# Wikipedia states that the relativistic Doppler effect is the same whether it is the source or the receiver that is stationary. Can this be explained?

According to Wikipedia, the relativistic Doppler effect is simply the classical Doppler effect for a stationary source, $$1-v/c$$, times the relativistic time dilation, $$1/\sqrt{1-v^2/c^2}$$ (where v is relative velocity) which simplifies nicely to $$\sqrt{\frac{1-v/c}{1+v/c}}.$$

It then goes on to say that assuming a stationary receiver gives the same result, but it doesn't, at least as far as its own derivation is concerned, because the classical Doppler effect for a stationary receiver is $$1/(1+v/c)$$. And this is not the same as $$1-v/c$$, so the two cannot be the same.

How is this resolved?

• Remember, all motion is relative. Commented Jul 15 at 16:49
• Precisely the reason for the question. Commented Jul 15 at 17:42
• @JohnHobson It's the answer to the question as well. "Stationary" is not a meaningful term in relativistic terms, as evidenced by the core definition of "relativity". Commented Jul 18 at 7:12
• I understand, but the issue is - do you use 1-v/c or 1/(1+v/c) when working with relative velocity.. The answers below do resolve the issue. Commented Jul 19 at 9:54
• Voting to re-open. Since when has Relativistic Doppler shift been a personal theory? It is a completely mainstream interpretation. See Wikipedia en.wikipedia.org/wiki/Relativistic_Doppler_effect
– KDP
Commented Jul 22 at 6:31

You have to include a factor of $$\gamma$$ in both effects. The terms that are the same are $$\gamma (1-\frac v c)$$ and $$\frac 1 {\gamma \left(1 + \frac v c \right)}$$. This is because

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

If there were not complete symmetry between receiver and source in the relativistic effect then we could detect absolute motion (we would know whether the receiver or the source was moving), which we cannot do in special relativity.

• Mathematically this makes sense but you seem to be saying that in the first case time dilation is gamma while in the second case time dilation is 1/gamma}. Why the difference? Commented Jul 15 at 17:56

The relativistic Doppler factor has to be the same whether it is the receiver or the source that is considered to be stationary, because in relativity there is no way to determine if it is the receiver or the source that is 'really' stationary or 'really' moving. When measuring the Doppler shift of sound in air it is possible to determine your own motion relative to the medium. When measuring the Doppler shift of light in a vacuum in relativity, there is no medium.

Mathematically this makes sense but ... in the first case time dilation is gamma while in the second case time dilation is 1/gamma}. Why the difference? – John Hobson

1. When the receiver is moving away, the the received frequency is reduced by the regular Doppler factor of $$(1-v/c)$$, but because the receiver's clock is considered to be running slow in this reference frame, it measures shorter intervals between the arriving peaks, increasing the classical received frequency by the gamma factor.

2. When the the source is going away from the receiver, the received frequency is reduced by the regular Doppler factor of $$1/(1+v/c)$$, but now it is the source that is considered to be time dilated and it emits photons at a slower rate, reducing the classical received frequency by a factor of 1/gamma.

The end result is the same whether it is the receiver or the source that is stationary as claimed by Wikipedia.

• I think I may have anticipated this but thanks for the clarification. But is there not another issue? Is the receiver the observer in both cases? And observers' own clocks never time dilate. Commented Jul 16 at 11:54
• @JohnHobson Dilation is always a relative notion. Saying the observer clock never time dilates is an absolute statement and thus does not really make sense. Indeed what we mean by this is that we fix our reference time to be the observers (as he is observing this ticking of his clock) and describe clocks ticking relative to this Commented Jul 16 at 12:45
• The comment answer by @ThomasTappeiner is spot on. Just wanted to add that while there are two observers, "THE observer" is traditionally the one that considers the other to be moving. If Anne is the emitter and Bob is the receiver, then $F_r/f_e$ is the ratio of what they individually measure and get when they compare notes at a later time. I could write the ratio as $f_{Anne}/f_{Bob}$ and they would both agree on this ratio. Clearly $f_{Anne}/f_{Bob}$ is not the measurement of a single observer. They use different calculations but get the same result.
– KDP
Commented Jul 16 at 15:48
• @MissUnderstands Thanks. Appreciate your comment very much.
– KDP
Commented Jul 16 at 15:55
• @JohnHobson We could even have a 3rd observer that considers both Anne and Bob to be moving in his reference frame. This 3rd observer would also agree on the ratio $f_{Anne}/f_{Bob}$ using his own calculations.
– KDP
Commented Jul 16 at 16:08

But your final two expressions are approximately the same in the limit $$v\ll c$$. Use the binomial expansion, \begin{align} (1+\epsilon)^n&= 1+n\epsilon+\frac{n(n-1)}{2!}\epsilon^2+\cdots \\&≈1+n\epsilon \end{align} where the approximation holds when the product $$n\epsilon\ll 1$$. This gives immediately $$(1-v/c)^{-1}≈1+v/c$$ in the nonrelativistic limit.

(This doesn't really answer your question, but it was too long for the comment box.)

• Approximately the same is no good, especially as the relativistic formula is only required when v is large compared to c. Commented Jul 15 at 17:41
• This is an interesting comment, but not a good answer. Commented Jul 16 at 20:27

For sound waves propagating in a medium, there is an obvious absolute frame of in which the sound wave exists with a frequency that does not depend on the motion of the source (Tx) or receiver (Rx). From there, the Tx and Rx processes are physically different. In their rest frames, one is transmitting in a moving fluid, and the other is receiving in a moving fluid--and the physical difference is reflected in the Doppler formula (e.g., the divergence at Mach 1).

Now if we imagine an electromagnetic plane wave just propagating through space: it doesn't have a frequency. It does have a 4-wave vector, which we write in a frame as:

$$k_{\mu} = (\omega/c, \vec k)$$

and all we know for sure is that:

$$k_{\mu}k^{\mu} = \frac{\omega^2}{c^2} -||\vec k||^2 = 0$$

in all frames (this is dispersion relation $$\omega=ck$$).

(Note, if you prefer to consider a photon and talk about momentum, then just use:

$$p_{\mu} = \hbar k_{\mu} = (E/c, \vec p)$$

it may be less abstract. The dispersion relation becomes $$E=pc$$ which means $$m_{\gamma}=0$$).

It is at this point where you have to fight the urge to associate $$\omega$$ with the source. Like the sound case, there appears to be a natural frame, but there isn't. There is no medium (aether) through which the wave propagates, and there is no way to assign a wavelength without considering the source's rest frame...

...or if the source is moving, then the frame defining the source's motion, but that is completely arbitrary.

Now on the receive end: the Rx has no idea what the source's motion is, and it has no effect on the result. There is just a frame in which the receiver is at rest, $$\omega$$ and $$\vec k$$ have values.

This is counterintuitive, since we generally associate EM radiation with source--the helium spectrum in the Sun, the 21 cm line, a 511 KeV annihilation $$\gamma$$...but really, that doesn't matter. The plane wave just exists when the Rx sees it. (remember, the imaginary wave has an arbitrary frequency, so it's not some physical process with a well defined frequency).

So what all that means is, if the source is radiating $$\omega_{Tx}$$, and the receiver sees $$\omega_{Rx}$$, the doppler shift can only depends on their relative velocity, and thus has to be symmetric.

$$v_T = 0$$ $$v_R \equiv v_0$$

is physically indistinguishable from:

$$v_T \equiv -v_0$$ $$v_R = 0$$

moreover, all cases in-between (or not) are indistinguishable, provided:

$$\frac{v_R + v_T}{1 + \frac{v_Rv_T}{c^2}} = v_0$$

So I hope that made sense. It really requires to understand the implications of:

1. All motion is relative.

2. There is no absolute rest frame.

3. Light always moves at $$c$$.

It's a hard step to take because relativity is introduced as "the effects of motion", but the core principle is that there is no such thing as "motion".

There is only relative motion, and even then the Tx and Rx don't have absolute velocities (hence the last formula above).

Now if you do take this to heart, you will not want to think of the relativistic Doppler shift as a classical doppler shift modified by length contraction or time dilation.

Nonetheless, it is instructive to think about time-dilated oscillations and length contracted distances between crest, that is, different 3 + 1 views of:

$$E(x) = E_0e^{i(\vec k \cdot \vec x - \omega t)}$$

the same way we do with, say, a pole moving through a barn with both doors "simultaneously" closing. Note that the pole-barn paradox is a confusion causing paradox, and similar pitfalls will occur when examining the crests and troughs of plane waves.

Also, if you look at the above plane wave, it can be written:

$$E(x) = E_0e^{-ik_{\mu}x^{\mu}}$$

where the phase is now a manifestly Lorentz invariant scalar:

$$\phi(x^{\mu}) = k_{\mu}x^{\mu} = -(\vec k \cdot \vec x - \omega t)$$

that is: all frames agree a crest is a crest, and a zero crossing is zero crossing.

In summary, the relativistic Doppler formulae are just two ends of Lorentz transforming a null 4-vector, which seems a too abstract an mathy.

There is a physical interpretation though. In the photon description where $$k_{\mu}$$ is replaced with the 4-momentum:

$$p_{\mu} = (E/c, \vec p)$$

and either the Tx or Rx velocity is defined by a 4-velocity:

$$u_{\mu} = (\gamma c, \gamma \vec v)$$

and a 4-momentum:

$$P_{\mu} = Mu_{\mu}$$

where $$M\gg E_{\gamma}/c^2$$ is the large mass of the apparatus, the requirement that the momentum transfer from the emission/absorption of the radiation is orthogonal to the 4-acceleration of the apparatus will reproduce the relativistic Doppler shift.

That is required, so that:

$$P^2/c^2 = M^2$$

before and after the emission or absorption events.

(That method, which I've posted somewhere on PSE, can also handle the Doppler shift with massive bosons, inelastic processes, and recoils effects).

This considers a more general case in which
both the source and the receiver are in motion in the lab frame.

This calculation (using a spacetime diagram and its geometry) will obtain the Doppler factor in special relativity for the situation with velocities $$0 ( so $$v_{RS}>0$$ ) after they met at event O (the receiver moves away from the source). Other variations can be considered using the same methods to be shown.

Consider the spacetime diagram below (with time running upwards)

• the source along OS (making rapidity angle $$\theta_{SL}$$ with the lab frame) and
• the receiver along OR (making rapidity angle $$\theta_{RL}$$ with the lab frame),
• so the relative rapidity of the receiver with respect to the source is $$\theta_{RS}=\theta_{RL}-\theta_{SL}$$.

The corresponding velocities [slopes] are equal to the hyperbolic-tangents of the rapidities.
For example, $$v_{SL}=c\ \tanh\theta_{SL}=\frac{MS}{OM}$$.

At event S, the source sends a signal (to be called SR) to be received by the receiver at event R.
With respect to the vertical time axis, the slope of SR is $$c=\frac{UR}{SU}$$.
At event O, when the source and receiver meet, the source sent a signal and was immediately received by the receiver [at event O].

OS is the period between emissions by the source.
OR is the period between receptions by the receiver.
The ratio of frequencies is $$\frac{f_R}{f_S}=\frac{1/(OR)}{1/(OS)}=\frac{OS}{OR}.$$ For convenience, trace back SR to meet the lab frame at event $$L$$ (before event $$M$$).

We will write OL in two ways, then eliminate OL to relate the expressions. \begin{align} OL=(OM-LM) &=(OM-(MS/c)) \mbox{ since c=\frac{MS}{LM}}\\ &=OM(1-(v_{SL}/c)) \mbox{ since v_{SL}=\frac{MS}{OM}} \end{align} and \begin{align} OL=(ON-LN) &=(ON-(NR/c)) \mbox{ since c=\frac{NR}{LN}}\\ &=ON(1-(v_{RL}/c)) \mbox{ since v_{RL}=\frac{NR}{ON}} \end{align}

Thinking non-relativistically (absolute time), we have $$OR=ON$$ and $$OS=OM$$ \begin{align} \frac{f_R}{f_S}&=\frac{OS}{OR}\\ &\stackrel{abs.time}{=}\frac{OM}{ON}\\ &\stackrel{abs.time}{=}\frac{1-v_{RL}/c}{1-v_{SL}/c}\\ \end{align} (The denominator would have a relative-plus sign, if the receiver were slower than the source [so the signal SR has a negative velocity]. The event $$L$$ would be after event $$M$$.)

Thinking relativistically, we have $$OR\ \gamma_{RL}=ON$$ and $$OS\ \gamma_{SL}=OM$$ \begin{align} \frac{f_R}{f_S} &=\frac{OS}{OR}\\ &=\frac{ \frac{OM}{\gamma_{SL}} }{ \frac{ON}{\gamma_{RL}} }\\ &=\frac{\gamma_{RL}}{\gamma_{SL}} \frac{ OM}{ON }\\ &=\frac{\gamma_{RL}}{\gamma_{SL}} \frac{1-v_{RL}/c}{1-v_{SL}/c}\\ &=\frac{\frac{1}{\sqrt{(1+v_{RL}/c)(1-v_{RL}/c)}}\ (1-v_{RL}/c)} {\frac{1}{\sqrt{(1+v_{SL}/c)(1-v_{SL}/c)}}\ (1-v_{SL}/c)}\\ &=\frac{ \sqrt{ \displaystyle\frac{ (1-v_{RL}/c)}{(1+v_{RL}/c)} } } { \sqrt{ \displaystyle \frac{ (1-v_{SL}/c)}{(1+v_{SL}/c)} } } =\sqrt{ \displaystyle\frac{ (1-v_{RS}/c)}{(1+v_{RS}/c)} } =\frac{1}{k_{RS}} \end{align} which depends only on the relative-velocity $$v_{RS}$$ between the receiver and source.
(In our example, $$v_{RS}>0$$ after they met at event O so, the receiver is moving away from the source and so the frequency of receptions is lower.)

In the last line, we did some algebra of the form \begin{align} \frac{\displaystyle\left(\frac{1-r}{1+r}\right)}{\displaystyle\left(\frac{1-s}{1+s}\right)} &=\displaystyle\frac{rs+r-s-1}{rs-r+s-1}\\ &=\displaystyle\frac{1-rs-(r-s)}{1-rs+(r-s)}\\ &=\displaystyle\frac{1-\frac{(r-s)}{1-rs} }{1+\frac{(r-s)}{1-rs}}\\ \end{align} where you hopefully recognize the relative-velocity formula.

Instead, Thinking trigonometrically, we write $$OL=(OM-LM)=(OS \cosh\theta_{SL} - OS\sinh\theta_{SL})\quad \mbox{since c=\frac{MS}{LM}}$$ $$OL=(ON-LN)=(OR \cosh\theta_{RL} - OR\sinh\theta_{RL})\quad \mbox{since c=\frac{NR}{LN}}$$ So, \begin{align} \frac{f_R}{f_S} =\frac{OS}{OR} &=\frac{\cosh\theta_{RL}-\sinh\theta_{RL}}{\cosh\theta_{SL}-\sinh\theta_{SL}}\\ &=\frac{\exp(-\theta_{RL})}{\exp(-\theta_{SL})}\\ &=\exp(-(\theta_{RL}-\theta_{SL}))\\ &=\exp(-\theta_{RS})=\frac{1}{k_{RS}}\\ \end{align}

Thinking in terms of the Bondi k-calculus,

$$OS=k_{SL}\ OL$$ $$OR=k_{RL}\ OL$$ $$OR=k_{RS}\ OS$$

So, $$\left(\frac{f_R}{f_S}\right)^{-1}=k_{RS} =\frac{OR}{OS}=\frac{ k_{RL}\ OL}{ k_{SL}\ OL}=\frac{ k_{RL}}{ k_{SL}}$$

Here is a numerical example (drawn on rotated graph paper so we can visualize the tickmarks along the worldlines):

$$v_{SL}=(3/5)c$$ and $$v_{RL}=(4/5)c$$

$$k_{SL}=\sqrt{\frac{1+v_{SL}/c}{1-v_{SL}/c}}= \sqrt{\frac{1+(3/5)}{1-(3/5)}}=\sqrt{\frac{8}{2}}=2$$ which agrees with $$(OS/OL)=4/2=2$$

$$k_{RL}=\sqrt{\frac{1+v_{RL}/c}{1-v_{RL}/c}}= \sqrt{\frac{1+(4/5)}{1-(4/5)}}=\sqrt{\frac{9}{1}}=3$$ which agrees with $$(OR/OL)=6/2=3$$

We expect $$k_{RS}=\displaystyle\frac{k_{RL}}{k_{SL}}=\frac{3}{2}$$ which agrees with $$(OR/OS)=6/4=3/2$$.

From \begin{align} v_{RS}=\displaystyle\frac{v_{RL}-v_{SL}}{1-v_{RL}v_{SL}/c^2} &=\frac{(4/5)-(3/5)}{1-(4/5)(3/5)}\\ &=\frac{(1/5)}{25/25-(12/25)}\\ &=\frac{5}{13} \end{align} we have $$k_{RS}=\sqrt{\frac{1+(5/13)}{1-(5/13)}}=\sqrt{\frac{18}{8}}=\sqrt{\frac{9}{4}}=\frac{3}{2}$$.

As a check, $$k_{RS}=$$$$\exp({\rm arctanh}(4/5)-{\rm arctanh}(3/5))=3/2$$.