# Talk:Relativistic Doppler effect

## Untitled

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)

## 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).

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 . 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  it says:
Template:Fontcolor
$\nu '=\nu {\frac {1-\cos \phi \cdot v/c}{\sqrt {1-v^{2}/c^{2}}}}.$ Template:Fontcolor
. . .
Template:Fontcolor
$\cos \phi '={\frac {\cos \phi -v/c}{1-\cos \phi \cdot v/c}}$ 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 (talkcontribs) 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). 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 . 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 $f_{s}/2$ 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 $\gamma$ 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 $\theta _{s}$ (i.e., the second formula in the article). Yevgeny Kats 16:29, 11 December 2006 (UTC)
Agreed Moroder 16:39, 11 December 2006 (UTC)
Outrageous. Lorentz contraction is not reflected in diagram 2, which is totally misleading. SJGooch (talk) 15:11, 11 December 2010 (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. 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 --TxAlien 18:12, 13 December 2006 (UTC)
This  looks correct as opposed to all your previous ones  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: . 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)

Did anyone can derive relativistic Dopper effect not from "time dilation" or "Lorentz transformation"? Because these two lack asysmetry. Especially "time dilation", it did not have direction. — Preceding unsigned comment added by Fsshl (talkcontribs) 01:20, 23 April 2012 (UTC)

Hey, not to go back to this whole deal, but are you sure the formulas are correct? Methinks the positive/negative on :$f_{o}={\frac {1}{t_{o}}}=\gamma (1-\beta )f_{s}={\sqrt {\frac {1-\beta }{1+\beta }}}\,f_{s}.$ is switched, since this expands to

$f_{o}={\frac {1}{t_{o}}}=\gamma (1-\beta )f_{s}={\sqrt {\frac {1-{\frac {v}{c}}}{1+{\frac {v}{c}}}}}\,f_{s}.$ . That, or the 7th edition of Physics for Scientists and Engineers by Serway and Jewett, page 1129 is wrong, which is very much a possibility. Can anyone confirm this? I put in a dubious just in case — Preceding unsigned comment added by 173.240.228.167 (talk) 11:17, 5 December 2012 (UTC)
No, it's correct. In this article, beta is negative if moving towards the observer, positive if moving away. Your textbook might be using the opposite sign convention. I'll remove the dubious.--BerFinelli (talk) 11:05, 26 December 2012 (UTC)

## Excuse me

Three formulas in the third "Motion in an arbitrary direction" section seem not to agree in all. namely,

$f_{o}={\frac {f_{s}}{\gamma \left(1+{\frac {v\cos \theta _{o}}{c}}\right)}}~.$ (1)
$\cos \theta _{o}={\frac {\cos \theta _{s}-{\frac {v}{c}}}{1-{\frac {v}{c}}\cos \theta _{s}}}\,.$ and

$f_{o}=\gamma \left(1-{\frac {v\cos \theta _{s}}{c}}\right)f_{s}.$ (2)

The first and second seem to yield the third but gamma to the power of 3.Like sushi (talk) 10:14, 7 May 2009 (UTC)

## I'm sorry. It was a mistake

The post just before was a simple mistake. Sorry. Gamma is simple gamma.Like sushi (talk) 13:47, 7 May 2009 (UTC)

## Visualization

I just edited the wording in the visualization section. I do not have a background in relativistic physics, I just reorganized the information already written. Please proofread my writing for technical accuracy. Also, as I was rewriting, I realized that I would like to see two different animations. The one that already exists shows an observer's velocity increasing against a stationary background. Another could show an emitter's velocity increasing relative to a stationary observer located somewhere on the grid (like the plot at the top of the article). Can someone please generate this plot? Also, the aberation of light doesn't appear anywhere in the article and has no context. Please provide a brief description of what causes the aberation of light. (I think it has to do with the time differnce between emitting and receiving the light, but I'm not sure.) David.hillshafer 21:31, 9 June 2009 (UTC)

I moved the visualization to the top, because a visual improves understanding and anticipates a formula. Also, it allows for direct comparison for the case of the observer and the emitter moving. David.hillshafer 17:22, 19 June 2009 (UTC)

## Analogy

I added an analogy to explain my understanding of the abberation of light. Again, I do not have a background in relativistic physics. Please check my analogy for technical accuracy. (My background is in engineering mechanics, so I'm comfortable with complicated classical physics equations.) I made this analogy in an attempt make the equations personal and within the scope of normal human experience. I intend to extend the analogy to compare the spin of the ball and the doppler shift, but I'm working on the most intuitive way to describe this. Also, I would like to make a few simple diagrams.David.hillshafer 17:36, 19 June 2009 (UTC)

Devising Conceptual and Mathematical Analogies are difficult chores. For example when discussing the second degree of freedom that makes an incompressible fluid (water) behave as a compressible one (a gas) in an open channel, the author of a text book that I studied "back in the day" compared the bow waves of a canal boat in a calm canal to the sonic booms generated by aircraft. I don't know how this works out I never did see the math.
However, the acoustic doppler effect has helped me to finally grasp the concept of the constant speed(limit) of mass and energy as being that of the speed of light - I think. Regardless of the speed of the train (horn) -(and it usually is a train which travels at 1/10 to 1/3 the speed of sound and sends long blasts approaching crossings and turns)- the speed of sound is determined soley by the mechanical and thermodynamic (pv^k etc.) properties of air. Simple. The only thing the speed of the source (relative to the observer) does is increase the frequency of the sound.
Therefore any analogy that uses projectiles to illustrate a concept has to be clear that the speed of the source does not add to nor subtract from the speed of the projectile. My tentative analogy would be a trolley about roll down a hill with a constant slope. The trolly can trigger the release of soccer balls from the trolley but because the balls quickly reach terminal velocity (determined but the slope and the aerodynamics of the balls) they travel at virtually the same speed.
So when the trolley is stationary it triggers the release of one ball every five seconds. The ball quickly reached terminal velocity limited by the slope of the hill - the speed of the trolley is limited to 1/3 of that of the balls. At the bottom of the hill a receiving gutter catches balls at a rate of 12/min. Once the trolley moves down the slope it continues to drop a ball every five seconds. Depending on the speed of the trolley the catcher receives more balls/min - BUT the balls still roll down the hill at the same speed.
Still an awkward analogy but I am working on it.
Great article. However this suggests more questions about analogies with the acoustical doppler phenomena and light transmission. For example since light does not travel through a medium (as once thought - "the ether") so what happens as the vehicle approaches the speed of light - not the same as sound(??) - Mach 1,2,3 vs, warp 1,2,3!! Further if the (apparent) frequency of light reaching the observer increases - does the spectrum go beyond visible and the source appears to disappear?? What about spectral analysis of starlight - does it have to adjusted to compensate for the speed that the star approaches us ["redshift", "blueshift" etc.]?
IMHO, That is is the bad and good of scientific analogies...... they don't always hold water, but they do a great job of stimulating questions in the minds of students until they better understand the precise mathematics of the phenomena.
Pete318 (talk) 17:05, 10 October 2012 (UTC)

## Discussion on Diagrams

I am curious about *Diagram 1* from the article page. This diagram depicts the redshift and blueshift of waves being emitted by a moving source. To my untrained eye, it looks like this picture does a good job of showing the relative shift in frequencies/wavelengths that would occur and in illustrating the Doppler effect in general terms. The note says that the source is moving at 0.7c. At this speed, I am thinking that the relativistic components of the Doppler effect would show up starkly in the picture. If this is the case, shouldn't waves emitted transverse (and even waves emitted up to some forward angle) appear redshifted in the "stationary" reference frame? When generating this image, was relativistic time dilation accounted for? If not, it would seem appropriate to regenerate this image taking all relativistic effects into account (considering that the wave emitter is moving at relativistic speeds and this article is about relativistic effects). Jsnydr (talk) 22:04, 23 October 2009 (UTC)

## Experimental confirmation of transverse Doppler effect

The following reference is mentioned in the Mossbauer spectroscopy article.

Y.-L. Chen, D.-P. Yang (2007). "Recoilless Fraction and Second-Order Doppler Effect". Mössbauer Effect in Lattice Dynamics. John Wiley & Sons. doi:10.1002/9783527611423.ch5. ISBN 9783527611423.

This may report another experimental confirmation of the transverse Doppler effect. Perhaps someone with access to the article could confirm this and determine whether the article also has references to other experiments confirming the effect. —Preceding unsigned comment added by 68.145.187.67 (talk) 21:44, 22 March 2010 (UTC)

## Wrong expression

Hallo, I am totally new to article editing, so I will only post here and not in the article. Please feel free to correct it yourself.

The expression of time dilation (second equation) is wrong, as the time between two clock ticks measured in the reference frame moving with respect to the clock is t_o = t * gamma, and not t_o = t / gamma. See for example the article on time dilation. Therefore the relation between observed and emitted frequency (fourth equation ) is wrong as well, since gamma should go on denominator and not on numerator. I believe the problem has to do with one wanting the (1 + beta) to simplify correctly in those formulas. This comes automatically as follows. Start the article by referring to the very first equation in the classic Doppler Effect article, setting v_r (observer's velocity) equal to zero, v_s (source velocity) equal to v, and v (wave velocity) equal to c. One then reads:

f_o (observed) = f_s (emitted) / (1 + beta), with beta= v/c.

Special relativity adds the time dilation effect gamma to times, so 1/gamma to frequencies, so that overall

f_o = (1 + beta) / gamma

The rest follows without problems. If you seek confirmation, see Rindler, Relativity, 2nd edition, page 79; and Weinberg, eq. (2.2.2) pag. 30. Bepibl (talk) 18:52, 8 November 2010 (UTC)

Moved new section to bottom per wp:TPG
I think you make a mistake. The two wave crest arrival events happen at the same place according to the observer, so per standard time dilation setup they appear longer in some frame in which the observer appears to be moving, in this case the frame of the source. So Δt = γ (Δt0 - v/c2 Δx0) and therefore (with Δx0 = 0), we have Δt = γ Δt0 and thus indeed Δt0 = Δt / γ.
DVdm (talk) 19:30, 8 November 2010 (UTC)

Hi DVdm, let me put it differently. The velocity appearing in the classic Doppler effect is the component of the relative velocity of source and observer in the direction of the line connecting them. The link between f_o and f_s in the section "Transverse Doppler effect" (eq. 2 of that section) states exactly this. So eq. 2 of transverse doppler section should reduce to eq. 4 from top if you take theta = pi/2. But it does not, as eq.3 of transverse doppler section states f_0 = f_s / gamma, not *times* gamma. As it is now, equation 4 from top and equation 3 of transverse doppler are inconsistent between them. Do you agree? Bepibl (talk) 20:43, 8 November 2010 (UTC)

Note that eq. 2 of transverse doppler section reduces to eq. 3 of same section (i.e. transverse) with θ=π/2, and should reduce to eq. 4 from top (i.e. longitudinal) if you take θ=0, (not π/2), which as you can verify, it does. In both cases we have fo< fs, giving redshift (since β>0), as it should, since in the longitudinal case the observer is moving away from the source. DVdm (talk) 21:09, 8 November 2010 (UTC)

Hi there, sorry, I meant of course theta = 0...! Eq.4 from top agrees with the general form (eq.2 of transverse doppler section) for theta=0 because gamma*(1-beta) = 1/(gamma*(1+beta)). However expressing eq.4 from top as it is now and not as eq. 2 of transverse doppler for theta=0 hides this equivalence. I think all would be much easier to understand if eq.4 from top were expressed as I stated above, i.e. in the reference frame of the observer. As is now, I find rather tortuous to follow, because in the first section the effect is "observed" by the source, whereas in the transverse section it is observed by the observer (in its own reference frame).Bepibl (talk) 22:31, 8 November 2010 (UTC)

Ah yes, I see what you mean. I have made a little change, replacing "the time observed between crests" with "the time (as measured in the reference frame of the source) between crest arrivals at the observer", so that should make it somewhat less tortuous. With this amendment, I wouldn't agree that in the first section the effect is "observed" by the source. As I read it now, it is observed by the observer, who can straightforwardly apply the standard Lorentz transformation to calculate the time between the two events (colocal from himself) of wave front reception in the frame of the source. Anyway, i.m.o. eq. 4 from top is indeed expressed in the reference frame of the observer, so to speak. And of course I don't think that it really makes sense to talk about such an equation as "expressed in the reference frame of the observer" or "expressed in the reference frame of the source". It is a an equation relating two quantities on equal footing, and the text explains (--a little bit better now--) where it comes from. DVdm (talk) 23:16, 8 November 2010 (UTC)

## What is this supposed to say?

In the section on Transverse Doppler Effect, it says, "If the predictions of special relativity are compared to those of a simple flat nonrelativistic light medium that is STATIONARY in the observer’s frame (“classical theory”), SR’s physical predictions of what an observer sees are always "redder", by the Lorentz factor..." Really? Stationary??? The Lorentz factor has no effect if the source is not moving relative to the observer. What is this supposed to say??? Thank you.— Preceding unsigned comment added by 98.212.132.146 (talkcontribs)

Please put new messages at the bottom, provide a header and edit summary, and sign with four tildes (~~~~)? Thanks.
It means that in SR the transverse Doppler effect predicts a redshift for a source that is instantanously moving transversally w.r.t. the observer. I.o.w. light from a source that is not approaching or receding from the observer, but that is moving w.r.t. the observer, is redshifted. In this case the source is moving w.r.t. to observer, but not approachin or receding: it could be moving in a circle with the observer at the centre, or it could be moving along a straigh line with the observer sitting at the point of shortest distance to the line — that's what "transverse (i.e. lateral)" means. (note: I have removed the unhelpful wikilink to Transversality). DVdm (talk) 11:45, 12 December 2010 (UTC)

## Accelerated motion

The article says:

For general accelerated motion, or when the motions of the source and receiver are analyzed in an arbitrary inertial frame, the distinction between source and emitter motion must again be taken into account.

The Doppler shift when observed from an arbitrary inertial frame:{{ safesubst:#invoke:Unsubst||date=__DATE__ |\$B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }}

$f_{o}={\frac {c+v_{o}\!\cdot \!cos(\theta _{o})}{c-v_{s}\!\cdot \!cos(\theta _{s})}}\cdot {\frac {\gamma _{o}}{\gamma _{s}}}\cdot f_{s}$ where:

$v_{s}$ is the speed of the source at the time of emission
$v_{o}$ is the speed of the receiver at the time of reception
$\gamma _{s}$ is the Lorentz factor of the source at the time of emission
$\gamma _{o}$ is the Lorentz factor of the receiver at the time of reception
$\theta _{s}$ is the angle between the light path and the velocity of the source
$\theta _{o}$ is the angle between the light path and the velocity of the receiver

Consider the situation where the observer and the source are moving in opposite directions at the same speed such that $v_{o}=v_{s}$ (where $v$ is the symbol for speed). In this case ${\frac {\gamma _{o}}{\gamma _{s}}}={\frac {\sqrt {1-v_{s}^{2}/c^{2}}}{\sqrt {1-v_{o}^{2}/c^{2}}}}=1$ , while $cos(\theta _{o})=-cos(\theta _{s})$ , and thus ${\frac {c+v_{o}\!\cdot \!cos(\theta _{o})}{c-v_{s}\!\cdot \!cos(\theta _{s})}}=1$ . The result would be that $f_{o}=f_{s}$ . Why is that? Kmarinas86 (Expert Sectioneer of Wikipedia) 19+9+14 + karma = 19+9+14 + talk = 86 16:11, 30 December 2010 (UTC)

Okay. I see:
Per above, if $cos(\theta _{o})=-cos(\theta _{s})$ when the velocities are equal and opposite of each other, then $f_{s}=f_{o}$ apparently resulting in no frequency shift. For the equation to work, $cos(\theta _{o})$ would have to equal $cos(\theta _{s})$ in this case. Yet this point is not clear, and in fact, made less clear with the following passage:
It appears that the angles $\theta _{s}$ and $\theta _{o}$ only make sense in connection with scalar products $\mathbf {c} _{s}\cdot \mathbf {v} _{s}$ and $\mathbf {c} _{o}\cdot \mathbf {v} _{o}$ , such that if the velocities of $s$ and $o$ go in opposite directions, then the angular directions would be opposite with respect to the arbitrary inertial frame and also opposite with respect to the light path. This would make the ratio ${\frac {c-v_{o}cos(\theta _{o})}{c-v_{s}cos(\theta _{s})}}$ , which can be represented as ${\frac {c-{\frac {1}{c}}\mathbf {c} _{o}\cdot \mathbf {v} _{o}}{c-{\frac {1}{c}}\mathbf {c} _{s}\cdot \mathbf {v} _{s}}}$ or ${\frac {\mathbf {c} \cdot \mathbf {c} -\mathbf {c} _{o}\cdot \mathbf {v} _{o}}{\mathbf {c} \cdot \mathbf {c} -\mathbf {c} _{s}\cdot \mathbf {v} _{s}}}$ .
Kmarinas86 (Expert Sectioneer of Wikipedia) 19+9+14 + karma = 19+9+14 + talk = 86 17:55, 30 December 2010 (UTC)

In the previous section, we have the following:

As you can see, the denominator here has a factor $\left(1+{\frac {v\cos \theta _{o}}{c}}\right)$ , whereas in the formula I quoted has the equivalent of $\left(1-{\frac {v\cos \theta _{s}}{c}}\right)$ if you factor out $c$ from both the numerator and the denominator. There would appear to me no justification for assuming that, in the general case, $cos(\theta _{o})=-cos(\theta _{s})$ .Kmarinas86 (Expert Sectioneer of Wikipedia) 19+9+14 + karma = 19+9+14 + talk = 86 16:45, 30 December 2010 (UTC)

I have removed both new sections added by Template:User-multi, since one was added without a source and the other was not to be found in the cited source. When we have proper sources some of it can be restored. DVdm (talk) 19:58, 30 December 2010 (UTC)

Note. I have left a second level wp:NOR warning on talk page of Template:User-multi. DVdm (talk) 17:14, 7 January 2011 (UTC)

Note. We have worked this out on (User NOrbeck's talk page). The equation is now taken over from the source. DVdm (talk) 12:43, 10 January 2011 (UTC)

## Refraction Outside Domain of Special Relativity?

The article says refraction is outside the domain of special relativity, and cites a web FAQ by Tom Roberts, but the cited page does not make any such claim, and as a matter of fact the claim is false. Special relativity has no trouble with refraction. Of course, one can't use vacuum equations when dealing with something other than vacuum, but this doesn't mean special relativity is inapplicable. As long as no gravitational fields are significant, special relativity applies. But the point is, the cited reference doesn't support the claim, so I'm boldly deleting it. Also, the statement in the article about the Doppler shift being being purely classical when viewed from the median reference frame between two objects is simply wrong, and there is no reference supporting that claim, so I'm boldly deleting it. To be helpful, I though I'd mention why it is false. The velocities appearing in the relativistic equation cannot be mapped to the classical velocities, so the whole statement is misleading. If people think this is too difficult to understand, I'd be happy to simply delete it. But definitely we shouldn't retain the false and unsourced claim.Cattlecall1 (talk) 21:24, 25 February 2011 (UTC)

Without a source for your claim, it is entirely useless. Replacing sourced content "because you think it is wrong" with unsourced content is one of the most definite no-no's of Wikipedia. I have reverted your edit and left a second level warning on your talk page. DVdm (talk) 22:43, 25 February 2011 (UTC)

No, you misunderstand. The content on the page was not sourced. It claimed that refraction is outside the domain of special relativity, and it referenced a web FAQ, but a review of the referenced page shows that it does not make any such statement. So it is not a well-sourced claim, and needs to be removed, per Wikipedia policy. Since there are multiple issues here, I'll just focus on this one specific point, and we can discuss the other unsourced claims in the article later. The most important and blatent one to fix is the "refraction outside the domain of special relativity" claim, which is completely unsourced. If you can provide a valid source for that claim, please do so. Thanks.Cattlecall1 (talk) 23:16, 25 February 2011 (UTC)

In support of Cattlecall1: The claim "Refractive media are explicitly outside the domain of Special Relativity, which applies only to light propagation in gravity-free vacuums" makes no sense here for several reasons:
• it's only gravity that lies outside the domain of SR, refractive media are certainly within it
• the cited source makes no mention of refraction so it's original research to claim that it does
• the section in which this sentence appears makes no mention of gravity, so why is it being brought up here?
However it is true to say that an equation derived for an inertial frame need not be true in a non-inertial frame, and I guess that is the point that is attempting to be made here. The cited source expresses the opinion that non-inertial frames (in the absence of gravity) are still within the domain of SR. -- 23:22, 25 February 2011 (UTC)
In further support of Cattlecall1 (or perhaps non-support of the statement that SR only applies to vacuum): has everybody forgotten the Fizeau experiment? The derivation of its odd result from Einstein's SR treatment of additive velocities at relativistic speeds (in this case, the 3/4 c speed of light in moving water, complete with use of water's refractive index of 4/3) was one of the first experimental tests of SR (albeit a "retrodictive one"). Lorentz had already come close to using Einstein's methods, but did not realize that he was onto a completely general principle of time and coordinate transformations that applied to all physics. SBHarris 02:05, 26 February 2011 (UTC)

Cattlecall1, sorry for my previous reverts. I hadn't looked closely and assumed that you just reverted, so my reverts were mistaken and my edit summaries and warnings inappropriate. My apologies. DVdm (talk) 15:21, 26 February 2011 (UTC)

## Inconsistent definitions of theta?

The formula for a 2D doppler shift appears twice on the page, once under "Transverse Doppler Effect" and once under "Motion in an arbitrary direction" but their definitions of $\theta _{o}$ are not consistent. The first definition has $\theta _{o}$ defined as, when seen from receiver's frame, the angle between the direction the emitter is traveling and the "observed direction of the light at reception" which to me translates as the vector from the emitter to the receiver. The second definition is the angle between the velocity and the direction from the receiver to the emitter. Assuming $v$ has the same definition in both equations (I didn't see anything to indicate otherwise), these definitions cannot both be true and using one or the other will change the sign of the $v\cos \theta _{o}$ . I think the second definition is the correct one and the first definition should be changed. — Preceding unsigned comment added by 198.129.105.67 (talk) 19:17, 22 December 2011 (UTC)

"The first definition has $\theta _{o}$ defined as, when seen from receiver's frame, the angle between the direction the emitter is traveling and the 'observed direction of the light at reception' which to me translates as the vector from the emitter to the receiver."
The direction the emitter is traveling is not the "vector from the emitter to the receiver", that would be the "observed direction of the light at reception".
"The second definition is the angle between the velocity and the direction from the receiver to the emitter."
That's the same thing as the first definition.siNkarma86—Expert Sectioneer of Wikipedia
86 = 19+9+14 + karma = 19+9+14 + talk
02:19, 23 December 2011 (UTC)
I think there is just a simple misunderstanding here, we agree that that the "observed direction of the light at reception" is the vector from the emitter to the receiver. The first definition has $\theta _{o}$ as the angle between the velocity and this vector, whereas the second definition has $\theta _{o}$ as the angle between the velocity and a vector from receiver to the emitter (different than vector from emitter to receiver). These two definitions are not consistent. — Preceding unsigned comment added by 198.129.105.67 (talk) 01:53, 30 December 2011 (UTC)

## Sources? Original Research?

The majority of this article doesn't list any sources at all. Here's a summary" by section:

1 Visualization NO SOURCES
2 Analogy NO SOURCES
3 Motion along the line of sight NO SOURCES
4 Systematic derivation for inertial observers 1 ref at very end, not clear what it refers to
5 Transverse Doppler effect NO SOURCES
5.1 Reciprocity NO SOURCES
5.2 Experimental verification 2 sources
6 Motion in an arbitrary direction NO SOURCES
7 Accelerated motion 2 sources (but mostly original research?)

I think it would be better to base the article more firmly on sources. At present it just seems to be a collection of essays written by various editors describing their personal (and in some cases non-standard) views of the subject.Flau98bert (talk) 14:39, 14 September 2012 (UTC)

## In addition to the classical Doppler effect...

Re Jordgette's edit and my amendment: actually, the original wording was, if properly interpreted, correct after all.

The classical Doppler effect is given by

$f_{o}={\frac {f_{s}}{1+{\frac {v\cos \theta _{o}}{c}}}}.$ The relativistic effect is

$f_{o}={\frac {f_{s}}{\gamma \left(1+{\frac {v\cos \theta _{o}}{c}}\right)}},$ and thus, for the transversal case —which is the subject of the section— this reduces to, classically

$f_{o}=f_{s},\,$ versus relativistic

$f_{o}={\frac {f_{s}}{\gamma }}.\,$ So, indeed, in addition to the classical Doppler effect —being null (aka factor 1) in the transversal case—, the received frequency is reduced by the Lorentz factor. Since we're dealing with factors here, the word "addition" was a bit awkward, so I propose we leave it out, specially since the section is about the transverse effect, which is really non-existent classically. Subtle. - DVdm (talk) 13:26, 10 January 2013 (UTC)

## C' or V'

Full disclaimer, I am not a physicist, but, under "Motion along the line of sight", shouldn't the last equation (the approximation) be v', not c'? How can there even be a c'? — Preceding unsigned comment added by 174.98.234.239 (talk) 16:35, 9 September 2013 (UTC)