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- 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?

##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?