I Isotropy of the speed of light

[etotheipi]

Okay, well I don't understand. To me it just looks like you wrote the time transformation. But it's quite possible I don't know enough.

 

[Sagittarius A-Star]

Okay, well I don't understand. To me it just looks like you wrote the time transformation. But it's quite possible I don't know enough.

I wrote the time transformation, yes, but I also concluded from the independence of the x'-coordinate: "There is no pseudo-gravitational time-dilation". And without a gradiend of a pseudo-gravitational potential, there can't be an inertial force.

@vanhees71
I think, your formulation in #233 "non-inertial reference frame" could be interpreted as:
A1) non-accelerated frame(with non-Einstein-synchronization) without test particle -> no inertial force
A2) non-accelerated frame (with non-Einstein-synchronization) with test particle -> no inertial force
B1) accelerated frame without test particle -> no inertial force
B2) accelerated frame with test particle -> inertial force.

So, as I said, it could imply the existence of fictuous forces (case B2), but which cannot appear in the discussed frame (cases A1, A2).

 

[PeterDonis]

I think, your formulation in #233 "non-inertial reference frame" could be interpreted as:
A1) non-accelerated frame(with non-Einstein-synchronization) without test particle -> no inertial force
A2) non-accelerated frame (with non-Einstein-synchronization) with test particle -> no inertial force
B1) accelerated frame without test particle -> no inertial force
B2) accelerated frame with test particle -> inertial force.

What do you mean by "without test particle" vs. "with test particle"? The definition of a frame does not have anything to do with test particles.

 

[vanhees71]

The equation of motion for a free particle in flat Minkowski space time of course always leads to a straight world line, i.e., uniform motion as it must be. That's of coarse also true in Newtonian mechanics. It's nothing else than the 1st Newtonian postulate.

As in Newtonian mechanics, it's however common slang for centuries that there are "inertial forces in a non-inertial reference frame". That's of course a bit misleading in the sense that there are not really "forces" but just "covariant derivatives of vector components" with respect to (in SR proper in Newtonian mechanics absolute) time.

It's also easy to derive from the action principle (here in the most simple "square form"). To that end one uses
$$L=-\frac{1}{2} \dot{x}^{\mu} \dot{x}_{\mu}=-\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu},$$
where the $x^{\mu}$ are arbitrary space-time coordinates and $g_{\mu \nu}$ the Minkowski-metric components wrt. the corresponding holonomic basis. The dot means derivative with respect to an arbitrary world-line parameter, $\lambda$, which is automatically an affine parameter, because $H=L=\text{const}$, because $L$ doesn't depend on $\lambda$ explicitly.

The equations of motion (leading to nothing else than the said straight lines in Minkowksi space) are
$$\frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial L}{\partial \dot{x}^{\mu}}-\frac{\partial L}{\partial x^{\mu}}=0.$$
This gives
$$\frac{\mathrm{d}}{\mathrm{d} \lambda} (g_{\mu \nu} \dot{x}^{\nu})-\frac{1}{2} \frac{\partial g_{\alpha \beta}}{\partial x^{\mu}} \dot{u}^{\alpha} \dot{u}^{\beta}=0.$$
This is nothing else than
$$\ddot{x}^{\mu}=-{\Gamma^{\mu}}_{\alpha beta} \dot{x}^{\alpha} \dot{x}^{\beta} \qquad (*)$$
with the standard Christoffel symbols
$${\Gamma^{\mu}}_{\alpha \beta}=\frac{1}{2} g^{\mu \gamma} (\partial_{\alpha} g_{\gamma \beta} + \partial_{\beta} g_{\alpha \gamma}-\partial_{\gamma} g_{\alpha \beta}).$$
Then you can say the right-hand side of (*) are the "fictitious/inertial forces" as in the analogous case in Newtonian mechanics.

That's of coarse just semantics and doesn't contain too much physics ;-).

 

[Sagittarius A-Star]

What do you mean by "without test particle" vs. "with test particle"? The definition of a frame does not have anything to do with test particles.

I added this distinction only because of etotheipi's concern in posting #239.

 

[PeterDonis]

I added this distinction only because of etotheipi's concern in posting #139.

Ah, I see. That concern was about the distinction between the metric and the geodesic equation; the latter is where "inertial forces" (which is really a misnomer) appear. But neither of those things are a frame, and defining a frame automatically gets you both an expression for the metric and an expression for the geodesic equation. There is no such thing as a "frame without test particle" where you only have the metric and no geodesic equation.

 

[Sagittarius A-Star]

Ah, I see. That concern was about the distinction between the metric and the geodesic equation; the latter is where "inertial forces" (which is really a misnomer) appear.

To my understanding, the geodesic equation follows automatically from the metric, because you can derive the geodesic equation by applying the principle of maximum proper time (which is compatible to the action principle) to a free test particle.

 

[PeterDonis]

To my understanding, the geodesic equation follows automatically from the metric

Yes, it does. But the two are still conceptually distinct: the metric describes the geometry of spacetime, the geodesic equation describes the curves that freely moving objects (objects that feel no force) follow.

 

[Dale]

The dot means derivative with respect to an arbitrary world-line parameter, λ, which is automatically an affine parameter, because H=L=const, because L doesn't depend on λ explicitly.

By the way, as an aside. Do you know of a particularly easy or convenient way to make sure that $\lambda$ is not just any affine parameter but is specifically the proper time (specifically for a timelike worldline). I know that I can take $\lambda$ and go back to the metric to convert it to proper time, but can I do anything to the Lagrangian or the Euler equations to make it happen automatically?

 

[PeterDonis]

Do you know of a particularly easy or convenient way to make sure that $\lambda$ is not just any affine parameter but is specifically the proper time (specifically for a timelike worldline)

Any affine parameter along a timelike worldline is proper time. Converting from one affine parameter to another is just rescaling the unit of time and/or picking a new "zero point". There is no one particular one that is "the" proper time; they all are.

 

[Dale]

Any affine parameter along a timelike worldline is proper time. Converting from one affine parameter to another is just rescaling the unit of time and/or picking a new "zero point". There is no one particular one that is "the" proper time; they all are.

Sorry, I should have been clear. I want the one that is the unit scaling in units where c=1. I.e. so that the affine parameter has the same units as the units of space.

 

[PeterDonis]

Sorry, I should have been clear. I want the one that is the unit scaling in units where c=1. I.e. so that the affine parameter has the same units as the units of space.

Isn't this going to be true automatically? I'm having trouble imagining how you would get into a situation where it wasn't true.

 

[vanhees71]

It's proper time, if you set the said conserved "Hamilton function" $H=L=-c^2/2$. The great advantage of this "square form" of the point-particle Hamiltonian is that it also applies to "massless particles", i.e., also to $H=L=0$. Then the affine parameter is arbitrary, but it then has no physical meaning anyway. It's just a parameter.

 

[vanhees71]

Sorry, I should have been clear. I want the one that is the unit scaling in units where c=1. I.e. so that the affine parameter has the same units as the units of space.

Then of course you have to set $H=-1/2$.

 

[DAH]

Have you carefully worked it out in detail? Draw up the two scenarios: incident beam comes in normal to grating and produces both a +1 and -1 diffraction max. The angular location of each will be the same even if the +x speed differs from the -x speed. This will be true because the overall diagonal speed of the diffracted beams +/-1 will also differ. This is a generic result as @Dale says

Check

That's an interesting video thanks.
In the video he say's that light from distant objects such as stars or galaxies could reach us instantaneously and we wouldn't know it. Surely that's not true, is it?

 

[Sagittarius A-Star]

That's an interesting video thanks.
In the video he say's that light from distant objects such as stars or galaxies could reach us instantaneously and we wouldn't know it. Surely that's not true, is it?

Of course that's true. But it is only formally the case. You can define the time-coordinate of your coordinate system in such a way, that the start time is formally equal to the arrival time. That's like if you drive for 1 hour and cross the border of a time zone. Then you can start at 10h00 and arrive at 10h00 of the other time zone.

 

[PeterDonis]

You can define the time-coordinate of your coordinate system in such a way, that the start time is formally equal to the arrival time.

Strictly speaking, you can't push things to this limit, because in this limit, surfaces of constant "time" are not spacelike, they're null. But that means there is no valid notion of "spatial distance" in these coordinates, which means there is also no valid notion of "speed".

I can't tell whether the person in the video is aware of this or not.

That's like if you drive for 1 hour and cross the border of a time zone. Then you can start at 10h00 and arrive at 10h00 of the other time zone.

No. The two time zones are two different coordinate systems with different time coordinates. They are not two parts of a single coordinate system. If you want a single coordinate system with a single time coordinate for the entire Earth, you need to look at something like UTC.

 

[Sagittarius A-Star]

Strictly speaking, you can't push things to this limit, because in this limit, surfaces of constant "time" are not spacelike, they're null. But that means there is no valid notion of "spatial distance" in these coordinates, which means there is also no valid notion of "speed".

"Spatial distance" is not changed by this coordinate system, because $x' = x$. Only light speed is infinite in this direction. Sound speed for example is then still finite.

 

[DAH]

Of course that's true. But it is only formally the case. You can define the time-coordinate of your coordinate system in such a way, that the start time is formally equal to the arrival time. That's like if you drive for 1 hour and cross the border of a time zone. Then you can start at 10h00 and arrive at 10h00 of the other time zone.

Yes, but obviously we know that the light from distant objects takes a certain amount of time to reach us. I was reading a paper the other day (don't have a link) about how scientists observed multiple images of a type 1a supernova caused by gravitational lensing, and there was a time delay between each image. They actually predicted when one of the images would appear using GR and the one way speed of light as c, and their calculations were correct. If light from the SN was instantaneous then those multiple images would have appeared at the same time.

 

[Sagittarius A-Star]

Yes, but obviously we know that the light from distant objects takes a certain amount of time to reach us. I was reading a paper the other day (don't have a link) about how scientists observed multiple images of a type 1a supernova caused by gravitational lensing, and there was a time delay between each image. They actually predicted when one of the images would appear using GR and the one way speed of light as c, and their calculations were correct. If light from the SN was instantaneous then those multiple images would have appeared at the same time.

The coordinate system I linked is for SR, not GR.

 

[PeterDonis]

"Spatial distance" is not changed by this coordinate system, because $x' = x$.

Yes, it is, because, as I said, the surfaces of constant time are null, not spacelike. In other words, the spatial metric changes because of the redefinition of the time coordinate, even though the spatial coordinates do not change.

One of the reasons videos like this one are not good sources is that they include no math. Actually doing the math for this coordinate system would force one to confront these issues.

Only light speed is infinite in this direction. Sound speed for example is then still finite.

The derivation you cite in that previous post is not valid for the limit case where $c$ becomes infinite in one direction; the equation for $c' / c$ is undefined for $\theta = 0$ (divide by zero error). So, as I said, this limit is not valid and so are any claims made using it.

 

[Sagittarius A-Star]

The derivation you cite in that previous post is not valid for the limit case where $c$ becomes infinite in one direction; the equation for $c' / c$ is undefined for $\theta = 0$ (divide by zero error). So, as I said, this limit is not valid and so are any claims made using it.

At mathpages, they allow also the "instantaneously" case at segments for calculating the round-trip-speed of light (k= +/- 1). They add traverse times (partly equal to zero) and avoid so deviding by zero:

Furthermore, this dependence is sufficient to ensure that the average speed of light over any closed loop path is precisely c (for any constant value of k in the range from −1 to +1).

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

 

[PeterDonis]

At mathpages, they allow also the "instantaneously" case at segments for calculating the round-trip-speed of light (k= +/- 1). They add traverse times (partly equal to zero) and avoid so deviding by zero:

No, the sums still have terms in them with zero in the denominator, so they're still not valid.

The fundamental issue, as I've already said, is that for the case $k = \pm 1$, surfaces of constant time are null, not spacelike. The issue is not that such a coordinate chart cannot be constructed; of course it can. The issue is whether such a chart justifies claims about "speed". It doesn't, for the reasons I have already given.

 

[PeterDonis]

for the case $k = \pm 1$, surfaces of constant time are null, not spacelike.

More precisely, they are null in whatever direction we choose as the "axis" of the anisotropy. For the coordinate transformation given in the mathpages article, this is the $x$ direction. The metric in the transformed chart is easily found to be (I use capital letters instead of primes for the transformed chart):

$$
ds^2 = - dT^2 + 2 k dT dX + \left( 1 - k^2 \right) dX^2 + dY^2 + dZ^2
$$

This makes it obvious that, for $k = \pm 1$, any line along the $X$ axis in a surface of constant $T$, i.e., with $dT = dY = dZ = 0$, is a null line. Therefore, the concepts of "distance" and "speed" make no sense along this line.

 

[cianfa72]

For your question, you take a system of moving objects, each with their own clock, radar, and accelerometers (6 degree of freedom type). You measure the object’s proper time, proper acceleration, and relative distance and speed (radar) to each of the other objects. Then you solve the resulting system of equations to determine if there exists an inertial frame that can describe the object’s motion. It may very well turn out that there is no solution to that system of equations. So the existence of such a solution is not a mere convention but a physical result.

With reference to this post, I take it as follows: take a system of N moving objects and starting from their proper time (measured by its own attached clock), proper acceleration and relative radar speed of all other N-1 objects (i.e. $N(N-1)$ total relative radar speeds) write down the resulting system of equations involving coordinate "labels" of an hypotetical inertial reference frame (i.e. assume as metric in those coordinates the standard diagonal SR metric in inertial coordinates). If that system of equations has or has not a solution is a physical result.

We know that if all N objects are freely moving (i.e. attached accelerometers measure zero proper acceleration) then it turns out that in a narrow region of spacetime a solution of the above system actually does exist.