I What is the meaning of constant speed of light?

[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?

 

[Herman Trivilino]

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.

 

[Ibix]

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

The "common explanation" is a best attempt at saying something that can't really be said well in natural language. The key point, in this context, is that you will never see these galaxies overtake a light pulse.

A loose analogy is this: turn around 360° on the spot. In your rotating frame, Alpha Centauri just traveled about 25 light years in a second or two. That's fine because the coordinate speed in this coordinate system isn't restricted to being less than 3×10⁸m/s. And Alpha Centauri does not overtake any light pulses.

Nothing is rotating in the cosmological case, but you are forced to use curved coordinate systems in curved spacetime. This has the same result that coordinate speeds do not really mean anything physical. Local measurement of the speed of light in those distant galaxies would show it to be $c$ in all inertial frames, just as here.

 

[vanhees71]

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?

In the standard Robertson-Walker metric (on the large-scale average) distant galaxies are at rest relative to each other. What changes with time is the scale $a(t)$. There's no "speed limit" for $\dot{a}$.

 

[PeterDonis]

In the standard Robertson-Walker metric (on the large-scale average) distant galaxies are at rest relative to each other.

This is an unusual use of the term "at rest". These galaxies certainly do not observe zero redshift in each other's light signals.

The more usual way of putting what you are saying here is that comoving objects in FRW spacetime have constant spatial coordinates in the standard FRW coordinate chart. But equating "constant spatial coordinates" with "at rest relative to each other" is problematic in non-stationary spacetimes.

 

[vanhees71]

Yes, and that's what I consider at "being at rest" relative to each other. Of course you have red-shift, because of the time-dependent scale factor. At least one should not say that the redshift is due to a Doppler shift, which is what leads to the usual misunderstandings about the "recession velocity", as can be seen in this thread. It's not a velocity about the relative motion between objects. For details, see

http://www.edu-observatory.org/physics-faq/Relativity/GR/hubble.html

 

[PeterDonis]

that's what I consider at "being at rest" relative to each other

As I said, this is an unusual use of the term "at rest". As far as I know, you are the only one that uses it.

At least one should not say that the redshift is due to a Doppler shift

I don't entirely agree. See below.

It's not a velocity about the relative motion between objects.

The reference you give doesn't agree with this. It says:

"the Doppler shift explanation is a linear approximation to the "stretched light" explanation. Switching from one viewpoint to the other amounts to a change of coordinate systems in (curved) spacetime."

In other words, calling it a "Doppler shift" is not wrong; it's just an approximation that breaks down for large enough separations between the objects (or more precisely becomes less appropriate for large enough separations between the objects). But that's not because the description "at rest relative to each other" becomes more appropriate as the separation between the objects gets larger.

 

[skanskan]

The "common explanation" is a best attempt at saying something that can't really be said well in natural language. The key point, in this context, is that you will never see these galaxies overtake a light pulse.

A loose analogy is this: turn around 360° on the spot. In your rotating frame, Alpha Centauri just traveled about 25 light years in a second or two. That's fine because the coordinate speed in this coordinate system isn't restricted to being less than 3×10⁸m/s. And Alpha Centauri does not overtake any light pulses.

Nothing is rotating in the cosmological case, but you are forced to use curved coordinate systems in curved spacetime. This has the same result that coordinate speeds do not really mean anything physical. Local measurement of the speed of light in those distant galaxies would show it to be cc in all inertial frames, just as here.

What would happen if I hooked one end of an undeformable string to one of those distant galaxies and I leave the other end near to us free? What would be the relative speed between us and that free end?
I understand that no massive object can be accelerated beyond the speed of light.

 

[PeterDonis]

an undeformable string

There is no such thing. "Undeformable" means "infinite tensile strength" and relativity sets a finite limit on the tensile strength of any material.

 

[Ibix]

What would happen if I hooked one end of an undeformable string to one of those distant galaxies and I leave the other end near to us free?

As Peter says, there's no such thing. Something that cannot be deformed requires an infinite speed of sound - which you can't have without contradicting relativity. So the string either stretches or breaks.

 

[vanhees71]

As I said, this is an unusual use of the term "at rest". As far as I know, you are the only one that uses it.
I don't entirely agree. See below.
The reference you give doesn't agree with this. It says:

"the Doppler shift explanation is a linear approximation to the "stretched light" explanation. Switching from one viewpoint to the other amounts to a change of coordinate systems in (curved) spacetime."

In other words, calling it a "Doppler shift" is not wrong; it's just an approximation that breaks down for large enough separations between the objects (or more precisely becomes less appropriate for large enough separations between the objects). But that's not because the description "at rest relative to each other" becomes more appropriate as the separation between the objects gets larger.

Ok, perhaps my interpretation of the Robertson-Walker metric is not appropriate though in calculations of the red shift you assume radial light trajectories between objects of fixed spatical coordinates.

I also think it's less confusing to stress that the Hubble-expansion redshift is a gravitational effect and not (entirely) due to Doppler shifts of light between source and observer that are moving with respect to each other.

 

[PeterDonis]

in calculations of the red shift you assume radial light trajectories between objects of fixed spatical coordinates

Coordinates are not physics. "At rest relative to each other", at least as I've always seen the term used in the literature, is a statement about physics, not a statement about coordinates. If you find the redshift confusing in this respect, think of round-trip light signals between two comoving observers in FRW spacetime. The round-trip travel times of these light signals will not be constant according to either observer; they will increase with each observer's proper time. That is a physical manifestation of not being "at rest relative to each other".

Another way to capture the same thing is to look at the expansion of the congruence of worldlines that describes the family of observers. "At rest relative to each other" means the expansion is zero. (Actually, strictly speaking, it means expansion and shear are zero--only vorticity can be nonzero. Another way to put it is that the congruence must be Born rigid.) The expansion of the congruence of comoving observer worldlines in FRW spacetime is not zero; it's positive.

 

[vanhees71]

Hm, I always considered the observers defined by the standard FLRW space time
$$\mathrm{d} s^2=\mathrm{d} t^2 - a^2(t) \left [\frac{\mathrm{d} r^2}{1-k r^2} + r^2 (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2) \right ],$$
as those observers at rest relative to the "cosmological substrate", because the energy-momentum tensor of the matter in these coordinates necessarily looks like that of an ideal fluid at rest, $T^{\mu \nu}=(\epsilon+P)u^{\mu} u^{\mu}-P g^{\mu \nu}$, $u^{mu}=(1,0,0,0)$. In this sense a "fundamental observer" defined as an observer whose worldline is given by $(r,\vartheta,\varphi)=\text{const}$ is comoving with the "cosmic substrate". It's of course also the (local) rest frame of the cosmic microwave background radiation (in the sense that for such an observer it's homogeneous and isotropic in his neighborhood).

Of course this has to be taken with some care, because it's a local notion of "being at rest" relative to the cosmic substrate. Of course you are right in saying that distant fundamental observers are not "at rest" relative to each other.

 

[PeterDonis]

Of course you are right in saying that distant fundamental observers are not "at rest" relative to each other.

Exactly: "at rest" relative to some "cosmic substrate" is not the same as being at rest relative to some other observer. They are two different notions of "at rest". If you use the expression "at rest relative to each other", which is the phrasing of yours that I originally objected to, you are using the second notion of "at rest", not the first, and you agree that the second notion does not apply to distant comoving observers (and hence not to "distant galaxies", which was the phrasing you originally used).

 

[vanhees71]

I agree with that, but I still do not think that the Hubble redshift is solely a Doppler effect but can only be understood as the total gravitational effect, described by the time-dependent scale factor of the RWFL spacetime. For sure there's no velocity-like quantity "faster than light" which is not allowed to be faster than light ;-)).

 

[timmdeeg]

I agree with that, but I still do not think that the Hubble redshift is solely a Doppler effect but can only be understood as the total gravitational effect, described by the time-dependent scale factor of the RWFL spacetime.

Solely Doppler effect would require that special relativity holds globally.

There are several views. Peacock's is:

https://aapt.scitation.org/doi/10.1119/1.3129103

A common belief about big-bang cosmology is that the cosmological redshift cannot be properly viewed as a Doppler shift (that is, as evidence for a recession velocity) but must be viewed in terms of the stretching of space. We argue that, contrary to this view, the most natural interpretation of the redshift is as a Doppler shift, or rather as the accumulation of many infinitesimal Doppler shifts.