Why Light Experienced a Doppler Shift?

[Mohammad Fajar]

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.

 

[Orodruin]

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).

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.

 

[Ibix]

But in this two case the wavelength remain constant

That is true for things like sound waves in a medium at non-relativistic speeds. Neither of those statements applies to light.

or experienced length contraction in the same amount without seeing whether it is receding or approaching)

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 traveling towards you is not it.

Your analysis is based on false premises.

 

[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, $c$.

 

[Vanadium 50]

Whoa! This is B-level!

 

[vanhees71]

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

 

[Orodruin]

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

 

[Dale]

But in this two case the wavelength remain constant

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

or experienced length contraction

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

 

[timmdeeg]

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.

Yes.

But in this two case the wavelength remain constant

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 .

 

[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

 

[timmdeeg]

To say it again. There's no way to understand relativity without a minimum of mathematics.

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.

 

[russ_watters]

Whoa! This is B-level!

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.

 

[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.

 

[russ_watters]

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...

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.

 

[SiennaTheGr8]

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.

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

 

[vanhees71]

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

 

[PAllen]

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.

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.

 

[Mohammad Fajar]

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.

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.

 

[Ibix]

light have a velocity equal c and frequency 1 Hz.

Light has a frequency of 10¹⁴-10¹⁵Hz, typically. But that's not particularly important here.

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.

Ok. So the received frequency is 1Hz.

But how about S,

S will measure the emitted frequency as $1/\sqrt 3\simeq 0.58$Hz, if my mental arithmetic is reliable.

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.

...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.

 

[russ_watters]

Sory, but light always travel with constant velocity.

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?

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.

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.

 

[Mohammad Fajar]

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.

I have read Doppler Effect for sound. When source (S) at rest and Observer (O) approaching, then speed of sound wave in medium (air) added by speed of O relative to the medium. So if speed of sound in medium is c then according to O it's speed become c + v, with v speed of O relative to medium. So in this case when frequency increased as O approaching then velocity of sound relative to him increased. Increasing in frequency makes velocity increasing, but wavelength constant. He will see more wave per second, but the wave still have the same wavelength.

For the case S approaching and O at rest, it is very different. For S velocity of sound relative to him become c - v, with v speed of S relative to medium. So in this case, as frequency Increased for observer O then wavelength decreased. If originally S emit a sound wave at frequency 1 Hz, then with v = 0.5 c, when one second passed the peak of wave he emit previously just a half original wavelength ahead of him and then he emit a second peak. But this two peak still travel with velocity of c relative to medium, so when arrive in O, he will see increasing in frequency that is two peak per second. He experience a shorter wavelength but the speed of sound not changed relative to him, because he at rest.

For the light it is very different. Light propagate without a medium. So there is no correction of speed of light when O approaching and S at rest, or S approaching and O at rest. Light always travel with c. So for S when he approaching there is no correction to velocity of light. S always see the same wavelength and the same frequency and the same velocity. For O it is also the same, like a sound wave, O does not have a role to generate the wave, so impossible the wavelength changed (except if we involving length contraction in calculation, so the wavelength become shorter for him). Wave generated by excitation in charger and it is have a definite energy, then why when we approaching that light the energy changed. If energy changed then the charge also change, but the charge is invariant quantity.

 

[Nugatory]

Light always travel with c.

Yes, and also...

So for S when he approaching there is no correction to velocity of light.

Yes, but...

S always see the same wavelength and the same frequency and the same velocity.

This does not follow. Try calculating it for yourself: if the crest of each wave leave the source $\Delta{t}$ seconds apart, and each crest travels at speed $c$, and the source is moving towards the receiver with speed $v$ so that each successive crest travels a distance that is less by $v\Delta{t}$... how much time elapsed between the arrival of each crest at the receiver? Don't guess or rely on the method of solution by emphatic assertion - do the calculation, it's just algebra.

Once you've done that, you can try correcting for the effects of time dilation: if the time between crests leaving the source is $\Delta{t}$ according to an observer at rest relative to the source, then the time between successive crests will be $\gamma\Delta{t}$ according to an observer at rest relative to the receiver. How does that affect the time between the arrival of successive crest and the distance between them?

 

[russ_watters]

...

You quoted my post but didn't actually respond to it. Do you have any response? Did you follow the logic?

Light propagate without a medium. So there is no correction of speed of light when O approaching and S at rest, or S approaching and O at rest. Light always travel with c. So for S when he approaching there is no correction to velocity of light. S always see the same wavelength and the same frequency and the same velocity. For O it is also the same, like a sound wave, O does not have a role to generate the wave, so impossible the wavelength changed (except if we involving length contraction in calculation, so the wavelength become shorter for him).

The way you put this is only true if the source and observer are not moving with respect to each other. You're basically saying that if C is the same with respect to each of them, then their velocity is zero with respect to each other. But it isn't. That's a contradiction.

And it is actually easier to visualize without the time dilation and length contraction considerations, just with the frame invariance of the speed light. The doppler shift at low speed is equal to the ratio of the speed of the source and the speed of light.
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/reldop3.html (last equation)

Wave generated by excitation in charger and it is have a definite energy, then why when we approaching that light the energy changed. If energy changed then the charge also change, but the charge is invariant quantity.

Energy has always been frame dependent, even from the time of Galileo.

You keep saying the same thing over and over with little variation and little indication you've read the responses. I'm not sure there is anything left to say if you won't address the responses.

 

[vanhees71]

Yes, and also...Yes, but...
This does not follow. Try calculating it for yourself: if the crest of each wave leave the source $\Delta{t}$ seconds apart, and each crest travels at speed $c$, and the source is moving towards the receiver with speed $v$ so that each successive crest travels a distance that is less by $v\Delta{t}$... how much time elapsed between the arrival of each crest at the receiver? Don't guess or rely on the method of solution by emphatic assertion - do the calculation, it's just algebra.

Indeed, and it is most simple to do it in a Lorentz-covariant way. The important quantity for the Doppler shift calculation (which also includes the calculation of aberration automatically) is the wave-four-vector,
$$(k^{\mu})=(\omega/c,\vec{k}),$$
which behaves like a four-vector under Lorentz transformations (particularly Lorentz boosts)
$$\tilde{k}^{\mu} = {\Lambda^{\mu}}_{\nu} k^{\nu}.$$
For a boost with relative velocity $\vec{v}$ it's
$$\tilde{k}^0=\gamma \left (k^0-\frac{\vec{v} \cdot \vec{k}}{c} \right), \quad \vec{\tilde{k}}=(\gamma-1) \frac{\vec{v} \cdot \vec{k}}{\vec{v}^2} \vec{v}+\vec{k}-\gamma k^0 \frac{\vec{v}}{c}.$$
Here, $\gamma=1/\sqrt{1-v^2/c^2}$ is the Lorentz factor. For details on Lorentz transformations etc. see my SRT FAQ article

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

 

[Mohammad Fajar]

You quoted my post but didn't actually respond to it. Do you have any response? Did you follow the logic?
The way you put this is only true if the source and observer are not moving with respect to each other. You're basically saying that if C is the same with respect to each of them, then their velocity is zero with respect to each other. But it isn't. That's a contradiction.

I just don't understand what mechanism have a role to make the wavelength changed for O. For example each beam of light separated by 1 meters then for observer O it is does't matter whether the S approaching or S receding the amount of changed in wavelength was given by Lorentz contraction, so it is shrink in the same amount. So when frequency changed and wavelength stay constant, then the velocity of light changed.

In this case I just see a light beam that emitted in definite frequency, not light as a continuous wave with very high frequency. Because this analogy was given in many book about derivation of relativistic Doppler shift. In many books there is no explanation about how the frequency of wave generated, it is only say that light beam emitted with definite frequency, and how this frequency changed when O moving in direction perpendicular to direction of light beam.

For classical Doppler Effect, there is no change in wavelength after it emitted by the source, it is always have that wavelength.

 

[Orodruin]

wavelength was given by Lorentz contraction

As you have already been told repeatedly, you cannot apply Lorentz contraction to the wavelength of a light signal. It does not satisfy the basic assumptions behind the derivation of Lorentz contraction.

 

[Mohammad Fajar]

As you have already been told repeatedly, you cannot apply Lorentz contraction to the wavelength of a light signal. It does not satisfy the basic assumptions behind the derivation of Lorentz contraction.

But they use time dilation to derive this Doppler Shift, then why I cannot use Lorentz contraction.
Sorry, but that picture I got from book Introduction of Modern Physics by Arthur Beiser.

 

[Orodruin]

Because they have, without telling you as is often the case in introductory literature, made sure that the assumptions behind the derivation of time dilation are satisfied. Unfortunately, many texts tend to focus on the flashy "Oh look! Time dilates and length contracts!" rather than methodically discussing what is actually assumed. This leads to many students trying to apply length contraction and time dilation in all manners of crazy ways that really are not allowed. In my experience with students, the ideas behind length contraction seem to be missed more often. Whenever in doubt, perform the full Lorentz transformation.

 

[PAllen]

@Mohammad Fajar , please, please just do exactly as @Nugatory described in post #22. It is elementary high school algebra. You don’t even need the Lorentz transform, nor do you need to know the 4 vector concepts needed for @vanhees71 calculation. Once you have done this, you have the relativistic Doppler period change. You already agree that c is invariant. If period changes and speed doesn’t, what must you conclude about wavelength?

 

[Mohammad Fajar]

Because they have, without telling you as is often the case in introductory literature, made sure that the assumptions behind the derivation of time dilation are satisfied. Unfortunately, many texts tend to focus on the flashy "Oh look! Time dilates and length contracts!" rather than methodically discussing what is actually assumed. This leads to many students trying to apply length contraction and time dilation in all manners of crazy ways that really are not allowed. In my experience with students, the ideas behind length contraction seem to be missed more often. Whenever in doubt, perform the full Lorentz transformation.

Sorry, is there any way to derive Doppler shift without using time dilation (and why I cannot use time dilation in the same place where I used Length contraction), or to derive that with full Lorentz tranformation?

If time dilation can I applied to every ticks of light, then why doesn't length contraction?
Are frequency of light not related to the time used to generate every peak of wave?
Are wavelength of light not related to the distance/space between every wave peak?

 

[Dale]

If time dilation can I applied to every ticks of light, then why doesn't length contraction?

Because the length contraction formula assumes that the object is at rest in one frame and compares the length in that test frame with the length in other frames where it is moving. Light is not at rest in any frame, so the length contraction cannot be applied to it.

The time dilation formula also does not apply to light for a similar reason. It applies to the emitter and the receiver.

My recommendation is to avoid the length contraction and time dilation formulas entirely. Use the Lorentz transform always. It will automatically simplify whenever appropriate.

 

[PAllen]

Sorry, is there any way to derive Doppler shift without using time dilation (and why I cannot use time dilation in the same place where I used Length contraction), or to derive that with full Lorentz tranformation?

If time dilation can I applied to every ticks of light, then why doesn't length contraction?
Are frequency of light not related to the time used to generate every peak of wave?
Are wavelength of light not related to the distance/space between every wave peak?

You can use the Lorentz transform. Just apply it for x = ct and x= ct + L, for example, getting the equations in primed coordinate. You will find that the speed is still c, but the L has become multiplied by a Doppler factor. This, of course also implies the frequency has changed.

As for length contraction, how many times do people need to repeat to you that this applies to length of something at rest in one frame, compared to how it is measured in a frame moving relative to that. Light is at test in no frame, so there is no rest length to contract.

 

[Mohammad Fajar]

Because the length contraction formula assumes that the object is at rest in one frame and compares the length in that test frame with the length in other frames where it is moving. Light is not at rest in any frame, so the length contraction cannot be applied to it.

The time dilation formula also does not apply to light for a similar reason. It applies to the emitter and the receiver.

My recommendation is to avoid the length contraction and time dilation formulas entirely. Use the Lorentz transform always. It will automatically simplify whenever appropriate.

Yes but they not consistent. They use time dilation when deriving Doppler shift. If light have frequency 10 Hz (for example), then period between generated peak is 0.1 second. And this is the time that attached to the light frame itself even though it is related to generator frequency at the source. So if the light have specific frequency in S frame, then it must have specific wavelength in S frame, because $\lambda \times f = c$.

 

[PeterDonis]

this is the time that attached to the light frame itself

No, it isn't. You are still missing a fundamental point: there is no "light frame". Light is not at rest in any frame; it moves at $c$ in all frames. That means there is no "time attached to the light frame" because there is no "light frame". The only relevant frames are the rest frame of the emitter and the rest frame of the receiver.

 

[robphy]

The Doppler Effect can be derived by using Bondi’s method. Radar methods are then used to connect the Doppler-Bondi k-factor with time and position displacements, and thus relative velocity measurements to get the Doppler formula: $k=\sqrt{\frac{1+v}{1-v}}$. Along the way, one can derive the Lorentz transformation... but it’s not necessary to obtain the doppler formula with Bondi’s method.

In addition, it might be useful to draw a Spacetime diagram of the wavefronts of a periodic source. Then consider the observations by other various inertial observers... in both the relativistic and nonrelativistic cases.

For special relativity, a key detail that is often neglected is that length contraction involves the separation between parallel timelike lines, whereas the Doppler effect as applied to wavelength applies to parallel lightlike lines. Thus the length contraction formula does not apply to wavelength as if it were a ruler-length.