I've been given the following scenario:
Observer $B$ is in the center of a train carriage which is moving at velocity $v$ with respect to an observer $A$. Two light signals are emitted from sources L at the left end and R at the right end of the carriage, such that they reach $B$ simultaneously. $B$ observes that the light signals were emitted at the same time. Show that $A$ doesn't agree.
I've no idea how to do this mathematically. I understand the concept; that as the carriage is moving to the right, $A$ will see that the light from the left will have to travel further than light traveling from the right, as $B$ is moving towards the light coming from the right.
What I've done so far (which I feel is wrong) is as follows:
Let the carriage be of length $2d$ from $A$'s point of view. Then the distance that light has travelled from the left is $d+v\Delta t_L=c\Delta t_L$. Similarly, the distance that light has travelled from the right to B is $d-v\Delta t_R=c\Delta t_R$. Can solve these to get $\Delta t_L=\dfrac{d}{c-v}$ and $\Delta t_R=\dfrac{d}{c+v}$.
If this is correct then I'm ok, as if $\Delta t_R \neq \Delta t_L$ then the light can't have been emmitted simultaneously according to $A$. However, I feel like having the denominator $c+v$ is wrong, as in some sense I'm adding velocity $v$ to the speed of light which is not allowed in SR.
Secondly, in B's frame of refence, the two times $\Delta t'_L=\Delta t'_R$. Surely two equal time differences will dilate the same and we get that $\gamma \Delta t_L= \gamma \Delta t_R$ and thus $\Delta t_L=\Delta t_R$?
Thanks for any replies!
Best Answer
It is permissible, in SR, to have 'non-physical speeds' in excess of c.
For example, you observe two trains speeding towards on another. You measure the speed of each train to be say, $0.9c$ relative to the track.
Then, according to you, their closing speed is $1.8c$, i.e., if at some time, the trains are separated by a distance $d$, the time to impact is
$$\frac{d}{0.9c + 0.9c}=\frac{d}{1.8c}$$
This is fine because there is no physical object with speed $1.8c$.
However, the speed of one train, according to an observer on the other train, must be less than $c$ and is given by the relativistic velocity addition formula.
So, the closing speeds, according to A, of B and the light beams are $c + v$ and $c - v$.
UPDATE to address the comments:
No. Regarding your second question, you seem to be trying to use time dilation inappropriately to understand the relativity of simultaneity.
First, it is true that, according to B's clocks, (one in the rear of the train, one in the center, and one in the front), the time it takes for light to travel from the rear to the center and from the front to the center are equal.
In fact, it's true by definition because this how B confirms that his clocks are synchronized; see Einstein synchronization.
But, according to A's (Einstein synchronized) clocks, B's clocks are not synchronized. In fact, according to A, the clock 'in front' is behind the clock 'in back' (though they both run at the same rate).
This is precisely the reason that $\Delta t'_L = \Delta t'_R$ while $\Delta t_L \gt \Delta t_R$; no time dilation is required to explain this - only the constancy of the speed of light and the Einstein synchronization convention.