Relativistic Accelerating Frame (non-inertial frame)

Question

One way to resolve the Twin paradox is to count in the effect of acceleration when velocity suddenly changes. I am therefore searching for a relativistic version of "accelerating frame" or non-inertial frame. Apparently it is much less straightforward than accelerating frame of Galilean relativity.

The key question is: if an observer stays at rest at the origin of an accelerating frame, how much (proper) time will she experience? In the laboratory frame, it certainly should not equal $T/\gamma(u),$ where $\gamma(u) = (1-u^2/c^2)^{-1/2},$ and $u$ is the instantaneous velocity $\frac{dx}{dt},$ because this does not resolve the Twin paradox.

Could anyone explain how the proper time changes with acceleration?

Answer 1

The issue is not so much that it is difficult in principle to construct a reference frame for a non-inertial object, but rather there is just no one way that is accepted as the standard way. So if Bob is inertial then the phrase "Bob's frame" has an accepted standard meaning. If Bob is non-inertial then the phrase "Bob's frame" can be defined any number of ways, subject only to the requirements that the coordinates must be smooth and invertible.

Since there is no one standard accepted way of defining a reference frame for a non-inertial object, in the end it is somewhat a matter of personal preference and taste. For me, my preferred method is Dolby and Gull's "radar coordinates". Essentially, you imagine the object sending radar pulses to every event in spacetime and listening for the return. If the time that the pulse was sent is $t_1$ and the time that the echo was received is $t_2$ then the time of the event is $(t_1+t_2)/2$ and the distance to the event is $c(t_2-t_1)/2$. The thing that I like about this definition is that it preserves the second postulate. Most non-inertial frames do not maintain the second postulate.

The proper time itself is an invariant, so its value does not depend on the coordinates at all. However, the formula for the proper time expressed in terms of the coordinates depends on the acceleration profile. For the usual twin's scenario it is given by: $$ds^2=\begin{cases} e^{-2\alpha}(d\tau^2-d\rho^2) & \text{region II} \\ d\tau^2-d\rho^2 & \text{regions F and P} \\ e^{2\alpha}(d\tau^2-d\rho^2) & \text{region I} \end{cases}$$ with all quantities and regions as defined in the paper. Notice in particular that the stay at home twin spends a considerable amount of time in region I, where the proper time runs fast. Thus the non-inertial frame correctly predicts that the home twin will age more than the traveling twin.

Answer 2

You only need a single frame, and an inertial one at that, to calculate the twin paradox exactly. Just focus on the fact that if two events are separated by $dt,dx,dy,dz$ in some inertial frame, then the proper time between them is $$ d\tau = \sqrt{ dt^2 - (dx^2 + dy^2 + dz^2)/c^2}. $$ For motion in one spatial dimension this simplifies to $$ d\tau = \sqrt{ dt^2 - dx^2 /c^2}. $$ It follows that the total proper time along a worldline $p$ between events $A$ and $B$ is $$ \tau = \int_{(p),A}^B \sqrt{ dt^2 - dx^2 /c^2} \; = \int_{(p),A}^B \sqrt{1 - v^2 /c^2} dt. $$

If we pick a frame where those events are at the same spatial location, then for the worldline going straight from $A$ to $B$ we have $v=0$ all the way, so the proper time for that worldline is $$ \tau_{v=0} = t(B) - t(A) $$ For some other worldline between the same two events, one will have $v^2 > 0$ for some or all of the worldline, so then the proper time is smaller. The twin who follows such a worldline will therefore age less, between $A$ and $B$, than the twin whose worldline was straight.

The above method of calculation is perfectly accurate for any worldline at all, including ones which first accelerate away and then turn around and come back. No approximations are introduced, and you do not need to assume a sudden turn-around or anything like that.

Answer 3

Basically in GR, all frames that are not subject to a force are inertial. This means that free-falling objects, which we would normally say are accelerating under the influence of gravity and so are not inertial frames in special relativity, are actually inertial frames.