# How do Doppler Effect and Time Dilation differ? Above, I have drawn a diagram showing Doppler Effect (here we are using space-time but in a non-relativistic sense. Time and distance are the same for A and B).

Edit: I am adding a relativistic space-time diagram below this with lines of simultaneity drawn. I am also editing the description to be more relavent to the updated diagram. The diagram shows the frame of a stationary observer B. A travels with velocity c/2. A emits flashes of light every second (according to the time of observer B - flashes of light are shown as dashed lines and are emitted where the lines of simultaneity meet the worldline of A and hence they are emitted every second in the time of B)

We see that in the frame of B, we begin to see the light one second after it is emitted. light is continuous and after the lag shown, light from a time delta t' (according to the time in frame B) is observed across time delta t (which is 2 x delta t').

This seems to indicate that what we OBSERVE in frame B seems to depend on only the slope of A. Is this indeed what we will see in B?

• A big difference is that the Doppler shift for inertial frame A relative to inertial frame B is directly observable, whereas the time dilation is not. To observe time dilation, you have to separate and then reunite two clocks, which involves moving at least one of them noninertially (as in the Hafele-Keating experiment en.wikipedia.org/wiki/Hafele-Keating_experiment ). – Ben Crowell Apr 8 '13 at 1:46
• I have added my answer as a document. Would appreciate your comments on it. I have used spacetime geometry and my analysis seems to point out that Relativistic Doppler shift is more than just a frequency shift, it's a visual shift of the whole of time. Of course, I could be wrong. – Physics n00b May 10 '13 at 18:08

One significant difference is that the doppler effect is dependent on the direction of the velocity, while time dilation is only dependent on the speed. This is why the doppler effect changes when A passes B, while the time dilation would be the same before and after. For this reason there is also no doppler effect when something moves perependicularly to you, while there is still time dilation.

Another difference is that the doppler effect is "stronger" than the time dilation effect when something moves towards you or from you. The doppler effect gives a factor $(1\pm\frac{v}{c})$ while the time dilation gives a factor:

$$\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$

Because of this, the combined effect, the relativistic doppler effect, is dominated by the doppler effect in these cases.

• I agree with your Doppler factor. I have tried to do some analysis on my own using space time geometry to combine time dilation and doppler effect. The results are presented as my own answer. I would appreciate you views/comments on it. According to my analysis, its not just frequency that shifts but the whole of time. I could be very wrong about this and I have no source of validation. So, I would appreciate if you took some time to look at my answer and see if and where I might have made a mistake. Thanks! – Physics n00b May 10 '13 at 18:12

I have given this question much thought. I am providing a link to what I think is the answer, as putting the answer here would be too long.

Please note that I am new to physics so my thinking may be wrong. I would appreciate your comments on whether you think this idea is correct or not.

Thanks!

Is important to realize time dilation and the Doppler effect are 2 frame-dependent parts of a single relativistic phenomenon: the constancy of the magnitude of four-velocity (which is always ||c||). Requiring that all inertial observers see a four-velocity with magnitude ||c|| leads to frame dependent observations of time dilation and Doppler effect.

For a Doppler radar, the relativistic Doppler effect can be derived by forcing the scattered radar's (read: photon's) four-acceleration to be orthogonal (in Minkowski space) to the targets 4-velocity, that is:

[(E, pc) - (E', p'c)]*(c, v) = 0.