How the Earth observer would see (time-ticks wise) a traveller rushing away or towards Earth at near-light speed?

There is an observer on the Earth.

1) There is a traveller moving away from the Earth at a speed approaching the speed of light.

2) There is a traveller moving towards the Earth at a speed approaching the speed of light.

In both cases there would be (exactly same) time difference of clocks - one tick of the traveller's clock would be very-very many ticks of the observer clock on earth.

I don't understand the following: what would the observer from earth see? Just that the traveller is moving away or towards at a great speed?

In case a traveller approaches a blackhole it is said that the earth observer would never see the traveller disappear behind the horizon (because one tick of travellers clock would be infinite time of observer's clock). Is this right?

But we have the same time dilution (caused by difference in relative speed, not gravity) in our first example, so it seems that the traveller shall "freeze" and seem not to move for the earth observer in both cases (1) and (2), even when the traveller is rushing towards Earth with tremendous speed!

I cannot find error in my reasoning.

So what the Earth observer (and the traveller) would see and why? I mean time difference (how many ticks of traveller per one tick of observer), not physical seeing (so that not to complicate things with doppler effect).

1 Answer

The Lorentz transformation doesn't describe what an observer sees. What we see is a matter of optics: how light rays reach our eyes. The symmetry you describe is not a symmetry of what the observer sees. The observer will see different Doppler shifts for the receding and approaching objects. (In this context, a Doppler shift is not just the shift in wavelength like blue to red, it also defines the rate at which signals reach us.)

So really the case of matter falling into a black hole is not so different from the case of an object just flying around in empty space. In optical measurements, there is no symmetry in either case.

Even without worrying about this distinction between the Lorentz transformation and the results of optical observations, there is no particular reason that we should expect a symmetry in the empty-space case to have a corresponding symmetry in the black hole case. They are different. There is no symmetry in the black hole case between the two observers, because one is at a different distance from the black hole compared to the other.

• I beliebe the OP meant this: time dilation is teh same for both, the approaching and the receding observer. Assume in the reference frame of the travelers the light emited is at 1Hz, and each peack of the light moves the tick of the clock. Now, if the dopler effect for the approaching observer increaes to frequency, why the observer on earth see the clock running slower if the frequency of the observed clock must speed up because it has to tick at a larger, not lower frequency? – Wolphram jonny Mar 31 '19 at 5:42