# Inconsistency in special relativity?

inconsistency in special relativity???

Hey i was just thinking about some of the stuff we have done in SR, and i thought of this situation, and i come up with an inconsistency.
the situation is, you have a rod moving at 0.76c relative to you (the observer). its moving to the right.
if the rest length of the ruler ( call it L') is 400m.
and then if a light signal is sent from one end of the rod to the other
if t' is the time it takes for the light signal to go from one one end of the rod to the other in the frame at which the rod is at rest. then
$$t' = \frac{L'}{c}$$
= 1.333 x 10^{-6} s
and if t is the time it takes for the light signal to go from one end to the other in the rest frame of the observer.
since the observer see the length of the rod to be $$\frac{L'}{\gamma}$$
then the time it takes for the light signal to reach the other end according to the observer is:
$$t = \frac{\frac{L'}{\gamma}}{c} + \frac{vt}{c}$$
= 3.611 x 10^{-6} s s
shouldnt these times be the same? or am i going crazy?
Thanks
-Sarah

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Pengwuino
Gold Member
I believe you did not incorporate time dilation... i THINK

could you show me what you would get for the each time?

this thing is driving me nuts!

Pengwuino
Gold Member
This really isnt my field of expertise (my phd is in sleeping and not doing work)... someone else should be along shortly to take your call. Please stay on the line. *elevator music plays*

ok well i am pretty sure that:

$$t = \frac{L'}{\gamma (c - v)}$$

and

$$t' = \frac{L'}{c}$$

but then how are t and t' related??????!?

i thought it was by:

$$t' = \frac{t}{\gamma}$$

but that doesnt work!?!

whats going on here?!? :(

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jtbell
Mentor
The two events (departure of the light from one end of the ruler, and arrival of the light at the other end of the ruler) are at different locations in the ruler's frame of reference. Therefore, in order to find the time between them according to a moving observer, you have to take into account two effects:

1. Time dilation

2. Relativity of simultaneity: if two events at different locations are simultaneous in the ruler's frame, they are not simultaneous in a relatively moving observer's frame. This also affects the time between events that are not simultaneous in the ruler's frame.

To include both effects properly, it's best to use the full Lorentz transformation equations, not the simple time-dilation equation. In your case, with the ruler frame being the "primed" frame and the observer's frame being the "unprimed" one, and $v$ being the velocity of the primed frame in the unprimed one, the equations are:

$$x = \gamma (x^\prime + vt^\prime)$$

$$t = \gamma (t^\prime + vx^\prime / c^2)$$

where the two frames are such that $x = x^\prime = 0$ at $t = t^\prime = 0$.

For your example, let the light be emitted at the left end of the ruler at $t_1 = t_1^\prime = 0$ and $x_1 = x_1^\prime = 0$ (in the ruler's reference frame); and let the light be received at the right end at $x_2^\prime = 400 m$ and $t_2^\prime = 1333 ns$. I'll leave it to you to calculate $x_2$ and $t_2$.

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jtbell said:
The two events (departure of the light from one end of the ruler, and arrival of the light at the other end of the ruler) are at different locations in the ruler's frame of reference. Therefore, in order to find the time between them according to a moving observer, you have to take into account two effects:

1. Time dilation

2. Relativity of simultaneity: if two events at different locations are simultaneous in the ruler's frame, they are not simultaneous in a relatively moving observer's frame. This also affects the time between events that are not simultaneous in the ruler's frame.

To include both effects properly, it's best to use the full Lorentz transformation equations, not the simple time-dilation equation. In your case, with the ruler frame being the "primed" frame and the observer's frame being the "unprimed" one, and $v$ being the velocity of the primed frame in the unprimed one, the equations are:

$$x = \gamma (x^\prime + vt^\prime)$$

$$t = \gamma (t^\prime + vx^\prime / c^2)$$

where the two frames are such that $x = x^\prime = 0$ at $t = t^\prime = 0$.

For your example, let the light be emitted at the left end of the ruler at $t_1 = t_1^\prime = 0$ and $x_1 = x_1^\prime = 0$ (in the ruler's reference frame); and let the light be received at the right end at $x_2^\prime = 400 m$ and $t_2^\prime = 1333 ns$. I'll leave it to you to calculate $x_2$ and $t_2$.

ok yep that makes sense.

i tryed to apply that approach to question (b) of this question:
http://img379.imageshack.us/img379/948/picture910js.th.png [Broken]

but it doesnt seem to work! :( would you be able to show me how you would do this part of the question using lorentz transformations?

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jtbell
Mentor
sarahisme said:
it doesnt seem to work! :(
What, precisely, doesn't seem to work? Show us what you did and I or someone else can probably point out the mistake.

don't worry about it, i think i got it! :)

Thanks for all your help though! :D