# Understanding Proper Time better

Wannabe Physicist
Homework Statement:
Imagine that space (not spacetime) is actually a finite box, or in more sophisticated
terms, a three-torus, of size ##L##. By this we mean that there is a coordinate system ##x^\mu = (t,x,y,z)## such that every point with coordinates ##(t,x,y,z)## is identified with every point
with coordinates ##(t,x+L,y,z)##, ##(t,x,y+L,z)##, and ##(t,x,y,z+L)##. Note that the time coordinate is the same. Now consider two observers; observer A is at rest in this coordinate system (constant spatial coordinates), while observer B moves in the x-direction with constant
velocity ##v##. A and B begin at the same event, and while A remains still, B moves once
around the universe and comes back to intersect the world line of A without having to
accelerate (since the universe is periodic). What are the relative proper times
experienced in this interval by A and B? Is this consistent with your understanding of
Lorentz invariance?
Relevant Equations:
##(\Delta \tau^2) = (\Delta t)^2-(\Delta x)^2-(\Delta y)^2-(\Delta z)^2##
Let us denote the events in spacetime before the trip has started by subscript 1 and those after the trip is over by subscript 2. So before the trip has begun, the coordinates in spacetime for A and B are

##A = (t_{A_1},x,y,z)## and ##B = (t_{B_1},x,y,z) = (t_{A_1},x,y,z)##.

After the trip is over,

##A = (t_{A_2},x,y,z)## and ##B = (t_{B_2},x+L,y,z)##.

Note that ##\Delta t_A = t_{A_2}-t_{A_1} = \frac{L}{v}## and ##\Delta t_B = t_{B_2}-t_{B_1}##. Now, by the time dilation formula, ##\Delta t_B = (\Delta t_A)/\gamma##

Thus if ##\Delta \tau_A## and ##\Delta \tau_B## are proper time intervals of A and V respectively
$$(\Delta \tau_A)^2 = (\Delta t_A)^2 - 0 - 0 -0 \implies \Delta \tau_A = \Delta t_A$$
$$(\Delta \tau_B)^2 = (\Delta t_B)^2 - L^2 - 0 -0 = \frac{(\Delta t_A)^2}{\gamma^2} - L^2 = \frac{(\Delta \tau_A)^2}{\gamma^2} - L^2$$

I feel that this must be right. I, with my naive eyes, am not able to see any flaw in my argument. But this blog post: https://petraaxolotl.wordpress.com/chapter-1-special-relativity-and-flat-spacetime/ seems to be telling a different story.

It says $$(\Delta \tau_B)^2 = (\Delta t_B)^2-(\Delta x)^2 = (\Delta \tau_B)^2(1-v^2) = (\Delta \tau_A)^2(1-v^2) = \frac{(\Delta \tau_A)^2}{\gamma^2}$$

Am I making any mistake?

Gold Member
I am skeptical in applying SR which is for observers in IFR to the torus 3D-space+time world which does not seem IFR to me globally.

Last edited:
docnet
Wannabe Physicist
I understand your point. But I am trying to teach myself GR and this exercise is in the first chapter of Carroll. He hasn't introduced curvature and stuff yet. Actually, I got the answer. It is ##\Delta \tau_B = \displaystyle\frac{L}{\gamma v}##. The mistake is in writing the coordinates and is very silly. A's and B's coordinates are in reference to A's and B's frames respectively. But in B's frame, the observer is obviously at rest. So we should have ##B = (t_{B_2}, x,y,z)##

Gold Member
Along the problem statement, say starting with zero time adjustment, when they meet again their own proper time are
$$\tau_A=\tau_B=\frac{L}{v}$$
and they see others time are by time dilation
$$t_A=t_B=\frac{L}{\gamma v}$$
We face contradiction because they have to be
$$\tau_A=t_A$$
$$\tau_B=t_B$$
One way to avoid the contradiction is to make L infinite or grow with rate of more than c to prohibit them meeting again.

docnet