How Does Uniform Acceleration Affect Time Perception in Special Relativity?

[James L. Stru]

I have some questions ion on Uniform Acceleration in SR, inspired by exercise 19 c) in Chapter 2 of
Schutz, A First Course in General Relativity.

Here is the exercise:

A body is uniformly accelerated [tex]\[
a = 10m/s^2
\] [/tex]
. Find the elapsed proper time for the body as a
function of t. (Integrate [tex]\[
d\tau
\] [/tex]
along its world line.) How much would a person accelerated like this
age on trip of [tex]\[
2 \times 10^{20} m
\] [/tex]
to the center of the Milky Way?

I can get the answer, they would age about 10 years.

My questions are:

1. Why does integrating [tex] \[
d\tau
\] [/tex]
along the world line give you the how much the person ages?

For an unaccelerated observer, [tex]\[
\tau
\] [/tex]
would be purely the time they experience. But this is observer is
accelerated. They know they are moving and [tex]\[
\tau
\] [/tex]
should be a mix of time and distance traveled.

2. If [tex]\[
\tau
\] [/tex]
is purely the time the accelerated observer experiences, how do you calculate the distance
they observe themselves to have traveled?

In this case, they ‘know’ they have traveled [tex]\[
2 \times 10^{20} m
\] [/tex]
because they are at the center of the Milky
Way, which they know was [tex] \[
2 \times 10^{20} m
\] [/tex]
away when they started out. But they are zooming along at
nearly the speed of light at this point. If they try to measure how far they have gone, its going to
be a lot less than [tex] \[
2 \times 10^{20} m
\] [/tex]
due to Lorentz Contraction.

3. You might think that you could calculate the time and distance observed by the accelerated
traveler by using the instantaneous velocity in the Lorentz Transformations. But that doesn’t
work. Regardless of where a uniformly accelerated observer is or how fast they are going,
the Lorentz Transformations transform you back to the starting point. This would seem to mean
that the uniformly accelerated observer isn’t moving at all.

Do the values you calculate from the Lorentz Transformations in this case have any physical
significance? Or is this whole business of commoving frames and invariant hyperbolas just a
way to obtain the relationship between the felt acceleration and the observables in the
unaccelerated frame?

Homework Equations

The Attempt at a Solution

 

[BigJDubs]

1. Integrating \[d\tau \] along the world line gives you the proper time elapsed because the proper time is related to the distance and velocity traveled by the observer. The integration of \[d\tau \] gives the total proper time experienced by the accelerated observer as a function of t.2. The distance that the accelerated observer observes themselves to have traveled can be calculated using the Lorentz Transformations. By transforming the coordinates of the starting point and ending point of the trip, you can calculate the distance that the accelerated observer would observe themselves to have traveled.3. The values you calculate from the Lorentz Transformations have physical significance as they can be used to calculate the distance that the accelerated observer would observe themselves to have traveled. The concept of commoving frames and invariant hyperbolas gives the relationship between the felt acceleration and the observables in the unaccelerated frame.