# B Why Light Experienced a Doppler Shift?

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1. Jan 2, 2018

It's already known in physical communities that the velocity of light always constant and not depend on motion of observer or source. But it is also teached to us that there are aslo Doppler shift experienced by light wave like that when measuring CMB radiation. But for me it is makes a non-sense. If I approaching the light source then the frequence of light I receive become increased, but when receding the light source the frequency of light I receive decreased. But in this two case the wavelength remain constant (or experienced length contraction in the same amount without seeing whether it is receding or approaching). Then the velocity of light not constant for this two case, becase velocity = wavelength x frequency.

2. Jan 2, 2018

### Orodruin

Staff Emeritus
No, this is not correct. You should learn actual relativity before you make blanket statements like this. The length contraction formula does not apply here in the way you think that it does.

3. Jan 2, 2018

### Ibix

That is true for things like sound waves in a medium at non-relativistic speeds. Neither of those statements applies to light.
This is not correct. Length contraction compares the length of an object as measured in a frame where it is moving to the length measured in its rest frame. Light has no rest frame; even if it did a frame where it is travelling towards you is not it.

Your analysis is based on false premises.

4. Jan 2, 2018

### vanhees71

Take a plane wave. It's characterized by the four-vector $(k^{\mu})=(\omega/c,\vec{k})$. The dispersion relation is covariantly written as $k_{\mu} k^{\mu}=0$, i.e., $\omega = c |\vec{k}|$.
Now in a frame, which moves with velocity $\vec{v}$ ($|\vec{v}|<c$) against the original frame the wave-vector components read
$$\tilde{k}^0=\gamma \left (k^0-\frac{\vec{v} \cdot \vec{k}}{c} \right), \quad \vec{\tilde{k}}=(\gamma-1) \frac{\vec{v} \cdot \vec{k}}{v^2} \vec{v}+\vec{k}, \quad \gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}},$$
which clearly shows that both the frequency $\omega$ and the wavelength change under Lorentz boosts. Of course the dispersion relation doesn't change since it's expressed as a manifestly covariant condition, and thus $\tilde{k}^{\mu} \tilde{k}_{\mu}=k^{\mu} k_{\mu}=0$. This of course implies that also in the new reference frame the speed of light is the same as in the old one, $c$.

5. Jan 2, 2018

Staff Emeritus
Whoa! This is B-level!

6. Jan 2, 2018

### vanhees71

Well, #4 is B-level. B-level cannot mean to make things simpler than possible!

7. Jan 2, 2018

### Orodruin

Staff Emeritus
Sometimes B-level is “I am sorry you need to learn things at a more advanced level to understand this properly”.

8. Jan 2, 2018

### Staff: Mentor

The wavelength does not remain constant. The wavelength changes inversely to the frequency so that the speed stays constant.

The simplified length contraction formula does not apply here. You need to use the full Lorentz transform.

9. Jan 2, 2018

### timmdeeg

Yes.
No. In addition to what has been said I would like to add that energy is frame dependent. The energy and hence the wavelength and the frequency of light remain constant (which means that there is zero Doppler shift then) only in case the source is at rest relative to the receiver .

10. Jan 3, 2018

### vanhees71

The wavelength cannot stay constant if the frequency changes, because free em. waves obey the dispersion relation (in other words the relation that tells us that the electromagnetic field is a massless field):
$$\omega=c |\vec{k}|, \quad |\vec{k}|=\frac{2 \pi}{\lambda}.$$
This is valid in any frame of reference since it can be written in a manifest covariant way by introducing the four-vector
$$(k^{\mu})=(\omega/c,\vec{k}) \; \Rightarrow \; k^{\mu} k_{\mu}=0.$$
To say it again. There's no way to understand relativity without a minimum of mathematics. For an introduction, see my SRT FAQ article

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

11. Jan 3, 2018

### timmdeeg

It depends on the level. For for a laymen it is already helpful if he can improve his notion and if wrong notions are disproved. To a certain extent this is possible without math. I suspect it is a challenge for experts to answer in ordinary language and that it would be easier for them to just write down the math.

It features PhysicsForums that there are experts who are willing and capable to talk to laymen.

Last edited: Jan 3, 2018
12. Jan 3, 2018

### Staff: Mentor

I agree. The OP has a conceptual problem with doppler shift that doesn't have anything directly to do with math or even Relativity.

Light, sound, water waves, tennis balls from a launcher; the concept is always the same: when the source and receiver are moving with respect to each other, the spacing of the peaks (tennis balls) changes.

That alone answers the question in the title. The math just tells you how much the spacing changes.

13. Jan 3, 2018

### vanhees71

Well, there's a great conceptual difference between the Doppler shift of light in a vacuum and sound waves. In the case of the light there's no preferred frame of reference (aka no aether), while for sound waves there is a preferred frame of reference, namely the one where the air is at rest, i.e., you have to distinguish between the cases whether the observer or the source are moving relative to the rest-frame of the air, while this is not the case for light, where you only need the relative velocity between source and observer. It's not clear to me how to comprehensively explain this without using the adequate mathematics.

14. Jan 4, 2018

### Staff: Mentor

Is that conceptual or quantification? I don't see how the OP's question - though he mentions light - is specific to light. But still: tennis balls fired from a launcher don't have a medium either.

15. Jan 4, 2018

### SiennaTheGr8

Nonetheless, redshift was observed and correctly recognized as a Doppler phenomenon decades before Einstein derived the right formula for light.

16. Jan 4, 2018

### vanhees71

Sure, most phenomena of electromagnetism and optics were known before RT, but with RT it's much better understood than before!

17. Jan 4, 2018

### PAllen

Actually, this shows a key difference between light and other cases modeled via Newtonian physics. Given any description of a periodic phenomenon in an emitter frame (doesn’t matter if a medium is motion relative to emitter, so anisotropy, or if a corpuscular - baseball - model is being used), then any receiver frame agrees on wavelengths, but disagrees on speed and frequency. This is trivially seen given that distance is a Galilean invariant while speed is frame dependent. Since the relationship of speed, frequency, and wavelength still holds, it is the frequency, speed pair which are frame variant in Galilean physics, while wavelength is invariant. SR actually changes this for all cases, but the difference is only significant for high speed waves and high speed relative motion.

18. Jan 4, 2018

Sory, but light always travel with constant velocity.

For example if source (S) emit a light beam at constant frequency one time a second, and there is observer O at distance 3 x 10^8 m receive this light he will agree with S that the light have a velocity equal c and frequency 1 Hz. But what if S approaching O with velocity 0.5 c at the same time when he start emitting the light, then when he arrive at half distance (1.5 x 10^8 m) the light he emitted previously arrive at O. Observer O will measure that the speed of light always c with frequency 1 Hz. But how about S, if he need to make sure that light always travel with velocity c, then when he arrive at point 1.5 x 10^8, the light must travel at distance 3 x 10^8 m surpassing O.

19. Jan 4, 2018

### Ibix

Light has a frequency of 1014-1015Hz, typically. But that's not particularly important here.
Ok. So the received frequency is 1Hz.
S will measure the emitted frequency as $1/\sqrt 3\simeq 0.58$Hz, if my mental arithmetic is reliable.
...what? This makes no sense to me. In the rest frame of S he isn't moving, O is. And neither of them agree on clock synchronisation or clock rate, so "when he arrive at point 1.5 x 10^8" is ambiguous because you have to say when according to which frame.

I suggest you look up the Lorentz transforms. These relate positions and times measured by one observer (e.g. S) to those measured by another moving at speed v relative to the first. You may also wish to google for the relativistic Doppler formula to see where I got the $1/\sqrt 3$Hz from.

20. Jan 4, 2018

### Staff: Mentor

Yeah, I get that. And your question is: shouldn't this new fact cause Doppler shift to not happen. And my point is that this new fact doesn't change the underlying mechanism of Doppler shift. And my approach is to start by making sure you understand the underlying mechanism of Doppler shift. Try this:

You have two people moving toward each other and one is firing a wave, stream of baseballs, whatever toward the other. At a certain moment in time, there is a certain number of baseballs wave peaks, etc., between them. At a later moment in time, when they are closer together, there is a smaller number of baseballs, wave peaks, etc. between them. Where did the missing baseballs/wave peaks go? And then why would the existence or lack thereof of a medium, invariant speed of the wave, etc., change that?
This is just a claim (an incorrect one); you haven't given a logical reason why. Merging with my logic from above, if you count the number of peaks between them, then after the first peak arrives, there is always a constant number of wave peaks between them, including at the time they meet - including after they pass! A nonsensical result. Because of the time delay between when the source starts emitting and the observer starts receiving, they are emitting and receiving waves for different amounts of time (from the start, until they meet), but they send/receive the same number of peaks.

This logic is the same for all of the various permutations of Doppler shift.

Last edited: Jan 4, 2018