How does SR explain constant light speed where the distance between observer and light source is increasing?

Question

Given are two cases of relative motion between an observer and a light source:

1. the speed of light is always observed to be the same regardless whether the light source is approaching the observer or the observer is approaching the light source, and

2. the speed of light is always observed to be the same regardless whether the light source is receding from the observer or the observer is receding from the light source.

Time dilation in Special Relativity resolves the first case in keeping light speed constant because the slowing of time (time dilation) keeps the same distance travelled per elapsed time on the observer's clock. But SR does not seem to resolve the second case because the observer's clock would have to speed up (not slow down) to maintain the same distance per elapsed time on the observer's clock. Please explain.

Answer 1

You are neglecting the relativity of simultaneity and length contraction. All three are needed to explain the invariance of $c$.

Specifically, the usual derivation is to assume the invariance of $c$ along with the principle of relativity as postulates. These two postulates imply the Lorentz transform which implies all three of time dilation, length contraction, and relativity of simultaneity, and vice versa.

However, it is also possible to reverse the proof. You can start with time dilation, length contraction, and the relativity of simultaneity and derive the Lorentz transform which then implies the invariance of $c$ and the principle of relativity. You cannot do the derivation from time dilation alone, you need all three (the relativity of simultaneity is the one usually neglected by new students).

Once you have the Lorentz transforms, deriving the invariance of $c$ in both directions (your two cases) is straightforward:

Suppose that you have a flash of light at $(t,x)=(0,0)$ then the part of the flash going to the right is given by $x_+=c t$ and the part of the flash going to the left is given by $x_-=-c t$.

If you plug those expressions into the Lorentz transform and simplify then you get $x'_+=ct'$ and $x'_-=-ct'$. So using the full Lorentz transform which includes the relativity of simultaneity immediately shows that the speed of light does not depend on the direction in either frame.

Answer 2

The key to understanding SR is to understand the relativity of simultaneity, which is the cause of effects such as time dilation and length contraction.

Suppose you walk East past a stationary person at a metre per second, and just as you pass, that person flashes a light that heads off East and West. After a second in the frame of the stationary person, the light will be 229,792,458 metres away in both directions, but in your frame you will have walked a metre towards the East, so when the light is equidistant from the stationary person it is 229,792,457 metres ahead of you and 229,792,459 metres behind you, ie it is nearer to you in the direction you are walking and further from you in the opposite direction. Given that the speed of light, for you, is the same in both directions, in your frame the time associated with the position of the light ahead of you must be just under a second from the time of the flash, while the time associated with the position of the light behind you must be just over a second from the time of the flash.

More generally, what happens is that when you are moving relative to someone else, your time axis in spacetime is tilted relative to theirs. That means a plane of constant time for them is a sloping slice through time for you, and vice versa. If you ponder on that for long enough, you should come to realise that it explains time dilation, and it also explains why time dilation is symmetrical in a particular way.