# Constant speed of light - in this specific scenario

Suppose there are $4$ points on a straight line from left to right - $A$, $B$, $C$, $D$

$A$ and $D$ are stationary wrt one another and so have same frame of reference. $A$ & $D$ are sufficiently far apart.

$B$ & $C$ are part of a lab which is moving from left to right (in the direction $A$ to $D$) at a significant fraction of $c$, say at $10$% of $c$ wrt $A$ & $D$.

Therefore $B$ and $C$ also have same frame of reference. Distance between $B$ and $C$ is fixed inside its frame.

Light pulse $P_1$ sent from $A$, reaches $B$ at $t_1$ and same pulse reaches $C$ at $t_2$. $t_1$ and $t_2$ are per lab's frame of reference.

Light pulse $P_2$ sent from $D$, reaches $C$ at $t_3$ and same pulse reaches $B$ at $t_4$. $t_3$ and $t_4$ are per lab's frame of reference.

As far as my understanding goes, per relativity, the lab should register $t_2-t_1 = t_4-t_3$.

Is this correct? If so, how to count for the motion of $B$ & $C$ - in direction of pulse $P_1$ and in opposite direction for pulse $P_2$?

As $B$ & $C$ are fixed part of the lab, they both should have same length contraction and same time dilation.

This kind of experiment seems very difficult but has something similar been tried.

I know about Michelson–Morley experiment but in there, the source of light is within the same frame of reference as is the measuring apparatus.

In the setup I described, the sources of light are outside the frame of reference of the lab and are on opposite sides of the relative motion.

• The answer to your question "How to account for the motion of B&C?" is that in the lab frame, there is no motion of B&C, hence nothing to account for. – WillO Apr 28 '17 at 15:46

You are correct. The two time intervals $t_2 - t_1$ and $t_4 - t_3$ will be equal. There is no need to account for the motion of $BC$ because they will always measure the speed of light to have the same value. Since their relative positions are fixed, light will always take the same amount of time to cross from one to the other. What will be different is the length of the time intervals as measured by the $AD$ system. Because they observe that $BC$ are closer together due to length contraction, they will measure a shorter time for the light pulse to go from $B$ to $C$ or vice versa.

• I believe you misread the question. You write "What will be different is the length of the time intervals as measured by the AD system". But those are exactly the time intervals that the OP asserts will be the same, not different. (His "lab frame" is the AD system ---unless I'm the one who misread.) – WillO Apr 28 '17 at 15:06
• Oops! My apologies! I am the one who misread! He is using "lab frame" to mean the BC system. You are totally correct. – WillO Apr 28 '17 at 15:07

How do we account for the motion of $B$ and $C$ in the direction of pulse $P_1$ and in the opposite direction of pulse $P_2$?

And in order to do this, I invite you to set it up in the A/D frame and then Lorentz-transform to the B/C frame of reference. This is the easiest way to really get a feel for what's happening.

# Quick review of Lorentz transforms

It's helpful to imagine that there's a bunch of lab assistants standing near clocks, at rest with respect to the A/D frame, and they're all synchronized and ticking at the same rate: when these things happen in the B/C lab, we imagine that someone standing at the nearest A/D clock who sees it, immediately writes down the A/D time that they saw this thing happen.

Now we write down their positions and times with little "primes", so this A/D observer is at some position $x'$ and they write down the time $t'$ for this event; inside the B/C lab the observers think the position is $x$ and the time is $t$. Assuming that the two frames choose a common event to be $x' = 0, t'=0, x=0, t=0$ then the proper way to transform between these coordinates is:$$x = \frac{x' - v~t'}{\sqrt{1-(v/c)^2}},~~~~~~~~ t = \frac{t' - v~x'/c^2}{\sqrt{1-(v/c)^2}}~~.\\$$Side note: usually it is nice to write $w = c~t,~~w'=c~t',~~\beta=v/c,~~\gamma=1/\sqrt{1 - \beta^2},$ and then these expressions become very pretty, $x = \gamma~(x' - \beta~w'),~~w=\gamma~(w' - \beta~x').$ If you are good at your linear algebra, you might try inverting this to find that the inverse relation simply looks the same but with $\beta\mapsto-\beta.$ I may occasionally lapse into using $\beta$ and $\gamma$ in the discussion that follows, I will try to stay away from using $w$.

# Setting up the problem, calculating some A/D-times

So now imagine that this B/C lab is cruising along at this speed $v$, and I'm going to name people at the different points Alice, Bob, Carol, Darren. First Carol passes Alice, then Bob passes Alice some time $\tau'$ thereafter, and they happen to all have synchronized all of their clocks in advance so that both frames agree that $t' = t = 0$ at that exact moment, and in the B/C lab therefore Bob is always at position $x=0$ while in the A/D reference frame Alice is always as $x'=0.$ Still with me? This time Alice sees, $\tau'$, is the apparent distance between Bob and Carol in her reference frame, if we multiply by $v$. It will turn out that that's not the distance Bob and Carol measure between themselves.

After this magic common-zero agreement of time and position when Bob passes Alice, at some new time $t_0',$ say 5 seconds later, Alice sends a light pulse towards Bob at $B$. It does not arrive precisely 5 seconds after that because Bob is moving forward in Alice's reference frame; instead we need to solve $v~t'=c~(t' - t_0'),$ the $v~t'$ being the $x'$ position of Bob, and $c~(t' - t_0')$ being the $x'$ position of the light pulse that started off at $x'=0$ when $t' = t_0'$. So we find that it is sent at $t_0'$ and then it passes Bob at $t_1' = t_0'/(1 - \beta),$ at which time he is at $x_1' = v~t_1'.$ Then the light pulse continues on to hit Carol at $C,$ which it does at time $t_2' = (t_0' + \beta~\tau')/(1 - \beta),$ at which point she is at position $x_2' = v~(t_2' + \tau').$ The time that Alice calculates between these two events is $\beta~\tau'/(1-\beta),$ perfectly consistent with the idea that the speed of light is constant in her frame of reference and the spaceship has length $v~\tau'$ and is moving away from her light pulse in her frame of reference.

# Transforming those to B/C times

But now let's plug these into our Lorentz transform: we find that this light passes Bob at position $x_1 = \gamma~(v~t_1' - v~t_1') = \gamma~0 = 0$ (of course, he must be at that position!) at his time $t_1 = \gamma~(t_1' - v^2 t_1'/c^2) = t_1'/\gamma.$ So Alice and Bob fundamentally disagree on what time this thing has hit Bob. This number $\gamma$ is always greater than 1 so from Alice's perspective, Bob's clocks are running slow. (From Bob's perspective, Alice's clocks are also running slow!) Actually we usually think of this the other way, that Bob is measuring a time between two events that are at the same position in his coordinates (Alice passes by him, light passes by him), and so he measures a special time between them called the "proper time" between them. For everyone else this amount of time gets "dilated" into a longer time interval than it should otherwise be.

Anyway, we have that $x_1 = 0$ and $t_1 = t_1'/\gamma = t_0'/(\gamma\cdot(1-\beta)).$ Now the light hits Carol, and our coordinate transform says that $x_2 = \gamma~(x_2' - v~t_2') = \gamma~(v~t_2' + v~\tau' - v~t_2') = \gamma~v~\tau'.$ So remember that Alice was measuring the length of the B/C lab as $v~\tau',$ that is a correct expression in her reference frame, but there is a "proper length" to the B/C lab which is actually $\gamma~v~\tau'$, a little bit longer than she saw, and what she saw was this proper length being "contracted" to a shorter distance. (The proper length between two points is a concept that only makes sense if those two points are moving at the same speed, otherwise everyone thinks they're getting closer/further apart. It is the distance between them as measured by someone who is moving along with them at the same speed.)

Finally, the time that it hits Carol is $t_2 = \gamma~(t_2' - v~x_2'/c^2),$ substituting in the easy expression $x_2' = v~(t_2' + \tau')$ gives simply $t_2 = t_2'/\gamma - \gamma~\beta^2~\tau'.$ Remember $c~\beta~\tau'$ is the length of the B/C lab as seen by Alice; we see the familiar time-dilation factor $t_2/\gamma$ but there is also a slightly different effect happening as well, which comes down to: according to Alice, the clocks that Bob and Carol are using are not just ticking slowly but they're not even in sync with each other. In fact, it may surprise you to know that this is the only interesting effect in relativity, the effects of time-dilation and length-contraction are just the aggregated results of it.

# The moment of truth

Now we already calculated $t_2' - t_1' = \tau'/(1-\beta).$ But what about $t_2 - t_1$? It is: $$t_2 - t_1 = \frac{t_0' + \beta\tau'}{\gamma~(1 - \beta)} - \gamma~\beta^2~\tau' - \frac{t_0'}{\gamma~(1-\beta)} = \gamma~\beta~\tau'~\left(\frac{1}{\gamma^2~(1-\beta)} - \beta\right).$$This is the moment of truth: we derive the Lorentz transform from the constancy of the speed of light and we know that in the B/C lab the light travels this distance $\gamma~v~\tau'$ and so if that right hand side in parentheses doesn't simplify to 1, we've reached self-contradiction.

But we know that the $1/\gamma^2$ factor on the left is $1 - \beta^2 = (1 - \beta)(1 + \beta),$ so that complicated expression in parentheses on the right turns out to simplify to exactly $1$. So the time is just $\gamma~\beta~\tau'$ and the distance it covers is just $\gamma~v~\tau'$ and therefore in the B/C frame the light packet travels also at speed $c$. It has the same speed in both reference frames.

You can work out the other direction if you want but I'll give you the punchline: the terms that were all $1/(1-\beta)$ become all $1/(1 + \beta)$ and at the end you get $\frac1{\gamma^2~(1 + \beta)} + \beta = 1$ the same way.