Accelerating a long "rigid" rod
From:        Bruce Bowen        
Date:        Mon, Mar 14 1994 5:16 pm
Subject: Re: Rigid Rod in Relativity
Newsgroups: sci.physics
References: <1994Mar11.180535.10371@oracorp.com>

>From article <1994Mar11.180535.10371@oracorp.com>, by daryl@oracorp.com (Daryl McCullough):

> Here's a relativity puzzle that seems to be difficult to solve. Any
> 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?


Ahhh, leaf hopper, you have stumbled upon one of the roads toward enlightenment in relativity!

The problem of a long accelerating rigid rod is essentially that of a "uniformly" accelerating coordinate system, also known as "Rindler Space". 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 refered to as "Born rigidity"). Also assume local proper acceleration is time independent (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 acceleration at the FRONT of the rod is g0.


(1)
The redshift due to acceleration is df/f = gdx/c^2. An individual needs an acceleration enhancement proportional to the redshift to maintain relative position on the rod, so we have:

dg/g = gdx/c^2   ---->   dg/g^2  = dx/c^2  ---->  g = g0/(1 - xg0/c^2)

We now 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 at the front of the rod, 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.

This subject also brings up the often debated issue of whether Lorentz contraction can break a string, and the answer is "Yes it can."

You have two heavy spaceships initially at rest, with identical inertial guidance systems. One is positioned many thousands of kilometers in from of the other one. The two ships are connected by a thread or fragile string that is light enough that its own mass is irrelevant to the problem.

Their guidance systems are synchonized and loaded with the same program to start uniform acceleration at a certain time. With the exception of the fragile thread (which is irrelevant) both ships are completely independent of the other.

Both ships start acceleration at the same time. In the inertial rest frame of the ships before they started, both ships have simultaneous velocities and constant separation throughout eternity (since they are independent of each other and have identical inertial guidance systems.)

In the frame of the trailing ship, he sees the lead ship pull ahead.

The lead ship sees the trailing ship fall behind and eventually fall thru his event horizon. In anycase, the thread breaks because it's Lorentz contracting and the separation between the ships isn't.

A way to view this is, if instead of constant acceleration, assume the guidance systems are programmed with uniformly timed impulses. Initially, both ships see their impulses synchronized, but as they pick up speed the impulses from the lead ship are seen to occur earlier in the frame of the trailing ship due to the de-synchonization in the Lorentz transformations. In the initial inertial rest frame the impulses of both ships remain synchronized throughout eternity.

If the trailing ship does not wish to fall behind the lead ship, he has to time his impulses at the OBSERVED rate he sees them coming in from the lead ship. These will be at a higher rate than what was initially programmed. The inhabitants of the trailing ship will then need feel a higher acceleration to keep up with the leader. With this latter case, the two ships are not independent of each other and the problem is now equivalent to our long rigid rod experiencing acceleration.

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.

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A very easy way to calculate the gravitational redshift in a weak field, using the principle of equivalence is as follows:

Assume you have an accelerating elevator, initially at rest, of vertical length dx, which is small compared to the gravitational gradient. Assume the gravitational acceleration is "g".

Light takes time dt = dx/c to go from the bottom to top of the elevator. The elevator's own change in position due to "g" is negligable in this calculation of dt.

In this time the elevator has accelerated to speed v = g*dt = g*dx/c.

Plug into doppler formula sqrt(c-v)/sqrt(c+v) and we have

f' = f*sqrt[ (c - g*dx/c) / (c + g*dx/c) ] ~~ f*(1 - g*dx/c^2)

df = f' - f ~~ f*g*dx/c^2, which, in the limit as dx -> 0 is an equality, ie. df = f*g*dx/c^2.

This gives us the differential equation for the redshift of light moving parallel to a static gravitational field for any value of proper acceleration "g".