What is the meaning of constant speed of light?

[Dale]

I think that though the wave front speed is 0 relative to the observer, he can still watch the concentric circles of the propagating wave extended in space and then he can still calculate the phase velocity.

This is very easy to calculate for a plane wave. E.g.

For a plane wave in the x direction we can write $A \cos (\omega t - k x)$. Then we can transform to a different frame (assuming v<<c) by substituting $x=x'+vt'$ and $t=t'$. This gives us $A \cos((\omega-kv)t' - kx')$. So the wavelength is $k$ and the frequency is $\omega-kv$ which leads to a different phase velocity.

 

[vanhees71]

For the acoustic Doppler effect treated in SR, see

https://aapt.scitation.org/doi/abs/10.1119/1.14430

It's of course purely academic, but of course it works relativistically too.

What's also very interesting is how poor Doppler was treated in connection with his effect:

https://physicstoday.scitation.org/doi/10.1063/PT.3.4429

 

[Mister T]

What is your meaning here?

I think you are confused by the notion that all measurements are invariant. For example, if I measure the speed of something to be 340 m/s all others will agree that that is what I got when I made the measurement.

However, they may not agree that the speed is 340 m/s relative to them.

This is different from a measurement of the speed $c$. If I measure the speed of something to be $c$ all others will agree that that is what I got when I made the measurement.

And, they will also agree that the speed is $c$ relative to them. This is what is meant by the declaration that the speed $c$ is invariant.

 

[thaiqi]

This is very easy to calculate for a plane wave. E.g.

For a plane wave in the x direction we can write $A \cos (\omega t - k x)$. Then we can transform to a different frame (assuming v<<c) by substituting $x=x'+vt'$ and $t=t'$. This gives us $A \cos((\omega-kv)t' - kx')$. So the wavelength is $k$ and the frequency is $\omega-kv$ which leads to a different phase velocity.

Imagine a water wave source, and its wavefront speed is v. Then one person(person 1) approaches it from the left with a velocity v.
Another person(person 2) leaves it away on the right side with a velocity v.
So the person 1 will measure both the wave front speed and the phase velocity v+v = 2v.
The person 2 will measure a wave front speed v-v = 0.
But what will the person 2 measure the phase velocity?

 

[Dale]

As long as they are far enough away that the plane wave approximation is valid then you can just use the formula above.

 

[jbriggs444]

Imagine a water wave source, and its wavefront speed is v.

Except that it is not.

The speed of every individual portion of the wavefront relative to the medium is v. If one were to measure the speed of the various portions of the wavefront from a different frame of reference then that speed would vary around the periphery of the front.

 

[thaiqi]

View attachment 258508
Imagine a water wave source, and its wavefront speed is v. Then one person(person 1) approaches it from the left with a velocity v.
Another person(person 2) leaves it away on the right side with a velocity v.
So the person 1 will measure both the wave front speed and the phase velocity v+v = 2v.
The person 2 will measure a wave front speed v-v = 0.
But what will the person 2 measure the phase velocity?

The person 2 will measure a wave front speed v-v = 0.
But what will the person 2 measure the phase velocity? Surely it is not 0, I think he will get the same wavelength and same time of period, so the phase velocity is v. Am I wrong?

 

[jbriggs444]

The person 2 will measure a wave front speed v-v = 0.
But what will the person 2 measure the phase velocity? Surely it is not 0,

Why not? If the person sees that he is keeping pace with a region of maximum pressure (for a sound wave) or maximum ripple height (for a water wave), surely he will measure the velocity of that phase as zero.

If you have him measuring wave length divided by period then since the period is Doppler shifted to infinity, that quotient will be zero also.

 

[Dale]

I think he will get the same wavelength and same time of period, so the phase velocity is v. Am I wrong?

Yes. Use the formula I posted. What does it tell you?

 

[thaiqi]

Why not? If the person sees that he is keeping pace with a region of maximum pressure (for a sound wave) or maximum ripple height (for a water wave), surely he will measure the velocity of that phase as zero.

If you have him measuring wave length divided by period then since the period is Doppler shifted to infinity, that quotient will be zero also.

The person 2 will measure the wave front speed as zero. But how could he regard the phase velocity as 0 while he saw and knew that wave is propagating as concentric circles? Shouldn't the phase velocity be of some particular value other than zero? (Shouldn't he regard so?)

 

[Ibix]

But how could he regard the phase velocity as 0 while he saw and knew that wave is propagating as concentric circles?

Remember that the source is moving in this person's frame. Naively, it's quite surprising that you get concentric circles since the successive wavecrests were not emitted from the same place.

But it turns out that the phase velocity, as measured in this frame, varies with direction - from zero to 2v. And it does it in such a way that the wavefronts remain circular and centred on an object at rest with respect to the medium.

 

[A.T.]

...person 2 will measure...But how could he regard ...

Physics only cares about the measure part. The regard part is irrelevant.

 

[thaiqi]

It's the phase velocity. In the vacuum it's also the group velocity:
$$\vec{v}_\text{g}=\frac{\partial \omega}{\partial \vec{k}}=\partial_{\vec{k}} c |\vec{k}|=c \hat{k}.$$
It's also the "speed of the wave front" in simple ("Drude like") models of dielectrica.

For a very deep understanding of the latter, see the now famous papers by Sommerfeld and Brillouin (I'm sure there are English translations of those):

It is said that there are cases where the phase velocity is larger than c. Why not consider it is the speed of the wave front in postulate 2 rather than phase velocity?

 

[A.T.]

What's also very interesting is how poor Doppler was treated in connection with his effect:

https://physicstoday.scitation.org/doi/10.1063/PT.3.4429

Oh dear...

"Petzval thought that no great science could come from a few simple lines of algebra: In his view, all natural phenomena were the manifestations of underlying differential equations."

 

[vanhees71]

It is said that there are cases where the phase velocity is larger than c. Why not consider it is the speed of the wave front in postulate 2 rather than phase velocity?

That's the point of the famous papers by Sommerfeld and Brillouin. In regions of anomalous dispersion the quantity you can formally calculate as $v_{\text{g}}$ can be larger than $c$, it can be even pointing in the opposite direction from what you'd expect. The reason is that this quantity in the region of anomalous dispersion, i.e., around a resonance frequency of the matter, has no longer the physical meaning of some wave-propagation velocity. In regions of normal dispersion it describes how a wave packet moves in the sense of e.g., the maximum of the wave packet moves, while the packet's overall shape changes only very slowly. In regions of anomalous dispersion the wave packet is rather deforming quite quickly, and you cannot make sense of how to define a center or peak of the packet anymore. That's why the group velocity looses its physical meaning in this case, because the corresponding approximation doesn't apply anymore.

The only thing that always must be $\leq c$ is the front velocity, and this is guaranteed by the relativistic in-medium Maxwell equations as long as the model for the medium is causal, and then the function theory (theorem of residues) takes care of the causality of the wave propagation. In the usual simple Drude models (which can also be motivated by in-medium QED calculations!) the front velocity is precisely $c$.

The reason is very intuitive: If the incoming wave first arrives at a point in the medium, the medium has not yet reacted to the corresponding em. field and also not emitted em. waves itself (the total em. field is the incoming field + the field emitted by the moving charges of the medium). That's why you can say, in the very first moment, the em. field is unaffected and moves as if would propagate in vacuo, i.e., with the vacuum velocity of light. But then you soon get quite fascinating transient states, the socalled Sommerfeld precursor followed by the Brilloui precursor.

The treatment of a semifinite wavetrain entering a dieelectric in Sommerfeld's way is among the most beautiful and elegant applications of dispersion theory, mathematically all based on complex-function theory applied to Fourier transformations. I highly recommend to read the treatment in

A. Sommerfeld, Lectures on Theoretical Physics vol. IV (optics)

It's boiled down to there to the most simple treatment possible. I'm not sure whether the two Annalen papers are translated to English, but I guess so, because they are real gems of the classical-field-theoretical literature.

 

[thaiqi]

I highly recommend to read the treatment in

A. Sommerfeld, Lectures on Theoretical Physics vol. IV (optics)

It's boiled down to there to the most simple treatment possible. I'm not sure whether the two Annalen papers are translated to English, but I guess so, because they are real gems of the classical-field-theoretical literature.

I take a look at Sommerfeld's book section 22 and find the contents in it are rather hard for me.

 

[jbriggs444]

View attachment 258515The person 2 will measure the wave front speed as zero. But how could he regard the phase velocity as 0 while he saw and knew that wave is propagating as concentric circles? Shouldn't the phase velocity be of some particular value other than zero? (Shouldn't he regard so?)

The center of the circle is moving. Have you never thrown a stone into a flowing river?

Or consider surfing on a standing wave. https://en.wikipedia.org/wiki/River_surfing The phase velocity of a standing wave is manifestly zero.

 

[PeroK]

The person 2 will measure the wave front speed as zero. But how could he regard the phase velocity as 0 while he saw and knew that wave is propagating as concentric circles? Shouldn't the phase velocity be of some particular value other than zero? (Shouldn't he regard so?)

In other words, all (wave) motion is absolute! Once you have decided on the most appropriate reference frame in which to describe a wave motion, then all measurements must be done in that frame. No one is allowed to use a different frame. Hence, all velocities are invariant!

For example: if a car is moving at $30m/s$ relative to a road, then all observers must use the reference frame of the road to measure the speed of the car. No one is allowed to use a reference frame in which the road is moving and in which the speed of the car is not $30m/s$.

As I said many posts ago, some people have a mental block about the whole concept of relative velocity. In your understanding of physics there are only absolute velocities. Nothing is relative!

 

[A.T.]

As I said many posts ago, some people have a mental block about the whole concept of relative velocity.

Much of the confusion about Special Relativity comes from not grasping Galilean Relativity in the first place.

 

[PeroK]

Much of the confusion about Special Relativity comes from not grasping Galilean Relativity in the first place.

We see that a lot on here. The basic concept of motion being different in different reference frames is the first major stumbling block.

 

[thaiqi]

The center of the circle is moving. Have you never thrown a stone into a flowing river?

I draw the image using the Earth as reference frame. The center of the circle is moving relative to person 2, while it is fixed on the earth.

The phase velocity of a standing wave is manifestly zero.

Now I admit the phase velocity is zero.

 

[jbriggs444]

I draw the image using the Earth as reference frame. The center of the circle is moving relative to person 2, while it is fixed on the earth.

Of course, this means that you have a choice of reference frames relative to which to measure the [phase/group/whatever] velocity.

[With reference to a surfer on a standing wave...]

Now I admit the phase velocity is zero.

What about relative to the flowing water?

 

[Dale]

Why not consider it is the speed of the wave front in postulate 2 rather than phase velocity?

The postulate doesn’t refer to any of this speeds. It refers to the one unique invariant speed.

 

[thaiqi]

[With reference to a surfer on a standing wave...]

What about relative to the flowing water?

Sorry, I don't catch what you mean here.

 

[jbriggs444]

Sorry, I don't catch what you mean here.

You've agreed that the phase velocity of a standing wave relative to a surfer on that wave is zero.

What about the phase velocity of the same standing wave relative to the water flowing through the wave?

 

[thaiqi]

You've agreed that the phase velocity of a standing wave relative to a surfer on that wave is zero.

What about the phase velocity of the same standing wave relative to the water flowing through the wave?

The wave source is at rest relative to the earth, the standing wave is not standing any more with respect to the water, it propagates as concentric circles in the water, I think.

 

[jbriggs444]

The wave source is at rest relative to the earth, the standing wave is not standing any more with respect to the water, it propagates as concentric circles in the water, I think.

Neither the wave source nor any hypothetical concentric circles that might be present in other circumstances are relevant. A particular phase of the wave (e.g. a crest, a trough or the midpoint on a leading edge) has a position as a function of time and, consequently, a velocity as a function of time.

You can express the position using coordinates from one frame of reference or coordinates from another. Depending on the choice, the velocity determined using those coordinates will be different.

Compared to a surfer on the wave, the crest is motionless a bit behind him, the trough is motionless a bit in front of him and the midpoint on the leading edge is motionless beneath his feet. All three have the same zero phase velocity.

Compared to a chip of wood floating on the water as it passes the surfer, the crest is rushing toward the chip, the trough is rushing away and the midpoint on the wave's leading edge is rushing past. All three have [nearly] the same non-zero phase velocity.

 

[cmb]

I confess I do struggle with the concept of doppler shift for light, when the speed is invariant. That means the photons carrying the EM energy have different frequencies, according to the observer. But why would a photon, with a given amount of energy at a given (invariant) speed have a different effect according to how fast an observer is traveling relative to wherever that photon came from?

I mean, imagine that photon, as a packet of energy, is traveling parallel to another photon of the same energy but that was emitted from a different source traveling faster than that of the first. These two photons, with the same energy, traveling at the same speed, are they then destined to interact with some distant observer in some different way to each other?

 

[Mister T]

But why would a photon, with a given amount of energy at a given (invariant) speed have a different effect according to how fast an observer is traveling relative to wherever that photon came from?

Because the effect depends on the energy, not the speed. The amount of energy is not invariant, but the speed is. Note that this is a purely relativistic notion, there is nothing like it in Newtonian physics.

 

[Ibix]

with the same energy

With the same energy as measured by who? Energy is frame dependent.

 

[PeroK]

I confess I do struggle with the concept of doppler shift for light, when the speed is invariant.

As speed is wavelength times frequency, the wavelength and frequency may not be frame-invariant; hence energy may be (and is) frame-dependent.

In other words, the speed of a wave does not determine the wavelength and frequency. If it did, there would only be one universal monochromatic light.

 

[vanhees71]

The Doppler effect within classical electromagnetism follows from the Lorentz transformation of the plane-wave modes of the electromagnetic field. It's sufficient to consider only the phase of the solution,
$$F_{\mu \nu}(x)=F_{\mu \nu}^{(0)} \exp[-\mathrm{i} (\omega t-\vec{k} \cdot \vec{x}).$$
The Lorentz transformation reads
$$\bar{F}^{\mu \nu}(\bar{x}) = {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(x) = {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(\hat{\Lambda}^{-1} \bar{x}).$$
Thus the phase is invariant, i.e.,
$$\bar{k} \cdot \bar{x}=\bar{k}_{\mu} \bar{x}^{\mu} = k_{\mu} x^{\mu}.$$
This implies that with $\bar{x}=\hat{\Lambda} x$ also $\bar{k} =\hat{\Lambda} k$.

Take a boost with velocity $\vec{v}=\vec{\beta} c$, $\gamma=1/\sqrt{1-\beta^2}$). Then
$$\bar{k}^0=\frac{1}{c} \bar{\omega}=\gamma (k^0-\vec{\beta} \cdot \vec{k}).$$
Since $k^0=|\vec{k}|$ due to $k \cdot k =0$ we get
$$\bar{k}^0=\gamma |\vec{k}| (1-\beta \cos \vartheta).$$
Here $\vartheta$ is the angle between $\vec{v}$ and $\vec{k}$. Especially for $\vartheta=0$ you get
$$\bar{k}^0(\vartheta=0)=|\vec{k}| \sqrt{\frac{1-\beta}{1+\beta}},$$
In this case the observer in $\bar{\Sigma}$ moves in the same direction as the wave relative to $\Sigma$, and thus you get the maximal possible red-shift (the observer moves away from the source, which is located at infinity).

For $\vartheta=\pi$ the observer moves towards the source, and you get the maximal possible blue-shift
$$\bar{k}^0(\vartheta=0)=|\vec{k}| \sqrt{\frac{1+\beta}{1-\beta}}.$$
Remarkably you also get a Doppler effect when the observer moves perpendicular to the wave-propagation direction, which is unknown to the non-relativistic Doppler effect (for sound):
$$\bar{k}^0(\vartheta=\pi/2)=\gamma |\vec{k}|.$$
This is of course caused by the well-known time-dilation effect.

If you evaluate the spatial part of the Lorentz transform of the four-vector $k$, you get the aberration formula for light.

 

[Nugatory]

I mean, imagine that photon, as a packet of energy, is traveling parallel to another photon of the same energy but that was emitted from a different source traveling faster than that of the first. These two photons, with the same energy, traveling at the same speed, are they then destined to interact with some distant observer in some different way to each other?

Your difficulty here comes from thinking that the electromagnetic wave is made up of photons traveling from the source to the destination, and that's not how it works.

But telling you that you're thinking about it all wrong isn't especially helpful... helpful would be telling you how it does work .

There's no substitute for learning the physics right: analyze light classically using Mawell's equations; understand how the fields transform from frames in which the source and/or receiver is at rest to ones in which they are moving; then move on to quantum mechanics; and finally encounter photons for the first time as you take on quantum electrodynamics. That's a lot more work than most non-specialists are willing to put in, and the mathematical price of admission is fairly steep.

Happily, there is a simpler and more layman-friendly way of thinking about light and photons. Light is electromagnetic radiation, the classical waves described by Maxwell's equations, so there's nothing surprising about Doppler. There are no photons except when the electromagnetic waves interact with matter. These interactions transfer energy between the oscillating electromagnetic fields and the matter; even though the wave is spread out in space the energy transfer always happens in discrete amounts delivered at a single point. When this happens we say "a photon happened at that point". From this point of view there's no problem understanding the Doppler shift: the light waves from a moving source are Doppler-shifted producing different electromagnetic fields at the receiver, and there's nothing surprising about different electromagnetic fields interacting with the receiver in different ways.
(But be warned that this explanation is oversimplified almost to the ragged edge of acceptability. As I said above, "There's no substitute for learning the physics right".)

 

[skanskan]

Things are more difficult than that.
Distant galaxies seem to be moving away at a speed much greater than c. (very high Z).
The common explanation is because that "extra" speed is due to the expansion of the universe.

Can anyone explain exactly how the maximum speed limit is taken into consideration here?

 

[PeroK]

Things are more difficult than that.
Distant galaxies seem to be moving away at a speed much greater than c. (very high Z).
The common explanation is because that "extra" speed is due to the expansion of the universe.

Can anyone explain exactly how the maximum speed limit is taken into consideration here?

You really ought to start your own thread on this question. In short:

In Special Relativity (flat spacetime) we have the invariant speed of light.

In General relativity (curved spacetime - including expanding space) we have that light moves locally at an invariant speed and, in general, on null geodesics.