I Isotropy of the speed of light

[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)?