What is the meaning of constant speed of light?

In summary, the speed of light in vacuum, also known as the invariant speed, is the same for all observers regardless of their relative motion. This is not the case for the speed of sound, which varies depending on the observer's motion relative to the medium through which the sound is propagating. While both the phase and group velocities may be constant relative to the medium, neither of them can be invariant. This is due to the concept of relative speed, where an object's speed can vary depending on the observer's motion.
  • #71
skanskan said:
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.
 
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  • #72
skanskan said:
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}##.
 
  • #73
vanhees71 said:
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.
 
  • #74
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
 
  • #75
vanhees71 said:
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.

vanhees71 said:
At least one should not say that the redshift is due to a Doppler shift

I don't entirely agree. See below.

vanhees71 said:
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.
 
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  • #76
Ibix said:
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.
 
  • #77
skanskan said:
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.
 
  • #78
skanskan said:
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.
 
  • #79
PeterDonis said:
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.
 
  • #80
vanhees71 said:
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.
 
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  • #81
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.
 
  • #82
vanhees71 said:
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).
 
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  • #83
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 ;-)).
 
  • #84
vanhees71 said:
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.
 
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