# What will the observer observe?

Consider the case where observer $A$ is at rest, and observer $B$ is moving with speed $\frac{c}{2}$ (where $c$ is the speed of light) propagating a wave with wave speed $c$. So my question is what will be the wave speed observed by observer $A$(which is at rest)?

I'm really curious to know the answer.

• The wave speed will again be c.For such relativistic speeds, the rule for transforming velocities is not the conceptual, Galilean one, i.e. adding the relative velocity to the velocity observed by one observer. You should include a relativistic factor predicted by Lorentz transformations. – Panos C. Jun 20 '18 at 16:11
• The speed of light is the same for every observer in an inertial frame of reference. – Andrei Geanta Jun 20 '18 at 16:15
• They will both see an EM wave to propagate with speed c. It is because you intuitively think that our speed are normal. The only speed is the speed of light, and everything is relative to that. In your case, A is traveling at 0c, B is traveling at c/2. They are obviously in this case seeing the EM wave propagate at speed c, if you see their speeds as being relative to c. This is because everything with no rest mass travels at speed c, and to slow down, you have to gain rest mass. Macro objects, like in our world, all have rest mass and all travel relative to c. – Árpád Szendrei Jun 20 '18 at 16:51

Observer A will see the wave propagating at speed $c$ (the same speed which Observer B sees). This is essentially one of the basic postulates of special relativity: the speed of light (in vacuum) has the same value as measured in all inertial frames.

Even if Observer B moved at speed $0.999c$ relative to Observer A, they would still both see the wave propagate at speed $c$ relative to themselves.

The relevant formula in special relativity should be the velocity-addition formula if you want to look at how you might compute this, but really the velocity-addition formula is constructed so as to reproduce the postulate mentioned above, so arguing from first principles perhaps makes more sense here.

Note that this is very different from Galilean velocity-addition, which would give the intuitive result that observer A sees the wave propagating at speed $\frac{3}{2} c$.

Also I assume here that all the motion is just in one dimension.

• But if you would consider the distance travelled by them in 1 second then it should be $AW=AB+BW$ which isn't the case here. – Sahil Silare Jun 20 '18 at 16:17
• @Grayscale The postulate of SR only applies to inertial frames, so it is not invariant "regardless of their motion." The proper speed in non-inertial frames remains $c$, but the coordinate speed of light is routinely changed by gravitational fields. – Zack Hutchens Jun 20 '18 at 16:18
• @zhutchens1 I think it should be fixed now. – Grayscale Jun 20 '18 at 16:22
• What's postulate of SR? – Sahil Silare Jun 20 '18 at 16:22
• @SahilSilare There are two main assumptions that SR makes. One is about the invariance of the speed of light (see above) and the other is that the laws of physics are the same in all inertial frames. See the linked article. – Grayscale Jun 20 '18 at 16:26

Einstein's two axioms for special relativity :

1. PRINCIPLE OF RELATIVITY: The laws of physics are identical in all inertial frames, or, equivalently, the outcome of any physical experiment is the same when performed with identical initial conditions relative to any inertial frame.

2. LAW OF LIGHT PROPAGATION: Light signals in vacuum are propagated rectilinearly, with the same speed c, at all times, in all directions, in all inertial frames.