs.
> So in a moving frame, measuring the total charge in a section of
> wire at any instant (in the moving frame) is analogous, in the
> stationary frame, to integrating the charge density in that section
> using different times for each slice of wire. This obviously doesn't
> have to stay zero in a transient situation.
So you agree that the total number of electrons is frame dependent? The
total number of protons in our finite section of wire is constant,
because we never add or remove any. So the net charge changes, even if
it was zero before.
Eventually, a net charge gets spread into a nonzero density.
> So far I understand that non-zero charge *density* is frame dependent,
> but don't yet quite see that the same holds for a *zero* charge density.
But if it starts out zero, and changes in one frame but not another,
then afterwards it must be zero in one frame, but not the other.
Ralph Hartley
Russell Blackadar
Dec29-04, 11:42 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Dushan Mitrovich wrote:\n\n[snip]\n\n> The contraction is determined by the relative frame motion, not by the\n> motions of the charges.\n\nThe relative frame motion of *which* frames?\n\nYou seem to think that the original measurement frame (in which\nthe particles have equal and opposite velocities) is somehow\nimportant, but why? We aren\'t using that frame any more.\n\nThe relevant frames for our measurement *now* are first of all\nthe new measurement frame, secondly the frame in which one set\nof particles is stationary (since that will be the frame in which\ntheir contraction is zero) and thirdly the frame in which the\nother set of particles is stationary (ditto). Since in our new\nmeasurement frame those two particle frames have different speeds,\nobviously they will have unequal Lorentz contractions in our new\nmeasurement frame. So Hartley is right.\n\nI think I understand where you went wrong -- yes, you *could*\nhave calculated the new particle-separation distances directly\nfrom their separation distances in the old measurement frame,\nvia a Lorentz transformation (which includes relativity of\nsimultaneity explicitly in the equations) without reference to\nthe particles\' rest frames. But that laborious calculation\nwould have worked out to exactly Hartley\'s result, not the one\nyou asserted. In other words, I think you confused the Lorentz\ntransformation -- which works in every case but can be cumbersome\nto apply -- with the Lorentz contraction formula. They are not\nthe same! The latter is a *shortcut* that gives the right answer\nonly under certain conditions; in particular, one of the frames\nmust be the rest frame of the thing measured.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Dushan Mitrovich wrote:
[snip]
> The contraction is determined by the relative frame motion, not by the
> motions of the charges.
The relative frame motion of *which* frames?
You seem to think that the original measurement frame (in which
the particles have equal and opposite velocities) is somehow
important, but why? We aren't using that frame any more.
The relevant frames for our measurement *now* are first of all
the new measurement frame, secondly the frame in which one set
of particles is stationary (since that will be the frame in which
their contraction is zero) and thirdly the frame in which the
other set of particles is stationary (ditto). Since in our new
measurement frame those two particle frames have different speeds,
obviously they will have unequal Lorentz contractions in our new
measurement frame. So Hartley is right.
I think I understand where you went wrong -- yes, you *could*
have calculated the new particle-separation distances directly
from their separation distances in the old measurement frame,
via a Lorentz transformation (which includes relativity of
simultaneity explicitly in the equations) without reference to
the particles' rest frames. But that laborious calculation
would have worked out to exactly Hartley's result, not the one
you asserted. In other words, I think you confused the Lorentz
transformation -- which works in every case but can be cumbersome
to apply -- with the Lorentz contraction formula. They are not
the same! The latter is a *shortcut* that gives the right answer
only under certain conditions; in particular, one of the frames
must be the rest frame of the thing measured.
Dushan Mitrovich
Dec30-04, 11:48 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:\n\n<snip>\n\n> You don\'t know *all* that.\n\nBy \'all\' I was referring to the items you enumerated. But anyway...\n\n> Assume you have a neutral wire carying a current (in some frame, since it\n> will turn out to matter).\n>\n> A charged test particle moving along the wire feels a purely magnetic\n> force, since there is no eclectic field. In the particle\'s rest frame,\n> there cannot be a magnetic force, but the total force is not zero (because\n> the total force being zero *is* the same in all frames). In its rest frame\n> there must be an electric field. So the wire must be charged.\n\nOkay, I understand this argument. In the \'neutral wire\' frame the JxB\nforce will push both positive and negative charges toward the center of the\nwire (the \'pinch\' force). In the rest frame of the positive charges the\nwire will look like it\'s negatively charged (to attract the positive\ncharges), and vice versa. So the charge density changes sign as the frame\nchanges between \'pos. at rest\' and \'neg. at rest\'.\n\n<snip>\n\n> Consider two lines of charge, positive charges moving to the right at v\n> and negative charges to the left, each with density 1 unit/m in its own\n> rest frame.\n>\n> In our frame they both have density slightly more (c/sqrt(c^2-v^2)), but\n> still the same, so the total charge density is zero.\n>\n> Now consider a new frame moving to the right with velocity v. The\n> density of the positive charges (which are at rest) is 1, no contraction\n> applies. But the negative charges are moving to the left with velocity\n> -2v/(1+v^2/c^2) (remember the formula for adding velocities) the\n> density in the new frame is (c^2+v^2)/(c^2-v*2) which is >1 if v!=0.\n>\n> We didn\'t have to choose the rest frame of one of the lines of charge,\n> but it is simpler.\n\nA good detailed explanation of how the change in charge density occurs.\nThanks.\n\n<snip>\n\n> Dushan Mitrovich wrote:\n>> But\n>> because the charge insertion/removal signal travels at finite speed in\n>> the wire, the charge density at any particular location will not remain\n>> zero.\n>\n> I don\'t quite understand what you mean.\n\nJust that the signal that pulls charges out of position travels at finite\nspeed down the wire. If the wire is two feet long, the charges in the\nmiddle don\'t know anything about the current flow for about 1 ns after you\nhave started to insert/remove charges from the ends (wire in rest frame).\n\n> Except when the current is changing, which we can assume happens very\n> slowly, everyone agrees that electrons enter and leave at the same rate.\n>\n> If I say the current took 100 seconds to increase from 0 to 1Amp\n> starting at t=0, any you agree about the left end of the wire, but say\n> that the current didn\'t start at the right end until t=0.0001, then we\n> will disagree about the total number of electrons in the wire at t=1000\n> by 0.0001 Coulombs.\n\nI wasn\'t quite sure what you were calculating here, but I\'m now convinced\nof your original assertion. Thanks for straightening me out on this point.\n\n- Dushan Mitrovich\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Ralph Hartley <hartley@aic.nrl.navy.mil> wrote:
<snip>
> You don't know *all* that.
By 'all' I was referring to the items you enumerated. But anyway...
> Assume you have a neutral wire carying a current (in some frame, since it
> will turn out to matter).
>
> A charged test particle moving along the wire feels a purely magnetic
> force, since there is no eclectic field. In the particle's rest frame,
> there cannot be a magnetic force, but the total force is not zero (because
> the total force being zero *is* the same in all frames). In its rest frame
> there must be an electric field. So the wire must be charged.
Okay, I understand this argument. In the 'neutral wire' frame the JxB
force will push both positive and negative charges toward the center of the
wire (the 'pinch' force). In the rest frame of the positive charges the
wire will look like it's negatively charged (to attract the positive
charges), and vice versa. So the charge density changes sign as the frame
changes between 'pos. at rest' and 'neg. at rest'.
<snip>
> Consider two lines of charge, positive charges moving to the right at v
> and negative charges to the left, each with density 1 unit/m in its own
> rest frame.
>
> In our frame they both have density slightly more (c/\sqrt(c^2-v^2)), but
> still the same, so the total charge density is zero.
>
> Now consider a new frame moving to the right with velocity v. The
> density of the positive charges (which are at rest) is 1, no contraction
> applies. But the negative charges are moving to the left with velocity
> -2v/(1+v^2/c^2) (remember the formula for adding velocities) the
> density in the new frame is (c^2+v^2)/(c^2-v*2) which is >1 if v!=0.
>
> We didn't have to choose the rest frame of one of the lines of charge,
> but it is simpler.
A good detailed explanation of how the change in charge density occurs.
Thanks.
<snip>
> Dushan Mitrovich wrote:
>> But
>> because the charge insertion/removal signal travels at finite speed in
>> the wire, the charge density at any particular location will not remain
>> zero.
>
> I don't quite understand what you mean.
Just that the signal that pulls charges out of position travels at finite
speed down the wire. If the wire is two feet long, the charges in the
middle don't know anything about the current flow for about 1 ns after you
have started to insert/remove charges from the ends (wire in rest frame).
> Except when the current is changing, which we can assume happens very
> slowly, everyone agrees that electrons enter and leave at the same rate.
>
> If I say the current took 100 seconds to increase from to 1Amp
> starting at t=0, any you agree about the left end of the wire, but say
> that the current didn't start at the right end until t=0.0001, then we
> will disagree about the total number of electrons in the wire at t=1000
> by .0001 Coulombs.
I wasn't quite sure what you were calculating here, but I'm now convinced
of your original assertion. Thanks for straightening me out on this point.
- Dushan Mitrovich
Dushan Mitrovich
Dec30-04, 11:48 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Russell Blackadar <russell@mdli.com> wrote:\n>Dushan Mitrovich wrote:\n>\n>[snip]\n>\n>> The contraction is determined by the relative frame motion, not by the\n>> motions of the charges.\n>\n> The relative frame motion of *which* frames?\n>\n> You seem to think that the original measurement frame (in which\n> the particles have equal and opposite velocities) is somehow\n> important, but why? We aren\'t using that frame any more.\n\nActually, no, I didn\'t think that frame was [more] important.\n\n> The relevant frames for our measurement *now* are first of all\n> the new measurement frame, secondly the frame in which one set\n> of particles is stationary (since that will be the frame in which\n> their contraction is zero) and thirdly the frame in which the\n> other set of particles is stationary (ditto). Since in our new\n> measurement frame those two particle frames have different speeds,\n> obviously they will have unequal Lorentz contractions in our new\n> measurement frame. So Hartley is right.\n\nYes, I was convinced by Hartley\'s explanations - see my response to him.\n\n> I think I understand where you went wrong -- yes, you *could*\n> have calculated the new particle-separation distances directly\n> from their separation distances in the old measurement frame,\n> via a Lorentz transformation (which includes relativity of\n> simultaneity explicitly in the equations) without reference to\n> the particles\' rest frames. But that laborious calculation\n> would have worked out to exactly Hartley\'s result, not the one\n> you asserted. In other words, I think you confused the Lorentz\n> transformation -- which works in every case but can be cumbersome\n> to apply -- with the Lorentz contraction formula. They are not\n> the same! The latter is a *shortcut* that gives the right answer\n> only under certain conditions; in particular, one of the frames\n> must be the rest frame of the thing measured.\n\n- Dushan Mitrovich\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Russell Blackadar <russell@mdli.com> wrote:
>Dushan Mitrovich wrote:
>
>[snip]
>
>> The contraction is determined by the relative frame motion, not by the
>> motions of the charges.
>
> The relative frame motion of *which* frames?
>
> You seem to think that the original measurement frame (in which
> the particles have equal and opposite velocities) is somehow
> important, but why? We aren't using that frame any more.
Actually, no, I didn't think that frame was [more] important.
> The relevant frames for our measurement *now* are first of all
> the new measurement frame, secondly the frame in which one set
> of particles is stationary (since that will be the frame in which
> their contraction is zero) and thirdly the frame in which the
> other set of particles is stationary (ditto). Since in our new
> measurement frame those two particle frames have different speeds,
> obviously they will have unequal Lorentz contractions in our new
> measurement frame. So Hartley is right.
Yes, I was convinced by Hartley's explanations - see my response to him.
> I think I understand where you went wrong -- yes, you *could*
> have calculated the new particle-separation distances directly
> from their separation distances in the old measurement frame,
> via a Lorentz transformation (which includes relativity of
> simultaneity explicitly in the equations) without reference to
> the particles' rest frames. But that laborious calculation
> would have worked out to exactly Hartley's result, not the one
> you asserted. In other words, I think you confused the Lorentz
> transformation -- which works in every case but can be cumbersome
> to apply -- with the Lorentz contraction formula. They are not
> the same! The latter is a *shortcut* that gives the right answer
> only under certain conditions; in particular, one of the frames
> must be the rest frame of the thing measured.
- Dushan Mitrovich
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