I want a good reference to a discussion of the meaning of "the speed of light is constant"

[gnnmartin]

... But one can make such variation appear as variation in the speed of light, or variation in the charge on the electron, or variation in Planck's constant, or any mixture of those, by an appropriate choice of units. The choice of units has nothing to do with the physics. ...

Thank you. I am only interested in the effect of any such variation on the metric. I set out my argument in 7 steps. It would help me if you can say which step is illogical or meaningless. Or even if you think it nonsense.

I agree with all your other comments, and apologise if I am careless with the distinction between a point and a neighbourhood.

 

[Ibix]

Thank you. I am only interested in the effect of any such variation on the metric. I set out my argument in 7 steps. It would help me if you can say which step is illogical or meaningless. Or even if you think it nonsense.

You can't change just the speed of light because of the fine structure constant, $\alpha$. You have to change one of the other values in the definition of $\alpha$.

If you also change one or more of the dimensionful constants it turns out that you are doing nothing more than a unit change - for example varying the electron charge varies the size of atoms, and if you double the speed of light and double $e^2$ then you halve the length of your rulers and all you've done is redefine 1m to mean what we used to call 50cm, so of course the speed of light in "new" metres is twice what it was in "old" meters.

It's only if you allow the dimensionless constant to change that you get a real physical change. But you could always choose to manifest that change as changes to $e$ or $\hbar$and keep $c$ unchanged, so it seems unlikely that messing around with $\alpha$ would change anything meaningful in the metric.

 

[PeterDonis]

I am only interested in the effect of any such variation on the metric.

A variation in the fine structure constant, by itself, does nothing to the geometry of spacetime. The geometry of spacetime is determined by the distribution of stress-energy.

I set out my argument in 7 steps. It would help me if you can say which step is illogical or meaningless.

I already told you what was wrong with your step 4.

 

[PeterDonis]

5: An illustration of the problem is the question often asked "What is the local effect of the expansion of the universe". The answer often given (including by me) is that the local effect is so small that it can be ignored

The correct answer is "none". From just local observations it is impossible to tell whether our universe is expanding or not.

6: I feel that at least we can say the speed of light must only vary extremely slowly along a line in space/time

I don't even know what this is supposed to mean.

7: The expansion of the universe, as it is generally described, assumes the speed of light varies along the time axis of a typical coordinate system representing the cosmic universe.

I have no idea where you are getting this from; it doesn't look like anything I've ever seen anyone say to describe the expansion of the universe.

 

[gnnmartin]

The correct answer is "none". From just local observations it is impossible to tell whether our universe is expanding or not.

Where does 'local' end?

We notice locally the expansion of the universe by noting the red shift of distant stars, isn't that incompatible with 'The correct answer is "none"'?

 

[gnnmartin]

...

Not the coordinate speed of light, which is what you appear to be interested. That is the ratio of a coordinate length to a coordinate time.

...

No, I'm definitely interested in the physical speed, so not (dx/dt), but Sqrt(-g(x,x)/g(t,t), where the coordinates are locally orthogonal and isotropic.

 

[Mordred]

Another source of papers regarding speed of light is the Lorentz invariance tests.

 

[Nugatory]

Where does 'local' end?

Where geodesic deviation becomes non-negligible.

We notice locally the expansion of the universe by noting the red shift of distant stars, isn't that incompatible with 'The correct answer is "none"'?

The local observation is the frequency and wavelength of the electromagnetic radiation right where we are. We note that this is different from the frequency and wavelength that would have been measured by a comoving observer present at the emission, so conclude that "local" does not cover the entire spacetime between the emission and reception events.
But our measurement is still completely local: there is electromagnetic radiation of a particular wavelength and frequency right where we are, obeying Maxwell's flat-spacetime laws and altogether unaffected by the expansion of the universe.

 

[Nugatory]

No, I'm definitely interested in the physical speed, so not (dx/dt), but Sqrt(-g(x,x)/g(t,t), where the coordinates are locally orthogonal and isotropic.

And the value of that quantity is determined solely by your choice of units... Not seeing how that is anything but a coordinate speed.

 

[PeterDonis]

Where does 'local' end?

When the region of spacetime necessary to cover all of the relevant objects can no longer be approximated as flat, i.e., when spacetime curvature can no longer be ignored.

We notice locally the expansion of the universe by noting the red shift of distant stars

No, we don't. Observations of redshifts of distant stars are not local because the stars are too far away for spacetime curvature to be ignored. Indeed, the observed redshifts are an effect of spacetime curvature.

 

[PeterDonis]

The local observation is the frequency and wavelength of the electromagnetic radiation right where we are.

But that observation, in itself, does not contain any "redshift".

our measurement is still completely local

But the determination of "redshift" is not, because it involves the properties of the light at emission as well as at reception, and the emission event is not "local"--it's far enough away that the effects of spacetime curvature cannot be ignored.

 

[gnnmartin]

And the value of that quantity is determined solely by your choice of units... Not seeing how that is anything but a coordinate speed.

Can you show how, please? Choose different coordinates (T,X,YZ) parallel to the first (t,x,y,z), and let dX/dT=a.dx/dt, then the value of Sqrt(-g(X,X)/g(T,T) is the same as the value of Sqrt(-g(x,x)/g(t,t) so the ratio that I am calling 'the speed of light' remains unchanged.

 

[PeterDonis]

I'm definitely interested in the physical speed, so not (dx/dt), but Sqrt(-g(x,x)/g(t,t), where the coordinates are locally orthogonal and isotropic.

You can't conclude anything about the "physical speed" of light at any place other than where you are from the ratios of metric coefficients where you are.

 

[gnnmartin]

When the region of spacetime necessary to cover all of the relevant objects can no longer be approximated as flat, i.e., when spacetime curvature can no longer be ignored.

Indeed, so you are giving the same answer as I am, but in different words. My answer was the difference is negligible. Your answer is because the expansion has a negligible effect on the local curvature. Both answers fail to acknowledge the underlying problem.

 

[PeterDonis]

My answer was the difference is negligible.

What "difference"? The redshift is certainly not negligible.

Your answer is because the expansion has a negligible effect on the local curvature.

But it does not have a negligible effect on the redshift.

Both answers fail to acknowledge the underlying problem.

What "underlying problem" are you talking about?

 

[PeterDonis]

Your answer is because the expansion has a negligible effect on the local curvature.

No, that was not my answer. My answer was that "local" means "over a small enough region of spacetime that the effects of curvature are negligible". That is not the same as saying "the expansion has a negligible effect on local curvature".

You are thinking of it backwards. The "expansion" is an "effect" of curvature; it's not a cause of curvature. The cause of the curvature is the distribution of matter and energy. Over a small enough region of spacetime (spacetime, note, not "space"), the effects of curvature are negligible--that includes the expansion itself. But redshift observations do not involve a small enough region of spacetime, so the effects of curvature, of which the redshift is one, are not negligible.

 

[gnnmartin]

What "difference"? The redshift is certainly not negligible.


But it does not have a negligible effect on the redshift.


What "underlying problem" are you talking about?

It is my assumption when asked 'what is the local effect of the expansion of the universe' is that the enquirer seeks to find some difference in local physics, due to (to use your words) the deviation from 'flat' of space time in our neighbourhood consistent with an expanding universe.

My answer is that the differences are negligible. Your answer is the same, but expressed in terms of curvature.

 

[Ibix]

My answer is that the differences are negligible. Your answer is the same, but expressed in terms of curvature.

The point is that this answer is a tautology. The definition of "local" in this context is that curvature is negligible over the "local" region. So the answer to the question in your point 5:

5: An illustration of the problem is the question often asked "What is the local effect of the expansion of the universe". The answer often given (including by me) is that the local effect is so small that it can be ignored, but that ducks the real issue.

should be "local is defined as the region where the answer to your question is 'nothing' - so either you are confirming the definition of local or you don't understand what you are asking". (Possibly phrased more tactfully.)

So I, at least, don't really understand what "underlying issue" you think is being ducked.

 

[PeterDonis]

It is my assumption when asked 'what is the local effect of the expansion of the universe' is that the enquirer seeks to find some difference in local physics, due to (to use your words) the deviation from 'flat' of space time in our neighbourhood consistent with an expanding universe.

My answer is that the differences are negligible. Your answer is the same

Not really. My answer is "none", because "local" by definition means "no effects from spacetime curvature".

 

[gnnmartin]

The point is that this answer is a tautology. The definition of "local" in this context is that curvature is negligible over the "local" region. So the answer to the question in your point 5:should be "local is defined as the region where the answer to your question is 'nothing' - so either you are confirming the definition of local or you don't understand what you are asking". (Possibly phrased more tactfully.)

So I, at least, don't really understand what "underlying issue" you think is being ducked.

You might assert here that 'local' means flat, but the person asking the question doubtless just uses 'local' in the everyday sense given in an English/American dictionary. We detect gravitational waves 'locally'. What you think of as flat becomes not-flat as your ability to measure improves. If you give the answer you suggest, then it will prompt the question 'How far out do you have to look in order to detect the difference between stars getting further away, and the space between us and the stars expanding?'. In other words, you will eventually say 'We don't know' or perhaps 'I don't know', or 'we can't detect any effect'.

 

[PeterDonis]

If you give the answer you suggest, then it will prompt the question 'How far out do you have to look in order to detect the difference between stars getting further away, and the space between us and the stars expanding?'.

Ok, then why don't you ask that question?

In other words, you will eventually say 'We don't know' or perhaps 'I don't know', or 'we can't detect any effect'.

How do you know without asking the question and seeing what responses you get?

 

[PeterDonis]

We detect gravitational waves 'locally'.

We detect anything we detect "locally" in the sense that we can only make measurements where we are. We can't directly measure what is happening in some galaxy far away.

However, many things we detect "locally" involve effects that are not "local" in the sense you appeared to be trying to use that term in your item 5; redshifts due to the expansion of the universe are an example.

If you're going to quibble over what "local" means, then let's stop using the word, and you rephrase your item 5. using some other words that more clearly express what you mean. Can you do that?

 

[gnnmartin]

Many thanks to all for your replies. I'm disappointed that nobody seems to have any sympathy with what I am trying to ask/say, but that in itself is useful to know. I'll continue to monitor this thread, and will respond to any post that adds something I find useful, but will otherwise retire from the thread.

Thanks again to all.

 

[PeterDonis]

I'm disappointed that nobody seems to have any sympathy with what I am trying to ask/say

I don't think it's a matter of "sympathy", it's a matter of not understanding what issue you think is being "ducked". None of us know of any such issue, so when we see someone saying they think there is, we want to know what that issue is.

 

[pervect]

It is my assumption when asked 'what is the local effect of the expansion of the universe' is that the enquirer seeks to find some difference in local physics, due to (to use your words) the deviation from 'flat' of space time in our neighbourhood consistent with an expanding universe.

My answer is that the differences are negligible. Your answer is the same, but expressed in terms of curvature.

Consider the Milne universe, if you're aware of it. https://en.wikipedia.org/wiki/Milne_model

It expands, but in the region of validity of the Milne universe, t>0, there is a diffeomorphism to the standard "Minkowskii" flat space-time of special relativity which does not expand.

My not-very precise resolution of this issue is to conclude that expansion, per se, does not cause physical effects. It's basically the result of a coordinate choice, though the expanding coordinates are lacking in complete coverage of the space-time.

However, acceleration or deacceleratio of the expansion DOES have physical effects. But these effects can be analyzed in terms of the effects of the matter, dark matter, energy, or dark energy that is causing the rate of expansion to accelerate or deaccelerate in the first place. The cause can ultimately be traced back to normal and dark matter/energy distributions.

Fore example, without dark energy, "gravity" causes the expansion of the universe to de-accelerate (slow down), this gravity has physical effects. Dark energy is the only one of the four that causes an accelerating expansion.

Orodruin wrote an article on some of this recently, with much more precision than what I wrote above, but I wasn't able to locate it.

 

[gnnmartin]

Consider the Milne universe, if you're aware of it. https://en.wikipedia.org/wiki/Milne_model

It expands, but in the region of validity of the Milne universe, t>0, there is a diffeomorphism to the standard "Minkowskii" flat space-time of special relativity which does not expand.

My not-very precise resolution of this issue is to conclude that expansion, per se, does not cause physical effects. It's basically the result of a coordinate choice, though the expanding coordinates are lacking in complete coverage of the space-time.

However, acceleration or deacceleratio of the expansion DOES have physical effects. But these effects can be analyzed in terms of the effects of the matter, dark matter, energy, or dark energy that is causing the rate of expansion to accelerate or deaccelerate in the first place. The cause can ultimately be traced back to normal and dark matter/energy distributions.

Fore example, without dark energy, "gravity" causes the expansion of the universe to de-accelerate (slow down), this gravity has physical effects. Dark energy is the only one of the four that causes an accelerating expansion.

Orodruin wrote an article on some of this recently, with much more precision than what I wrote above, but I wasn't able to locate it.

Thanks, very helpful.