# What effect does direction have on special relativity?

Okay, so here's what I'm stuck on. First imagine there is an observer (observor A) on a train heading west at half the speed of light. It is chasing a beam of light. There is another observer (observer B) on the platform.

Both observers see the light heading away from them at speed c. The observer on the platform says that the person on the train 'should' see the light moving at c/2, but he doesn't, he sees it moving at c, leading observer B to conclude that observer A's clocks are running slow.

However, from the point of view of observor A, observer B is moving east at c/2, and so 'should' see the light beam moving away at 1.5 times the speed of light, but observer B insists it is moving away at c, leading observer A to conclude that Observer B's clocks are running fast. Right?

The problem is everything I've ever read about SR says that both observers should claim the other's time is running slow, this seems to contradict it.(eg. I've approached the situation with light clocks).

Can somebody help clear this up please.

• You are not using any Special Relativity in this example, so it cannot tell you anything about Special Relativity. – m4r35n357 Jul 17 '15 at 19:06

## 4 Answers

"The observer on the platform says that the person on the train 'should' see the light moving at c/2". Your logic is flawed ,you are using gallilean transformation instead of lorentzian

• I used the quotation marks to indicate that this was the tradition (Galilean) view. ie. that is what common sense would lead the observer to believe, though I know this is not the reality – Padraig Stapleton Jul 17 '15 at 18:32
• Dear friend common sense is often a bad guide when entering into the realm of modern physics we must not rely on it – Sahil Chadha Jul 18 '15 at 10:40

There are three crucial components to a good answer to this question: what's the correct thing to think, why is it the correct thing to think, and why is the reasoning given wrong.

# Both think the others' clocks are running slow

This is a deep requirement from the principle of relativity; that is to say the principle that there is no "correct" reference frame but the laws of physics work well in all of them. As a consequence, if I say "you're moving relative to me, therefore your clocks are moving slowly!" you must say the same thing to me.

So what you've identified would not be classified as a "derivation" but rather a "paradox."

# Why that's correct and anything else leads to a preferred frame

Suppose anyone is travelling relative to anyone else in the $\hat x$ direction: Alice thinks she's at rest but Bob sees her as moving with velocity $v \hat x$. Alice fires a laser pulse a finite distance $L$ in the $\hat y$ direction and times it to take time $\tau = L/c$. Suppose that both Alice and Bob agree on the speed of light and on the perpendicular measurement $L$, as the Lorentz transformation demands. Then Bob thinks that it took a time $t$ given by the Pythagorean theorem, $c^2 t^2 = v^2 t^2 + L^2$. The math works out to $t = \gamma ~ \tau$ where $\gamma = 1/\sqrt{1 - (v/c)^2}$. If the event at $y=L$ is a reflection back to the emitter, then one of Alice's clocks reports a time $2\tau$ whereas Bob sees $2\gamma~\tau$, therefore he sees her clock moving "slow".

But if there is no privileged frame of reference, Bob can also fire a laser pulse in this same way to one of his clocks, and Alice must see that clock moving slowly: there can be no difference.

# Resolving this fast/slow paradox in SR

The first crucial fact here is that I don't see someone who is moving past me at speed $c/2$ with their clocks ticking slow by a factor of 2 but rather with a factor of $1/\sqrt{1 - (1/2)^2} = \sqrt{4/3}$. So we're going to have to be very clear about what we mean here, because all three of the aggregate effects of relativity -- time dilation, length contraction, and simultaneity shifts -- can potentially play a huge role here, depending on how you set up the problem.

So let's review length contraction. Alice measures a distance to a block in the $+\hat x$ direction by putting a mirror on it and measuring a round-trip pulse time of $\tau$; she therefore knows that the object is of length $\lambda = \tau/(2 c)$. We already know that Bob measures this time as $t = \gamma \tau = 2 \gamma \lambda/c$, the problem is, for him this light travels a different distance. Let's say he measures the distance between Alice's emitter and mirror as $\ell$, then Bob thinks that this took the time $t = \frac\ell{c + v} + \frac\ell{c - v}$, as he measures the relative velocities of these things (the light and the reflector/absorber) to be $c - v$ and $c + v$ in the two cases. If you remember how to manipulate fractions, this is $t = (2\ell / c)/(1 - (v/c)^2) = 2\gamma^2\ell / c$. The only way these two can be the same is if $\ell = \lambda / \gamma$.

We'll skip simultaneity shifting.

Now we've almost got all the parts we need for your example. Suppose that Bob shines a light in the $-\hat x$ direction, with $v = c/2$. How does Alice measure how fast the light is moving away from her? How does she measure that it's traveling away at speed $c$? Here is one way: suppose in her reference frame, at rest, there are some simple clouds of dust or powder in the path of the light, separated by distance (in her frame) $\lambda$. Now Bob's light beam scatters some light at each cloud, some of which comes back to her. So she now sees this beam travelling away from her as a set of light flashes in the clouds, with period $\tau = 2 \lambda / c$.

Bob of course thinks that these clouds are closer-spaced by $\ell = \lambda / \gamma$. So let's say he fires the light just as one is passing his laser; they are then a grid $x_n = n \ell$, and he receives a pulse from each of these back to him after a time $\frac {2 x_n c}{c^2 - v^2}$, so he measures the period as $$\frac {2 \ell c}{c^2 - v^2} = 2 \gamma^2 \frac{\lambda}{\gamma} c = \gamma \tau.$$ If he immediately forwards this signal with period $\gamma \tau$ onwards to Alice, his light pulses will be in sync with the pulses from the clouds, as all he's done is to absorb and immediately re-emit light. So it must be the case that in Bob's perspective her clocks are running slow by a factor $\gamma$, even though her velocity is directly opposed to the light beam.

If we think non-relativistically for relativity then we'll have certainly false results.

"... ; and asking the right question is frequently more than halfway to the solution of the problem."

Werner Heisenberg in 'Physics and Philosophy' Penguin Edition 1958.

This answer is a non-relativistic dialogue :

A to B : You see $1.5c$ for the speed of light beam, do you ???

B to A : No, I see exactly $c$ !!!

A to B : That's impossible. May be your time is running faster, say by a factor $f=1.5$

B to A : Yes, my time is running faster by a factor $f=1.5$ with respect to your time. But may be I measure distances greater than yours by the same factor $f=1.5$ , so dividing distances by time periods I see the same speed $c$ as you.

A to B : Marvelous, now I understood !!!

You're correct. Two observers moving at some relative velocity to each other such that $v \neq 0$ will always see a clock at the other observer as running slower. However, this is not a problem since at least one of them must accelerate in order for the observers to compare their clocks with each other. In this process, length contraction and time dilation will always make sure that they both agree on what the clocks shows.

Using your example, in A's frame of reference A sees B moving eastwards at a velocity of $v = 0.5 c$ and the light beam westwards with velocity of $v = c$. In B's frame of reference B sees A moving eastwards at a velocity of $v = 0.5 c$ and the light beam eastwards with velocity of $v = c$.