# Does the Doppler effect disprove the constancy of $c$? [closed]

Why doesn't the Doppler effect disprove the relativity postulate that $c$ is constant relative to all sources or observers?

It seems it does because of the following example:

Suppose a ship of length $L$ faces an isotropic emitter (at the left in the diagram) emitting a constant frequency $\omega$ of wave crests a distance $\ell$ from the ship's stern:

Definitions:
$x_0 = 0, t_0$: emitter (origin)
$x_1$, $t_1$: stern
$x_2$, $t_2$: bow
$c$: speed of light (same constant in all frames)

# Case 1: ship at rest

## stern

$$|x_1-0|=c(t_1-t_0)$$$$x_1=\ell$$$$\implies\boxed{t_1=t_0+\ell/c}$$

## bow

$$|x_2-0|=c(t_2-t_0)$$$$x_2=\ell+L$$$$\implies\boxed{t_2=t_0+(\ell+L)/c}$$ Propagation time from bow to stern:$$\implies\boxed{t_2-t_1=L/c}$$

# Case 2: ship moving

(Note: Prime variables here do not mean another frame; all measurements are done in the ship's frame)
Ship moving at speed $v$ toward emitter:

## stern

$$|x_1'-0|=c(t_1'-t_0')$$$$x_1'=\ell'-v(t_1'-t_0')$$$$\implies\boxed{t_1'=t_0'+\ell'/(c+v)}$$

## bow

$$|x_2'-0|=c(t_2'-t_0')$$$$x_2'=\ell'+L-v(t_2'-t_0')$$$$\implies\boxed{t_2'=t_0'+(\ell'+L)/(c+v)}$$ Propagation time from bow to stern:$$\implies\boxed{t_2'-t_1'=L/(c+v)}$$

Thus, the bow and stern observers together measure different values for $c$ ($c$ in case 1 and $c+v$ in case 2). This contradicts the assumption that $c$ is constant.

The time difference$$\Delta t=(t_2-t_1)-(t_2'-t_1')=L/c-L/(c+v)\approx vL/c^2$$is still the same regardless if the clocks are biased or run at different rates. Even if time dilation came into play for the primed times, it would be a 2nd order effect in $v/c$. Thus, relativity of simultaneity or time dilation has no effect. Lorentz contraction, even if it made a difference in the observers' frame (the ship frame), is also a second-order effect: $$\Delta L\approx L(v/c)^2/2.$$ But the $\Delta t$ above is 1st order: $$\Delta t\propto (v/c)^1.$$

## closed as unclear what you're asking by ACuriousMind♦, Colin McFaul, Danu, Brandon Enright, Kyle KanosOct 17 '14 at 0:41

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• Why do you think that it would? – Sean Oct 16 '14 at 18:39
• No it doesn't. Why do you think it does? – Kyle Kanos Oct 16 '14 at 18:40
• Doppler effect changes frequencies, not speed. – jinawee Oct 16 '14 at 19:37
• The relativistic Doppler shift (linked in my previous comment) requires the Lorentz transformation, so it kinda depends on $c$ being an invariant quantity hence it cannot disprove the postulate of $c$'s invariance. – Kyle Kanos Oct 16 '14 at 19:42
• Good grief, all would be clear if you would draw a spacetime diagram. I don't have the time or inclination to decipher your derivation but I suspect you haven't taken into account that the clocks at the bow and stern are not synchronized in the emitter frame of reference. Again, this would be clear in a spacetime diagram. – Alfred Centauri Oct 17 '14 at 0:33

As stated in the comments: The Doppler shift is a change in observed frequency due to relative speed difference. However, the speed with which the signal propagates is the speed of light.

The speed of a photon, and all massless particles, is essentially always $c = 299792458\space m/s$. The Doppler effect is entirely contingent on the fact that the velocity is the same throughout. We have a source, $S$, emitting light at a certain speed: $$|\space\space\space\space\space|\space\space\space\space\space|S|\space\space\space\space\space|\space\space\space\space\space|$$ moving at velocity, $v$: $$v\rightarrow\\|\space\space\space\space\space\space\space|\space\space\space\space|S|\space\space|\space|$$ Where $v=f\lambda$ $$z = \frac{\lambda_{\mathrm{obsv}} - \lambda_{\mathrm{emit}}}{\lambda_{\mathrm{emit}}}$$ Or, with respect to frequency: $$z = \frac{f_{\mathrm{emit}} - f_{\mathrm{obsv}}}{f_{\mathrm{obsv}}}$$ At no point does the velocity change.

• You're onto something good with the diagrams, but you should probably define $z$ and explain where the heck those formulae came from. – DanielSank Oct 16 '14 at 21:24
• But you've got to show there is not a contradiction. How am I supposed to know whether your final equation implies a contradiction or not to your assumption $c=$const.? – Geremia Oct 17 '14 at 0:14

It appears from your edit that you added the velocity of the ship and the speed of light classically. However, the velocity of the ship and the speed of light need to be added relativistically, using a Lorentz Transformation.

Here's a quick way to know something is amiss with your moving ship: You state that the stern observer sees a wave propagation speed of $c+v$ However, one consequence of Special Relativity is that information can not be transferred faster than the speed of light. Because $c+v>c$, you know something is up. If you apply a Lorentz Transformation you will observe a constant speed of light, $c$.

• That's not the issue, since Geremia is analyzing everything from the perspective of the frame of the source. If the ship's length is L and its speed is v in this frame, then it is indeed true that in this frame, the time for the light to get from the stern to the bow is L/(c+v). The real issue is forgetting to take into account the relativity of simultaneity when calculating what clocks on board the ship will measure, see my series of comments about this. – Hypnosifl Oct 17 '14 at 0:59
• …it is indeed true that in this frame, the time for the light to get from the stern to the bow is L/(c+v). But it should be $c$ in all frames. – Geremia Oct 17 '14 at 2:53
• @Geremia - It is. (c+v) is not the speed of the light signal itself in any frame, rather it is the "closing speed" between the light signal and the bow of the ship, i.e. the rate that the distance between them is shrinking, as seen in the frame of the source of the light. In this frame the light itself is moving at c, the bow itself is moving at v, and so the distance between them shrinks at (c+v). And because of the relativity of simultaneity, the speed of this same light signal is c in the ship's rest frame--see my series of three comments in reply to your original question. – Hypnosifl Oct 17 '14 at 5:08
• For more on the notion of the "closing speed" between two things in some frame, and how it is distinct from the individual speed of either one in that frame, see en.wikipedia.org/wiki/Faster-than-light#Closing_speeds – Hypnosifl Oct 17 '14 at 5:11