Debunking a thought experiment [Archive]

Kefka G

Oct27-04, 10:56 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>\n\n\nHi.\n\nI decided to take another look at an old thought experiment that I\'d thought\nabout in high school a little bit (I\'m almost out of college now), and\nunfortunately I\'m not coming up with much more than I could then. I have a\nfeeling it\'s got to be wrong somewhere, since the conclusion is a bit absurd,\nbut I can\'t really put my finger on the problem. Hopefully someone on the\ngroup can point out my misstep to me.\n\nLet\'s assume that we\'re in D+1 spacetime dimensions, and we\'re working with the\nappropriate D+1 dimensional generalization of Maxwell\'s equations (preview -\nwe\'re going to try to solve for D). Don\'t worry, we don\'t need much about\nhigher dimensional electrodynamics here, just the static Coulomb force.\n\nHere\'s what to do - start with a "dumbbell" of positive charge, with a rest\nseparation L. Calculate the energy in the rest frame. Transform to a moving\nframe (in the direction of the separation), and calculate the energy. Note that\nthe dumbbell will "appear" shorter in the moving frame due to Lorentz\ncontraction. Stop the dumbbell by applying an impulse to each end, thus\nenabling us to trap it inside a "barn" of length L / gamma. Now calculate the\nfinal energy.\n\nNote that this is just the pole/barn experiment except we catch the pole in the\nbarn (yes, I know, the barn or pole will probably just explode - assume the\nbarn is really strong and the "pole" is just two electrons) and worry about\nwhere the energy comes from.\n\nHere\'s how we do it: forget the energy in the rest frame. Just convert it to\nan "effective mass" M, which we don\'t specify yet. Then in the moving frame,\nthe energy (i.e. 0 component of energy/momentum vector) will be\n\nEmov = gamma * M c^2\n\nNote that we\'ve already taken into account the stress addition to the energy\nfrom the bar piece of the dumbbell, since from Poincare we know that the whole\nobject transforms as a four vector, hence we treat it as such (with an\neffective mass M) rather than worrying about the fact that electromagnetic\nenergy transforms anomalously when sources are present. But now once we stop\nit, the ends of the dumbbell will be a distance L/gamma from each other. This\nwill lead to an electrostatic energy (in D space dimensions) of\n\nEelec = k q^2 (gamma / L)^(D-2)\n\nThis is just the Coulomb energy in the appropriate dimensionality, as can be\nobtained quite easily from the stress/energy tensor (plus an integration).\n\nThere will be other energy in the final situation, of course, such as\nradiation, heat, etc (not to mention the kinetic energy lost from the system\nwhen it was slowed down, or the rest masses of the particles involved!). So\nEelec should be a lower bound on the outgoing energy, and hence should be less\nthan or equal to the incoming energy:\n\nEelec < Emov\nor\nk q^2 (gamma/L)^(D-2) < gamma * M c^2\n\nFor (D-2) > 1,\n\n(gamma/L)^(D-3) < M c^2 / k q^2\n\nThis should hold regardless of velocity - but gamma goes to infinity as v->c,\nand the right hand side is a constant, hence there\'s no way this can hold for\narbitrary v < c. Unless, of course, D = 3 so that the left hand side is a\nconstant, too.\n\nLuckily this is the way things really work out, since we live in three\ndimensions. But it seems like there must be some sleight of hand involved in\norder to "derive" this fact without assuming it to start. Can anyone figure\nout where the error is? This is really bugging me, and so far nobody I\'ve\nasked can see exactly where it\'s wrong (although there\'s almost universal\nagreement that it is).\n\n-Eric\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>Hi.

I decided to take another look at an old thought experiment that I'd thought
about in high school a little bit (I'm almost out of college now), and
unfortunately I'm not coming up with much more than I could then. I have a
feeling it's got to be wrong somewhere, since the conclusion is a bit absurd,
but I can't really put my finger on the problem. Hopefully someone on the
group can point out my misstep to me.

Let's assume that we're in D+1 spacetime dimensions, and we're working with the
appropriate D+1 dimensional generalization of Maxwell's equations (preview -
we're going to try to solve for D). Don't worry, we don't need much about
higher dimensional electrodynamics here, just the static Coulomb force.

Here's what to do - start with a "dumbbell" of positive charge, with a rest
separation L. Calculate the energy in the rest frame. Transform to a moving
frame (in the direction of the separation), and calculate the energy. Note that
the dumbbell will "appear" shorter in the moving frame due to Lorentz
contraction. Stop the dumbbell by applying an impulse to each end, thus
enabling us to trap it inside a "barn" of length L / \gamma. Now calculate the
final energy.

Note that this is just the pole/barn experiment except we catch the pole in the
barn (yes, I know, the barn or pole will probably just explode - assume the
barn is really strong and the "pole" is just two electrons) and worry about
where the energy comes from.

Here's how we do it: forget the energy in the rest frame. Just convert it to
an "effective mass" M, which we don't specify yet. Then in the moving frame,
the energy (i.e. component of energy/momentum vector) will be

Emov = \gamma * M c^2

Note that we've already taken into account the stress addition to the energy
from the bar piece of the dumbbell, since from Poincare we know that the whole
object transforms as a four vector, hence we treat it as such (with an
effective mass M) rather than worrying about the fact that electromagnetic
energy transforms anomalously when sources are present. But now once we stop
it, the ends of the dumbbell will be a distance L/\gamma from each other. This
will lead to an electrostatic energy (in D space dimensions) of

Eelec = k q^2 (\gamma / L)^(D-2)

This is just the Coulomb energy in the appropriate dimensionality, as can be
obtained quite easily from the stress/energy tensor (plus an integration).

There will be other energy in the final situation, of course, such as
radiation, heat, etc (not to mention the kinetic energy lost from the system
when it was slowed down, or the rest masses of the particles involved!). So
Eelec should be a lower bound on the outgoing energy, and hence should be less
than or equal to the incoming energy:

Eelec < Emov
or
k q^2 (\gamma/L)^(D-2) < \gamma * M c^2

For (D-2) > 1,(\gamma/L)^(D-3) < M c^2 / k q^2

This should hold regardless of velocity - but \gamma goes to infinity as v->c,
and the right hand side is a constant, hence there's no way this can hold for
arbitrary v < c. Unless, of course, D = 3 so that the left hand side is a
constant, too.

Luckily this is the way things really work out, since we live in three
dimensions. But it seems like there must be some sleight of hand involved in
order to "derive" this fact without assuming it to start. Can anyone figure
out where the error is? This is really bugging me, and so far nobody I've
asked can see exactly where it's wrong (although there's almost universal
agreement that it is).

-Eric

Mark Palenik

Oct30-04, 08:10 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>\n\n"Kefka G" <kefkag@aol.com> wrote in message\nnews:20041025171210.22988.00002614@mb-m10.aol.com...\n>\n>\n>\n<snip>\n> Let\'s assume that we\'re in D+1 spacetime dimensions, and we\'re working\nwith the\n> appropriate D+1 dimensional generalization of Maxwell\'s equations\n(preview -\n> we\'re going to try to solve for D). Don\'t worry, we don\'t need much about\n> higher dimensional electrodynamics here, just the static Coulomb force.\n>\n> Here\'s what to do - start with a "dumbbell" of positive charge, with a\nrest\n> separation L. Calculate the energy in the rest frame. Transform to a\nmoving\n> frame (in the direction of the separation), and calculate the energy. Note\nthat\n> the dumbbell will "appear" shorter in the moving frame due to Lorentz\n> contraction. Stop the dumbbell by applying an impulse to each end, thus\n> enabling us to trap it inside a "barn" of length L / gamma. Now calculate\nthe\n> final energy.\n>\n> Note that this is just the pole/barn experiment except we catch the pole\nin the\n> barn (yes, I know, the barn or pole will probably just explode - assume\nthe\n> barn is really strong and the "pole" is just two electrons) and worry\nabout\n> where the energy comes from.\n>\n> Here\'s how we do it: forget the energy in the rest frame. Just convert it\nto\n> an "effective mass" M, which we don\'t specify yet. Then in the moving\nframe,\n> the energy (i.e. 0 component of energy/momentum vector) will be\n>\n> Emov = gamma * M c^2\n\nIf M is the effective mass, isn\'t M = m*gamma, which makes you\'re statement\nEmov = gamma^2*Mc^2. Shouldn\'t this be Mc^2 - Mc^2/gamma?\n\n>\n> Note that we\'ve already taken into account the stress addition to the\nenergy\n> from the bar piece of the dumbbell, since from Poincare we know that the\nwhole\n> object transforms as a four vector, hence we treat it as such (with an\n> effective mass M) rather than worrying about the fact that electromagnetic\n> energy transforms anomalously when sources are present. But now once we\nstop\n> it, the ends of the dumbbell will be a distance L/gamma from each other.\nThis\n> will lead to an electrostatic energy (in D space dimensions) of\n>\n> Eelec = k q^2 (gamma / L)^(D-2)\n>\n> This is just the Coulomb energy in the appropriate dimensionality, as can\nbe\n> obtained quite easily from the stress/energy tensor (plus an integration).\n\nI\'m taking physics 435, which mainly deals with electrostatics and magnetic\nfields (some more electrodynamics at the end), so I don\'t really know much\nabout dynamic electric fields, but Eelec = kq^2/(gamma/L)^(D-2) seems\nreasonable to me, although I don\'t know why the stress/energy tensor needs\nto come into play. I would think the flux of the field would remain\nconstant, giving an electrostatic force proportional to 1/r^(D-1), and thus,\na potential proportional to 1/r^(D-2), so that\'s about all I can say with\nregards to the validity of that statement.\n\n>\n> There will be other energy in the final situation, of course, such as\n> radiation, heat, etc (not to mention the kinetic energy lost from the\nsystem\n> when it was slowed down, or the rest masses of the particles involved!).\nSo\n> Eelec should be a lower bound on the outgoing energy, and hence should be\nless\n> than or equal to the incoming energy:\n>\n> Eelec < Emov\n> or\n> k q^2 (gamma/L)^(D-2) < gamma * M c^2\n\n\nWhat about your inital potential? I believe this should be Eelec2 <= Emov +\nEelec1\n\nAlso, we need to change the expression for Kinetic energy\n\nso kq^2(gamma/L)^(D-2) <= Mc^2 - Mc^2/gamma + kq^2/L^(D-2)\n\n>\n> For (D-2) > 1,\n>\n> (gamma/L)^(D-3) < M c^2 / k q^2\n\nActually, the power of L doesn\'t change, since all you did was divide by\ngamma.\n\nNote that in you\'re equation, however, both sides are actually constant.\n\ngamma/L = Lnaut = constant, so Lnaut^(D-3) is constant. If you hadn\'t\nmysteriously decreased the power of L when dividing by gamma, however, it\nwouldn\'t be.\n\nBut given our new equation (an I\'ll change M to m, to denote "rest" mass)\n\nkq^2(gamma/L)^(D-2) <= m*gamma*c^2 - mc^2 + kq^2/L^(D-2)\n\nkq^2(gamma/L)^(D-2) - m*gamma*c^2 <= m*c^2 + kq^2/L^(D-2)\n\nwhere the right hand side is ***NOT*** a constant.\n\nRemember, L is really Lnaut*gamma, at least, as you\'ve defined it.\n\n>\n> This should hold regardless of velocity - but gamma goes to infinity as\nv->c,\n> and the right hand side is a constant, hence there\'s no way this can hold\nfor\n> arbitrary v < c. Unless, of course, D = 3 so that the left hand side is a\n> constant, too.\n>\n\nIf I change all my Ls to Lnauts, which are constant,\n\nkq^2*/Lnaut^(D-2) - m*gamma <= m*c^2 + kq^2*gamma/Lnaut^(D-2)\n\nOh, and I just realized, we should probably flip our signs on the electric\npotentials, although at this point, it doesn\'t really matter for the\noriginal problem.\n\n-kq^2*/Lnaut^(D-2) - m*gamma <= m*c^2 - kq^2*gamma/Lnaut^(D-2)\n\nand now, we have a function of gamma to the first power on each side, and a\nconstant on each side. No specific number of dimensions required.\n\nI\'m almost certain I\'ve made some error somewhere in my calculations, but\nthe point is, since you made L = Lnaut*gamma, L/gamma is actually constant,\nand not a function of gamma, so it all works out - but not in what your\noriginal equation should have been, unless you make that one particular\nalgebra mistake, which lead to the actual original equation you wrote.\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>"Kefka G" <kefkag@aol.com> wrote in message
news:20041025171210.22988.00002614@mb-m10.aol.com...
>
>
>
<snip>
> Let's assume that we're in D+1 spacetime dimensions, and we're working
with the
> appropriate D+1 dimensional generalization of Maxwell's equations
(preview -
> we're going to try to solve for D). Don't worry, we don't need much about
> higher dimensional electrodynamics here, just the static Coulomb force.
>
> Here's what to do - start with a "dumbbell" of positive charge, with a
rest
> separation L. Calculate the energy in the rest frame. Transform to a
moving
> frame (in the direction of the separation), and calculate the energy. Note
that
> the dumbbell will "appear" shorter in the moving frame due to Lorentz
> contraction. Stop the dumbbell by applying an impulse to each end, thus
> enabling us to trap it inside a "barn" of length L / \gamma. Now calculate
the
> final energy.
>
> Note that this is just the pole/barn experiment except we catch the pole
in the
> barn (yes, I know, the barn or pole will probably just explode - assume
the
> barn is really strong and the "pole" is just two electrons) and worry
about
> where the energy comes from.
>
> Here's how we do it: forget the energy in the rest frame. Just convert it
to
> an "effective mass" M, which we don't specify yet. Then in the moving
frame,
> the energy (i.e. component of energy/momentum vector) will be
>
> Emov = \gamma * M c^2

If M is the effective mass, isn't M = m*\gamma, which makes you're statement
Emov = \gamma^2*Mc^2. Shouldn't this be Mc^2 - Mc^2/\gamma?

>
> Note that we've already taken into account the stress addition to the
energy
> from the bar piece of the dumbbell, since from Poincare we know that the
whole
> object transforms as a four vector, hence we treat it as such (with an
> effective mass M) rather than worrying about the fact that electromagnetic
> energy transforms anomalously when sources are present. But now once we
stop
> it, the ends of the dumbbell will be a distance L/\gamma from each other.
This
> will lead to an electrostatic energy (in D space dimensions) of
>
> Eelec = k q^2 (\gamma / L)^(D-2)
>
> This is just the Coulomb energy in the appropriate dimensionality, as can
be
> obtained quite easily from the stress/energy tensor (plus an integration).

I'm taking physics 435, which mainly deals with electrostatics and magnetic
fields (some more electrodynamics at the end), so I don't really know much
about dynamic electric fields, but Eelec = kq^2/(\gamma/L)^(D-2) seems
reasonable to me, although I don't know why the stress/energy tensor needs
to come into play. I would think the flux of the field would remain
constant, giving an electrostatic force proportional to 1/r^(D-1), and thus,
a potential proportional to 1/r^(D-2), so that's about all I can say with
regards to the validity of that statement.

>
> There will be other energy in the final situation, of course, such as
> radiation, heat, etc (not to mention the kinetic energy lost from the
system
> when it was slowed down, or the rest masses of the particles involved!).
So
> Eelec should be a lower bound on the outgoing energy, and hence should be
less
> than or equal to the incoming energy:
>
> Eelec < Emov
> or
> k q^2 (\gamma/L)^(D-2) < \gamma * M c^2

What about your inital potential? I believe this should be Eelec2 <= Emov +
Eelec1

Also, we need to change the expression for Kinetic energy

so kq^2(\gamma/L)^(D-2) <= Mc^2 - Mc^2/\gamma + kq^2/L^(D-2)

>
> For (D-2) > 1,
>
> (\gamma/L)^(D-3) < M c^2 / k q^2

Actually, the power of L doesn't change, since all you did was divide by
\gamma.

Note that in you're equation, however, both sides are actually constant.

\gamma/L =[/itex] Lnaut = constant, so Lnaut^(D-3) is constant. If you hadn't
mysteriously decreased the power of L when dividing by \gamma, however, it
wouldn't be.

But given our new equation (an I'll change M to m, to denote "rest" mass)

kq^2(\gamma/L)^(D-2) <= m*\gamma*c^2 - mc^2 + kq^2/L^(D-2)kq^2(\gamma/L)^(D-2) - m*\gamma*c^2 <= m*c^2 + kq^2/L^(D-2)

where the right hand side is ***NOT*** a constant.

Remember, L is really Lnaut*\gamma, at least, as you've defined it.

>
> This should hold regardless of velocity - but \gamma goes to infinity as
v->c,
> and the right hand side is a constant, hence there's no way this can hold
for
> arbitrary v < c. Unless, of course, [itex]D = 3 so that the left hand side is a
> constant, too.
>

If I change all my Ls to Lnauts, which are constant,

kq^2*/Lnaut^(D-2) - m*\gamma <= m*c^2 + kq^2*\gamma/Lnaut^(D-2)

Oh, and I just realized, we should probably flip our signs on the electric
potentials, although at this point, it doesn't really matter for the
original problem.

-kq^2*/Lnaut^(D-2) - m*\gamma <= m*c^2 - kq^2*\gamma/Lnaut^(D-2)

and now, we have a function of \gamma to the first power on each side, and a
constant on each side. No specific number of dimensions required.

I'm almost certain I've made some error somewhere in my calculations, but
the point is, since you made L = Lnaut*\gamma, L/\gamma is actually constant,
and not a function of \gamma, so it all works out - but not in what your
original equation should have been, unless you make that one particular
algebra mistake, which lead to the actual original equation you wrote.

Kefka G

Nov2-04, 12:38 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\nMark Palenik wrote:\n\n[sorry, I\'m going to start clipping since this is getting long]\n\n>> Here\'s how we do it: forget the energy in the rest frame. Just convert it\n>to\n>> an "effective mass" M, which we don\'t specify yet. Then in the moving\n>frame,\n>> the energy (i.e. 0 component of energy/momentum vector) will be\n>>\n>> Emov = gamma * M c^2\n>\n>If M is the effective mass, isn\'t M = m*gamma, which makes you\'re statement\n>Emov = gamma^2*Mc^2. Shouldn\'t this be Mc^2 - Mc^2/gamma?\n>\n\nWatch out - I\'m not using "effective mass" to mean "relativistic mass," as is\noften done (and in my opinion leads to confusion). I\'m simply saying, hey, we\ndon\'t really want to worry about breaking the incoming energy into parts, so\nlet\'s just throw the damn thing on a scale and see how much it weighs - by the\nequivalence principle, this is also the inertial mass. We can try to come up\nwith an appropriate expression for mass - it will contain a few pieces, most of\nwhich are easy to tackle, but without doing that we avoid several possible\nmistakes. I\'m not sure where you got Mc^2 - Mc^2/gamma from, though - that\ndoesn\'t follow from anything I said, unless we\'re crossing wires somewhere...\n\n>>\n>> Note that we\'ve already taken into account the stress addition to the\n>energy\n>> from the bar piece of the dumbbell, since from Poincare we know that the\n>whole\n>> object transforms as a four vector, hence we treat it as such (with an\n>> effective mass M) rather than worrying about the fact that electromagnetic\n>> energy transforms anomalously when sources are present. But now once we\n>stop\n>> it, the ends of the dumbbell will be a distance L/gamma from each other.\n>This\n>> will lead to an electrostatic energy (in D space dimensions) of\n>>\n>> Eelec = k q^2 (gamma / L)^(D-2)\n>>\n>> This is just the Coulomb energy in the appropriate dimensionality, as can\n>be\n>> obtained quite easily from the stress/energy tensor (plus an integration).\n>\n>I\'m taking physics 435, which mainly deals with electrostatics and magnetic\n>fields (some more electrodynamics at the end), so I don\'t really know much\n>about dynamic electric fields, but Eelec = kq^2/(gamma/L)^(D-2) seems\n>reasonable to me, although I don\'t know why the stress/energy tensor needs\n>to come into play. I would think the flux of the field would remain\n>constant, giving an electrostatic force proportional to 1/r^(D-1), and thus,\n>a potential proportional to 1/r^(D-2), so that\'s about all I can say with\n>regards to the validity of that statement.\n>\n\nFor a stationary electric field in any dimensionality, the total enery in the\nfield will be proportional to the integral of |E|^2 over all space. For two\npoint charges, we can perform an integration by parts to break this integral\ninto two pieces, 1) a divergent piece corresponding to the infinite\nelectrostatic energy of a point charge, and 2) a distance dependent piece which\ngoes as 1/r^(D-2). It only takes a few lines, but I hate ascii so I\'ll leave\nit as an exercise...\n\n>>\n>> There will be other energy in the final situation, of course, such as\n>> radiation, heat, etc (not to mention the kinetic energy lost from the\n>system\n>> when it was slowed down, or the rest masses of the particles involved!).\n>So\n>> Eelec should be a lower bound on the outgoing energy, and hence should be\n>less\n>> than or equal to the incoming energy:\n>>\n>> Eelec < Emov\n>> or\n>> k q^2 (gamma/L)^(D-2) < gamma * M c^2\n>\n>\n>What about your inital potential? I believe this should be Eelec2 <= Emov +\n>Eelec1\n>\n>Also, we need to change the expression for Kinetic energy\n>\n>so kq^2(gamma/L)^(D-2) <= Mc^2 - Mc^2/gamma + kq^2/L^(D-2)\n>\n\nThe point of creating an effective mass M was so that we wouldn\'t have to worry\nabout the incoming, kinetic energy. If we DID want to do this (which I didn\'t\nbecause it\'s kind of a pain), we would use the following. The energy in the\nrest frame will be\n\nErest = 2mc^2 + kq^2 / L^(D-2)\n\nas expected (assuming both ends of the dumbbell have mass m), breaking it into\nrest energy and potential energy. The moving energy will be\n\nEmov = gamma * Erest + anomalous terms\n= gamma (2mc^2 + kq^2 / L^(D-2)) + anomalous terms\n\nThe anomalous terms exist because unfortunately electromagnetic energy doesn\'t\ntransform exactly like a four vector when we have constraints (i.e. the bar\nholding the charges in place). There will be a contribution to the energy from\nthe stress caused by the bar - as an example, try Lorentz transforming a second\nrank tensor (i.e. stress/energy tensor) - the 00 component picks up pieces from\nits neighbors, including the 11 component, which will be nonzero in this case.\nIf we integrate a second rank tensor that only has a 00 component, then in\nfact, thanks to the Lorentz transformation, the INTEGRAL will transform with\njust one factor of gamma. As it turns out, in the thought experiment here, the\nstresses will be essentially negligible (they carry a factor of v^2/c^2, hence\nmax out with a factor of unity when v = c, hardly the several powers of gamma\nthat would be necessary to get rid of the conclusion). This is why often we\nget "close" to the right answers by using an energy/momentum 4-vector. But\nwhen we have stresses, this is not the right way to do it. I should mention\nfor completeness that in 3-D these anomalous pieces are necessary for exact\nconservation - if you\'re interested, I can go through that proof or give you a\nreference.\n\nOf course, consistency of the theory requires that this "all works out" when we\ndeal with electromagnetism - the combination of stresses, rest energies, and\nelectromagnetic energies, when integrated over space, must transform like a\nfour vector. Otherwise our mechanics would be somewhat pathological (to some\nextent, the theory would probably be classically nonrenormalizable). See\nRohrlich\'s electrodynamics book for the explanation of this and how it arises\nin 3-D - it\'s not entirely trivial. But the point is, if we have a rigid\nobject, and the theory is consistent, then we can always just weigh it to\ndetermine the mass, and then say that\n\nErest = Mc^2\nEmov = gamma * Erest = gamma Mc^2\n\nThis seems to be much simpler, and of much greater generality since we don\'t\neven care what types of stresses are present.\n\n>>\n>> For (D-2) > 1,\n>>\n>> (gamma/L)^(D-3) < M c^2 / k q^2\n>\n>Actually, the power of L doesn\'t change, since all you did was divide by\n>gamma.\n\nThis is correct, of course - I apologize, the equation should read:\n\ngamma^(D-3) < M L^(D-2) c^2 / k q^2\n\n>\n>Note that in you\'re equation, however, both sides are actually constant.\n\nNot true - in this thought experiment we hold everything except v (and hence\ngamma) constant, thus while the right side is constant, the left is variable.\nI\'ll read on, though - perhaps we\'re arguing apples and oranges.\n\n>\n>gamma/L = Lnaut = constant, so Lnaut^(D-3) is constant. If you hadn\'t\n>mysteriously decreased the power of L when dividing by gamma, however, it\n>wouldn\'t be.\n>\n\nOkay, I think this is where we\'re mixed up - here, I\'ve taken L to be the rest\nlength of the dumbbell, so the contracted length will be L / gamma. No matter\nwhat, L should NOT be a variable.\n\n>But given our new equation (an I\'ll change M to m, to denote "rest" mass)\n>\n>kq^2(gamma/L)^(D-2) <= m*gamma*c^2 - mc^2 + kq^2/L^(D-2)\n>\n>kq^2(gamma/L)^(D-2) - m*gamma*c^2 <= m*c^2 + kq^2/L^(D-2)\n>\n>where the right hand side is ***NOT*** a constant.\n>\n>Remember, L is really Lnaut*gamma, at least, as you\'ve defined it.\n>\n\nYes, I believe that the mistake you pointed out above led to this - taking that\nmistake into account, along with what I said about foregoing the process of\ntrying to explicitly define M (which should be constant, not variable), the\nfinal equation should be\n\ngamma^(D-3) < M L^(D-2) c^2 / k q^2\n\nEvery piece of the right hand side is constant (wrt velocity) but the left goes\nto infinity as v->c. Hence as far as I can see, the paradox still persists.\n\n>>\n>> This should hold regardless of velocity - but gamma goes to infinity as\n>v->c,\n>> and the right hand side is a constant, hence there\'s no way this can hold\n>for\n>> arbitrary v < c. Unless, of course, D = 3 so that the left hand side is a\n>> constant, too.\n>>\n>\n>If I change all my Ls to Lnauts, which are constant,\n>\n>kq^2*/Lnaut^(D-2) - m*gamma <= m*c^2 + kq^2*gamma/Lnaut^(D-2)\n>\n>Oh, and I just realized, we should probably flip our signs on the electric\n>potentials, although at this point, it doesn\'t really matter for the\n>original problem.\n>\n>-kq^2*/Lnaut^(D-2) - m*gamma <= m*c^2 - kq^2*gamma/Lnaut^(D-2)\n>\n>and now, we have a function of gamma to the first power on each side, and a\n>constant on each side. No specific number of dimensions required.\n>\n>I\'m almost certain I\'ve made some error somewhere in my calculations, but\n>the point is, since you made L = Lnaut*gamma, L/gamma is actually constant,\n>and not a function of gamma, so it all works out - but not in what your\n>original equation should have been, unless you make that one particular\n>algebra mistake, which lead to the actual original equation you wrote.\n>\n\nJust to reiterate, L should be a constant, and is defined as the rest length\nbetween the two charges on the dumbbell. I mussed up the algebra a little, but\nI\'m pretty sure I have it okay now. See the comments above about why to use M\n(which is also a constant, and has no powers of gamma) instead of m.\n\nThanks, though - I missed that algebra glitch before.\n\nIntuitively, though, think about it this way - when we physically compress\nsomething by a power of gamma, we need to do some amount of work on it that\ndepends on how the charges interact (i.e. on the potential function, which is\ndependent upon dimensionality). But when we use a Lorentz boost to compress\nsomething, we need to add a very specific amount of energy to it - the kinetic\nenergy, or, for rest mass M, (gamma - 1) * M. We can then slow the thing down\nand have an effective physical compression. It would take a monstrous\ncoincidence for the energy to balance exactly for any potential function U(r) -\nin fact, arguments such as this are used to show that rigid bodies can\'t exist\nin relativity.\n\nIf a rigid body existed, the potential energy as a function of length would\nlook something like a step graph:\n\nU(r) = constant for r > r0\n= 0 for r = r0\n= constant for r < r0\n\nIf we started with the body at size r0, then accelerated it (or perhaps\nswitched reference frames) by just a little bit, we would add negligible\nkinetic energy, since in the classical limit the kinetic energy goes as v^2.\nBut ANY velocity at all is enough to catch it in a barn at some size less than\nr0, and hence the body would pick up a bit of energy for free. Clearly this is\na violation of energy conservation. Incidentally, we\'ve also proven that NO\ndiscontinuous jumps (at least positive jumps, as we compress an object) are\nallowed in "relativistically acceptable potential functions."\n\nThis gives me a little more confidence that there is SOME result from the\ndumbbell thought experiment. I\'m just not positive I\'ve gotten the best\nresult, or that I\'m presenting it convincingly enough. Any thoughts are\nwelcome.\n\n-Eric\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>Mark Palenik wrote:

[sorry, I'm going to start clipping since this is getting long]

>> Here's how we do it: forget the energy in the rest frame. Just convert it
>to
>> an "effective mass" M, which we don't specify yet. Then in the moving
>frame,
>> the energy (i.e. component of energy/momentum vector) will be
>>
>> Emov = \gamma * M c^2
>
>If M is the effective mass, isn't M = m*\gamma, which makes you're statement
>Emov = \gamma^2*Mc^2. Shouldn't this be Mc^2 - Mc^2/\gamma?
>

Watch out - I'm not using "effective mass" to mean "relativistic mass," as is
often done (and in my opinion leads to confusion). I'm simply saying, hey, we
don't really want to worry about breaking the incoming energy into parts, so
let's just throw the damn thing on a scale and see how much it weighs - by the
equivalence principle, this is also the inertial mass. We can try to come up
with an appropriate expression for mass - it will contain a few pieces, most of
which are easy to tackle, but without doing that we avoid several possible
mistakes. I'm not sure where you got Mc^2 - Mc^2/\gamma from, though - that
doesn't follow from anything I said, unless we're crossing wires somewhere...

>>
>> Note that we've already taken into account the stress addition to the
>energy
>> from the bar piece of the dumbbell, since from Poincare we know that the
>whole
>> object transforms as a four vector, hence we treat it as such (with an
>> effective mass M) rather than worrying about the fact that electromagnetic
>> energy transforms anomalously when sources are present. But now once we
>stop
>> it, the ends of the dumbbell will be a distance L/\gamma from each other.
>This
>> will lead to an electrostatic energy (in D space dimensions) of
>>
>> Eelec = k q^2 (\gamma / L)^(D-2)
>>
>> This is just the Coulomb energy in the appropriate dimensionality, as can
>be
>> obtained quite easily from the stress/energy tensor (plus an integration).
>
>I'm taking physics 435, which mainly deals with electrostatics and magnetic
>fields (some more electrodynamics at the end), so I don't really know much
>about dynamic electric fields, but Eelec = kq^2/(\gamma/L)^(D-2) seems
>reasonable to me, although I don't know why the stress/energy tensor needs
>to come into play. I would think the flux of the field would remain
>constant, giving an electrostatic force proportional to 1/r^(D-1), and thus,
>a potential proportional to 1/r^(D-2), so that's about all I can say with
>regards to the validity of that statement.
>

For a stationary electric field in any dimensionality, the total enery in the
field will be proportional to the integral of |E|^2 over all space. For two
point charges, we can perform an integration by parts to break this integral
into two pieces, 1) a divergent piece corresponding to the infinite
electrostatic energy of a point charge, and 2) a distance dependent piece which
goes as 1/r^(D-2). It only takes a few lines, but I hate ascii so I'll leave
it as an exercise...

>>
>> There will be other energy in the final situation, of course, such as
>> radiation, heat, etc (not to mention the kinetic energy lost from the
>system
>> when it was slowed down, or the rest masses of the particles involved!).
>So
>> Eelec should be a lower bound on the outgoing energy, and hence should be
>less
>> than or equal to the incoming energy:
>>
>> Eelec < Emov
>> or
>> k q^2 (\gamma/L)^(D-2) < \gamma * M c^2
>
>
>What about your inital potential? I believe this should be Eelec2 <= Emov +
>Eelec1
>
>Also, we need to change the expression for Kinetic energy
>
>so kq^2(\gamma/L)^(D-2) <= Mc^2 - Mc^2/\gamma + kq^2/L^(D-2)
>

The point of creating an effective mass M was so that we wouldn't have to worry
about the incoming, kinetic energy. If we DID want to do this (which I didn't
because it's kind of a pain), we would use the following. The energy in the
rest frame will be

Erest = 2mc^2 + kq^2 / L^(D-2)

as expected (assuming both ends of the dumbbell have mass m), breaking it into
rest energy and potential energy. The moving energy will be

Emov = \gamma * Erest + anomalous terms
= \gamma (2mc^2 + kq^2 / L^(D-2)) + anomalous terms

The anomalous terms exist because unfortunately electromagnetic energy doesn't
transform exactly like a four vector when we have constraints (i.e. the bar
holding the charges in place). There will be a contribution to the energy from
the stress caused by the bar - as an example, try Lorentz transforming a second
rank tensor (i.e. stress/energy tensor) - the 00 component picks up pieces from
its neighbors, including the 11 component, which will be nonzero in this case.
If we integrate a second rank tensor that only has a 00 component, then in
fact, thanks to the Lorentz transformation, the INTEGRAL will transform with
just one factor of \gamma. As it turns out, in the thought experiment here, the
stresses will be essentially negligible (they carry a factor of v^2/c^2, hence
max out with a factor of unity when v = c, hardly the several powers of \gamma
that would be necessary to get rid of the conclusion). This is why often we
get "close" to the right answers by using an energy/momentum 4-vector. But
when we have stresses, this is not the right way to do it. I should mention
for completeness that in 3-D these anomalous pieces are necessary for exact
conservation - if you're interested, I can go through that proof or give you a
reference.

Of course, consistency of the theory requires that this "all works out" when we
deal with electromagnetism - the combination of stresses, rest energies, and
electromagnetic energies, when integrated over space, must transform like a
four vector. Otherwise our mechanics would be somewhat pathological (to some
extent, the theory would probably be classically nonrenormalizable). See
Rohrlich's electrodynamics book for the explanation of this and how it arises
in 3-D - it's not entirely trivial. But the point is, if we have a rigid
object, and the theory is consistent, then we can always just weigh it to
determine the mass, and then say that

Erest = Mc^2
Emov = \gamma * Erest = \gamma Mc^2

This seems to be much simpler, and of much greater generality since we don't
even care what types of stresses are present.

>>
>> For (D-2) > 1,
>>
>> (\gamma/L)^(D-3) < M c^2 / k q^2
>
>Actually, the power of L doesn't change, since all you did was divide by
>\gamma.

This is correct, of course - I apologize, the equation should read:

\gamma^(D-3) < M L^(D-2) c^2 / k q^2

>
>Note that in you're equation, however, both sides are actually constant.

Not true - in this thought experiment we hold everything except v (and hence
\gamma) constant, thus while the right side is constant, the left is variable.
I'll read on, though - perhaps we're arguing apples and oranges.

>
>\gamma/L = Lnaut = constant, so Lnaut^(D-3) is constant. If you hadn't
>mysteriously decreased the power of L when dividing by \gamma, however, it
>wouldn't be.
>

Okay, I think this is where we're mixed up - here, I've taken L to be the rest
length of the dumbbell, so the contracted length will be L / \gamma. No matter
what, L should NOT be a variable.

>But given our new equation (an I'll change M to m, to denote "rest" mass)
>
>kq^2(\gamma/L)^(D-2) <= m*\gamma*c^2 - mc^2 + kq^2/L^(D-2)
>
>kq^2(\gamma/L)^(D-2) - m*\gamma*c^2 <= m*c^2 + kq^2/L^(D-2)
>
>where the right hand side is ***NOT*** a constant.
>
>Remember, L is really Lnaut*\gamma, at least, as you've defined it.
>

Yes, I believe that the mistake you pointed out above led to this - taking that
mistake into account, along with what I said about foregoing the process of
trying to explicitly define M (which should be constant, not variable), the
final equation should be

\gamma^(D-3) < M L^(D-2) c^2 / k q^2

Every piece of the right hand side is constant (wrt velocity) but the left goes
to infinity as v->c. Hence as far as I can see, the paradox still persists.

>>
>> This should hold regardless of velocity - but \gamma goes to infinity as
>v->c,
>> and the right hand side is a constant, hence there's no way this can hold
>for
>> arbitrary v < c. Unless, of course, D = 3 so that the left hand side is a
>> constant, too.
>>
>
>If I change all my Ls to Lnauts, which are constant,
>
>kq^2*/Lnaut^(D-2) - m*\gamma <= m*c^2 + kq^2*\gamma/Lnaut^(D-2)
>
>Oh, and I just realized, we should probably flip our signs on the electric
>potentials, although at this point, it doesn't really matter for the
>original problem.
>
>-kq^2*/Lnaut^(D-2) - m*\gamma <= m*c^2 - kq^2*\gamma/Lnaut^(D-2)
>
>and now, we have a function of \gamma to the first power on each side, and a
>constant on each side. No specific number of dimensions required.
>
>I'm almost certain I've made some error somewhere in my calculations, but
>the point is, since you made L = Lnaut*\gamma, L/\gamma is actually constant,
>and not a function of \gamma, so it all works out - but not in what your
>original equation should have been, unless you make that one particular
>algebra mistake, which lead to the actual original equation you wrote.
>

Just to reiterate, L should be a constant, and is defined as the rest length
between the two charges on the dumbbell. I mussed up the algebra a little, but
I'm pretty sure I have it okay now. See the comments above about why to use M
(which is also a constant, and has no powers of \gamma) instead of m.

Thanks, though - I missed that algebra glitch before.

Intuitively, though, think about it this way - when we physically compress
something by a power of \gamma, we need to do some amount of work on it that
depends on how the charges interact (i.e. on the potential function, which is
dependent upon dimensionality). But when we use a Lorentz boost to compress
something, we need to add a very specific amount of energy to it - the kinetic
energy, or, for rest mass M, (\gamma - 1) * M. We can then slow the thing down
and have an effective physical compression. It would take a monstrous
coincidence for the energy to balance exactly for any potential function U(r) -
in fact, arguments such as this are used to show that rigid bodies can't exist
in relativity.

If a rigid body existed, the potential energy as a function of length would
look something like a step graph:

U(r) = constant for r > r0
= for r = r0
= constant for r < r0

If we started with the body at size r0, then accelerated it (or perhaps
switched reference frames) by just a little bit, we would add negligible
kinetic energy, since in the classical limit the kinetic energy goes as v^2.
But ANY velocity at all is enough to catch it in a barn at some size less than
r0, and hence the body would pick up a bit of energy for free. Clearly this is
a violation of energy conservation. Incidentally, we've also proven that NO
discontinuous jumps (at least positive jumps, as we compress an object) are
allowed in "relativistically acceptable potential functions."

This gives me a little more confidence that there is SOME result from the
dumbbell thought experiment. I'm just not positive I've gotten the best
result, or that I'm presenting it convincingly enough. Any thoughts are
welcome.

-Eric

Mark Palenik

Nov2-04, 12:38 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>\nI don\'t know what is wrong with me - the equation in the last message for KE\nshould have been m*c^2/gamma - mc^2. For some reason I keep wanting to\nwrite m*gamma.\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 don't know what is wrong with me - the equation in the last message for KE
should have been m*c^2/\gamma - mc^2. For some reason I keep wanting to
write m*\gamma.

Mark Palenik

Nov2-04, 12:38 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\n"Mark Palenik" <markpalenik@wideopenwest.com> wrote in message\nnews:_eKdnf-ORvGfOx_cRVn-iQ@wideopenwest.com...\n>\n> -kq^2*/Lnaut^(D-2) - m*gamma <= m*c^2 - kq^2*gamma/Lnaut^(D-2)\n\nYeah, I definitely made some mistakes here. One thing that confused me was\nthe way you wrote L and M as "relativistic" quantities, then used gamma to\nwrite their values in the rest frame - which gave the appearance of\nconstants not being constants. I made a mistake with my conversion of M to\nm, because I wasn\'t thinking clearly about the way you had defined you\'re\nvariables.\n\nAlso, since I don\'t have much experience with electrodynamics yet, I wasn\'t\nthinking about magnetism.\n\nTo get the correct equation you must\n\nFind the energy stored in the electric field when the charges are moving\nFind the energy stored in the magnetic field\nUse the *correct* formula for kinetic energy, m*gamma*c^2 - m*c^2\n\nAnd set the sum of those things equal to (or greater than) the energy stored\nin the electric field at rest\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>"Mark Palenik" <markpalenik@wideopenwest.com> wrote in message
news:_eKdnf-ORvGfOx_cRVn-iQ@wideopenwest.com...
>
> -kq^2*/Lnaut^(D-2) - m*\gamma <= m*c^2 - kq^2*\gamma/Lnaut^(D-2)

Yeah, I definitely made some mistakes here. One thing that confused me was
the way you wrote L and M as "relativistic" quantities, then used \gamma to
write their values in the rest frame - which gave the appearance of
constants not being constants. I made a mistake with my conversion of M to
m, because I wasn't thinking clearly about the way you had defined you're
variables.

Also, since I don't have much experience with electrodynamics yet, I wasn't
thinking about magnetism.

To get the correct equation you must

Find the energy stored in the electric field when the charges are moving
Find the energy stored in the magnetic field
Use the *correct* formula for kinetic energy, m*\gamma*c^2 - m*c^2

And set the sum of those things equal to (or greater than) the energy stored
in the electric field at rest

Kefka G

Nov5-04, 08:20 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>\n"Mark Palenik" <markpalenik@wideopenwest.com> wrote:\n>>\n>> -kq^2*/Lnaut^(D-2) - m*gamma <= m*c^2 - kq^2*gamma/Lnaut^(D-2)\n>\n>Yeah, I definitely made some mistakes here. One thing that confused me was\n>the way you wrote L and M as "relativistic" quantities, then used gamma to\n>write their values in the rest frame - which gave the appearance of\n>constants not being constants. I made a mistake with my conversion of M to\n>m, because I wasn\'t thinking clearly about the way you had defined you\'re\n>variables.\n\nSee my other message, but just to clarify - L and M are constants, and contain\nno powers of gamma. L is the rest length, and M is the effective rest mass.\nThey would both be measured in the rest frame.\n\n>\n>Also, since I don\'t have much experience with electrodynamics yet, I wasn\'t\n>thinking about magnetism.\n>\n>To get the correct equation you must\n>\n>Find the energy stored in the electric field when the charges are moving\n>Find the energy stored in the magnetic field\n>Use the *correct* formula for kinetic energy, m*gamma*c^2 - m*c^2\n\nJust a note - this matches up with the way that I calculated energy if you also\ninclude the rest energy, mc^2, to get E = gamma mc^2.\n\n>\n>And set the sum of those things equal to (or greater than) the energy stored\n>in the electric field at rest\n>\n\nI think this message was sent before my reply to your previous one appeared on\nthe group - in any case, these questions are a little tricky so it\'s probably\nworthwhile to go through why this method won\'t quite work in a bit of detail.\n\nFirst of all, we need to discuss higher dimensional electrodynamics a bit. If\nwe start out with Maxwell\'s equations in 3D, we have:\n\ndiv E = rho / eps0\ndiv B = 0\ncurl E = -dB/dt\nc^2 curl B = j/eps0 + dE/dt\n\nNow, how does one generalize this to higher dimensions? One way would be to\ntry and generalize the div/curl operators to act on higher dimensional vectors.\nWe can bump gradient up to any dimensionality, and div is just a dot product\nof grad with a vector, so that works fine. But think about the curl - this is\nan intrinsically 3-D operator, since it relies upon a mapping from 2-forms\n(essentially oriented area elements) to vectors - recall that you need to use a\nright hand rule to think about cross products. This will not exist in higher\ndimensions.\n\nSo what do we do now? It appears that the problem is really that the division\nof the electromagnetic field into E and B is not possible in general. We\nrecall that we can, instead, speak of the potentials A_u, which form a 4-vector\nfield. If we fix the gauge, we can then simply obtain D+1 copies of the wave\nequation (D spatial dimensions),\n\ndalembertian (A_u) = (d^2/dx_0^2 - ... - d^2/dx_D^2) A_u = j_u\n\n(up to choice of units)\n\nIn fact, much of electrodynamics goes through to higher dimensions "as usual"\nif we just do it in relativistic notation. In any case, the point is that we\ncan\'t separate electric and magnetic components of the field quite so simply in\nhigher dimensions.\n\nOf course, then we need to express the energy content of the fields in terms of\nthe potentials. To do this, we can use Noether\'s theorem applied to the\nelectromagnetic Lagrangian to get the stress energy tensor of the field - I\'m a\nbit lazy, and I\'m not going to go through this now. Maybe another time...in\nany case, though, we can deal with static situations fairly easily. I\'ll leave\nit to you to show that if you use the four force expression\n\nK^u = q n_v F^uv\n\nwhere q is the charge of the particle, n^v is the proper velocity, and F^uv =\ndA^v/dx_u - dA^u/dx_v then we can find that the work involved in moving two\npoint charges from infinity to a distance r will be proportional to 1/r^(D-2) -\na very plausible result, in my opinion.\n\nThis neatly skirts the infinite field energy issues, which are really the\nproblem with the method you mentioned above. For instance - step 1, find the\nenergy stored in the electric field when the charges are moving, gives us an\ninfinite answer, since the integral will go all the way down to the point\ncharge. Thus it\'s a bit more difficult than that. We\'d have to figure out how\nto cut off the integral at some radius rmin, which we can certainly do. BUT -\ncutting off an integral at rmin in the rest frame of the particle means that we\nhave to cut it off at an ellipsoid in the moving frame (thanks to Lorentz\ncontraction), which can be a bit tricky. There are ways to do this, but even\nthen, things don\'t necessarily go smoothly. I\'m sure you\'ve heard of the 4/3\nproblem (and other related ones) - these are related to the fact that\nelectromagnetic energy has some strange transformation properties. Part of the\nproblem is that when we have sources, the stress energy tensor of the fields\ndoes not have zero divergence - in particular, this means that we can\'t\nintegrate components over space to get four vectors.\n\nSo you can see that it\'s a lot more difficult than it would initially seem to\nget a good expression for the energy of a moving dumbbell in higher dimensions.\nSo instead, we can try to assume that the entire energy/momentum content\ntransforms like a four vector. If this isn\'t true, it\'s almost as bad for the\ntheory as loss of conservation.\n\nFor a first approximation, where we neglect the stresses of the connecting rod,\nwe can do the following. First we figure out the energy in the rest frame:\n\nErest = 2mc^2 + kq^2 / L^(D-2)\n\nwhere L is the rest length of the dumbbell. The total energy in the moving\nframe will be (neglecting stresses)\n\nEmov = gamma * Erest = gamma (2mc^2 + kq^2 / L^(D-2))\n\nThe energy when we\'ve stopped it in the "barn" will be\n\nEbarn = 2mc^2 + kq^2 / (L/gamma)^(D-2)\n\nwhere we\'ve now lost quite a bit of energy to radiation and from the kinetic\nenergy of the particles. Really, Emov should contain a piece from the\ntransformation of the stresses, but for now let\'s just see what happens.\nSetting Emov >= Ebarn, we get\n\ngamma (2mc^2 + kq^2 / L^(D-2)) >= 2mc^2 + kq^2 / (L/gamma)^(D-2)\n\n(gamma -1)2mc^2 >= kq^2/L^(D-2) * (gamma^(D-2) - gamma)\n\nor\n\n2mc^2 L^(D-2) / kq^2 >= (gamma^(D-2) - gamma) / (gamma - 1)\n\nand again, we have a constant LHS (wrt v) and a variable RHS which goes to\ninfinity as v->c, so we have a major problem. To see this, just do a\npolynomial division of the rhs - you\'ll get gamma times a partial sum of a\ngeometric series in gamma, as long as D>3.\n\nHad we taken stresses into account, this would have been altered - if I\nremember correctly, the final equation will become\n\n2mc^2 L^(D-2) / kq^2 >= [gamma^(D-2) - (1+v^2/c^2)gamma] / (gamma - 1)\n\ninstead of the above equation. The only difference is that (1+v^2/c^2) factor,\nwhich doesn\'t essentially alter the conclusion since it just goes to 2 as v->c,\nand it can\'t compete with the gamma^(D-2) term. I should mention that in 3-D,\nwe do need that term for conservation to work right (although this particular\nanalysis does not exhibit that fact).\n\nIf you\'d like the energy inequality for an arbitrary potential energy U(x),\nwhich leads to a force F(x) = -dU/dx, here it is, without proof (this includes\nthe stress energy):\n\n(gamma - 1) 2 mc^2 > U(x/gamma) - gamma U(x) - gamma (v^2/c^2) * F(x)\n\nwhere the incoming energy is\n\nEmov = gamma (2mc^2 + U(x) + (v^2/c^2) * F(x))\n\nand the outgoing energy is the same as before.\n\nThese are derived by assuming that the stress tensor of the rod is that of a\nperfect fluid with the appropriate pressure to keep the particles at constant\ndistance. If you\'d like the details, let me know and I\'ll send you the\nderivation - I\'ve gone long on this message, and it\'s got some tensor stuff\nwhich is tricky in ascii.\n\nThanks for your comments, again.\n\n-Eric\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>"Mark Palenik" <markpalenik@wideopenwest.com> wrote:
>>
>> -kq^2*/Lnaut^(D-2) - m*\gamma <= m*c^2 - kq^2*\gamma/Lnaut^(D-2)
>
>Yeah, I definitely made some mistakes here. One thing that confused me was
>the way you wrote L and M as "relativistic" quantities, then used \gamma to
>write their values in the rest frame - which gave the appearance of
>constants not being constants. I made a mistake with my conversion of M to
>m, because I wasn't thinking clearly about the way you had defined you're
>variables.

See my other message, but just to clarify - L and M are constants, and contain
no powers of \gamma. L is the rest length, and M is the effective rest mass.
They would both be measured in the rest frame.

>
>Also, since I don't have much experience with electrodynamics yet, I wasn't
>thinking about magnetism.
>
>To get the correct equation you must
>
>Find the energy stored in the electric field when the charges are moving
>Find the energy stored in the magnetic field
>Use the *correct* formula for kinetic energy, m*\gamma*c^2 - m*c^2

Just a note - this matches up with the way that I calculated energy if you also
include the rest energy, mc^2, to get E = \gamma mc^2.

>
>And set the sum of those things equal to (or greater than) the energy stored
>in the electric field at rest
>

I think this message was sent before my reply to your previous one appeared on
the group - in any case, these questions are a little tricky so it's probably
worthwhile to go through why this method won't quite work in a bit of detail.

First of all, we need to discuss higher dimensional electrodynamics a bit. If
we start out with Maxwell's equations in 3D, we have:

div E = \rho /[/itex] eps0
div B =
curl E = -dB/dtc^2 curl [itex]B = j/eps0 + dE/dt

Now, how does one generalize this to higher dimensions? One way would be to
try and generalize the div/curl operators to act on higher dimensional vectors.
We can bump gradient up to any dimensionality, and div is just a dot product
of grad with a vector, so that works fine. But think about the curl - this is
an intrinsically 3-D operator, since it relies upon a mapping from 2-forms
(essentially oriented area elements) to vectors - recall that you need to use a
right hand rule to think about cross products. This will not exist in higher
dimensions.

So what do we do now? It appears that the problem is really that the division
of the electromagnetic field into E and B is not possible in general. We
recall that we can, instead, speak of the potentials A_u, which form a 4-vector
field. If we fix the gauge, we can then simply obtain D+1 copies of the wave
equation (D spatial dimensions),

dalembertian (A_u) = (d^2/dx_0^2 - ... - d^2/dx_D^2) A_u = j_u

(up to choice of units)

In fact, much of electrodynamics goes through to higher dimensions "as usual"
if we just do it in relativistic notation. In any case, the point is that we
can't separate electric and magnetic components of the field quite so simply in
higher dimensions.

Of course, then we need to express the energy content of the fields in terms of
the potentials. To do this, we can use Noether's theorem applied to the
electromagnetic Lagrangian to get the stress energy tensor of the field - I'm a
bit lazy, and I'm not going to go through this now. Maybe another time...in
any case, though, we can deal with static situations fairly easily. I'll leave
it to you to show that if you use the four force expression

K^u = q n_v F^{uv}

where q is the charge of the particle, n^v is the proper velocity, and F^{uv} =dA^v/dx_u - dA^u/dx_v then we can find that the work involved in moving two
point charges from infinity to a distance r will be proportional to 1/r^(D-2) -
a very plausible result, in my opinion.

This neatly skirts the infinite field energy issues, which are really the
problem with the method you mentioned above. For instance - step 1, find the
energy stored in the electric field when the charges are moving, gives us an
infinite answer, since the integral will go all the way down to the point
charge. Thus it's a bit more difficult than that. We'd have to figure out how
to cut off the integral at some radius rmin, which we can certainly do. BUT -
cutting off an integral at rmin in the rest frame of the particle means that we
have to cut it off at an ellipsoid in the moving frame (thanks to Lorentz
contraction), which can be a bit tricky. There are ways to do this, but even
then, things don't necessarily go smoothly. I'm sure you've heard of the 4/3
problem (and other related ones) - these are related to the fact that
electromagnetic energy has some strange transformation properties. Part of the
problem is that when we have sources, the stress energy tensor of the fields
does not have zero divergence - in particular, this means that we can't
integrate components over space to get four vectors.

So you can see that it's a lot more difficult than it would initially seem to
get a good expression for the energy of a moving dumbbell in higher dimensions.
So instead, we can try to assume that the entire energy/momentum content
transforms like a four vector. If this isn't true, it's almost as bad for the
theory as loss of conservation.

For a first approximation, where we neglect the stresses of the connecting rod,
we can do the following. First we figure out the energy in the rest frame:

Erest = 2mc^2 + kq^2 / L^(D-2)

where L is the rest length of the dumbbell. The total energy in the moving
frame will be (neglecting stresses)

Emov = \gamma * Erest = \gamma (2mc^2 + kq^2 / L^(D-2))

The energy when we've stopped it in the "barn" will be

Ebarn = 2mc^2 + kq^2 / (L/\gamma)^(D-2)

where we've now lost quite a bit of energy to radiation and from the kinetic
energy of the particles. Really, Emov should contain a piece from the
transformation of the stresses, but for now let's just see what happens.
Setting Emov >= Ebarn, we get

\gamma (2mc^2 + kq^2 / L^(D-2)) >= 2mc^2 + kq^2 / (L/\gamma)^(D-2)(\gamma -1)2mc^2 >= kq^2/L^(D-2) * (\gamma^(D-2) - \gamma)

or

2mc^2 L^(D-2) / kq^2 >= (\gamma^(D-2) - \gamma) / (\gamma - 1)

and again, we have a constant LHS (wrt v) and a variable RHS which goes to
infinity as v->c, so we have a major problem. To see this, just do a
polynomial division of the rhs - you'll get \gamma times a partial sum of a
geometric series in \gamma, as long as D>3.

Had we taken stresses into account, this would have been altered - if I
remember correctly, the final equation will become

2mc^2 L^(D-2) / kq^2 >= [\gamma^(D-2) - (1+v^2/c^2)\gamma] / (\gamma - 1)

instead of the above equation. The only difference is that (1+v^2/c^2) factor,
which doesn't essentially alter the conclusion since it just goes to 2 as v->c,
and it can't compete with the \gamma^(D-2) term. I should mention that in 3-D,
we do need that term for conservation to work right (although this particular
analysis does not exhibit that fact).

If you'd like the energy inequality for an arbitrary potential energy U(x),
which leads to a force F(x) = -dU/dx, here it is, without proof (this includes
the stress energy):

(\gamma - 1) 2 mc^2 > U(x/\gamma) - \gamma U(x) - \gamma (v^2/c^2) * F(x)

where the incoming energy is

Emov = \gamma (2mc^2 + U(x) + (v^2/c^2) * F(x))

and the outgoing energy is the same as before.

These are derived by assuming that the stress tensor of the rod is that of a
perfect fluid with the appropriate pressure to keep the particles at constant
distance. If you'd like the details, let me know and I'll send you the
derivation - I've gone long on this message, and it's got some tensor stuff
which is tricky in ascii.

Thanks for your comments, again.

-Eric

Kefka G

Dec3-04, 04:55 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>Mark Palenik wrote:\n>>[kefkag wrote:]\n>> Note that we\'ve already taken into account the stress addition to the energy\n>> from the bar piece of the dumbbell, since from Poincare we know that the\nwhole\n>> object transforms as a four vector, hence we treat it as such (with an\n>> effective mass M) rather than worrying about the fact that electromagnetic\n>> energy transforms anomalously when sources are present. But now once we\nstop\n>> it, the ends of the dumbbell will be a distance L/gamma from each other.\nThis\n>> will lead to an electrostatic energy (in D space dimensions) of\n>>\n>> Eelec = k q^2 (gamma / L)^(D-2)\n>>\n>> This is just the Coulomb energy in the appropriate dimensionality, as can be\n>> obtained quite easily from the stress/energy tensor (plus an integration).\n>\n>I\'m taking physics 435, which mainly deals with electrostatics and magnetic\n>fields (some more electrodynamics at the end), so I don\'t really know much\n>about dynamic electric fields, but Eelec = kq^2/(gamma/L)^(D-2) seems\n>reasonable to me, although I don\'t know why the stress/energy tensor needs\n>to come into play. I would think the flux of the field would remain\n>constant, giving an electrostatic force proportional to 1/r^(D-1), and thus,\n>a potential proportional to 1/r^(D-2), so that\'s about all I can say with\n>regards to the validity of that statement.\n\nThis thread is old, but I saw this and figured I should clarify these points to\navoid confusion. First of all, we should try to see exactly why a higher\ndimensional potential goes as 1/r^(D-2), as claimed. There are several ways to\ndo this. One is to note that a generalized Maxwell theory would still probably\nsatisfy Div E = 0 in empty space (this can be verified quite generally if you\ndefine E correctly), and that if we set E proportional to grad U, then U\nproportional to 1/r^(D-2) is a solution to the equations. This is a bit\nsloppy, however, since we have not shown that U is in any way related to the\nenergy content of the fields. In effect, we\'re _assuming_ things about the\nforce on a charged particle, namely that E is proportional to the force on a\nresting particle and that whatever energy goes into the particle comes from the\nfield (and vice versa) - again, this can be verified more carefully if desired.\n\nWe can also use a least action approach. Up to a constant, we seek to\nextremize the integral of F^uv F_uv over spacetime, taking F_uv = d_u A_v - d_v\nA_u where d denotes partial integration. We treat A_u as the (set of)\nfunctions to vary, and apply a similar approach to minimizing the integral as\nwhen we\'re dealing with particles - I don\'t have time to go through this method\nin detail, and I\'m sure it\'s explained in many places (Wald\'s GR book covers\nthis in appendix E). In any case, when we do this we obtain field equations\nfor the field tensor F. But what is also nice is that by taking a variational\nderivative w.r.t. g_uv, we can actually obtain the conserved stress tensor for\nthe fields (up to a factor, after dividing out a volume factor dependent upon\ng) - notice that to obtain the stress tensor in _flat_ space, we go temporarily\nto _curved_ space, only to set g_uv = eta_uv at the end. In any case, after\nsome calculation we obtain (up to a possible sign error which I\'m too lazy to\ncheck at the moment...it depends on what signature you use)\n\nT_uv = F^a_u F_av - 1/4 eta_uv F^cd F_cd\n\nThe energy density in the fields is given by the 00 component, and using the\ndefinition of F, it\'s fairly easy to check that for purely electric fields\n(i.e. only field components that are nonzero are the "electric" ones along the\ntop row, plus the side row by antisymmetry) the energy density is simply |E|^2\n(up to a factor, of course), or at least it should be for the right sign choice\n(there\'s a balancing act between the two terms). For two particles separated\nby some distance r at rest, we write (where the integrals are over the t = 0\nsurface in the rest frame of the particles)\n\nEnergy = integral |E|^2\n= integral (grad U) dot (grad U) (U is the time component of\nthe potential A_u)\n\nIntegrating by parts and dropping a boundary term,\n\n= - integral U (del^2 U)\n\nNow, if we recognize that del^2 U is proportional to the charge density, we can\ninsert delta functions for sources, and integrate. We must, of course,\nsubtract off infinite constants corresponding to the infinite self energy of\nthe point charges, but this is nothing new. We end up with a potential energy\nproportional to 1/r^(D-2), as promised, since that is a time independent radial\nsolution of the field equations for A. Notice that we\'ve obtained this by\ndirectly calculating the energy in the _fields_, NOT from any considerations of\nforces on particles. In fact, we can essentially derive the static forces on\nthe particles this way, if we assume that our theory is consistent.\n\nAs for the stress energy question, when we have two particles at rest at some\nseparation, we do not have a static situation. As such, we must have some\nother form of matter involved to hold the particles in place. This affects the\ntransformation properties of the energy in the system - in particular, the\nenergy of the fields AND whatever is holding the particles in place must\ntransform like a four vector, but neither part alone will. This can be\nverified by directly computing the field energy involved and transforming the\nexpression (note that the stress energy tensor DOES transform like a tensor -\nin fact, it is exactly those terms in that tensor other than the 00 one that\nadd to the anomalous transformation energy when integrated). It\'s a bit easier\nto compute the anomalous energy transformation of a rod holding the charges in\nplace than of the fields themselves, although either approach ultimately works\n(one contribution is the negative of the other).\n\nIn any case, the point is, if we have two particles separated by r, we can\'t\nsimply assume that the field energy when we go to a moving frame is gamma times\nthe energy in the rest frame - we must also pick up an extra term, dependent\nupon the derivative of the potential in the rest frame (i.e. on the force\nrequired to hold the charges in place, which is proportional to the required\npressure).\n\nIf we go through all of this, we end up finding that the energy contained in\nthe fields (after subtracting the infinite constants mentioned before) of two\npositive particles moving with speed v and separated (in their rest frame) by a\nlength r is\n\ngamma[2mc^2 + U(r) - r (v/c)^2 U\'(r)]\n\nwhere U\' denotes the derivative wrt r of U. In higher dimensional\nelectrodynamics, what bothers me is the fact that the field energy of two point\ncharges a distance r/gamma away, at rest, is not always less than the field\nenergy of the same two charges while moving (while at that same separation, in\nthe lab frame). Pictorially:\n\n1)\nO O\n(separation r / gamma)\nvs\n\n2)\nO--> O-->\n(separation r / gamma)\n\nFor some separations and velocities, the second situation above actually has\nless field energy than the first.\n\nIn such a theory, adding velocity to a distribution of charge does NOT\nnecessarily increase its energy, at least if we trust these results (which do,\nin fact, give us perfect energy conservation in 3-D). This seems like a\nparticularly nasty quality to have in a theory, and I\'m not sure what exactly\nit means - possibly instability of some sort. The real problem seems to be\nthat the energy in 1) is only dependent upon U(r/gamma), but the energy in 2)\nis only dependent upon U(r) and U\'(r). It is a lucky coincidence that in 3-D\nthe two expressions happen to balance nicely - the r/gamma brings one power of\ngamma to the expression in 1), and there is already one factor of gamma in 2),\nso they have similar dependences.\n\nAnyone have any idea how to resolve this in higher dimensions where the U = 1 /\nr^(D-2) potential will bring in D-2 powers of gamma to the energy for situation\n1), without bringing in any (other than the one there from the energy\ntransformation) powers of gamma for situation 2)?\n\n-Eric\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>Mark Palenik wrote:
>>[kefkag wrote:]
>> Note that we've already taken into account the stress addition to the energy
>> from the bar piece of the dumbbell, since from Poincare we know that the
whole
>> object transforms as a four vector, hence we treat it as such (with an
>> effective mass M) rather than worrying about the fact that electromagnetic
>> energy transforms anomalously when sources are present. But now once we
stop
>> it, the ends of the dumbbell will be a distance L/\gamma from each other.
This
>> will lead to an electrostatic energy (in D space dimensions) of
>>
>> Eelec = k q^2 (\gamma / L)^(D-2)
>>
>> This is just the Coulomb energy in the appropriate dimensionality, as can be
>> obtained quite easily from the stress/energy tensor (plus an integration).
>
>I'm taking physics 435, which mainly deals with electrostatics and magnetic
>fields (some more electrodynamics at the end), so I don't really know much
>about dynamic electric fields, but Eelec = kq^2/(\gamma/L)^(D-2) seems
>reasonable to me, although I don't know why the stress/energy tensor needs
>to come into play. I would think the flux of the field would remain
>constant, giving an electrostatic force proportional to 1/r^(D-1), and thus,
>a potential proportional to 1/r^(D-2), so that's about all I can say with
>regards to the validity of that statement.

This thread is old, but I saw this and figured I should clarify these points to
avoid confusion. First of all, we should try to see exactly why a higher
dimensional potential goes as 1/r^(D-2), as claimed. There are several ways to
do this. One is to note that a generalized Maxwell theory would still probably
satisfy Div E = in empty space (this can be verified quite generally if you
define E correctly), and that if we set E proportional to grad U, then U
proportional to 1/r^(D-2) is a solution to the equations. This is a bit
sloppy, however, since we have not shown that U is in any way related to the
energy content of the fields. In effect, we're _assuming_ things about the
force on a charged particle, namely that E is proportional to the force on a
resting particle and that whatever energy goes into the particle comes from the
field (and vice versa) - again, this can be verified more carefully if desired.

We can also use a least action approach. Up to a constant, we seek to
extremize the integral of F^{uv} F_{uv} over spacetime, taking F_{uv} = d_u A_v - d_vA_u where d denotes partial integration. We treat A_u as the (set of)
functions to vary, and apply a similar approach to minimizing the integral as
when we're dealing with particles - I don't have time to go through this method
in detail, and I'm sure it's explained in many places (Wald's GR book covers
this in appendix E). In any case, when we do this we obtain field equations
for the field tensor F. But what is also nice is that by taking a variational
derivative w.r.t. g_{uv}, we can actually obtain the conserved stress tensor for
the fields (up to a factor, after dividing out a volume factor dependent upon
g) - notice that to obtain the stress tensor in _flat_ space, we go temporarily
to _curved_ space, only to set g_{uv} = \eta_uv at the end. In any case, after
some calculation we obtain (up to a possible sign error which I'm too lazy to
check at the moment...it depends on what signature you use)

T_{uv} = F^{a_u} F_{av} - 1/4 \eta_uv F^{cd} F_{cd}

The energy density in the fields is given by the 00 component, and using the
definition of F, it's fairly easy to check that for purely electric fields
(i.e. only field components that are nonzero are the "electric" ones along the
top row, plus the side row by antisymmetry) the energy density is simply |E|^2(up to a factor, of course), or at least it should be for the right sign choice
(there's a balancing act between the two terms). For two particles separated
by some distance r at rest, we write (where the integrals are over the t =
surface in the rest frame of the particles)

Energy = integral |E|^2
= integral (grad U) dot (grad U) (U is the time component of
the potential A_u)

Integrating by parts and dropping a boundary term,

= - integral U (del^2 U)

Now, if we recognize that del^2 U is proportional to the charge density, we can
insert \delta functions for sources, and integrate. We must, of course,
subtract off infinite constants corresponding to the infinite self energy of
the point charges, but this is nothing new. We end up with a potential energy
proportional to 1/r^(D-2), as promised, since that is a time independent radial
solution of the field equations for A. Notice that we've obtained this by
directly calculating the energy in the _fields_, NOT from any considerations of
forces on particles. In fact, we can essentially derive the static forces on
the particles this way, if we assume that our theory is consistent.

As for the stress energy question, when we have two particles at rest at some
separation, we do not have a static situation. As such, we must have some
other form of matter involved to hold the particles in place. This affects the
transformation properties of the energy in the system - in particular, the
energy of the fields AND whatever is holding the particles in place must
transform like a four vector, but neither part alone will. This can be
verified by directly computing the field energy involved and transforming the
expression (note that the stress energy tensor DOES transform like a tensor -
in fact, it is exactly those terms in that tensor other than the 00 one that
add to the anomalous transformation energy when integrated). It's a bit easier
to compute the anomalous energy transformation of a rod holding the charges in
place than of the fields themselves, although either approach ultimately works
(one contribution is the negative of the other).

In any case, the point is, if we have two particles separated by r, we can't
simply assume that the field energy when we go to a moving frame is \gamma times
the energy in the rest frame - we must also pick up an extra term, dependent
upon the derivative of the potential in the rest frame (i.e. on the force
required to hold the charges in place, which is proportional to the required
pressure).

If we go through all of this, we end up finding that the energy contained in
the fields (after subtracting the infinite constants mentioned before) of two
positive particles moving with speed v and separated (in their rest frame) by a
length r is

\gamma[2mc^2 + U(r) - r (v/c)^2 U'(r)]

where U' denotes the derivative wrt r of U. In higher dimensional
electrodynamics, what bothers me is the fact that the field energy of two point
charges a distance r/\gamma away, at rest, is not always less than the field
energy of the same two charges while moving (while at that same separation, in
the lab frame). Pictorially:

1)
O O
(separation r / \gamma)
vs

2)
O--> O-->
(separation r / \gamma)

For some separations and velocities, the second situation above actually has
less field energy than the first.

In such a theory, adding velocity to a distribution of charge does NOT
necessarily increase its energy, at least if we trust these results (which do,
in fact, give us perfect energy conservation in 3-D). This seems like a
particularly nasty quality to have in a theory, and I'm not sure what exactly
it means - possibly instability of some sort. The real problem seems to be
that the energy in 1) is only dependent upon U(r/\gamma), but the energy in 2)
is only dependent upon U(r) and U'(r). It is a lucky coincidence that in 3-D
the two expressions happen to balance nicely - the r/\gamma brings one power of
\gamma to the expression in 1), and there is already one factor of \gamma in 2),
so they have similar dependences.

Anyone have any idea how to resolve this in higher dimensions where the U = 1 /r^(D-2) potential will bring in D-2 powers of \gamma to the energy for situation
1), without bringing in any (other than the one there from the energy
transformation) powers of \gamma for situation 2)?

-Eric

Neil

Dec10-04, 05:02 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>"Kefka G" <kefkag@aol.com> wrote in message\nnews:20041203035204.11961.00001181@mb-m16.aol.com...\n> Mark Palenik wrote:\n> >>[kefkag wrote:]\n> >> Note that we\'ve already taken into account the stress addition to the energy\n> >> from the bar piece of the dumbbell, since from Poincare we know that the\n> whole\n> >> object transforms as a four vector, hence we treat it as such (with an\n> >> effective mass M) rather than worrying about the fact that electromagnetic\n> >> energy transforms anomalously when sources are present. But now once we\n> stop\n> >> it, the ends of the dumbbell will be a distance L/gamma from each other.\n> This\n> >> will lead to an electrostatic energy (in D space dimensions) of\n> >>\n> >> Eelec = k q^2 (gamma / L)^(D-2)\n> >>\n<snip>\n> 1)\n> O O\n> (separation r / gamma)\n> vs\n>\n> 2)\n> O--> O-->\n> (separation r / gamma)\n>\n> For some separations and velocities, the second situation above actually has\n> less field energy than the first.\n>\n> In such a theory, adding velocity to a distribution of charge does NOT\n> necessarily increase its energy, at least if we trust these results (which do,\n> in fact, give us perfect energy conservation in 3-D). This seems like a\n> particularly nasty quality to have in a theory, and I\'m not sure what exactly\n> it means - possibly instability of some sort. The real problem seems to be\n> that the energy in 1) is only dependent upon U(r/gamma), but the energy in 2)\n> is only dependent upon U(r) and U\'(r). It is a lucky coincidence that in 3-D\n> the two expressions happen to balance nicely - the r/gamma brings one power of\n> gamma to the expression in 1), and there is already one factor of gamma in 2),\n> so they have similar dependences.\n>\n> Anyone have any idea how to resolve this in higher dimensions where the U = 1 /\n> r^(D-2) potential will bring in D-2 powers of gamma to the energy for situation\n> 1), without bringing in any (other than the one there from the energy\n> transformation) powers of gamma for situation 2)?\n>\n> -Eric\n>\n\nI think there is a way to show that the electromagnetic mass (ie, inertia)\nproblem *cannot* be resolved when D is not three. Consider the\ndifferently-argued point below, using "N" for space dimnesionality, based on\nmy previous posts to sci.physics etc.:\n\nI argue that electromagnetic mass, calculated in terms of the work done\naccelerating charges to a given velocity (versus calculating the field\ntransformation at constant velocity) gives consistent results only in\nthree-dimensional space. My approach bypasses the messy issue of dealing\nwith the stresses in a constraint. It\'s about the question, do consistency\nrequirements force space to have three dimensions? Many have claimed so in\nthe past (e.g., H. Weyl back in 1918), but now the consensus is: although\nthere may be reasons why space did turn out to be 3-D (e.g., the\nBrandenberger-Vafa process in string theory), it did not have to, and\nother spaces "could exist."\n\nEM mass has been a messy boondoggle in physics for a long time (the infamous\n"4/3 paradox") because the EM mass (or, presumed inertia) of a system moving\nat v and naively calculated from field transformation gives the wrong answer\ninstead of U\' = gamma U_0. (It is 4/3 of the correct value for a spherical\ncharge, but twice too much in the "barbell" case.) The usual reply is: the\nstress in the constraint (e.g., the connecting rod between two charges)\ncompensates and gives a correct total answer. As noted by Palenik and Kefka\nG in this thread, it is controversial whether this works out in general for\nN-D space, partly because covariant formulations are often used that seem\n(to me, at least) to force a circularly-reasoned correct answer.\n\nWhat should EM mass equal per se, aside from issues of velocity\ntransformation? It must be consistent with the work done/given bringing\ncharges together from infinity, regardless of what field transformation\ngives. Otherwise we can violate energy conservation by trading energy\nbetween a forcing motor and the charges, then take advantage of the\neffective mass difference (such as raising the system when it\'s lighter,\nlowering it when it\'s heavier, etc.) Using Gaussian-style units, we assume\n\nf = q1q2r^(1-N). Hence, m_EM =\n\n(c^-2) q1q2 Integral(inf - r) 2r^(1-N) dr\n\nWhen N < 3, the integral diverges and such spaces are ruled out, though many\nseem to overlook this. With N > 2 we have the result\n\n(c^-2)[q1q2r^(2-N)]/(N-2).\n\nSome writers bypass calculating field energy by finding the actual forces\nbetween *accelerating* charges, which are not the simple N-D Coulomb force\nf = q1q2r^(1-N) because of how each charge moves towards or away from\nthe field "projected" by the other charge in the past. They combine that\nwith the stress correction (which alters the force required to accelerate\nthe system) to get a final result for the effective electromagnetic mass in\nterms f = m_effective * a. The result is correct in our space, although it\nis not (?) clear whether this applies correctly when N is not three.\n\nI tried a new way to find consistency for EM mass: the work done as\nforce*distance while accelerating two charges when forces are simply applied\nto them, *without* a connecting rod. We must take into account the relative\nmotion of the charges during the travel time of light between them. We can\nfind the forces by calculating the force change over the increments Delta r\n= at^2 /2, where t = r/c, and r is the base proper distance between charges\nq1 and q2 (i.e. measured in the co-moving inertial frame that the charges\nare accelerating away from.) In N-D, the simple Coulomb force between\ncharges is presumed to be f = q1q2r^(1-N). The derivative\ndf/dr = (1-N) q1q2r^(-N). The charges undergo constant proper\nacceleration a0 (constant relative to their own standards, not dv/dt),\nand will maintain constant "proper separation," i.e. will be in hyperbolic\nmotion. Hence, their accelerations will *not* equal each other\'s but will\nvary w.r.t. their Rindler coordinates as follows: a0 = c^2/X. Note that\nDelta r is negative for q1, which moves toward q2 during the travel of\nEM signals, but positive for q2, which travels away from q1 during\nDelta t. If we use f to show applied reaction forces along the X coordinate,\nwe can write\n\nf1 = q1q2r^(1-N) - (df1/dr)[(a01*t^2)]/2\n\nf2 = -q1q2r^(1-N) + (df2/dr)[(a02*t^2)]/2.\n\nHence\n\nf1 = q1q2r^(1-N) + (N-1)q1q2[r^(2-N)](a01/2c^2)\n\nf2 = -q1q2r^(1-N) + (N-1)q1q2[r^(2-N)](a02/2c^2)\n\nNow for the total work W = W1 + W2 accelerating both charges to a velocity,\nin terms of gamma factor: The distance moved s, is given by\ns = (gamma - 1)c^2/a0 = (gamma - 1)X. (Because of relativity of\nsimultaneity, the leading charge stops accelerating later for given v and\nmoves farther than the trailing charge despite having lower acceleration.)\nMultiplying by the respective forces:\n\nW1 + W2 = f1*s1 + f2*s2 =\n\n(gamma - 1)[q1q2]{[r^(1-N)]X1 + [(N-1)r^(2-N)]/2\n\n- [r^(1-N)]X2 + [(N-1)r^(2-N)]/2}\n\nSince X2 = X1 + r, and combining terms we have:\n\n(gamma - 1)[q1q2][(N-1)r^(2-N) - r^(2-N)] =\n\n(gamma - 1)[q1q2] [(N-2)r^(2-N)]\n\nThe EM mass should be W(c^2)/ (gamma - 1), since the work corresponds\nto kinetic energy.\n\nNow, solve for N in q1q2r^(2-N)]/(N-2) =\n\n[q1q2] [(N-2)r^(2-N)], and we find N = 1 or 3. Since N = 1 is ruled out for\nother reasons (infinite potential energy), N = 3 is the only permitted space\ndimensionality in terms of consistent treatment of electromagnetic mass.\n\n\nNeil Bates\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>"Kefka G" <kefkag@aol.com> wrote in message
news:20041203035204.11961.00001181@mb-m16.aol.com...
> Mark Palenik wrote:
> >>[kefkag wrote:]
> >> Note that we've already taken into account the stress addition to the energy
> >> from the bar piece of the dumbbell, since from Poincare we know that the
> whole
> >> object transforms as a four vector, hence we treat it as such (with an
> >> effective mass M) rather than worrying about the fact that electromagnetic
> >> energy transforms anomalously when sources are present. But now once we
> stop
> >> it, the ends of the dumbbell will be a distance L/\gamma from each other.
> This
> >> will lead to an electrostatic energy (in D space dimensions) of
> >>
> >> Eelec = k q^2 (\gamma / L)^(D-2)
> >>
<snip>
> 1)
> O O
> (separation r / \gamma)
> vs
>
> 2)
> O--> O-->
> (separation r / \gamma)
>
> For some separations and velocities, the second situation above actually has
> less field energy than the first.
>
> In such a theory, adding velocity to a distribution of charge does NOT
> necessarily increase its energy, at least if we trust these results (which do,
> in fact, give us perfect energy conservation in 3-D). This seems like a
> particularly nasty quality to have in a theory, and I'm not sure what exactly
> it means - possibly instability of some sort. The real problem seems to be
> that the energy in 1) is only dependent upon U(r/\gamma), but the energy in 2)
> is only dependent upon U(r) and U'(r). It is a lucky coincidence that in 3-D
> the two expressions happen to balance nicely - the r/\gamma brings one power of
> \gamma to the expression in 1), and there is already one factor of \gamma in 2),
> so they have similar dependences.
>
> Anyone have any idea how to resolve this in higher dimensions where the U = 1 /
> r^(D-2) potential will bring in D-2 powers of \gamma to the energy for situation
> 1), without bringing in any (other than the one there from the energy
> transformation) powers of \gamma for situation 2)?
>
> -Eric
>

I think there is a way to show that the electromagnetic mass (ie, inertia)
problem *cannot* be resolved when D is not three. Consider the
differently-argued point below, using "N" for space dimnesionality, based on
my previous posts to sci.physics etc.:

I argue that electromagnetic mass, calculated in terms of the work done
accelerating charges to a given velocity (versus calculating the field
transformation at constant velocity) gives consistent results only in
three-dimensional space. My approach bypasses the messy issue of dealing
with the stresses in a constraint. It's about the question, do consistency
requirements force space to have three dimensions? Many have claimed so in
the past (e.g., H. Weyl back in 1918), but now the consensus is: although
there may be reasons why space did turn out to be 3-D (e.g., the
Brandenberger-Vafa process in string theory), it did not have to, and
other spaces "could exist."

EM mass has been a messy boondoggle in physics for a long time (the infamous
"4/3 paradox") because the EM mass (or, presumed inertia) of a system moving
at v and naively calculated from field transformation gives the wrong answer
instead of U' = \gamma U_0. (It is 4/3 of the correct value for a spherical
charge, but twice too much in the "barbell" case.) The usual reply is: the
stress in the constraint (e.g., the connecting rod between two charges)
compensates and gives a correct total answer. As noted by Palenik and Kefka
G in this thread, it is controversial whether this works out in general for
N-D space, partly because covariant formulations are often used that seem
(to me, at least) to force a circularly-reasoned correct answer.

What should EM mass equal per se, aside from issues of velocity
transformation? It must be consistent with the work done/given bringing
charges together from infinity, regardless of what field transformation
gives. Otherwise we can violate energy conservation by trading energy
between a forcing motor and the charges, then take advantage of the
effective mass difference (such as raising the system when it's lighter,
lowering it when it's heavier, etc.) Using Gaussian-style units, we assume

f = q1q2r^(1-N)[/itex]. Hence, m_{EM} =(c^-2) q1q2 Integral(inf - r) 2r^(1-N) dr

When N < 3, the integral diverges and such spaces are ruled out, though many
seem to overlook this. With N > 2 we have the result

(c^-2)[q1q2r^(2-N)]/(N-2).

Some writers bypass calculating field energy by finding the actual forces
between *accelerating* charges, which are not the simple N-D Coulomb force
f = q1q2r^(1-N) because of how each charge moves towards or away from
the field "projected" by the other charge in the past. They combine that
with the stress correction (which alters the force required to accelerate
the system) to get a final result for the effective electromagnetic mass in
terms f = m_{effective} * a. The result is correct in our space, although it
is not (?) clear whether this applies correctly when N is not three.

I tried a new way to find consistency for EM mass: the work done as
force*distance while accelerating two charges when forces are simply applied
to them, *without* a connecting rod. We must take into account the relative
motion of the charges during the travel time of light between them. We can
find the forces by calculating the force change over the increments \Delta r= at^2 /2, where t = r/c, and r is the base proper distance between charges
q1 and q2 (i.e. measured in the co-moving inertial frame that the charges
are accelerating away from.) In N-D, the simple Coulomb force between
charges is presumed to be f = q1q2r^(1-N). The derivative
df/dr = (1-N) q1q2r^(-N). The charges undergo constant proper
acceleration a0 (constant relative to their own standards, not dv/dt),
and will maintain constant "proper separation," i.e. will be in hyperbolic
motion. Hence, their accelerations will *not* equal each other's but will
vary w.r.t. their Rindler coordinates as follows: a0 = c^2/X. Note that
\Delta r is negative for q1, which moves toward q2 during the travel of
EM signals, but positive for q2, which travels away from q1 during
\Delta t. If we use f to show applied reaction forces along the X coordinate,
we can write

f1 = q1q2r^(1-N) - (df1/dr)[(a01*t^2)]/2f2 = -q1q2r^(1-N) + (df2/dr)[(a02*t^2)]/2.

Hence

f1 = q1q2r^(1-N) + (N-1)q1q2[r^(2-N)](a01/2c^2)f2 = -q1q2r^(1-N) + (N-1)q1q2[r^(2-N)](a02/2c^2)

Now for the total work W = W1 + W2 accelerating both charges to a velocity,
in terms of \gamma factor: The distance moved s, is given by
s = (\gamma - 1)c^2/a0 = (\gamma - 1)X. (Because of relativity of
simultaneity, the leading charge stops accelerating later for given v and
moves farther than the trailing charge despite having lower acceleration.)
Multiplying by the respective forces:

W1 + W2 = f1*s1 + f2*s2 =(\gamma - 1)[q1q2]{[r^(1-N)]X1 + [(N-1)r^(2-N)]/2- [r^(1-N)]X2 + [(N-1)r^(2-N)]/2}

Since X2 = X1 + r, and combining terms we have:

(\gamma - 1)[q1q2][(N-1)r^(2-N) - r^(2-N)] =(\gamma - 1)[q1q2] [itex][(N-2)r^(2-N)]

The EM mass should be W(c^2)/ (\gamma - 1), since the work corresponds
to kinetic energy.

Now, solve for N in q1q2r^(2-N)]/(N-2) =

[q1q2] [(N-2)r^(2-N)], and we find N = 1 or 3. Since N = 1 is ruled out for
other reasons (infinite potential energy), N = 3 is the only permitted space
dimensionality in terms of consistent treatment of electromagnetic mass.

Neil Bates

Kefka G

Dec12-04, 03:52 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>>\n>I think there is a way to show that the electromagnetic mass (ie, inertia)\n>problem *cannot* be resolved when D is not three. Consider the\n>differently-argued point below, using "N" for space dimnesionality, based on\n>my previous posts to sci.physics etc.:\n\n<snip>\n\n>Now, solve for N in q1q2r^(2-N)]/(N-2) =\n>\n>[q1q2] [(N-2)r^(2-N)], and we find N = 1 or 3. Since N = 1 is ruled out for\n>other reasons (infinite potential energy), N = 3 is the only permitted space\n>dimensionality in terms of consistent treatment of electromagnetic mass.\n>\n>\n>Neil Bates\n\nI\'m going to have to go through that derivation in more detail a little\nlater - right now I just don\'t have the time. One thing that I wanted\nto note, however, was that we need to be very careful about what our\nassumptions are when we claim that a theory is inconsistent. I will\nagree 100% that for a whole slew of reasons 3+1 is the "optimal"\ndimensionality for Maxwell e/m, because a lot of things work out pretty\neasily there. For one, the energy conditions I derived earlier, and for\nanother, the conditions which you derived (assuming they are correct -\nI\'ll get back to you after I\'ve worked it out myself) do seem to point\nquite convincingly to three dimensional space. But I am bothered by the\nfact that (classically, NOT in QFT), it has been shown fairly\nconvincingly that the self action problem is resolvable in any\ndimensionality through additions of appropriate counterterms to the\nlagrangian. We need to perform more than just a usual mass\nrenormalization, however, and I fear that the thought experiment you\'ve\noutlined assumes that only a mass renormalization is taking place (in\nwhich case, you are correct - 3+1 is the only dimensionality that can be\nconsistently done in). I\'m just not sure exactly what the other\nrenormalizations mean physically - to some extent, I\'m convincing myself\nthat they mean that a moving charge has less energy than a stationary\none in a higher dimensional e/m theory, for reasons I\'ll go into in\nanother post (i.e. after I smooth out my factors of 2pi, etc.). This\nhas drawbacks as well, though, and so far I\'m really on the fence as to\nhow these theories must work.\n\nIn any case, I\'d be careful about your conclusion - to my knowledge,\nalthough I haven\'t understood the proofs, higher dimensional\nelectrodynamics with point charges IS indeed consistent; whether it is\nanything at all like 3+1 e/m is an altogether different question, and\nI\'d think that the answer is a big fat no. My guess is that the higher\ndimensional point particle has no analog in 3+1 theory, and this is why\nwe\'re obtaining counterintuitive results.\n\nI want to point out also that the thought experiment I outlined\npreviously was specifically designed to avoid making any assumptions\nabout the field theory itself, relying only upon the resultant "Coulomb"\nresting potential. I can\'t see if the one you just went through has the\nsame property, although I\'m definitely interested enough to investigate.\nI don\'t think I went through it in detail, but in another post I\'ll\noutline how we can extract a bit of info on the radiative aspects of a\nfield theory from nothing but its "Coulomb" potential between two\ncharges. I\'m still not certain, but I think we may not even have to\nassume linearity in the theory for the conclusion to hold. Anyway, I\'m\npromising a lot of info and providing little, so I\'ll just finish this\nup when I actually have the time (I\'m late for a train right now).\n\n-Eric\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 think there is a way to show that the electromagnetic mass (ie, inertia)
>problem *cannot* be resolved when D is not three. Consider the
>differently-argued point below, using "N" for space dimnesionality, based on
>my previous posts to sci.physics etc.:

<snip>

>Now, solve for N in q1q2r^(2-N)]/(N-2) =
>
>[q1q2] [(N-2)r^(2-N)], and we find N = 1 or 3. Since N = 1 is ruled out for
>other reasons (infinite potential energy), N = 3 is the only permitted space
>dimensionality in terms of consistent treatment of electromagnetic mass.
>
>
>Neil Bates

I'm going to have to go through that derivation in more detail a little
later - right now I just don't have the time. One thing that I wanted
to note, however, was that we need to be very careful about what our
assumptions are when we claim that a theory is inconsistent. I will
agree 100% that for a whole slew of reasons 3+1 is the "optimal"
dimensionality for Maxwell e/m, because a lot of things work out pretty
easily there. For one, the energy conditions I derived earlier, and for
another, the conditions which you derived (assuming they are correct -
I'll get back to you after I've worked it out myself) do seem to point
quite convincingly to three dimensional space. But I am bothered by the
fact that (classically, NOT in QFT), it has been shown fairly
convincingly that the self action problem is resolvable in any
dimensionality through additions of appropriate counterterms to the
lagrangian. We need to perform more than just a usual mass
renormalization, however, and I fear that the thought experiment you've
outlined assumes that only a mass renormalization is taking place (in
which case, you are correct - 3+1 is the only dimensionality that can be
consistently done in). I'm just not sure exactly what the other
renormalizations mean physically - to some extent, I'm convincing myself
that they mean that a moving charge has less energy than a stationary
one in a higher dimensional e/m theory, for reasons I'll go into in
another post (i.e. after I smooth out my factors of 2pi, etc.). This
has drawbacks as well, though, and so far I'm really on the fence as to
how these theories must work.

In any case, I'd be careful about your conclusion - to my knowledge,
although I haven't understood the proofs, higher dimensional
electrodynamics with point charges IS indeed consistent; whether it is
anything at all like 3+1 e/m is an altogether different question, and
I'd think that the answer is a big fat no. My guess is that the higher
dimensional point particle has no analog in 3+1 theory, and this is why
we're obtaining counterintuitive results.

I want to point out also that the thought experiment I outlined
previously was specifically designed to avoid making any assumptions
about the field theory itself, relying only upon the resultant "Coulomb"
resting potential. I can't see if the one you just went through has the
same property, although I'm definitely interested enough to investigate.
I don't think I went through it in detail, but in another post I'll
outline how we can extract a bit of info on the radiative aspects of a
field theory from nothing but its "Coulomb" potential between two
charges. I'm still not certain, but I think we may not even have to
assume linearity in the theory for the conclusion to hold. Anyway, I'm
promising a lot of info and providing little, so I'll just finish this
up when I actually have the time (I'm late for a train right now).

-Eric

Kefka G

Dec14-04, 01:13 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>other. . .Individual\nproblems, sins or even crimes, were not generally cause for wider\nsocial concern. . ."But the impact of the twin movements to the city\nand to the factory, both just gathering force in 1835, had a\nprogressive effect on personal behavior throughout the 19th century\nand into the 20th. The factory demanded regularity of behavior, a life\ngoverned by obedience to the rhythms of clock and calendar, the\ndemands of foreman and supervisor. In the city or town, the needs of\nliving in closely packed neighborhoods inhibited many actions\npreviously unobjectionable.\n\nBoth blue- and white-collar employees in larger establishments were\nmutually dependent on their fellows. as one man\'s work fit into\nanother\'s, so one man\'s business was no longer his own. "The results\nof the new organization of life and work were apparent by 1900, when\nsome 76 percent of the 2,805,346 inhabitants of Massachusetts were\nclassified as urbanites. Much violent or irregular behavior which had\nbeen tolerable in a casual, independent society was no longer\nacceptable in the more formalized, cooperative atmosphere of the later\nperiod. . .The move to the cities had, in short, produced a more\ntractable, more socialized, more \'civilized\' generation than its\npredecessors."\n\n17. (Paragraph 117) Apologists for the system are fond of citing cases\nin which elections have been decided by one or two votes, but such\ncases are rare.\n\n18. (Paragraph 119) "Today, in technologically advanced lands, men\nlive very similar lives in spite of geographical, religious and\npolitical differences. The daily lives of a Christian bank clerk in\nChicago, a\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>other. . .Individual
problems, sins or even crimes, were not generally cause for wider
social concern. . ."But the impact of the twin movements to the city
and to the factory, both just gathering force in 1835, had a
progressive effect on personal behavior throughout the 19th century
and into the 20th. The factory demanded regularity of behavior, a life
governed by obedience to the rhythms of clock and calendar, the
demands of foreman and supervisor. In the city or town, the needs of
living in closely packed neighborhoods inhibited many actions
previously unobjectionable.

Both blue- and white-collar employees in larger establishments were
mutually dependent on their fellows. as one man's work fit into
another's, so one man's business was no longer his own. "The results
of the new organization of life and work were apparent by 1900, when
some 76 percent of the 2,805,346 inhabitants of Massachusetts were
classified as urbanites. Much violent or irregular behavior which had
been tolerable in a casual, independent society was no longer
acceptable in the more formalized, cooperative atmosphere of the later
period. . .The move to the cities had, in short, produced a more
tractable, more socialized, more 'civilized' generation than its
predecessors."

17. (Paragraph 117) Apologists for the system are fond of citing cases
in which elections have been decided by one or two votes, but such
cases are rare.

18. (Paragraph 119) "Today, in technologically advanced lands, men
live very similar lives in spite of geographical, religious and
political differences. The daily lives of a Christian bank clerk in
Chicago, a

Neil

Dec14-04, 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>so that the trend is not altered\nbut only pushed a step ahead.\n\n101. The first principle is almost a tautology. If a trend were not\nstable with respect to small changes, it would wander at random rather\nthan following a definite direction; in other words it would not be a\nlong-term trend at all.\n\n102. SECOND PRINCIPLE. If a change is made that is sufficiently large\nto alter permanently a long-term historical trend, than it will alter\nthe society as a whole. In other words, a society is a system in which\nall parts are interrelated, and you can\'t permanently change any\nimportant part without change all the other parts as well.\n\n103. THIRD PRINCIPLE. If a change is made that is large enough to\nalter permanently a long-term trend, then the consequences for the\nsociety as a whole cannot be predicted in advance. (Unless various\nother societies have passed through the same change and have all\nexperienced the same consequences, in which case one can predict on\nempirical grounds that another society that passes through the same\nchange will be like to experience similar consequences.)\n\n104. FOURTH PRINCIPLE. A new kind of society cannot be designed on\npaper. That is, you cannot plan out a new form of society in advance,\nthen set it up and expect it to function as it was designed to.\n\n105. The third and fourth principles result from the complexity of\nhuman societies. A change in human behavior will affect the economy of\na society and its physical environment; the economy will affect the\nenvironment and vice versa, and the changes in the economy and the\nenvironment will affect human behavior in complex, unpredictable ways;\nand so forth. The network of causes and effects is far too complex to\nbe untangled and understood.\n\n106. FIFTH PRINCIPLE. People do not consciously and rationally choose\nthe form of their soc\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>so that the trend is not altered
but only pushed a step ahead.

101. The first principle is almost a tautology. If a trend were not
stable with respect to small changes, it would wander at random rather
than following a definite direction; in other words it would not be a
long-term trend at all.

102. SECOND PRINCIPLE. If a change is made that is sufficiently large
to alter permanently a long-term historical trend, than it will alter
the society as a whole. In other words, a society is a system in which
all parts are interrelated, and you can't permanently change any
important part without change all the other parts as well.

103. THIRD PRINCIPLE. If a change is made that is large enough to
alter permanently a long-term trend, then the consequences for the
society as a whole cannot be predicted in advance. (Unless various
other societies have passed through the same change and have all
experienced the same consequences, in which case one can predict on
empirical grounds that another society that passes through the same
change will be like to experience similar consequences.)

104. FOURTH PRINCIPLE. A new kind of society cannot be designed on
paper. That is, you cannot plan out a new form of society in advance,
then set it up and expect it to function as it was designed to.

105. The third and fourth principles result from the complexity of
human societies. A change in human behavior will affect the economy of
a society and its physical environment; the economy will affect the
environment and vice versa, and the changes in the economy and the
environment will affect human behavior in complex, unpredictable ways;
and so forth. The network of causes and effects is far too complex to
be untangled and understood.

106. FIFTH PRINCIPLE. People do not consciously and rationally choose
the form of their soc