Rigid Rod in Relativity
Daryl McCullough
suggestions?
Assume that we have a rigid rod and we accelerate it in the direction
of its length. (Well, no rod can be perfectly rigid, but let me assume
that it as rigid as is possible in Relativity). How, then, does its
length change as a function of time (as measured in an inertial
reference frame)? If the rod is moving at a constant velocity v, its
length would be given by L' = L/gamma, where L is the proper length
and gamma = 1/square-root(1-v^2/c^2). However, this formula can't be
correct for an accelerated rod for the following reason: If the rod is
accelerating, then gamma is increasing, which means that the rod is
contracting, which means that the front end of the rod is moving
slower than the back end (relative to an inertial reference frame).
Therefore, there is no single value v to use: the velocity of the back
will be different from the velocity of the front.
I assume that the length of the rod will be given by some differential
equation, but which one? Does the answer depend on the precise nature
of the forces inside the rod?
Daryl McCullough
ORA corp.
Ithaca, NY
Steven Sharp
>Assume that we have a rigid rod and we accelerate it in the direction
>of its length. (Well, no rod can be perfectly rigid, but let me assume
>that it as rigid as is possible in Relativity). How, then, does its
>length change as a function of time (as measured in an inertial
>reference frame)?
You can start by making this more rigorous. Replace the statement of
rigidity with the velocity profile of the endpoints. One possible
definition of "rigid" might be that the length of the rod is always
the same in the instantaneous frame of reference of the rod itself.
This means that the velocities of the endpoints are always equal in
the frame of reference of the rod. As you point out, this then makes
getting the length in an inertial frame difficult, since the endpoints
are not moving at the same velocity simultaneously in that frame.
Given a time in the inertial frame, getting the length will require
solving for both the coordinates of the endpoints and the velocities of
the endpoints together, since the coordinate transformation depends on
the velocity, and the velocities of the endpoints depend on the times in
the accelerating frame, which are themselves dependent on the results of
the coordinate transformation. Simultaneous nonlinear equations. Fun.
Ron Maimon
|>
|> I assume that the length of the rod will be given by some differential
|> equation, but which one? Does the answer depend on the precise nature
|> of the forces inside the rod?
I don't know, but I do know that the rod will, in the instantanious
comoving frame, look the same as if it was suspended in a gravitational
field, amd that gives you a differential equation for the stretching of the
rod, which depends on the elasticity constants.
Once you know what the rod looks like in a static gravitational field, you
can boost to any speed you want to see what it would look like under
uniform acceleration.
Ron Maimon
Bruce Bowen
The problem of a long accelerating rigid rod is essentially that of
a "uniformly" accelerating coordinate system. This coordinate system
has an event horizon. I put uniformly in quotes because if local
proper length is maintained, then acceleration is not a constant
function of position, but varies as a function of position parallel to
the direction of acceleration.
To see this assume we are accelerating in flat space and LOCAL
PROPER LENGTH IS MAINTAINED. Also assume local proper acceleration is
time independant (it may vary as a function of position though).
Assume the acceleration at any point "x" along the length of the rod is
g(x), g for short. Assume the acc. at the nose of the rod is g0.
The redshift due to acceleration is df/f = gdx/c^2. An individual feels
an acceleration "enhancement" proportional to the redshift, so we have:
dg/g = gdx/c^2 <---> dg/g^2 = dx/c^2 <---> g = g0/(1 - xg0/c^2)
So we see there is a characteristic length of x = c^2 / g0, where the
acceleration diverges. The length of the rod cannot be longer than
this and still stay together. This is the position of the event
horizon. You can think of this as the "back end of the rod having to
move at speed "c" in order to keep up with the Lorentz contraction of
this long rod as it accelerates." The tail end of this long rod
must experience infinite acceleration. For one earth gravity this
length is approximately 9e15 meters, or slightly under one light year.
It would not be a comfortable trip for people in the tail end of a
long rocket.
The above model of a long rocket is convenient for qualitatively
understanding the event horizon of a black hole, although in this
latter case, space time is curved so that you "accelerate" just by
"standing still". A person released into freefall at the nose end of
our accelerating rocket would eventually be passed by the tail end
moving at speed c. He falls thru the rocket's event horizon. If the
rocket continues to accelerate, there is nothing he can do to reach
it. Even if he shines a laser at it it will never catch it. This
points out that you can outrun light forever if you have a head start.
If we send a person out in a spacecraft, who continuously accelerates
at one g. Once he gets more than a light year away there is nothing
we can do to communicate with him or call him back.
-Bruce bbo...@megatest.com