(This is probably a clone of some other question, but I haven't been able to find a satisfactory answer, so I hope this is okay.)

So, I know that in special relativity, there's no such thing as a privileged reference frame: If you say you're moving one speed, it's equally valid to say you're moving another.

I've seen setups for measuring absolute velocity wrt light, and I've seen explanations of why they don't work. But I don't understand them. Usually, the setup is a line with a clock and a light detector at each end. Light goes from A to B, measure the time, and since the distance is known, you compare the measured time to the calculated time. Usually, the explanation is that A and B end up desynchronized.

I can understand that the experiment doesn't work because it's essentially expecting the speed of light relative to them to change. If I'm understanding correctly, A and B would measure the distance between them shrinking, right? From a lab frame, the arc length the light travels ends up the same, correct? I'll make a small diagram here (in side view from the lab frame):

A ~>~>~> B
at rest, light travels the distance shown

  A ~>~> B
Once moving, the distance between A and B shrinks, so light ends up travelling
the same distance in the lab frame.

So what exactly happens in A and B's frames of reference? What do they see?

If this is incorrect, here is another setup which I think might help me understand why. (This is presented in top-down view, again from the lab frame.)

At rest, A sends a signal to B. The distance between A and B is very large.
Note also that A and B move together.

  B    |     B   
       |   ^
       |   |
  ^    |
  |    |
  A    |     A   
-----> |   ----->
The whole frame moves sideways, and A sends a signal to B. By the time the light
reaches the place that B was, B is no longer there, and the signal misses.

Since whether they are moving or not only depends on what frame you view them from,
this is a paradox.

What happened? Does the signal reach B or not?
  • 1
    $\begingroup$ In the moving frame the light moves upwards and to the right in such a way that it reaches B. This has nothing to do with special relativity per se; the same thing would happen if instead of light A threw a ball at B. $\endgroup$
    – Javier
    Feb 1, 2016 at 15:06

2 Answers 2


This is a very good question. For your second example, the light does actually reach B, but the paths it takes to get there are different for each frame of reference. If you are in A and B's reference frame, you could argue to be at rest and the light travels a direct path. Looking from the labs perspective however, the light moves along with the frame AB. It has to take a longer diagonal path to reach B. This means from the lab perspective time in AB appears to have slowed.

Light paths in the two frames

It is important to note that in special relativity everything inside your own reference frame seems normal. So it could be argued that you are always in a resting frame as long as $delta$v is zero.

So in all your examples A and B would observe a normal light beam moving between them as if they were always at rest.

Let us take a different example (from Brian Greene's The elegant Universe). Imagine you had a train moving at a constant velocity with a light source exactly at its centre. Now if two people stood on the ends of the train, and the light was flashed, they would both see it at the same time because the light is moving at speed c in relation to them and they are in relation to one another static. Someone on the platform that the train passes will say however that the person in the back saw the light first because they were moving towards it decreasing the length of the lights path. Who is right? Both observers are right within their frames of reference.

Speed cannot be absolute because the way we measure it, through time and distance is not absolute.

You can think about it this way: Even in normal newtonian mechanics, the velocity is relative but not because of properties of time and space but because the earth too is moving. However when you throw a ball across a field you never claim that it travels at thousands of kilometers per hour. This is because your classical newtonian frame of reference is different. If you are on something thats moving and you dont know it you might as well not be moving.

It is the equivalence between any state of uniform motion that makes it impossible for velocity to be absolute. If you can't tell whether you're moving or not, how are you supposed to tell an absolute velocity. There simply isn't a static background or frame to relate it to.

  • $\begingroup$ From the lab frame of reference, what causes the light's path to "tilt"? Wouldn't the light have a higher total speed if that happened? (i.e. √((c^2)+(x^2)) where x is the horizontal speed, seen from the lab frame) $\endgroup$
    – Aieou
    Feb 2, 2016 at 10:34
  • $\begingroup$ No the light travels a longer distance with the same speed so this means time must be slower $\endgroup$
    – Jaywalker
    Feb 2, 2016 at 12:02
  • $\begingroup$ Why does the light travel on a diagonal? Imagine you were on a train and jumped, you would move straight up and down in your frame of reference but someone outside the train would see you travelling in a parabola. This is the same effect with the light but while your relative speed does change, the light's speed remains invariant. So it has to result in a time dilation. $\endgroup$
    – Jaywalker
    Feb 2, 2016 at 12:05

I think that there's two things which you're leaving out:

First, slanty lines. This was adequately covered by @Jaywalker but if you need a further synopsis: suppose in reference frame $R_1$ both $A$ and $B$ are at rest and $A$ emits a laser pulse at $B$, describing the trajectory $x = 0, y = c \tau.$ We transform to a reference frame $R_2$ moving in the $x$-direction with velocity $-v$ relative to $R_1$, meaning that both of these points $A$ and $B$ are moving forward in $R_2$ with velocity $+v~\hat x.$

The only way that relativity can prescribe the crucial property that "all reference frames agree on what events have happened" is if the light pulse is now moving somewhat forwards in the $\hat x$ direction too. This line has to become slanty, otherwise one reference frame will say "$B$ received the pulse" and the other will say "$B$ did not receive the pulse." And that would be enough to conclude "we can no longer do physics." So if we're going to keep doing physics, this line must become slanted.

In fact, let's imagine that $A$ emits this pulse in a "circle" in the $yz$-plane perpendicular to $\hat x$. As it expands it describes a "disk" shape over time. Well in $R_2$ that "disk" must all tilt forward and become a "cone." If you now imagine a uniform radiation in all directions, half is on one side of the disk and half is on the other. So in $R_2$, half must be within this cone and half must be outside of it. If you stare at that fact long enough you will derive a known effect called relativistic beaming: if a particle is uniformly radiating in its rest frame, then in a frame $R$ where it is travelling near the speed of light it "beams" almost all of its radiation in the direction that it's going. That's the formal name for these slanty lines. (If you actually understood this: congrats, most undergraduates struggle for a long time through some spinor mathematics in their senior year or their first Master's year to get this important effect.)

If you accept these slanty lines then you can derive the Lorentz transformation from those examples, just beware of one thing...

Second, getting the light back where it started. This is extremely important. You can view all of the effects of one big relativistic boost from $R_1$ to $R_2$ as coming from compounding tons of tiny "mini-boosts" which are much simpler. This mini-boost by a tiny speed $\delta v$ in the $x$-direction maps the tuple $$(w,\, x,\, y,\, z) ~\mapsto~ \left(w - x~\frac{\delta v}{c},\;x - w~\frac{\delta v}{c},\; y,\; z\right),$$where $w = ct$ is our new geometric view of time in relativity. Notice that the mapping $x \mapsto x - w~\delta v/c$ is exactly the $x \mapsto x - t~\delta v$ that you always see with Newtonian reference frames, and the only new effect is that this symmetrically happens with the time coordinate $w$ as well.

You can view this as saying: if two clocks in $R_1$ are in-sync but are separated in the $\hat x$-direction, then in $R_2$ they will inevitably be out of sync. In fact whenever you accelerate, you see in-sync clocks come out of sync if they are spaced in the direction that you are accelerating. Length contraction and time dilation are just the summed-up effects of this de-synchronization.

Therefore it does not matter just yet when the light is going perpendicular to the direction that we are boosting, but it will become massively important if you want to calculate length contraction that you fire the light pulse to the front of the spaceship, reflect it off of a mirror, and then have it get detected by a detector at the back of the spaceship. Because this detector is in the same place as the source of the light, we do not have to care about how the clocks at the emitter and absorber have desynchronized in $R_2$: they simply haven't, they are the same clock! This means that your time expression in $R_2$ will look like $L'/(c + v) + L'/(c - v) = 2\gamma^2 L'/c$ whereas your time expression in $R_1$ is $2L/c$, even after including time dilation we get $L' = L/\gamma.$ If you do not get the light back where it started, then you will not get this result because you will be assuming that the clocks stay synchronized between $R_1$ and $R_2$, whereas the whole point is that the clocks desynchronize.

  • $\begingroup$ Okay, once I accept "slanty lines", the rest follows. But I don't understand how that happens. Wouldn't that mean that the light is speeding up? i.e. it has speed c in the y direction, but speed whatever in the x direction? Isn't that impossible? And since this most likely doesn't happen, what causes the light to tilt in the lab frame's point of view? (Aside from the consistency that requires that the light from A reaches B, no matter what.) $\endgroup$
    – Aieou
    Feb 1, 2016 at 22:38
  • 1
    $\begingroup$ @Aieou The light maintains a uniform speed that is why time must dilate for it to travel different distances in two frames of reference $\endgroup$
    – Jaywalker
    Feb 2, 2016 at 12:10
  • $\begingroup$ @Aieou I'll skip the question of whether light "speeds up" until we understand the maths, ok? Instead, first I want you to appreciate that this happens with non-relativistic balls, too. If I roll a ball width-wise across a train, someone on the ground will see that ball moving in a slanty line forwards as it goes from one side of the track to the other. That's entirely from the $x - w \delta v/c$ part: that takes the constant-x coordinate and turns it into a changing-x coordinate, so $(0,0,0,0)\mapsto(0,0,0,0)$ but $(w,0,\epsilon,0)\mapsto(w,w\delta v/c,\epsilon,0).$ $\endgroup$
    – CR Drost
    Feb 2, 2016 at 15:57
  • 1
    $\begingroup$ In other words, that's the "classical part" of the transformation: if you want to understand "how that happens", you have to understand that this is fundamental to all coordinate transforms and is embedded in this $x - v t$ formulation, which slants the lines forwards. It's the $t - \delta v x/c^2$ part in relativity that's going to keep things from going faster than the speed of light, so that's where the "cool stuff" happens, but the "slanty lines" are part of the "boring stuff" that's always happened. It's just the crucial adequacy condition, "all reference frames agree on what happened." $\endgroup$
    – CR Drost
    Feb 2, 2016 at 16:03

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