# SR Doppler Redshift & SR Time Dilation

1. Dec 19, 2008

### nutgeb

Light sources A & B begin at the same location. Then, source A begins moving directly (and soon inertially) away from Observer at 0.8c, while source B begins moving directly (and soon inertially) toward Observer at 0.8c.

SR Doppler redshift causes the light received by Observer from A to be redshifted, and the light from B to be blueshifted. Light from A is redshifted because the wavecrests look stretched apart, and light from B is blueshifted because the wavecrests look compressed together.

However, SR time dilation causes Observer to perceive that BOTH A's and B's clock run slower than Observer's clock. In general, in its rest frame an observer perceives an inertially moving object's clock to run slower than the observer's own clock, regardless of the direction of movement. Light originating from a slower clocked source, when viewed in a reference frame with a faster clock, creates a redshifting effect.

Does the Lorentz tranformation time dilation cancel out the Lorentz transformation blueshift from source B, and double the Lorentz transformation SR Doppler redshift from source A, as perceived by Observer?

Last edited: Dec 20, 2008
2. Dec 20, 2008

### JesseM

If by "perceives" you mean what Observer actually sees visually, that's incorrect. The time dilation equation tells you the coordinate time between ticks of a moving clock in Observer's rest frame, but this is different than the rate that Observer sees the clock ticking with his own eyes. This is because if the clock is moving, then the light from successive ticks is emitted at different distances from him, and therefore has a different amount of time to reach his eyes.

For example, take clock B moving towards the observer at 0.8c. Say at t=10 seconds in Observer's frame, clock B is 50 light-seconds away, and it shows a time of T=0 seconds. Then at t=20 seconds in Observer's frame, clock B is 42 light-seconds away (because at 0.8c it will have gotten 8 light-seconds closer in those 10 seconds), and it shows a time of T=6 seconds (this follows from the time dilation equation, which says that in 10 seconds a clock moving at 0.8c will elapse $$10*\sqrt{1 - 0.8^2}$$ = 6 seconds).

Now, because the clock read T=0 seconds when it was 50 light-seconds away at t=10 seconds, it will take 50 seconds for the light from this event to reach Observer's position, so he'll see the clock reading T=0 seconds at t = 10 + 50 = 60 seconds. And because the clock read T=6 seconds when it was 42 light-seconds away at t=20 seconds, it will take 42 seconds for the light from this event to reach Observer's eyes, so he'll see it at t = 20 + 42 = 62 seconds, just 2 seconds later. So to him the moving clock will appear sped up rather than slowed down, advancing forward by 6 seconds in only 2 seconds of his own time, so visually it looks to be sped up by a factor of 3! Indeed, this is exactly what you'd predict if you just plug v=0.8c into the equation for relativistic Doppler effect, which tells you the visual frequency of clock ticks will be greater then the frequency of ticks in the clock's own rest frame (1 tick per second) by a factor of $$\sqrt{\frac{1 + 0.8}{1 - 0.8}}$$ = 3 (and this is also the factor by which the frequency of the individual photons is blueshifted).

If you didn't understand the difference between time dilation and visual appearances, perhaps this partly accounts for your confusion in the twin paradox thread? You might want to take a look at the Doppler shift analysis from the twin paradox page on John Baez's site, which explains how each twin sees the other one aging faster as they are approaching one another after the turnaround, but the twin who accelerates sees the approaching phase lasting just as long as the receding phase (assuming the relative speed is the same in both phases), while the inertial twin sees the receding phase take up a greater proportion of the total trip time than the approaching phase. There's also a good spacetime diagram of this on the too many explanations page:

Last edited: Dec 20, 2008
3. Dec 21, 2008

### nutgeb

Thanks JesseM.

John Baez gives a very simple answer to my question: The SR Doppler Redshift formula is a combination of two (and only two) discrete elements: The classical (nonrelativistic) Doppler Effect, and the time dilation as between the inertial reference frames of the source and observer.

For some reason I had the misimpression that Doppler Redshift at relativistic speeds introduces a redshift adjustment which is separate from (and additional to) the inertial time dilation. But that's not so.

It's reassuring that this subject is simpler than I was making it out to be.