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According to my understanding of SR, if I travel at 0.8c relative to a line of clocks, I should see the clocks in front of me going 3 times faster than my own, and those behind me going 3 times slower than my own (Doppler effect). OK, so what happens at my exact location? I reckon that as I look nearer and nearer to my origin, I would see a discontinuity between the forwards and backwards directions.

That's bad enough, but at my origin there is no light travel delay, so my local time is my proper time, and the speed of the clocks should then be in accordance with gamma (4/3) at 0.8c.

So, my question is: how do I reconcile the 3 contradictory speeds that I should observe for a clock at my origin? I think it should be the value given by gamma, but I can't explain the discontinuity resulting from the Doppler effect both forward and back in close proximity to the origin.

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I'm guessing you "really" see nothing more than a blur! (Kidding. This is actually a pretty interesting question.) –  Kyle Strand Aug 1 at 23:42

5 Answers 5

up vote 42 down vote accepted

I assume you used the formulae $f_o = fs\sqrt{\frac{1+v/c}{1-v/c}}$ for the clocks ahead of you and $f_o = fs\sqrt{\frac{1-v/c}{1+v/c}}$ for the clocks behind you. Those formulae do imply a singularity for the clock that is closest to you. Which equation to use?

The answer is neither. Those expressions assume the travel is along the line of sight to the source. There is a singularity because collisions are singularities. Your spaceship is plowing through the line of clocks. What you'll see in front of you are a series of clocks ticking faster than yours. Behind you, you'll see a cloud of pulverized clocks. Your spaceship had better have some very good forward shields.

Your spacecraft presumably isn't doing that. Instead, you are flying parallel to the line of clocks, with some constant, non-zero distance between the spacecraft and the line of clocks. You need to use the more generic expession $$f_o = fs \frac{\sqrt{1-\left(\frac v c\right)^2}} {1-\frac v c \cos \theta_o}$$ where $\theta_o$ is the angle between the clock in question and your line of travel, as observed by you. The sign convention here is that $\theta_o$ is positive for clocks in front of you, negative for clocks behind you.

The above expression reduces to the simpler expressions at the start of my answer for clocks very far in front of you and very far behind you. In between, you'll get a nice continuous change from faster in front, slower behind. The clock right next to you? It's a bit redshifted, and hence slower. Here $\cos \theta_0=0$, so in this case $fo=fs\sqrt{1-(v/c)^2}$. This is called the transverse Doppler redshift. This means that there's a clock just slightly ahead of you that is ticking at your own clock's rate.

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+1 for going the extra mile on the math. And for the "line of pulverized clocks". Good to be reminded of the effect of some thought experiments...! –  Floris Aug 1 at 12:07
Thanks for the answer David, and yes I am well aware of the collision aspect as I've simplified my question from 2+1 (where there is still a discontinuity) to 1+1 spacetime for clarity. So what you seem to be saying (in terms of my original question) is that yes, there is a discontinuity, but there's no point trying to assign a value at my origin. Furthermore, in light of your answer, I would like to discount the idea that this is a two-stage process as others have suggested. Would you concur? –  m4r35n357 Aug 1 at 12:32
When you look at David's equation, the top line is step 1 (Lorentz transformation of the time in the frame of the observer) and the bottom line is the rate of arrival of the clock ticks (step 2). There is no discrepancy between this answer and mine. –  Floris Aug 1 at 12:52
@m4r35n357 - No matter how you look at it, there are two effects to consider, time dilation and a changing distance between ticks. Time dilation says a traveling clock appears to tick slower than a clock stationary wrt the observer. The changing distance because of velocity means that the time needed for tick n versus that needed for tick n + 1 to reach you are different. The relativistic Doppler effect takes both of these into account. Whether you want to call that two stages or one is a bit pedantic. One way or the other, both effects do need to be taken into account. –  David Hammen Aug 1 at 15:18

The signal from the clock moving towards you is the Doppler shifted version of the value you "know" it to be - that is, first slow it down by gamma (clock moving relative to your frame of reference), then speed it up by Doppler shift.

Ditto, with sign reversed, for clocks behind you. Now the clock moving at right angles shows what you expect and there is no discontinuity.

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OK, this is an interesting answer to a subject I have found it hard to find useful information on. I have a slight issue in my mind that I calculated the Doppler effect using the Lorentz Transform in (t', x') by setting x = t, so what you have suggested sounds to me like applying the LT twice. What have I misunderstood? –  m4r35n357 Aug 1 at 11:17
Floris can you point me to any online references to the composition process that you have described? –  m4r35n357 Aug 1 at 11:40
See this earlier answer - it shows the two step approach, and how these end up giving you the observed frequency shift. –  Floris Aug 1 at 12:02
My point was - first establish how fast the clocks go in your frame of reference; next look at the rate at which these signals arrive at the observer. These are two separate things - in other words the clocks are always slower in your frame of reference, even though you see them going faster due to Doppler shift. David Hammen explains it very well in his answer. –  Floris Aug 1 at 12:29
Yes you understand what I meant to say now! –  Floris Aug 1 at 12:41

If you consider a straight-on trajectory, there really will be a discontinuity. That is the same as with the audible Doppler effect.

There is a smooth drop in frequency of a fire truck's siren passing you on the street. The reason is, that there is a certain distance between you and the truck at the closest point. If that were not the case, i.e. the siren moves directly through you without the truck smashing you, you would perceive a sudden drop in frequency.

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How does your answer relate to that given by Floris above? He seems to be describing a "composition" of LT then Doppler, whereas you seem to be saying that my original analysis is correct. –  m4r35n357 Aug 1 at 11:39
Floris is right as well. When I say "faster" or "slower" I mean the speed of clocks in front of you relative to the clocks behind you. Both suffer the effect of time dilation due to the relative speed between you and them. –  M.Herzkamp Aug 1 at 11:44
OK, then I would ask you the same question as I did Floris; can you give me a reference to the composition procedure? In my studies (mostly in 2+1 spacetime) I have been doing just one LT with t set to sqrt(x^2 + y^2) in the "RHS" frame. This has reproduced all the visual effects that I expected to see, so if my method is wrong I'd like to correct it. –  m4r35n357 Aug 1 at 11:55

I am assuming we are talking about the one dimensional case in which we can "move through" clocks.

Assuming you synchronise the clocks in your own frame, those further away from you will show an older time than those closer because it takes longer for light to reach you from them.

On top of that, all clocks will be equally time dilated since they are all moving in the same frame.

The doppler effect enters because it will change the color of the clock - those ahead will appear blue and those behind red. The doppler effect does not change the actual time on the clock, though.

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I don't have the reputation to comment, so I'll comment about David Hammen's (accepted) answer here.

His conclusion is correct, but mentions "Those formulae do imply a singularity for the clock that is closest to you" in reference to the forumale he thought you were referring to. He also mentions "In between, you'll get a nice continuous change from faster in front, slower behind" in reference to his latter formula involving a cosine.

There is no singularity present in your original formulae that is not also present in David's. He has compared your equations (using a head-on speed) to his (using the proper, 'pythagorean' relative speed).

In particular, if you are headed straight for a clock the angle $\theta_0$ will instantaneously flip $180^\circ$ when you pass it. At the same time $v$'s magnitude will flip. The 'singularity' is still there!

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