Re: Debunking a thought experiment
From: Mark Palenik (markpalenik_at_wideopenwest.com)
Date: 10/30/04
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Date: Sat, 30 Oct 2004 13:10:54 +0000 (UTC)
"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. 0 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 = 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, 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.
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