I Isotropy of the speed of light

[Vanadium 50]

t's like the question of parity symmetry.

Martin Deutsch (discoverer of positronium) told the story that in the early 1950's a grad student working on a positron experiment came up to him and said that there were more events in the left side of the detector. Marty told him that was was ridiculous. That would mean that the weak interaction was parity violating. He told the student to go fix things. Which he did.

 

[Tendex]

No, they are not outside the empirical scope. If you assume isotropy of space for (locally) inertial observers you can test this empirically. In principle nothing prevents you from finding a dependence of the speed of light on the orientation of your measurement device, and then this specific symmetry assumption would be disproven. Of course, today no such violation is known.

Maybe I'm not getting my point across, I think Vanadium 50 got It right. It's the inertial frame what cannot be empirically discarded without mathematical contradiction. So inertial frames are postulated together with the 2 postulates of SR, and then, as Einstein underlined several times no contradiction will be found in the theory. So inertial motion existence is a primitive definition/postulate of the theory just in order to not enter into contradiction and it is not subject to empirical test as an assumption. In other words its presence guarantees the theory is not contradictory, therefore its absence leads to contradiction. This is known as the Law of the excluded middle. If a physical model that claims to be based on mathematics failed it it wouldn't be taken too seriously.

 

[Sagittarius A-Star]

No, they are not outside the empirical scope. If you assume isotropy of space for (locally) inertial observers you can test this empirically. In principle nothing prevents you from finding a dependence of the speed of light on the orientation of your measurement device, and then this specific symmetry assumption would be disproven. Of course, today no such violation is known.

But that is only valid for the 2-way-speed of light. The isotropy of the one-way speed of light in an inertial (= not accelerated) frame is not an assumption. It is a definition. If you synchonize clocks at the light source and at the light detector differently from the Einstein-synchonization, then you will measure a non-isotropic one-way-speed of light.

 

[hutchphd]

The isotropy of the one-way speed of light in an inertial (= not accelerated) frame is not an assumption. It is a definition.

It is an assumption of convenience. As is the standard method of synchronization. It is not a definition because the theory is consistent without it being specified.

 

[Sagittarius A-Star]

It is an assumption of convenience. As is the standard method of synchronization. It is not a definition because the theory is consistent without it being specified.

No, it is a definition:

you declare: "I maintain my previous definition nevertheless, because in reality it assumes absolutely nothing about light. There is only one demand to be made of the definition of simultaneity, ...

Source

 

[Dale]

I think that there is a good point that @Tendex is making. There are assumptions that we make that are purely conventional and are not subject to empirical testing. That includes all definitions of terms and many mathematical conventions, like using the right hand rule, and other conventions like positive charge for protons or units. None of those assumptions are subject to testing.

There are other assumptions that we make which are empirically testable, like isotropy of the two-way speed of light, or conservation of momentum. So we need to test these testable assumptions and we need to distinguish when an assumption is testable or not so that we don't waste time trying to test untestable assumptions and so that we don't mentally elevate our convention assumptions to the status of facts about nature.

 

[Sagittarius A-Star]

like isotropy of the two-way speed of light, or conservation of momentum.

I think, the conservation of "one-way momentum" also depends on the definition of simultaneity, so it should be: "like isotropy of the two-way speed of light, or conservation of two-way momentum."

 

[PeterDonis]

I think, the conservation of "one-way momentum" also depends on the definition of simultaneity, so it should be: "like isotropy of the two-way speed of light, or conservation of two-way momentum."

There is no such thing as "one-way momentum" vs. "two-way momentum". There is just momentum.

 

[Sagittarius A-Star]

There is no such thing as "one-way momentum" vs. "two-way momentum". There is just momentum.

Yes. But consider a coordinate chart, that creates an anisotropy of the one-way light speed in (+/-) x-direction and describe in this coordinate chart an explosion. The momentum will not be conserved.

 

[PeterDonis]

Yes. But consider a coordinate chart, that creates an anisotropy of the one-way light speed in (+/-) x-direction and describe in this coordinate chart an explosion. The momentum will not be conserved.

This is not correct. Conservation of momentum is not a coordinate-dependent law. In the coordinate chart you describe, the one-way light speed won't be the only thing that is different from a standard inertial chart. Other things will be different as well, in just the right way to preserve momentum conservation.

 

[Sagittarius A-Star]

Conservation of momentum is not a coordinate-dependent law.

I think that is only the case, as long you also use the Einstein definition of simultaneity:

Salmon (1977, 273) argues, however, that the standard formulation of the law of conservation of momentum makes use of the concept of one-way velocities, which cannot be measured without the use of (something equivalent to) synchronized clocks at the two ends of the spatial interval that is traversed; thus, it is a circular argument to use conservation of momentum to define simultaneity.

Source:
https://plato.stanford.edu/entries/spacetime-convensimul/

 

[vanhees71]

I think that there is a good point that @Tendex is making. There are assumptions that we make that are purely conventional and are not subject to empirical testing. That includes all definitions of terms and many mathematical conventions, like using the right hand rule, and other conventions like positive charge for protons or units. None of those assumptions are subject to testing.

There are other assumptions that we make which are empirically testable, like isotropy of the two-way speed of light, or conservation of momentum. So we need to test these testable assumptions and we need to distinguish when an assumption is testable or not so that we don't waste time trying to test untestable assumptions and so that we don't mentally elevate our convention assumptions to the status of facts about nature.

That's true. You always need both theory and operational definitions that relate the theoretical (mathematical) elements to the phenomena you measure, i.e., make quantifiable and describable by abstract mathematical models (e.g., to use the real numbers to measure the distance between points in Euclidean geometry, which is a pretty modern finding by Hilbert).

Nevertheless even conventions like the definition of units are in principle subject to experimental test, i.e., when there accumulates evidence of inconsistencies between measurement results of a certain phenomenon using theory to define these units, it may be that the theory is not (precisely) correct. E.g., if the fine structure constant is not really a constant our definition of the Ampere (or Coulomb) in the SI is not consistent, and indeed a possible time dependence of the fine structure constant is indeed considered (with no hint so far that this may really be the case).

 

[vanhees71]

I think that is only the case, as long you also use the Einstein definition of simultaneity:

Source:
https://plato.stanford.edu/entries/spacetime-convensimul/

Momentum conservation is a direct consequence of the assumption of homogeneity of space for any inertial observer. As such it does not depend on any choice of coordinates, because everything observable is independent of the choice of coordinates by construction.

 

[Sagittarius A-Star]

Momentum conservation is a direct consequence of the assumption of homogeneity of space for any inertial observer. As such it does not depend on any choice of coordinates, because everything observable is independent of the choice of coordinates by construction.

Is this also true in an anisotropic inertial frame?

Salmon argued that momentum conservation in its standard form assumes isotropic one-way speed of moving bodies from the outset.
...
In addition, Iyer and Prabhu distinguished between "isotropic inertial frames" with standard synchrony and "anisotropic inertial frames" with non-standard synchrony.[25]

Source:
https://en.wikipedia.org/wiki/One-way_speed_of_light#Inertial_frames_and_dynamics

 

[vanhees71]

What do you mean by "anisotropic inertial frame"? If the space of an inertial observer is not isotropic you change the standard space-time model itself. This cannot be done by simply choosing some coordinates in Minkowski space. All geometrical properties are independent of the choice of coordinates.

It's as in Euclidean space: Only because you use spherical coordinates, implying to choose an arbitrary point as the origin and a direction as the polar axis, you don't destroy isotropy and homogeneity of Euclidean (affine) space.

 

[Sagittarius A-Star]

What do you mean by "anisotropic inertial frame"?

I mean the primed frame in:
$$x' = x \ \ \ \ \ y' = y \ \ \ \ \ z' = z \ \ \ \ \ t' = t + \frac{kx}{c}$$
Source:
https://www.mathpages.com/home/kmath229/kmath229.htm

 

[Tendex]

I mean the primed frame in:
$$x' = x \ \ \ \ \ y' = y \ \ \ \ \ z' = z \ \ \ \ \ t' = t + \frac{kx}{c}$$
Source:
https://www.mathpages.com/home/kmath229/kmath229.htm

That's using a synchronization convention different than Einstein's that makes you use non-inertial coordinates that are more contrived. Doesn't affect what vanhees said.

 

[Sagittarius A-Star]

That's using a synchronization convention different than Einstein's that makes you use non-inertial coordinates that are more contrived. Doesn't affect what vanhees said.

That are inertial coordinates. Reason: This frame is not accelerated. Therefore, it does not contain fictitious forces.

 

[PeterDonis]

I mean the primed frame in:
$$x' = x \ \ \ \ \ y' = y \ \ \ \ \ z' = z \ \ \ \ \ t' = t + \frac{kx}{c}$$
Source:
https://www.mathpages.com/home/kmath229/kmath229.htm

This is not an inertial frame. The article you reference does not say that it is.

 

[PeterDonis]

That are inertial coordinates. Reason: This frame is not accelerated. Therefore, it does not contain fictitious forces.

"Not accelerated" is a necessary condition for "no fictitious forces", but not a sufficient one. Try describing the motion of a free body (i.e., one that would have zero coordinate acceleration in a standard isotropic inertial frame) in your anisotropic frame. What is its coordinate acceleration?

 

[Dale]

Yes. But consider a coordinate chart, that creates an anisotropy of the one-way light speed in (+/-) x-direction and describe in this coordinate chart an explosion. The momentum will not be conserved.

I am not sure that is correct. I don't see why momentum would not be conserved. I could possibly see angular momentum not being conserved, but even that I would want a derivation to show.

 

[Sagittarius A-Star]

This is not an inertial frame. The article you reference does not say that it is.

It does say this for system x,y,z,t:

Given any inertial coordinate system x,y,z,t, we are free to apply a coordinate transformation of the form

The coordinate transformation does not change this. It changes constant velocity components in x-direction to a different constant velocity.

 

[PeterDonis]

The coordinate transformation does not change this.

The fact that you are free to apply any coordinate transformation does not mean that any coordinate transformation you apply will result in an inertial frame.

 

[Sagittarius A-Star]

The fact that you are free to apply any coordinate transformation does not mean that any coordinate transformation you apply will result in an inertial frame.

In this case it does. See my example calculation for one-way sound speed:
https://www.physicsforums.com/threads/measuring-the-one-way-speed-of-light.995539/post-6413191

 

[PeterDonis]

It changes constant velocity components in x-direction to a different constant velocity.

No, it doesn't. Check your math.

 

[PeterDonis]

In this case it does. See my example calculation for one-way sound speed:
https://www.physicsforums.com/threads/measuring-the-one-way-speed-of-light.995539/post-6413191

That calculation has nothing whatever to do with your claims that your anisotropic frame is "inertial".

 

[Sagittarius A-Star]

No, it doesn't. Check your math.

It does because $t'$ depends linear on $x$.

 

[vanhees71]

This is not an inertial frame. The article you reference does not say that it is.

But if you'd use the correct transformation from one set of coordinates to another the speed of light cannot change just by construction. A light-like vector just stays a light-like vector no matter which (holonomous) coordinate basis you use to define its components.

 

[Sagittarius A-Star]

I don't see why momentum would not be conserved.

For that reason:

Salmon argued that momentum conservation in its standard form assumes isotropic one-way speed of moving bodies from the outset.

Source:
https://en.wikipedia.org/wiki/One-way_speed_of_light#Inertial_frames_and_dynamics

 

[Dale]

For that reason:

The key there is "in its standard form". Per Noether's theorem a homogenous but anisotropic speed of light should be compatible with conservation of momentum. But I can easily believe that it would not be "in its standard form".

 

[Sagittarius A-Star]

The key there is "in its standard form". Per Noether's theorem a homogenous but anisotropic speed of light should be compatible with conservation of momentum. But I can easily believe that it would not be "in its standard form".

A different definition of simultaneity than that of Einstein has the effect, that the symmetry of nature is not reflected in the math. You must then replace Minkowsi spacetime by something elso to describe the same physics. That may become more complicated, including non-conservation of momentum "in its standard form" (in the complicated mathematical model).

 

[PeterDonis]

You must then replace Minkowsi spacetime by something elso to describe the same physics.

No. Minkowski spacetime is an invariant geometric object. It is the physics. You can't change it to something else by changing coordinates.

 

[Sagittarius A-Star]

No. Minkowski spacetime is an invariant geometric object. It is the physics. You can't change it to something else by changing coordinates.

But in the mathematial model of Minkowski spacetime, the one-way-speed of light is isotropic, which is only a definition.

 

[DrGreg]

I mean the primed frame in:
$$x' = x \ \ \ \ \ y' = y \ \ \ \ \ z' = z \ \ \ \ \ t' = t + \frac{kx}{c}$$
Source:
https://www.mathpages.com/home/kmath229/kmath229.htm

In this coordinate system you wouldn't define momentum as
$$ \frac{m \textbf{v}'} {\sqrt{1 - |\textbf{v}'|^2 / c^2}} $$
(where $\textbf{v}' = \rm{d}\textbf{x}' / \rm{d}t'$) because that isn't conserved.

You'd define it as
$$ m \frac{ \rm{d} \textbf{x}'} {\rm{d} \tau}$$
(where $\tau$ is proper time).

 

[PeterDonis]

in the mathematial model of Minkowski spacetime

There is no single mathematical model of Minkowski spacetime, the geometric object. There are an infinite number of possible coordinate charts you can use to describe Minkowski spacetime. None of them change its geometry. Nor are all of them "inertial frames" just because they all describe Minkowski spacetime.

 

[Sagittarius A-Star]

You'd define it as
$$ m \frac{ \rm{d} \textbf{x}'} {\rm{d} \tau}$$
(where $\tau$ is proper time).

I think, then it would be possible to synchonize distant stationary clocks equivalently to an Einstein synchronization without defining, that the one way-speed of light is isotropic. I could shoot from the middle between the clocks 2 equal cannon balls (with built-in clocks) with equal momentum in both directions (by an explosion between them). The stationary clocks are then synchronized to the built-in clocks of the cannon balls, when they are reached.

 

[PeterDonis]

I think, then it would be possible to synchonize distant stationary clocks equivalently to an Einstein synchronization without defining, that the one way-speed of light is isotropic.

As I have already said, you can choose whatever coordinates you want; it won't change any actual physics. The process of actually doing Einstein clock synchronization is a physical process; its results are the same no matter what your choice of coordinates is.

I could shoot from the middle between the clocks 2 equal cannon balls (with built-in clocks) with equal momentum in both directions. The stationary clocks are then synchronized to the built-in clocks of the cannon balls, when they are reached.

Yes, this would just be "Einstein synchronization" with cannon ball clocks instead of light signals.

What I do not see is how any of this has anything to do with your claim that an anisotropic coordinate system is an "inertial frame".

 

[Sagittarius A-Star]

Yes, this would just be "Einstein synchronization" with cannon ball clocks instead of light signals.

But, as I said, without defining, that the one way-speed of light is isotropic, what would be a requirement for an Einstein synchronization with light.

What I do not see is how any of this has anything to do with your claim that an anisotropic coordinate system is an "inertial frame".

I did not claim that it has to do anything with it.

 

[Dale]

That may become more complicated, including non-conservation of momentum "in its standard form" (in the complicated mathematical model)

But I wouldn’t call not “in its standard form” non-conservation. After all, the conservation of momentum in relativity is not “in its standard form” either, but we still say that momentum (in its relativistic form) is conserved.

Edit: actually, now that I think of it this is just a coordinate transform so all covariant laws remain. So the conservation of four-momentum definitely still works.

 

[Sagittarius A-Star]

Edit: actually, now that I think of it this is just a coordinate transform so all covariant laws remain. So the conservation of four-momentum definitely still works.

In SR, "convariant" relates to Lorentz transformation. But doesn't Lorentz transformation rely on Einstein synchronization (one way-speed of light is isotropic)?

 

[Dale]

In SR, "convariant" relates to Lorentz transformation. But doesn't Lorentz transformation rely on Einstein synchronization (one way-speed of light is isotropic)?

No, covariant means that the law holds under any arbitrary coordinate transform. It is not just limited to Lorentz transforms. At least that is how I have always seen the term used.

 

[Sagittarius A-Star]

No, covariant means that the law holds under any arbitrary coordinate transform. It is not just limited to Lorentz transforms. At least that is how I have always seen the term used.

Is then my above statement in posting #86 correct? (Einstein synchonization possible without definition of isotropy of one-way light speed, replaced by assumption of momentum conservation)

Then the following would be wrong:

Salmon (1977, 273) argues, however, that the standard formulation of the law of conservation of momentum makes use of the concept of one-way velocities, which cannot be measured without the use of (something equivalent to) synchronized clocks at the two ends of the spatial interval that is traversed; thus, it is a circular argument to use conservation of momentum to define simultaneity.

Source:
https://plato.stanford.edu/entries/spacetime-convensimul/

 

[Dale]

Is then my above statement in posting #86 correct? (Einstein synchonization possible without definition of isotropy of one-way light speed, replaced by assumption of momentum conservation)

Then the following would be wrong:Source:
https://plato.stanford.edu/entries/spacetime-convensimul/

I think that it is not correct unless you define “the standard formulation” in a very narrow way.

 

[Sagittarius A-Star]

I think that it is not correct unless you define “the standard formulation” in a very narrow way.

Is then my above statement in posting #86 correct?

I think, then it would be possible to synchonize distant stationary clocks equivalently to an Einstein synchronization without defining, that the one way-speed of light is isotropic. I could shoot from the middle between the clocks 2 equal cannon balls (with built-in clocks) with equal momentum in both directions (by an explosion between them). The stationary clocks are then synchronized to the built-in clocks of the cannon balls, when they are reached.

 

[Dale]

Is then my above statement in posting #86 correct?

I doubt it. The one way speed of light is coordinate dependent. The conservation of four momentum is covariant. So I am skeptical that the four momentum can be used to select a particular coordinate system.

 

[bhobba]

Is this also true in an anisotropic inertial frame?

It follows from Noethers Theorem if all points are equivalent as far as the laws of physics go, momentum is conserved. As long as that is the case, then yes. This is one of the issues with moving away from SR as a consequence of the symmetry properties of an inertial frame. It is possible for certain symmetries to fail which we know from Noether causes problems. In fact that is the reason Hilbert gave Noether the problem of non-energy conservation in GR to sort out - this was very troubling. The answer - energy conservation is a consequence of all instants of time being equivalent, and that does not necessarily apply to curved space-time, was of course startling, and one of the greatest discoveries ever of physics - as well as one of the most useful and beautiful.

SR, in inertial frames, actually has nothing to do with light. If follows directly from the symmetries of its definition except for a constant c that must be determined experimentally (of course experiment shows that c is the speed of light - but does not have to be determined by actually measuring the speed of light - one way or otherwise). I often give the following derivation, but for those that have not seen it:
http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

Mathematically during the 19th century it was discovered there is a strong connection between symmetries and geometries (eg the Erlangen program). So it is no surprise they determine SR. In modern times many textbooks ignore Einsteins original musings on SR such as what would happen if you caught up to a beam of light, and just give a derivation like the above. Rindler and Morin do it that way, as well as a discussion of its relation to Einstein's original thinking. Ohanian is the 'odd' man out:
https://www.physicscurriculum.com/specialrelativity

My view is those interested in SR should be aware of both approaches. I prefer Rindler and Morin rather than Ohanian, but that is just a personal preference. The 'beauty' of physics is what attracts me to it. That's probably because my background is math. Others more into experiment likely see it differently.

As has been emphasised here, correctly, science is based on experiment, not aesthetics. Many books have been written on what science is, but I sum it up in one word - doubt. The only 'truth' is experiment - not beauty - even though in the hands of masters like Dirac it can take us a long way.

Thanks
Bill

 

[bhobba]

That calculation has nothing whatever to do with your claims that your anisotropic frame is "inertial".

By definition an inertial frame is isotropic. The issue is do inertial frames actually exist. We know they, strictly speaking, do not. But deep in interstellar space they are very very close - at least as far as we can tell today.

Thanks
Bill

 

[Tendex]

The issue is do inertial frames actually exist. We know they, strictly speaking, do not.

As I said we can certainly check empirically that nature is compatible with inertial frames to a certain order of approximation but this way, by the nature of measurements we can never afirm their strict existence.
I'm not sure if this is what makes you say that we know they actually don't exist strictly. But that is a claim that I've tried to explain that is not only incompatible with SR and all theories derived from it but incompatible with any geometric theory of motion since it would lead to contradiction. So even if we know some of the frames we use as approximately inertial for the purposes needed are non inertial(like earth's) we cannot seriously say that inertial frames don't exist without contradiction, and this is the sense in which they are not empirical but a conventional assumption.

 

[Sagittarius A-Star]

By definition an inertial frame is isotropic.

I think, the following primed frame is isotropic in a physical sense (conservasion of 4-momentum) and anisotropic only in a coordinate sense (non-isotropic one way-speed of light). Therefore, it can be inertial.

Edit: Another argument: The frame (x',y',z',t') is moving with constant velocity $\vec v = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ relative to the inertial frame (x,y,z,t). Therefore, it must be also inertial.

Given any inertial coordinate system x,y,z,t, we are free to apply a coordinate transformation of the form
$$x' = x \ \ \ \ \ y' = y \ \ \ \ \ z' = z \ \ \ \ \ t' = t + \frac{kx}{c}$$

Source:
https://www.mathpages.com/home/kmath229/kmath229.htm

 

[Tendex]

I think, the following primed frame is isotropic in a physical sense (conservasion of 4-momentum) and anisotropic only in a coordinate sense (non-isotropic one way-speed of light). Therefore, it can be inertial.

Edit: Another argument: The frame (x',y',z',t') is moving with constant velocity $\vec v = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ relative to the inertial frame (x,y,z,t). Therefore, it must be also inertial.Source:
https://www.mathpages.com/home/kmath229/kmath229.htm

Sure, changing the simultaneity convention to one more contrived that uses non inertial coordinates doesn't change anything about the physics, i.e. about the general isotropy of light, you are just expressing it in the more contrived coordinates that don't apply the Einstein convention but some other convention anisotropic in the one-way direction.