# Derivation of the Relativistic Doppler Effect

1. Jan 11, 2007

### NanakiXIII

I'm working through the derivation found on this site:

http://maths.dur.ac.uk/~dma0cvj/mathphys/supplements/supplement5/supplement5.html [Broken]

And though most of it is clear to me, I stumbled upon a little problem.

My question is about the section that starts with "This is all according to A. What about B?", it states that x'_r = 0 when the light is received. I'm not sure I understand that. Does this simply say that the observer doesn't move relative to himself?

Secondly, could this entire section not just be substituted by an explanation involving time dilation? Or even, isn't it just time dilation? It comes to the same result, doesn't it?

Well, even posting this, it's starting to dawn just a little more, but still, if someone could confirm my thoughts on this matter, I'd appreciate to know for sure.

Last edited by a moderator: May 2, 2017
2. Jan 11, 2007

### Hootenanny

Staff Emeritus
This just means that the observer observes the light at the origin of the S' frame (i.e. there is no displacement in his frame of reference, he is stood on the origin we he sees the light if you like).
Yes, the Doppler effect is simply a re-statement of time dilation; T = 1\f

3. Jan 11, 2007

### NanakiXIII

4. Jan 11, 2007

### robphy

Actually it's not.
Their spacetime-geometrical problems [and, therefore, their physical interpretations] are differrent.

In the simplest cases,...
The [Longitudinal] Doppler effect deals with a triangle with two timelike legs and one lightlike one.
Time-dilation deals with a [Minkowski-]right triangle with two timelike legs and one spacelike one.

While the two effects are related, neither is a re-statement of the other.

5. Jan 11, 2007

### Hootenanny

Staff Emeritus
Yes, related is a better word, my bad. Time dilation is the difference between the relativistic and non-relativistic Doppler effects. My apologies

6. Jan 11, 2007

### NanakiXIII

I have another question which is also likely to be a simple one. I know this thread isn't in the right forum for the question I'm about to ask, but I hope you'll forgive me for posting it in this thread, since it's still about the derivation.

In one of the images on the site, it says this:

http://maths.dur.ac.uk/~dma0cvj/mathphys/supplements/supplement5/img19.png [Broken]

Well, this is just an algebra question (apparently I'm not very good at algebra anymore). How does one get from the third part to the fourth part of the equation? (A part being everything between equal-signs, I can't think of the proper English name.) I know what the gamma represents, that's not part of my problem.

Last edited by a moderator: May 2, 2017
7. Jan 11, 2007

### robphy

write (1-v/c) as sqrt(1-v/c)*sqrt(1-v/c).
use the formula for gamma in terms of v and factor.

If you use rapidities, where gamma=cosh theta and v/c=tanh theta, you will be more naturally guided to simplify the expressions.

Last edited: Jan 11, 2007
8. Jan 11, 2007

### bernhard.rothenstein

Doppler effect

Yes, the Doppler effect is simply a re-statement of time dilation; T = 1\f[/QUOTE]
Special relativity teaches us that time dilation formula relates a proper time interval to a coordinate time interval, that the Doppler formula relates two proper time intervals and that the Lorentz transformation for the time coordinate of the same event relates two coordinate time intervals.
It is advisible to have all that in mind and to be able to distinguish between the concepts of proper time and coordinate time intervals.
You can have an illuminating look at
Thomas A. Moore, Six Ideas that shaped Physics. Unit R: The laws of physics are frame independent. McGraw-Hill 1998 pp.33-46

9. Jan 12, 2007

### Jorrie

You should rather have stated: "The relativistic Doppler effect is simply a re-statement of the Newtonian Doppler effect with time dilation included."

Time dilation and Doppler shift are different animals, as has been said before in this thread.

Jorrie

10. Jan 12, 2007

### bernhard.rothenstein

Doppler Effect

I think you quoted me incorrectly. What I had to say was

Special relativity teaches us that time dilation formula relates a proper time interval to a coordinate time interval, that the Doppler formula relates two proper time intervals and that the Lorentz transformation for the time coordinate of the same event relates two coordinate time intervals.
It is advisible to have all that in mind and to be able to distinguish between the concepts of proper time and coordinate time intervals.
You can have an illuminating look at
Thomas A. Moore, Six Ideas that shaped Physics. Unit R: The laws of physics are frame independent. McGraw-Hill 1998 pp.33-46

I use "animals" in zoologie not in physics.

11. Jan 12, 2007

### robphy

Technically speaking, in the time-dilation formula you are also comparing the two proper-time intervals, OS and OT, where O is the meeting event and S and T are simultaneous according to the inertial observer (along OS) measuring the time-dilation of the moving inertial clock (along OT). That is, the 4-vector ST is orthogonal to OS (but not to OT).

In the longitudinal Doppler effect, S and T are lightlike-related... if S is a source emission event, then T, a reception event, is on the future light-cone of S. Assuming an emission at O, then OS is the proper period of the source and OT is the proper period of the receiver.

12. Jan 12, 2007

### Jorrie

Wrong quote - sorry!

Oops, sorry! The quoted part from your post #8 looked like you stated it! I think you left out the opening bracketed quote command.

I now noticed it was Hootenanny who originally stated it, was corrected by Robhy and was acknowledged.

Jorrie

Last edited: Jan 12, 2007
13. Jan 12, 2007

### bernhard.rothenstein

doppler effect and time dilation

I do not understand what you mean by technically speaking. As far as I know in the case of time dilation we compare a time interval measured by two distant clocks located where the involved events take place respectively(coordinate time interval) with the same time interval measured by a single clock present where the two events take place (proper time interval). The measurement of the coordinate time interval involves clock synchronization whereas the measurement of the proper time interval does not. What we have to do is to ensure that the stationary clock where from the moving clock starts to move and the moving clock read a zero time when they are related at the same point in space (initialization). I have learned all that long time ago from Rosser.
In the case of the longitudinal Doppler effect you say in a differtent wording the same thing as I did and we should not use synchronized clocks in the involved inertial reference frames.
I think the correct understanding of all that is essential. If I am wrong please let me know where.
Thanks for discussing special relativity with me.

14. Jan 12, 2007

### robphy

By "technically speaking", I mean that: "while there may be different ways to describe something, at the essence of it is..."

The triangle characterizations of time-dilation and the Doppler-effect are based on "Flat and Curved Space-Times" by Eliis & Williams. They are based on the following spacetime diagrams:

Time-Dilation compares timelike-intervals OS and OT (as measured by inertial observers along those worldlines) where ST is Minkowski-orthogonal to OS (and is thus purely spatial to the observer along OS). That is, S and T are simultaneous according to OS. Then, $$OS=(\gamma)OT$$, or more suggestively, $$OS=(\cosh\theta)OT$$, where $$\theta$$ is the rapidity [Minkowski-angle], measured at event O. (Here, in the declaration of S and T being simultaneous to OS, the notion of clock synchronization is involved.)

$$\begin{picture}(100,130)(0,0) \qbezier(0,0)(0,100)(0,100)\put(0,-5){O} \qbezier(0,0)(80,100)(80,100)\put(85,100){T} \qbezier(0,100)(80,100)(80,100)\put(-5,100){S} \end{picture}$$

Doppler-Effect compares timelike-intervals OS and OT (as measured by inertial observers along those worldlines) where ST is lightlike. That is, S is on the past-lightcone of T. Assuming for simplicity that the first emission occurred at event O, OS represents the proper-period of the source's emissions and OT represents the proper-period of the receiver's receptions.
Then, $$OS=\frac{1}{k}OT$$, where $$k=\exp(\theta)$$.

$$\] \begin{picture}(100,130)(0,0) \qbezier(0,0)(0,20)(0,20)\put(0,-5){O} \qbezier(0,0)(80,100)(80,100)\put(85,100){T} \qbezier[525](0,20)(40,60)(80,100)\put(-5,20){S} \end{picture} \[$$

Last edited: Jan 12, 2007
15. Jan 12, 2007

### NanakiXIII

I've been doing some more puzzling, but I'm just not seeing it. I also don't fully know what rapidities are and thus would rather not use them just yet. Would anyone care to type up a more extensive explanation, just using the necessary algebra?

16. Jan 12, 2007

### Hootenanny

Staff Emeritus
There's really nothing more extensive to it consider this;

$$\sqrt{a}\cdot\sqrt{a} = \left(a\right)^{\frac{1}{2}}\cdot \left(a\right)^{\frac{1}{2}} = \left(a\right)^{\left(\frac{1}{2}+\frac{1}{2}\right)} = \left(a\right)^1 = a$$

Do you follow? Now, if a:=1-v/c, then;

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

Last edited: Jan 12, 2007
17. Jan 12, 2007

### NanakiXIII

I understand that, just not how to use it in this matter. Perhaps I've made a mistake somewhere, but you get this I believe (sorry for the lack of Latex, I don't know how to use it):

t'_r = t_s * sqrt(1-v^2/c^2)/(1-v/c)

So you have a squared term above the division line and a regular one below. I don't know how to work around that. From the result it seems that the following should be true:

sqrt(1-v^2/c^2)/sqrt(1-v/c) = sqrt(1+v/c)

I don't see how.

EDIT: Alright, it just dawned on me right after I posted that.

(1-v/c)(1+v/c)=1-v^2/c^2

I understand that now. I never learned how to do that the other way around, though, that's why I was having trouble seeing the link. Thanks for pointing me in the right direction.

Last edited: Jan 12, 2007
18. Jan 12, 2007

### NanakiXIII

I apologize for the double post, I hope to be forgiven on the account that this post is about a somewhat different problem and I felt that editing my previous post for a second time wouldn't be favorable to the clarity of the topic.

Well, hoping not to have stepped on anyone's toes, I'll get down to business. The derivation on the site states the following:

http://maths.dur.ac.uk/~dma0cvj/mathphys/supplements/supplement5/img26.png [Broken]

Now using the factor sqrt((1+v/c)/(1-v/c)) in the previous equation with t'_r and t_s, that I understand. [0, t_s] and [0', t'_r] (I hope you'll understand my improvised method of describing the intervals) are entirely different intervals, even if you do not take the relativistic effects into account. But T and T' are essentially the same interval, just seen from different reference frames. That being so, should not the Lorentz transformations be applied to this interval? Why use the factor I mentioned above?

Last edited by a moderator: May 2, 2017
19. Jan 18, 2007

### NanakiXIII

Ah, dreadful, isn't it? A triple post. Yes, I'm shamelessly using my new question to bump my thread. It's just that I wonder why no one has replied in a while. Some reply would be appreciated, even if just to scold me for some stupidity.

At any rate, I laid off the Doppler effect for a while and only just took another peek at it. It took me about three seconds to realize I was quite obviously wrong in my previous post. T and T' are of course not the same interval. What still bothers me, though, is that the derivation states

$$T' = T \sqrt{\frac{1+v/c}{1-v/c}}$$

as a result of the similar equation using $$t'_r$$ and $$t_s$$. I don't fully understand, however, how this is derived. If anyone could explain what's in between, or why this is correct, I'd much appreciate it.

20. Jan 19, 2007

### Jorrie

It's not clear what exactly you are still unclear about - is it how the $$t'_r$$, $$t_s$$ relationship was derived or is it how one gets from that to the $$T'$$, $$T$$ relationship?

21. Jan 19, 2007

### NanakiXIII

Thanks for your reply, Jorrie. My problem is the latter you mentioned. I understand how the $$t'_r$$, $$t_s$$ relationship was derived, but not how the $$T'$$, $$T$$ relationship is proven by that.

22. Jan 19, 2007

### robphy

Let $$k=\sqrt{\frac{1+v/c}{1-v/c}}$$.

So, $$t_r'=k t_s$$, and by similarity [or repeating the derivation] $$t_r'+T'=k (t_s+T)$$. By subtraction, $$T'=k T$$.

23. Jan 20, 2007

### NanakiXIII

Alright, I feel stupid now. Thanks, robphy.

24. Jan 20, 2007

### robphy

In a simplified derivation [as seen in some Bondi k-calculus derivations of the Doppler Effect], where an emission [and reception] occurs at the meeting event, ts coincides with T and tr' coincides with T'.

25. Jan 20, 2007

### Jorrie

At the risk of being hand-slapped by the science advisers, I still like this wavelength and Lorentz contraction view of the relativistic Doppler effect.

Let's first view it purely Galilean: a stationary receiver (sub-scripted $${re}$$) will measure the transmitted wavelength of a moving transmitter (sub-scripted $${tr}$$) as:

$$\lambda_{re} = \lambda_{tr}(1+v/c),$$

where $${v}$$ is an opening (radial) velocity and $${c}$$ has its usual meaning.

Now, in a relativistic scenario, the transmitter's wavelength is Lorentz contracted by a factor $$\sqrt{1-(v/c)^2}$$, so we must divide by the same factor to get the rest frame's (receiver's) wavelength:

$$\lambda_{re} = \frac{\lambda_{tr}(1+v/c)}{\sqrt{1-(v/c)^2}} = \lambda_{tr}\sqrt{\frac{ 1+v/c }{1-v/c} }.$$

Periods ($$T$$) are just $$\lambda/c$$, so the relationship stays identical.

This is essentially the same thing as the Lorentz transformation, but I find it somewhat more comprehensible, even if it is perhaps not rigorous.

Jorrie

PS: I hope the TEX shows correctly; I have difficulty previewing it on my computer, for reasons unknown to me. Somehow "Reload this page" does not do the trick.

Last edited: Jan 20, 2007