Talk:Relativistic Doppler effect

From Wikipedia, the free encyclopedia

WikiProject Physics This article is within the scope of WikiProject Physics, which collaborates on articles related to physics.
??? This article has not yet received a rating on the assessment scale. [FAQ]
??? This article has not yet received an importance rating within physics.

Help with this template Please rate this article, and then leave comments to explain the ratings and/or to identify its strengths and weaknesses.

This article makes no sense. Just to start, it could be better explained in what way speed is a "rotation". Nickptar 21:07, 16 Apr 2005 (UTC)

Moreover, the discussion here doesn't really belong in this page. It should probably be in special relativity if anywhere. 67.165.197.242 06:58, 23 Apr 2005 (UTC)

The basic equation seems to be mis-labled. As the velocity approaches c, the observed wavelength goes to infinity, *not* the frequency, or the convention that recession is positive velocity is mis-stated.

I agree that this presentation is weird; it seems to have been made up by someone with a particular background. The standard way of presenting Doppler equations presents the ratio between measured frequencies, as can also be found in the 1905 paper of Einstein -- html link in Special relativity.

Harald88 22:08, 9 January 2006 (UTC)

Hi guys, I agree with your comments, and I revised the article. Hope you like it. Yevgeny Kats 05:46, 23 January 2006 (UTC)

IMO it looks better although you made several new unwarrented claims. Anyway, thanks for cleaning up the mess, it now has a much better look. Harald88 20:43, 23 January 2006 (UTC)
PS Your simple derivation to illustrate the connection to true Doppler is just what I had in mind to do myself. Thanks again! Harald88 12:14, 24 January 2006 (UTC)
Thanks, Harald, but why did you revert my correction of the centrifugal force article? What they have there now is a complete nonsense (based on someone's misunderstanding of the Principia or something). (Sorry, I don't feel like contributing to their 100-page talk page :) Yevgeny Kats 04:09, 25 January 2006 (UTC)
Sorry, I had not noticed that the intro had already been messed up; I had not reverted far enough (problem with my watchlist). Harald88 11:13, 26 January 2006 (UTC)


[edit] Formulas

Comparison of the formulas on this page with those on Doppler effect is very confusing. On this page, fo (letter o) stands for the frequency observed, but on the other page, f0 (number 0) stands for the actual frequency (i.e. frequency of the source). These should be clarified/reconciled.

No, it doesn't. The formulas are consistent. Moroder 02:31, 8 December 2006 (UTC)

===Incorrect Plots===--TxAlien 21:45, 10 December 2006 (UTC)--TxAlien 21:45, 10 December 2006 (UTC)

The plots are incorrect because the follow the first set of formulas instead of the second. The error is easy to tell since the plots show an INCORRECT redshift at 90 degrees instead of the correct blueshift. Note to the authors: could you please redo the plots for the correct formulas:

f_\mathrm{detected} = f_\mathrm{rest}{\left(1 - \frac{v}{c}\ cos\phi\right)/\sqrt{1 - \frac{v^2}{c^2}} }
as deduced by Einstein (1905)[1].

Thank you Moroder 16:36, 6 December 2006 (UTC)

I just added a note explaining that the two plots represent the wrong formula, the plots need to be regenerated in order to represent the right formula:
f_\mathrm{detected} = f_\mathrm{rest}{\left(1 - \frac{v}{c}\ cos\phi\right)/\sqrt{1 - \frac{v^2}{c^2}} }

Moroder 16:05, 7 December 2006 (UTC)

Well, according to your formulas, unstable particles in cyclotron should live shorter then the same particles at rest. But it is wrong.. so, I will restore the old version of the article.--TxAlien 18:10, 7 December 2006 (UTC)
These are not my formulas, they belong to Einstein. See here [2]. And your plots, pretty as they are are still dead wrong. Please read the Einstein paper (paragraph 7), perhaps you will understand why. Moroder 19:38, 7 December 2006 (UTC)
In [3] it says:
From the equation for \omega ' \; it follows that if an observer is moving with velocity v relatively to an infinitely distant source of light of frequency \nu \;, in such a way that the connecting line "source-observer" makes the angle \phi \; with the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light, the frequency \nu ' \; of the light perceived by the observer is given by the equation
\nu ' = \nu \frac{1-\cos \phi \cdot v/c}{\sqrt{1 - v^2/c^2}}.
This is Doppler's principle for any velocities whatever.
. . .
If we call the angle between the wave-normal (direction of the ray) in the moving system and the connecting line "source-observer" \phi ' \;, the equation for \phi ' \; [*6] assumes the form
\cos \phi '= \frac{\cos \phi - v/c}{1 - \cos \phi \cdot v/c}
or
\nu ' = \nu \frac{\sqrt{1 - v^2/c^2}}{1 + \cos \phi ' \cdot v/c}
We talk about that angle \phi ' \, in this article . And it was used in the images.--TxAlien 22:57, 7 December 2006 (UTC)


This is what I've been trying to tell you for the last 3 posts. Only the first formula for the Doppler effect shows up in the Einstein paper, the second one (the one that you deduce from the first one using the aberration transformation) does not. Wonder why? If you look at the RHS you can see that it mixes variables from the source and the observer frame (the frequency from one frame and the angle from the other one). This is a "no-no" in relativity. Frankly, the formula that you deduced has no place in wiki. It simply confuses things. So now, would you please regenerate the plots for the original Einstein formula? :-) —The preceding unsigned comment was added by Moroder (talk • contribs) 00:50, 8 December 2006 (UTC).
Dear Moroder, your claim that the formula
f_o = \frac{f_s}{\gamma\left(1+\frac{v\cos\theta_o}{c}\right)}
is problematic because the RHS "mixes variables from the source and the observer frame" doesn't make any sense. The whole purpose of this formula is to convert quantities from one frame to the other. Would you be happier, for example, if I rearranged the formula as
\gamma\left(1+\frac{v\cos\theta_o}{c}\right) f_o = f_s
Now it doesn't mix quantities from different frames in the same side of the equation, but it's still exactly the same formula!! Yevgeny Kats 01:25, 8 December 2006 (UTC)


Yes Yevgeny, I would be happy, this is the formula that I kept suggesting for my last 3 posts. Now, can you convince User:TxAlien to regenerate his colored plots? They show an incorrect red shift at 90 degrees , when in reality the Ives-Stilwell experiment shows a blue shift. We have come full circle to my original complaint, the colored plots are WRONG. Moroder 01:59, 8 December 2006 (UTC)
Dear Moroder, please read the article and my response above more carefully. The formula that makes you happy (that I wrote above, which is equivalent to the first formula in the article) is very different from the second formula in the article, which is
 :f_o = \gamma\left(1-\frac{v\cos\theta_s}{c}\right)f_s
Both formulas are correct, and there is no reason to change anything in the text of the article. Yevgeny Kats 05:51, 8 December 2006 (UTC)


Dang, I missed what you wrote. One more time, the correct formula is :

f_\mathrm{observed} = f_\mathrm{source}{\left(1 - \frac{v}{c}\ cos\phi\right)/\sqrt{1 - \frac{v^2}{c^2}} }
as deduced by Einstein (1905)[4]. This is what Einstein wrote, this is what is used in the Ives-Stilwell experiment. This is exactly what Einstein wrote:
\nu ' = \nu \frac{1-\cos \phi \cdot v/c}{\sqrt{1 - v^2/c^2}}.

Why not use his exact formula? This is also the formula that should drive the correct(ed) plots.Moroder 06:45, 8 December 2006 (UTC)

Both formulas that appear in the article are correct. You can use either of them to convert between the emitted frequency and the received frequency and vice versa. Actually, the analogs of both of them appear at the bottom of p. 6 of your reference [5]. Which formula to use depends on what angle you know: whether you know the direction of the velocity in the frame of the observer (\theta\, in that reference, our \theta_o\,) or the direction of the velocity in the frame of the source (\phi\, in that reference, our \theta_s\,). If you're an astronomer, it would be more natural to you to know the angle \theta\,, i.e. \theta_o\,, and so use the first formula in our article. In the Ives-Stilwell experiment, it's also more natural to use the angle \theta\,, and it is indeed what is used there in the same reference of yours - see top of p. 7. (The signs are different here and there because we define angle = 0 when the two are moving away from each other, while they define in the opposite way in the case of \theta\, - see picture on p. 6). Yevgeny Kats 16:56, 8 December 2006 (UTC)
On the other hand, I agree with Moroder that the plots are incorrect. The article uses the convention that v is positive when the source is moving away from the observer (angle 0), and then there should be a redshift, while the plots show a blueshift. Another problem with the plots is that they don't say whether the angle is measured in the frame of the observer or in the frame of the source. Therefore, I remove the plots from the article for now. Yevgeny Kats 05:51, 8 December 2006 (UTC)
If this new plot is good enough to the article what should I change then? --TxAlien 20:49, 10 December 2006 (UTC)
Something is still not quite right. The way I know it is that at 90 degrees you should get a blueshift that increases as v/c ->1. In the diagram, the 90 degree line is imbedded in a yellow domain, coresponding to f_o/f_s=1. This cannot be right. Moroder 07:53, 11 December 2006 (UTC)
I see the error, you insist on not plotting the formula \nu ' = \nu \frac{1-\cos \phi \cdot v/c}{\sqrt{1 - v^2/c^2}}. . Why? Moroder 16:18, 11 December 2006 (UTC)
What do these diagrams represent? I guess Diagram 1 corresponds to the first formula in the article, and then it looks fine. But I don't understand what Diagram 2 represents: the frequency is independent of the velocity when the angle is 90; the frequency approaches fs / 2 for large velocities when the angle is 0 - what is this? Yevgeny Kats 21:16, 10 December 2006 (UTC)
Yes, the first diagram represents the first formula. And second plot represents the same formula without \gamma \;. Actually it is classic case. So, I thought that it will be useful to see the difference. You are welcome to add any comments, and I can change the plots if it is necessarily. Or, we can forget about those plots if you think so.--TxAlien 21:45, 10 December 2006 (UTC)
The "classical" case isn't something universal because the classical result also depends on the assumption of whether the medium in which the light propagates is moving with the observer, with the source, or at a completely different velocity. I don't think that simply ingnoring the factor of γ has any universal "classical" meaning. So I would suggest not including the classical case. On the other hand, I would suggest including a second plot in terms of the angle θs (i.e., the second formula in the article). Yevgeny Kats 16:29, 11 December 2006 (UTC)
Agreed Moroder 16:39, 11 December 2006 (UTC)

Well, it is not universal classic case, of course, but the simplest one. On the other hand the second formula describes not easy case. It should have a better explanatory. It might be a short way to show how these formulas were found. Something like this:

assume the source of waves moves along the trajectory (ct(s),\vec{r}(s)) \;, where s \; is the proper time of that object and
\dfrac{d}{ds}(ct(s),\vec{r}(s)) = \left( c\dfrac{dt}{ds},\dfrac{d \vec{r}}{ds}\right) = \gamma \left( c,\vec{v}\right)
where \left( c\dfrac{dt}{ds},\dfrac{d \vec{r}}{ds}\right) is velocity four-vector. Due to the finite velocity of light, the frequency at the point of observation (T,0) \; is determined by state of source at the earlier time t(s) = T-r(s)/c \;.
Differentiating this relation with respect to s \;, we get
\dfrac{\omega _{s}}{\omega _{o}} = \dfrac{dT}{ds} = \dfrac{dt}{ds} + \dfrac{1}{cr} \left( \vec{r}\cdot \dfrac{d \vec{r}}{ds}\right) = \gamma \left( 1 + \dfrac{1}{cr}\left( \vec{r}\cdot \vec{v}\right) \right) = \gamma \left( 1 + \dfrac{v}{c}\cos \varphi_{o}\right)
or
\omega _{o} = \omega _{s}\dfrac{1}{\gamma \left( 1+\dfrac{v}{c}\cos \varphi_{o}\right) }
where \varphi _{o} is angle, relative to the direction from the observer to the source at the time when the light is emitted.
Differentiating relation t = T(s_{o})-r(s_{o})/c \; (as it seen from the reference of the source) with respect to the proper time s_{o} \; of the observer, we get
\omega _{o} = \omega _{s}\gamma \left( 1-\dfrac{v}{c}\cos \varphi_{s}\right)
where angle \varphi _{s} is measured in the reference frame of the source at the time when the light is received by the observer. It is not the easiest way, but it helps to find out how moving objects looks like in theory of relativity.
Anyway, the new image will be ready in a few minutes.--TxAlien 03:55, 12 December 2006 (UTC)
The images are clearly still wrong.Is this because you insist in plotting the first formula instead of the second one? Moroder 07:44, 12 December 2006 (UTC)
Diagram 1 is the first formula and Diagram 2 is the second one. Can you tell me exactly what you think is wrong whith these images?--TxAlien 01:29, 13 December 2006 (UTC)
I've been telling you the same thing over and over: at 90 degrees you should see a clear blue shift in formula 2 because of \omega _{o} = \omega _{s}\gamma \, . Your picture no 2 shows yellow, which is incorrect. Moroder 01:49, 13 December 2006 (UTC)
I did not expect this kind of misunderstanding. This image might help
http://img242.imageshack.us/img242/3837/dopplerextk9.jpg --TxAlien 02:53, 13 December 2006 (UTC)
Yes, it is a surface F(x,y)=F(v/c, cos(phi)). Somehow your plot misses the fact that at cos(phi)=0 you need to get a blue curve embedded in the surface , actually a narrow blue band on both sides of the curve. I don't know how you are getting the yellow but I do know it is wrong.I think that there is a clear error in your second surface plot, for example, at 90 degrees all the surface points should be above 1 (at altitude \gamma\,). Your plot shows the points in the 0.25 altitude range which is clearly wrong. I think that you are continuing to plot ::\omega _{o} = \omega _{s}\dfrac{1}{\gamma \left( 1+\dfrac{v}{c}\cos \varphi_{o}\right) } instead of \omega _{o} = \omega _{s}\gamma \left( 1-\dfrac{v}{c}\cos \varphi_{s}\right) Moroder 04:24, 13 December 2006 (UTC)
As you can see on Diagram 2 all values at the angle = 90^\circ are greater then 1 and -> \infty . 3d surface was scaled (to make a nice colors, it was just one of many simple ways), I did not care about its true values. I've made this image for fun, but after this discussion it is not a fun anymore.--TxAlien 16:20, 13 December 2006 (UTC)
Not really, in the surface representation you can see z coordinate looming around 0.25[6]. This explains the incorrect yellow coloring as well (it should be greenish/blue). Moroder 16:29, 13 December 2006 (UTC)
If you scaled your surface plot, did you scale the color bar accordingly? This may be a partial explanation for the color representation error Moroder 16:59, 13 December 2006 (UTC)
It is my last try to explain this to you [7]--TxAlien 18:12, 13 December 2006 (UTC)
This [8] looks correct as opposed to all your previous ones [9] that were incorrect. So , it looks like my criticism turned into something positive, you fixed your plots. They are very nice and correct now, I suggest that you reinsert them in the main aricle. The 3D representation is also much better than the previous 2D ones in terms of clarity. Moroder 19:43, 13 December 2006 (UTC)
Yevgeny Kats made big contribution to this article. So, let him decide the fate of these images. (3D is too fancy)--TxAlien 04:10, 14 December 2006 (UTC)
I think the 3D plots are great. So what was the error in your previous plots? How did you correct it? Moroder 04:27, 14 December 2006 (UTC)
I don't think past contributions give the person extra rights in future decisions :) However, I agree with TxAlien that the two-dimensional plots are better. Yevgeny Kats 05:29, 14 December 2006 (UTC)
Difference between old and new images: [10]. It is only scale, but I did not intend to put 3d images in this article. So, there were all correct from my point of view.--TxAlien 05:10, 14 December 2006 (UTC)
The old plots (left column) are clearly wrong. The new plots (right column) are correct. Moroder 07:14, 14 December 2006 (UTC)