View Full Version : 4-magnitudes in curved space?
Jon Absoul
Jun6-04, 04:27 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>I thought I knew this stuff, but now I feel stupid!\n\nIn flat spacetime, a particle with rest mass m moving along one\ncoordinate axis has\n(1) 4-velocity \\gamma(1, v, 0, 0), and\n(2) 4-momentum m \\gamma(1, v, 0, 0),\nwhere \\gamma = \\sqrt{1-v^2}.\nAnd (massless) radiation with energy-density \\rho moving along one\ncoordinate axis has\n(3) 4-velocity (1, 1, 0, 0) (YES…?), and\n(4) Energy-momentum tensor T^{mn}=\\rho\n((1,1,0,0),(1,1,0,0),(0,0,0,0),(0,0, 0,0))\n\nBut what about in curved spacetime?\nFor example, let\'s take Schwarzshild spacetime:\ng_{ab} = ((A,0,0,0),(0,-B,0,0),(0,0,-r^2,0,),(0,0,0,-r^2\nsin^2\\theta)),\nwhere A=(1-2M/r) and B=1/A.\nWhat is the\n(1) 4-velocity (contravariant) of a particle of rest mass m sitting\nstill at r=R,\n(2) 4-momentum (contravariant) of same,\n(3) 4-velocity of outgoing (massless) radiation ("sunlight"), and\n(4) the energy-momentum tensor of same?\n\nI can\'t even figure out if the 4-magnitude of p should be m or\nm*\\sqrt{A}!\n\nNo, this is not homework. I\'m self-studying from several textbooks.\n\n-jon\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>I thought I knew this stuff, but now I feel stupid!
In flat spacetime, a particle with rest mass m moving along one
coordinate axis has
(1) 4-velocity \gamma(1, v, 0, 0), and
(2) 4-momentum m \gamma(1, v, 0, 0),
where \gamma = \sqrt{1-v^2}.
And (massless) radiation with energy-density \rho moving along one
coordinate axis has
(3) 4-velocity (1, 1, 0, 0) (YES…?), and
(4) Energy-momentum tensor T^{mn}=\rho((1,1,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0 ))
But what about in curved spacetime?
For example, let's take Schwarzshild spacetime:
g_{ab} = ((A,0,0,0),(0,-B,0,0),(0,0,-r^2,0,),(0,0,0,-r^2sin^2\theta)),
where A=(1-2M/r) and B=1/A.
What is the
(1) 4-velocity (contravariant) of a particle of rest mass m sitting
still at r=R,
(2) 4-momentum (contravariant) of same,
(3) 4-velocity of outgoing (massless) radiation ("sunlight"), and
(4) the energy-momentum tensor of same?
I can't even figure out if the 4-magnitude of p should be m orm*\sqrt{A}!
No, this is not homework. I'm self-studying from several textbooks.
-jon
Greg Egan
Jun8-04, 04:03 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>\nIn article <e55764f0.0406051648.32a0913c@posting.google.com>, \njonabsoul@yahoo.com (Jon Absoul) wrote:\n\n[snip]\n\n> I can\'t even figure out if the 4-magnitude of p should be m or\n> m*\\sqrt{A}!\n\nThe magnitude of the energy-momentum 4-vector p always equals the rest\nmass, m. (Some people, like me, choose a -+++ signature for the metric,\nso g(p,p)=-m^2, but with your choice of signature, g(p,p)=m^2).\nSimilarly, if the 4-velocity is u, g(u,u)=1 with your choice of signature.\n\nJust remember that the tangent space at a given event, where these\nvectors live, is indistinguishable from that of flat-spacetime SR. So\nfor localised stuff -- vectors and tensors at a single event -- the only\nreal complication is that in general a coordinate system such as the\nSchwarzschild one will give rise to a non-orthonormal basis for the\ntangent space. If you take it as axiomatic that in an orthonormal basis\neverything (local) looks just as it does in SR, then all the results in a\ngeneral spacetime can be seen to be change-of-basis mathematics.\n\nThere are some calculations of a few things for the Schwarzschild metric\nthat you might find of interest at:\n\nhttp://gregegan.customer.netspace.net.au/PLANCK/Tour/TourNotes.html\n\nNo energy-momentum tensors, though.\n\nGreg Egan\n\nEmail address (remove name of animal and add standard punctuation):\ngregegan netspace zebra net au\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 <e55764f0.0406051648.32a0913c@posting.google.com>,
jonabsoul@yahoo.com (Jon Absoul) wrote:
[snip]
> I can't even figure out if the 4-magnitude of p should be m or
> m*\sqrt{A}!
The magnitude of the energy-momentum 4-vector p always equals the rest
mass, m. (Some people, like me, choose a -+++ signature for the metric,
so g(p,p)=-m^2, but with your choice of signature, g(p,p)=m^2).
Similarly, if the 4-velocity is u, g(u,u)=1 with your choice of signature.
Just remember that the tangent space at a given event, where these
vectors live, is indistinguishable from that of flat-spacetime SR. So
for localised stuff -- vectors and tensors at a single event -- the only
real complication is that in general a coordinate system such as the
Schwarzschild one will give rise to a non-orthonormal basis for the
tangent space. If you take it as axiomatic that in an orthonormal basis
everything (local) looks just as it does in SR, then all the results in a
general spacetime can be seen to be change-of-basis mathematics.
There are some calculations of a few things for the Schwarzschild metric
that you might find of interest at:
http://gregegan.customer.netspace.net.au/PLANCK/Tour/TourNotes.html
No energy-momentum tensors, though.
Greg Egan
Email address (remove name of animal and add standard punctuation):
gregegan netspace zebra net au
tessel@tum.bot
Jun12-04, 07:19 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>On Sun, 6 Jun 2004, Jon Absoul wrote:\n\n> I thought I knew this stuff, but now I feel stupid!\n>\n> In flat spacetime, a particle with rest mass m moving along one\n> coordinate axis has\n> (1) 4-velocity \\gamma(1, v, 0, 0), and\n> (2) 4-momentum m \\gamma(1, v, 0, 0),\n> where \\gamma =3D \\sqrt{1-v^2}.\n> And (massless) radiation with energy-density \\rho moving along one\n> coordinate axis has\n> (3) 4-velocity (1, 1, 0, 0) (YES=85?), and\n> (4) Energy-momentum tensor T^{mn}=3D\\rho\n> ((1,1,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0))\n\nI\'d highly recommend learning to think of vector fields as first order\n(homogeneous) linear operators. For example, your (3) is\n\nX =3D @/@t + @/@x\n\nFor example, to change to a null chart\n\np =3D (t+x)/2, q =3D (t-x)/2\n\nwe just apply X to p, q and find that\n\nX =3D @/@p\n\nSo your vector field is the "flow" which translates E^(1,1) along the nul=\nl\nvector @/@p. I know you knew that, but the point is that this notation i=\ns\nfar superious to the mysterious (1,1,0,0), or even worse, to something\nwritten with "delta functions" (which really play a completely different\nrole in this subject).\n\nSee the archived post "What is a Vector Field?" on Baez\'s webpages. See\nalso my recent posts discussing "canonical energy-momentum tensors" for\nthe wave equation, Klein-Gordon equation, sine-Gordon equation.\n\n> But what about in curved spacetime?\n> For example, let\'s take Schwarzshild spacetime:\n> g_{ab} =3D ((A,0,0,0),(0,-B,0,0),(0,0,-r^2,0,),(0,0,0,-r^2\n> sin^2\\theta)),\n> where A=3D(1-2M/r) and B=3D1/A.\n\nThis is the Schwarzschild chart for the exterior region, and you should\nalways include the intended range of the coordinates, e.g.\n\nds^2 =3D -(1-2M/r) dt^2 + dr^2/(1-2M/r) + r^2 (du^2 + sin(u)^2 dv^2),\n\n-infty < t < infty, 2M < r < infty, 0 < u < pi, -pi < v < pi\n\n> What is the\n>\n> (1) 4-velocity (contravariant) of a particle of rest mass m sitting\n> still at r=3DR,\n\nConsider the coordinate vector @/@t. This is the timelike irrotational\nKilling vector which expresses the fact that the spacetime is -static-.\nWe can normalize it to get the unit length velocity vector field or flow\nwhose integral curves are the world lines of the family of static\nobservers:\n\ne_1 =3D 1/sqrt(1-2M/r) @/@t ~ (1 + M/r) @/@t\n\nNotice that as r -> infty, this -> @/@t, as we would expect for an\nasympotically flat spacetime.\n\n> (2) 4-momentum (contravariant) of same,\n\nThe momentum is most properly treated as a covector field (although it ha=\ns\na dual vector field), which we obtain from the coordinate covector dt the\nsame way as in flat spacetime. If we now pretend that our observers are\ntest particles each with mass m << M, then\n\no^1 =3D -sqrt(1-2M/r) dt\n\nso\n\npi =3D m o^1 =3D -m sqrt(1-2M/r) dt\n\nwith dual vector field\n\nP =3D m/sqrt(1-2M/r) @/@t\n\nThis is a timelike vector field with uniform length ||P|| =3D m.\n\n(The minus sign arises from demanding metric duality between covectors\n[aka one-forms] and vectors; use here the Minkowski metric, which is vali=\nd\nin the tangent space to each event. See the figure in the chapter in MTW\nwhich explains why the covectors which are dual to -forward pointing-\ntimelike vectors must be thought of as little spacelike surface elements\nwhich are marching into the -past-!)\n\n> (3) 4-velocity of outgoing (massless) radiation ("sunlight"),\n\np =3D -E (o^1 + e^2) =3D -E sqrt(1-2M/r) dt - E/sqrt(1-2M/r) dr\n\n> (4) the energy-momentum tensor of same?\n\nFirst see an archived post on "Relativity on the World Wide Web"\n\nhttp://math.ucr.edu/home/baez/PUB/einsteinmaxwell\n\nwhich explains how to use Noether\'s theorem to find the canonical energy\nmomentum tensor for EM fields on curved spacetimes.\n\nFollowing the trick there, let me answer an easier but related question\nfirst:\n\nTo obtain the energy momentum tensor of a spherically symmetric\n"electrostatic test field" (one whose field energy density is too small t=\no\ndisturb the Schwarzschild geometry), we can use a vector potential which\nis an undetermined spherically symmetric scalar multiple of the flow whic=\nh\ngenerates time translation (our timelike Killing vector):\n\nA =3D f(r) @/@t\n\nIn the nontrivial Maxwell equation we demand that the current vanish (onl=\ny\n"test charges" allowed in our "electrovacuum" region), which gives (up to\nadditive constant)\n\nf(r) =3D q/(r-2M)\n\nSo, to obtain a spherically symmetric static EM field we use the vector\npotential\n\nA =3D 1/(1-2M/r) q/r @/@t\n\nwhich has the dual covector -q/r dt, just as for a spherically symmetric\nstatic EM field in Minkowski vacuum. Then, proceeding as for flat\nspacetime, the EM field measured by our static observers is the\nspherically symmetric electrostatic field\n\nE =3D -q/r^2 e_2 =3D -q/r^2\n\nB =3D 0\n\nand the energy-momentum tensor is\n\nT^(ab) =3D 8 pi q^2/r^4 diag(1,-1,1,1)\n\nThe latter is exactly what we\'d get in flat spacetime! Notice we have a\nnonnull EM field with Lorentz invariants\n\n1/4 F_(ab) F^(ab) =3D q^2/r^4 =3D |B|^2 - |E|^2\n\n1/2 F_(ab) *F^(ab) =3D 0 =3D E dot B\n\nJust as in flat spacetime, we could "Wick rotate" to get a spherically\nsymmetric magnetostatic field with the same EM stress-energy tensor.\n\nVery important: we have expanded T^(ab) wrt the the -frame- (aka\northonormal basis) read off from the coordinate basis for our coordinate\nchart! Sometimes this is called "computing the physical components". See\nMTW or other archived posts on RWWW. Once we have this, the coordinate\nbasis components are easily found:\n\nG^(tt) =3D 1(1-2m/r) 8 pi q^2/r^4\n\nG^(rr) =3D -(1-2m/r) 8 pi q^2/r^4\n\nG^uu =3D 1/r^2 8 pi q^2/r^4\n\nG^vv =3D 1/r^2/sin(u)^2 8 pi q^2/r^4\n\nBut the frame components are the components which would be -measured- by\nour static observers! The coordinate basis components have no physical\nmeaning; they are just what we obtain using curvilinear coordinate\nmethods; to obtain the physical components we\'d need to use the usual\nindex gymnastics conversion, in this case, from the coordinate basis to\nour frame (our orthonormal basis of vectors).\n\nNote that it would have been no harder to have found the\nReissner-Nordstrom electrovacuum in this way (see the archived post).\n\nNext, reverting for a moment to flat spacetime, to obtain a -null- EM tes=\nt\nfield, we should take for our vector potential an undetermined scalar\nmultiple of a flow generating a -parabolic- Lorentz transformation fixing\nthe x axis:\n\nA =3D f(r) [z @/@t + t @/@z + z @/@x - x @/@z]\n--------------- -----------------\nboost rotate\nalong z axis around y axis\n\n(the point here is that the z axis and y axis are not the x axis--- there\nare of course other choices of vector potentials which will do the job.)\n\nDemanding that the current computed from the nontrivial Maxwell equation\nshould vanish, we find that we must take\n\nA =3D q/2 [z @/@t + z @/@x + (z- x) @/@z]\n\nThen we obtain vanishing current, as we should, and we find the EM fields\n\nE =3D q e_4\n\nB =3D -q e_3\n\n(notice that these are transverse to the direction of propagation e_1 =3D\n@/@x; only in flat spacetime can a coordinate basis also be a frame). Th=\ne\nLorentz invariants of the EM field tensor F_(ab) are\n\n1/4 F_(ab) F^(ab) =3D 1/4 F_(ab) *F^(ab) =3D 0\n\nThis is the EM field of an EM wave propagating in the x direction, as\ndesired. The EM stress-energy tensor is:\n\n[ 1 1 0 0 ]\nT^(ab) =3D 16 pi q^2 [ 1 1 0 0 ]\n[ 0 0 0 0 ]\n[ 0 0 0 0 ]\n\nas expected.\n\nExercise: try to follow the analogous procedure to obtain spherically\noutgoing EM radiation in the polar spherical chart for Minkowksi vacuum.\nWhat about the outgoing Eddington chart for Minkowksi vacuum? (Let m =3D=\n0\nin the usual outgoing Eddington chart for Schwarzschild vacuum.)\n\nExercise: this suggests that to obtain a spherically symmetric test field\nin the Schwarzschild vacuum, corresponding to outgoing EM radiation, we\nshould reuse the Ansatz which succeeded in the previous exercise.\n\nWhat about an -exact- solution? Well, the outgoing radiation must carry\noff energy, so it should reduce the mass of the star which we regard\n(together with the energy density of the EM field itself) as the source o=\nf\nthe gravitational field. So we must look for a time varying but\nspherically symmetric solution whose Einstein tensor has the right form t=\no\nmatch the EM stress-energy tensor of a null EM field for radially outgoin=\ng\nradiation. This can be obtained using the same trick!\n\nIt turns out that the prescription for obtaining the Vaidya null dust is\nvery simple:\n\n1. change to the outgoing Eddington chart for the Schwarzschild vacuum, i=\nn\nwhich dr becomes a null covector,\n\n2. let the mass be a function of the new time coordinate.\n\nFor details see\n\nhttp://math.ucr.edu/home/baez/PUB/vaidya\n\nfor the "Vaidya null dust", an exact solution of the EFE which models a\nspherically symmetric outflux of massless radiation, say from a\nspherically symmetric star, which can vary in time. Time reversing this,\nwe can construct an extremely valuable thought experiment in which a\nspherically symmetric shell of massless radiation (e.g. EM radiation)\ncollapses to form a black hole. C.f. the discussion in Feynman\'s\nLectures.\n\n> I\'m self-studying from several textbooks.\n\nCool! Which ones?\n\n"T. Essel" (a mathbookworm hiding somewhere in cyberspace)\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 Sun, 6 Jun 2004, Jon Absoul wrote:
> I thought I knew this stuff, but now I feel stupid!
>
> In flat spacetime, a particle with rest mass m moving along one
> coordinate axis has
> (1) 4-velocity \gamma(1, v, 0, 0), and
> (2) 4-momentum m \gamma(1, v, 0, 0),
> where \gamma =3D \sqrt{1-v^2}.
> And (massless) radiation with energy-density \rho moving along one
> coordinate axis has
> (3) 4-velocity (1, 1, 0, 0) (YES=85?), and
> (4) Energy-momentum tensor T^{mn}=3D\rho
> ((1,1,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0))
I'd highly recommend learning to think of vector fields as first order
(homogeneous) linear operators. For example, your (3) is
X =3D @/@t + @/@x
For example, to change to a null chart
p =3D (t+x)/2, q =3D (t-x)/2
we just apply X to p, q and find that
X =3D @/@p
So your vector field is the "flow" which translates E^(1,1) along the nul=
l
vector @/@p. I know you knew that, but the point is that this notation i=
s
far superious to the mysterious (1,1,0,0), or even worse, to something
written with "\delta functions" (which really play a completely different
role in this subject).
See the archived post "What is a Vector Field?" on Baez's webpages. See
also my recent posts discussing "canonical energy-momentum tensors" for
the wave equation, Klein-Gordon equation, sine-Gordon equation.
> But what about in curved spacetime?
> For example, let's take Schwarzshild spacetime:
> g_{ab} =3D ((A,0,0,0),(0,-B,0,0),(0,0,-r^2,0,),(0,0,0,-r^2
> sin^2\theta)),
> where A=3D(1-2M/r) and B=3D1/A.
This is the Schwarzschild chart for the exterior region, and you should
always include the intended range of the coordinates, e.g.
ds^2 =3D -(1-2M/r) dt^2 + dr^2/(1-2M/r) + r^2 (du^2 + sin(u)^2 dv^2),-\infty < t < \infty, 2M < r < \infty,< u < \pi, -\pi < v < \pi
> What is the
>
> (1) 4-velocity (contravariant) of a particle of rest mass m sitting
> still at r=3DR,
Consider the coordinate vector @/@t. This is the timelike irrotational
Killing vector which expresses the fact that the spacetime is -static-.
We can normalize it to get the unit length velocity vector field or flow
whose integral curves are the world lines of the family of static
observers:
e_1 =3D 1/\sqrt(1-2M/r) @/@t ~ (1 + M/r) @/@t
Notice that as r -> \infty, this -> @/@t, as we would expect for an
asympotically flat spacetime.
> (2) 4-momentum (contravariant) of same,
The momentum is most properly treated as a covector field (although it ha=
s
a dual vector field), which we obtain from the coordinate covector dt the
same way as in flat spacetime. If we now pretend that our observers are
test particles each with mass m << M, then
o^1 =3D -\sqrt(1-2M/r) dt
so
\pi =3D m o^1 =3D -m \sqrt(1-2M/r) dt
with dual vector field
P =3D m/\sqrt(1-2M/r) @/@t
This is a timelike vector field with uniform length ||P|| =3D m.
(The minus sign arises from demanding metric duality between covectors
[aka one-forms] and vectors; use here the Minkowski metric, which is vali=
d
in the tangent space to each event. See the figure in the chapter in MTW
which explains why the covectors which are dual to -forward pointing-
timelike vectors must be thought of as little spacelike surface elements
which are marching into the -past-!)
> (3) 4-velocity of outgoing (massless) radiation ("sunlight"),
p =3D -E (o^1 + e^2) =3D -E \sqrt(1-2M/r) dt - E/\sqrt(1-2M/r) dr
> (4) the energy-momentum tensor of same?
First see an archived post on "Relativity on the World Wide Web"
http://math.ucr.edu/home/baez/PUB/einsteinmaxwell
which explains how to use Noether's theorem to find the canonical energy
momentum tensor for EM fields on curved spacetimes.
Following the trick there, let me answer an easier but related question
first:
To obtain the energy momentum tensor of a spherically symmetric
"electrostatic test field" (one whose field energy density is too small t=
o
disturb the Schwarzschild geometry), we can use a vector potential which
is an undetermined spherically symmetric scalar multiple of the flow whic=
h
generates time translation (our timelike Killing vector):
A =3D f(r) @/@t
In the nontrivial Maxwell equation we demand that the current vanish (onl=
y
"test charges" allowed in our "electrovacuum" region), which gives (up to
additive constant)
f(r) =3D q/(r-2M)
So, to obtain a spherically symmetric static EM field we use the vector
potential
A =3D 1/(1-2M/r) q/r @/@t
which has the dual covector -q/r dt, just as for a spherically symmetric
static EM field in Minkowski vacuum. Then, proceeding as for flat
spacetime, the EM field measured by our static observers is the
spherically symmetric electrostatic field
E =3D -q/r^2 e_2 =3D -q/r^2B =3D
and the energy-momentum tensor is
T^(ab) =3D 8 \pi q^2/r^4 diag(1,-1,1,1)
The latter is exactly what we'd get in flat spacetime! Notice we have a
nonnull EM field with Lorentz invariants
1/4 F_(ab) F^(ab) =3D q^2/r^4 =3D |B|^2 - |E|^21/2 F_(ab) *F^(ab) =3D=3D E[/itex] dot B
Just as in flat spacetime, we could "Wick rotate" to get a spherically
symmetric magnetostatic field with the same EM stress-energy tensor.
Very important: we have expanded T^(ab) wrt the the -frame- (aka
orthonormal basis) read off from the coordinate basis for our coordinate
chart! Sometimes this is called "computing the physical components". See
MTW or other archived posts on RWWW. Once we have this, the coordinate
basis components are easily found:
G^(tt) =3D 1(1-2m/r) 8 \pi q^2/r^4G^(rr) =3D -(1-2m/r) 8 \pi q^2/r^4G^{uu} =3D 1/r^2 8 \pi q^2/r^4G^{vv} =3D 1/r^2/sin(u)^2 8 \pi q^2/r^4
But the frame components are the components which would be -measured- by
our static observers! The coordinate basis components have no physical
meaning; they are just what we obtain using curvilinear coordinate
methods; to obtain the physical components we'd need to use the usual
index gymnastics conversion, in this case, from the coordinate basis to
our frame (our orthonormal basis of vectors).
Note that it would have been no harder to have found the
Reissner-Nordstrom electrovacuum in this way (see the archived post).
Next, reverting for a moment to flat spacetime, to obtain a -null- EM tes=
t
field, we should take for our vector potential an undetermined scalar
multiple of a flow generating a -parabolic- Lorentz transformation fixing
the x axis:
A =3D f(r) [z @/@t + t @/@z + z @/@x - x @/@z]
--------------- -----------------
boost rotate
along z axis around y axis
(the point here is that the z axis and y axis are not the x axis--- there
are of course other choices of vector potentials which will do the job.)
Demanding that the current computed from the nontrivial Maxwell equation
should vanish, we find that we must take
[itex]A =3D q/2 [z @/@t + z @/@x + (z- x) @/@z]
Then we obtain vanishing current, as we should, and we find the EM fields
E =3D q e_4B =3D -q e_3
(notice that these are transverse to the direction of propagation e_1 =3D@/@x; only in flat spacetime can a coordinate basis also be a frame). Th=
e
Lorentz invariants of the EM field tensor F_(ab) are
1/4 F_(ab) F^(ab) =3D 1/4 F_(ab) *F^(ab) =3D
This is the EM field of an EM wave propagating in the x direction, as
desired. The EM stress-energy tensor is:
[ 1 1 ]
T^(ab) =3D 16 \pi q^2 [ 1 1 ]
[ ]
[ ]
as expected.
Exercise: try to follow the analogous procedure to obtain spherically
outgoing EM radiation in the polar spherical chart for Minkowksi vacuum.
What about the outgoing Eddington chart for Minkowksi vacuum? (Let m =3D=
in the usual outgoing Eddington chart for Schwarzschild vacuum.)
Exercise: this suggests that to obtain a spherically symmetric test field
in the Schwarzschild vacuum, corresponding to outgoing EM radiation, we
should reuse the Ansatz which succeeded in the previous exercise.
What about an -exact- solution? Well, the outgoing radiation must carry
off energy, so it should reduce the mass of the star which we regard
(together with the energy density of the EM field itself) as the source o=
f
the gravitational field. So we must look for a time varying but
spherically symmetric solution whose Einstein tensor has the right form t=
o
match the EM stress-energy tensor of a null EM field for radially outgoin=
g
radiation. This can be obtained using the same trick!
It turns out that the prescription for obtaining the Vaidya null dust is
very simple:
1. change to the outgoing Eddington chart for the Schwarzschild vacuum, i=
n
which dr becomes a null covector,
2. let the mass be a function of the new time coordinate.
For details see
http://math.ucr.edu/home/baez/PUB/vaidya
for the "Vaidya null dust", an exact solution of the EFE which models a
spherically symmetric outflux of massless radiation, say from a
spherically symmetric star, which can vary in time. Time reversing this,
we can construct an extremely valuable thought experiment in which a
spherically symmetric shell of massless radiation (e.g. EM radiation)
collapses to form a black hole. C.f. the discussion in Feynman's
Lectures.
> I'm self-studying from several textbooks.
Cool! Which ones?
"T. Essel" (a mathbookworm hiding somewhere in cyberspace)
<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>The magnitude of the energy-momentum 4-vector p always equals the rest\n>mass, m. (Some people, like me, choose a -+++ signature for the metric,\n>so g(p,p)=-m^2, but with your choice of signature, g(p,p)=m^2).\n>Similarly, if the 4-velocity is u, g(u,u)=1 with your choice of\n>signature.\n\nHuh? So the lorentz transform (4-d) is, whatever, \'unitary\'?\nAt least in some sense or other.\n\nOk, that\'s not so implausible now I think about it.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<\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
>The magnitude of the energy-momentum 4-vector p always equals the rest
>mass, m. (Some people, like me, choose a -+++ signature for the metric,
>so g(p,p)=-m^2, but with your choice of signature, g(p,p)=m^2).
>Similarly, if the 4-velocity is u, g(u,u)=1 with your choice of
>signature.
Huh? So the lorentz transform (4-d) is, whatever, 'unitary'?
At least in some sense or other.
Ok, that's not so implausible now I think about it.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<
Frank Hellmann
Jun14-04, 02:50 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\nOz <oz@farmeroz.port995.com> wrote in message news:<XAYKo1gsYbxAFwAU@farmeroz.port995.com>...\n> Greg Egan <gregegan@netspace.zebra.net.au> writes\n> >The magnitude of the energy-momentum 4-vector p always equals the rest\n> >mass, m. (Some people, like me, choose a -+++ signature for the metric,\n> >so g(p,p)=-m^2, but with your choice of signature, g(p,p)=m^2).\n> >Similarly, if the 4-velocity is u, g(u,u)=1 with your choice of\n> >signature.\n>\n> Huh? So the lorentz transform (4-d) is, whatever, \'unitary\'?\n> At least in some sense or other.\n>\n> Ok, that\'s not so implausible now I think about it.\n\nIt\'s orthogonal, or rather, pseudo orthogonal. The Lorentz Group is\nSO(3,1) Where the 3,1 refers to the signature of the metric.\n\nGallileo transformations are replace by rotation and translation in\nMinkowski space, there\'s nothing else. That\'s why SR is much simpler\nthen Newtonian Mechanics. ;)\n\n\n---\nfrank\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:<XAYKo1gsYbxAFwAU@farmeroz.port995.com>...
> Greg Egan <gregegan@netspace.zebra.net.au> writes
> >The magnitude of the energy-momentum 4-vector p always equals the rest
> >mass, m. (Some people, like me, choose a -+++ signature for the metric,
> >so g(p,p)=-m^2, but with your choice of signature, g(p,p)=m^2).
> >Similarly, if the 4-velocity is u, g(u,u)=1 with your choice of
> >signature.
>
> Huh? So the lorentz transform (4-d) is, whatever, 'unitary'?
> At least in some sense or other.
>
> Ok, that's not so implausible now I think about it.
It's orthogonal, or rather, pseudo orthogonal. The Lorentz Group is
SO(3,1) Where the 3,1 refers to the signature of the metric.
Gallileo transformations are replace by rotation and translation in
Minkowski space, there's nothing else. That's why SR is much simpler
then Newtonian Mechanics. ;)
---
frank
<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>Frank Hellmann <C.i.m@gmx.net> writes\n>It\'s orthogonal, or rather, pseudo orthogonal. The Lorentz Group is\n>SO(3,1) Where the 3,1 refers to the signature of the metric.\n>\n>Gallileo transformations are replace by rotation and translation in\n>Minkowski space, there\'s nothing else. That\'s why SR is much simpler\n>then Newtonian Mechanics. ;)\n\nOk. So if I want to describe how an object looks in another frame of\nreference, all I need to do is multiply it by the appropriate LT?\n\nSo if I have some object, stationary in one frame, I would represent its\nmomentum as\n\nm[c,0,0,0] = M\n\nIn another frame it would be ML, but the magnitude in both frames would\nbe the same, as mentioned by greg egan.\n\nNow if we wanted to add a force, it seems to me (in my infinite\nignorance) that we would quite like an infinitesimal - dL.\n\nThen we could have an equation of motion like\n\nM\' = MdL.\n\nAll we would be doing would be moving momentum from one direction, and\ntransferring it to another, since the magnitude of the 4-momentum always\nremains unchanged.\n\nHmm. This is a bit profound. I need to think about this.\n\nHey, that means, I think, that a photon should have zero magnitude of\n4-momentum. Ok, so it has equal momentum in both time and a spatial\ndimension so that equals zero.\n\nMy head\'s spinning. That doesn\'t make sense. If I bounce a photon off a\nparticle it will change momentum. Oh, but it won\'t. Its 4-momentum will\nremain unchanged.\n\nBu**er.....\n\nOK, now I have a very basic question. I imagine that if a particle M\nabsorbed a photon B then it would work simply additively?\n\nThat is M\' = M + B\n\nI think I should stop there.....\n\nIts a whole new way of looking at things.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<\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>Frank Hellmann <C.i.m@gmx.net> writes
>It's orthogonal, or rather, pseudo orthogonal. The Lorentz Group is
>SO(3,1) Where the 3,1 refers to the signature of the metric.
>
>Gallileo transformations are replace by rotation and translation in
>Minkowski space, there's nothing else. That's why SR is much simpler
>then Newtonian Mechanics. ;)
Ok. So if I want to describe how an object looks in another frame of
reference, all I need to do is multiply it by the appropriate LT?
So if I have some object, stationary in one frame, I would represent its
momentum as
m[c,0,0,0] = M
In another frame it would be ML, but the magnitude in both frames would
be the same, as mentioned by greg egan.
Now if we wanted to add a force, it seems to me (in my infinite
ignorance) that we would quite like an infinitesimal - dL.
Then we could have an equation of motion like
M' =[/itex] MdL.
All we would be doing would be moving momentum from one direction, and
transferring it to another, since the magnitude of the 4-momentum always
remains unchanged.
Hmm. This is a bit profound. I need to think about this.
Hey, that means, I think, that a photon should have zero magnitude of
4-momentum. Ok, so it has equal momentum in both time and a spatial
dimension so that equals zero.
My head's spinning. That doesn't make sense. If I bounce a photon off a
particle it will change momentum. Oh, but it won't. Its 4-momentum will
remain unchanged.
Bu**er.....
OK, now I have a very basic question. I imagine that if a particle M
absorbed a photon B then it would work simply additively?
That is [itex]M' = M + B
I think I should stop there.....
Its a whole new way of looking at things.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<
Frank Hellmann
Jun18-04, 03:21 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>> In another frame it would be ML, but the magnitude in both frames would\n> be the same, as mentioned by greg egan.\n\n> OK, now I have a very basic question. I imagine that if a particle M\n> absorbed a photon B then it would work simply additively?\n>\n> That is M\' = M + B\n>\n\nYep, conservation of four momentum dictates it. If we have\ntranslational invariance in all four directions all four components of\nfour momentum are conserved individually. (Important to remember that\nthis is energy and momentum rolled into one vector, ordinary momentum\nis the spatial part of it).\n\n> I think I should stop there.....\n>\n\nYou are spot on with everything. Absolutely right. Of course in GR\nthings become a bit more complicated since you need parallel transport\nto actually add stuff and it might be ambigious.\n\n> Its a whole new way of looking at things.\n\nWelcome to a whole new world.\nIt\'s beautifully simple isn\'t it?\n\nTake the four momentum:\n(p,p) = m^2\n\ndifferentiate:\n\n(p,dp/dtau) = 0\n\nThe four force is always orthogonal to the four momentum (four\nacceleration always orthogonal to the four velocity).\n(implicit assumption: that the force acting does not change the rest\nmass! But such a force would hardly be relativistic since the rest\nmass is a scalar.)\n\nWhere to go from here?\nAll we put into our mechanics is that they be invariant under SO(3,1).\nNow for QM we need a complex representation of this group. Fiddle\naround a bit and you have Weyls and Diracs equation as the most simple\nnon trivial SO(3,1) invariant equations falling out immidiately,\nwithout conjuring up any vague chants of quantization.\n\nOf course maths spoils the fun and everything isn\'t all that nice and\ndandy, we are actually looking at the poincare group, thus we get\ntranslations as well. This is neither compact nor simple. And a lot of\nthe maths get a bit messy. You are right now at the spot that during\nmy studying has been the most beautifull bit of physics I encountered,\nI\'ll soon be going into some more GR and I hope for a similar beauty\nthere as well...\n\n---\nfrank\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 frame it would be ML, but the magnitude in both frames would
> be the same, as mentioned by greg egan.
> OK, now I have a very basic question. I imagine that if a particle M
> absorbed a photon B then it would work simply additively?
>
> That is M' = M + B
>
Yep, conservation of four momentum dictates it. If we have
translational invariance in all four directions all four components of
four momentum are conserved individually. (Important to remember that
this is energy and momentum rolled into one vector, ordinary momentum
is the spatial part of it).
> I think I should stop there.....
>
You are spot on with everything. Absolutely right. Of course in GR
things become a bit more complicated since you need parallel transport
to actually add stuff and it might be ambigious.
> Its a whole new way of looking at things.
Welcome to a whole new world.
It's beautifully simple isn't it?
Take the four momentum:
(p,p) = m^2
differentiate:
(p,dp/dtau) =
The four force is always orthogonal to the four momentum (four
acceleration always orthogonal to the four velocity).
(implicit assumption: that the force acting does not change the rest
mass! But such a force would hardly be relativistic since the rest
mass is a scalar.)
Where to go from here?
All we put into our mechanics is that they be invariant under SO(3,1).
Now for QM we need a complex representation of this group. Fiddle
around a bit and you have Weyls and Diracs equation as the most simple
non trivial SO(3,1) invariant equations falling out immidiately,
without conjuring up any vague chants of quantization.
Of course maths spoils the fun and everything isn't all that nice and
dandy, we are actually looking at the poincare group, thus we get
translations as well. This is neither compact nor simple. And a lot of
the maths get a bit messy. You are right now at the spot that during
my studying has been the most beautifull bit of physics I encountered,
I'll soon be going into some more GR and I hope for a similar beauty
there as well...
---
frank
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.