I Synchronizing clocks in an inertial frame if light is anisotropic

[Dale]

I would agree with that view. There are no non-zero Christoffel symbols in an Anderson frame, hence no fictitious forces. So Newton’s first law is satisfied.

But I would still not consider it to be an inertial frame for that reason. There are Anderson-frame expressions for conserved energy and momentum, but they differ enough that I don’t think the second and third laws are useful in such frames.

 

[Nugatory]

A calorimeter doesn’t measure KE.

If it is at rest in the center of mass frame and we have arranged for a fully inelastic collision to happen inside it?

 

[Ibix]

Been preoccupied with other matters this week, but is it not the case that the Anderson velocity, $\tilde{v}_x=dX/dT$, depends on the synchronization convention? Hence, is it not the case that two particles with what a Lorentz frame would call equal and opposite velocities would not have equal and opposite Anderson velocities? Then we should not expect the energies for particles with equal and opposite Anderson velocities to be equal, because equal and opposite Anderson velocities doesn't mean what our physicist reflexes say equal and opposite velocities ought to mean.

To put it another way, if we take two identical particles with equal and opposite Lorentz frame velocities, not equal and opposite Anderson frame velocities, are their Anderson energies equal?

Or to put it yet another way, I think $\tilde{E}$ is an explicit algebraic expression for the rank-0 tensor $p_au^a$ in terms of vector components. In an anisotropic frame we shouldn't be surprised if the functional form depends on that anisotropy to correct for the anisotropy in the components.

 

[PeterDonis]

is it not the case that the Anderson velocity, $\tilde{v}_x=dX/dT$, depends on the synchronization convention?

I believe it depends on the Anderson parameter $\kappa$, yes. If I've done the math right, a pair of Lorentz frame velocities $v$ and $-v$ (equal and opposite) becomes a pair of Anderson frame velocities $v / \left(1 + \kappa \right)$ and $- v / \left(1 - \kappa \right)$, whose magnitudes are obviously not equal.

 

[Sagittarius A-Star]

I still see no explanation of how different synchronization convention can cause otherwise identical sand beds to result in different crater depth.

The crater depth will not be different.

 

[PAllen]

The crater depth will not be different.

Then my point is proven: KE is a directly measurable local quantity that does not depend on clock synch (you don't even need one clock, let alone two). It depends only only on the relative motion of detector and test body. The formula for it may be synchronization dependent, taking the 'simplest' form only with standard clock synch.

 

[Dale]

KE is a directly measurable local quantity that does not depend on clock synch

I still disagree with this. KE in a specified frame is measurable. Not KE.

 

[PAllen]

I still disagree with this. KE in a specified frame is measurable. Not KE.

What matters is the relative motion of detector and body, not the details of frame construction (in particular, clock synchronization plays no role).

 

[Dale]

Just to clarify - this is the geodesic Lagrangian of a free particle, not the relativistic Lagrangian of a free particle, is that correct?

Sorry it took me so long to respond to this. I had to dig back through my notes more carefully.

Perhaps it's best to ask if the line element for Anderson's coordinates (in geometric units) is:

$$-dT^2 - 2 \kappa\, dX \, dT + (1 - \kappa^2) dX^2 + dY^2 + dZ^2$$

as I can write the geodesic equations from there.

What I have in my notes is indeed $$ds^2=-dT^2-2\kappa \ dT \ dX+(1-\kappa^2) \ dX^2 + dY^2 + dZ^2$$

Just to clarify - this is the geodesic Lagrangian of a free particle, not the relativistic Lagrangian of a free particle, is that correct? As I review the material, it seems that the geodesic Lagrangian corresponds to the square of the proper time (sometimes I've also seen it multiplied by a constant factor of 1/2), while the relativistic Lagrangian is not squared, it is just the proper time.

It looks like my software produced the Lagrangian without a square root (proper time squared), and I manually multiplied by mass and took the square root. So the Lagrangian I used was $$L=m\sqrt{\left( \dot T + (\kappa-1) \dot X \right) \left( \dot T + (\kappa +1) \dot X \right)- \dot Y^2 - \dot Z^2}$$

I've also realized I want another space-like Killing vector, as a different way of getting the momentum conservation law in the new coordinates X directio . I think I might be able to puzzle it out, but if you happen to know both Killing vectors, it'd be helpful.

Using that Lagrangian I get the conserved quantities I wrote above. The conserved energy (conjugate to $T$) is $$E=\frac{m(1+\kappa \ V_X)}{\sqrt{1-V^2+2 \kappa \ V_X + \kappa^2 \ V_X^2}} \approx m + \frac{1}{2} m V^2 - \kappa m \ V_X^3$$

 

[PAllen]

I'll sum up my position:
A given detector can measure KE of a test body (relative to that detector, of course) with no recourse to coordinates, clocks or conventions. Among coordinate systems which place that detector 'at coordinate rest', the formulas for predicting that measurement in terms of coordinate velocity, of course depend on the coordinates. If the coordinates are anisotropic (e.g. the one way speed of light is anisotropic), then big surprise, the KE formula will be anisotropic.

Newtonian physics in 'standard expression' picks isotropic coordinates because its formulas are isotropic.

 

[Dale]

A given detector can measure KE of a test body (relative to that detector, of course)

And my position is that it is wrong to call that quantity just "KE". That should be identified as "KE relative to the detector" or "KE in the frame of the detector". That quantity is an invariant and "KE" is not invariant while "KE in the frame of the detector" is invariant.

 

[PeterDonis]

And my position is that it is wrong to call that quantity just "KE". That should be identified as "KE relative to the detector" or "KE in the frame of the detector". That quantity is an invariant and "KE" is not invariant while "KE in the frame of the detector" is invariant.

Terminology for things like this is problematic, because it invites confusion both ways. "KE relative to the detector" sounds like it's frame-dependent, when in fact it's an invariant; while "KE" without qualification sounds like it should be an invariant, but in fact it's frame-dependent. Unfortunately I don't know of any useful terminology that avoids this. Once you understand relativity well enough, the reason for the terminology makes sense, but it's still confusing.

 

[A.T.]

"KE relative to the detector" sounds like it's frame-dependent, when in fact it's an invariant; while "KE" without qualification sounds like it should be an invariant, but in fact it's frame-dependent.

It sounds backwards, but makes perfect sense. Similarly to this:

Why do we call an inertial frame one that does not have inertial forces but a non-inertial frame is one that does?

 

[pervect]

This means for proper time squared:
$(d\tau)^2 = (dT + \kappa dX)^2 - dX^2 - dY^2 - dZ^2\ \ \ \ \ (1)$

Multiplying equation $(1)$ with $(m/d\tau)^2$ yields the Anderson energy-momentum equation:$$m^2 = (\tilde E + \kappa \tilde p_x)^2 - \tilde {p_x}^2 - \tilde {p_y}^2 - \tilde {p_z}^2\ \ \ \ \ (2)$$Multiplying equation $(1)$ with $(1/dT)^2$ yields the inverse squared Anderson gamma-factor:$$1/\tilde \gamma^2 = ({d\tau \over dT})^2 = (1 + \kappa \tilde v_x)^2 - {\tilde v_x}^2 - {\tilde v_y}^2 - {\tilde v_z}^2\ \ \ \ \ (3)$$Equation $(2)$ is fulfilled by $(4)$ and $(5)$:$$\tilde E = \tilde \gamma m = {m \over \sqrt {(1 + \kappa \tilde v_x)^2 - {\tilde v_x}^2 - {\tilde v_y}^2 - {\tilde v_z}^2}}\ \ \ \ \ = \gamma m- \kappa p_x\ \ \ \ \ (4)$$ $$ \vec {\tilde p} =\tilde \gamma m \vec {\tilde v}\ \ \ \ \ \ ={dT \over d \tau} m {d \over d T} (X, Y, Z)=m {d \over d \tau} (x, y, z)= \gamma m \vec { v}\ \ \ \ \ (5)$$
The one-way kinetic energy $m (\tilde \gamma -1)$ depends significantly on the clock-synchronization scheme. It can't be measured without clock synchronization.
Equations $(5)$ prove, that the relativistic 3-momentum $\vec {\tilde p}$ does not depend on the clock-synchronization scheme.

Source (see sentence before chapter 1.5.3 - containing relativistic mass - and sentence after equation 33):
https://ui.adsabs.harvard.edu/abs/1998PhR...295...93A/abstract

I'd agree that the momentum is some constant times the (1, beta,0,0), where beta is the normalized coordinante velocity. (in this context, it's normalized to the average speed of light for a round trip). This is because for an object moving at some velocity beta,we want x = beta * t by definition.

Thus, I would disagree that that constant is gamma. The constant should be chosen so the length of the 4 velocity is, with your sign convention, 1. This is the expression I gave for the four-velocity in my post, I won't repeat it unless there is interest (and I have the time to respond).

I would disagree that the energy is the 0 component of the 4-velocity, and the momentum is the 1 component. That's true in the standard metric, but the coordinate independent version I'm proposing is that the energy is given by the inner product of the 4-velocity and a killing vector k_e, which has components (1,0,0,0). Because the metric is non diagonal, this is NOT the zeroth component of the 4-velocity as it is in the Minkowskii metric, though it reduces to that in the Minkowskii case.

This is assuming I am correct in my conclusion that (1,0,0,0) and (0,1,0,0) are killing vectors of the metric, but I believe this is correct. The intuitive argument is that the x-axis and t-axis of any reference frame at any velocity is a Killing vector of the Minkowskii metric, meaning any constant velocity is a Killing vector of a flat space-time. Which should have been obvious, but wasn't to me at first.

Also, interestingly enough, I found that all the Christoffel symbols vanished, which was initially surprising to me but probably also shouldn't have been. But I digress.

Similarly, I would say that the momentum is given by the inner product of the 4-velocity (appropriately normalized by the factor that replaces gamma), and the killing vector (0,1,0,0). Which is again different from the 1 component of the 4-velocity because of the nondiagonal metric.

We can and probably will want to add constants to the values we compute for energy and momentum. Following Newtonian conventions, we can make the energy and momentum of a particle "at rest" equal to zero.

I've thought about this a bit off and on, but haven't taken the time to really run through and re-check my calculations. But I'm thinking maybe things aren maybe not so awful for these coordinates if we normalize energy and momentum to zero for a stationary particle. This is a bit unusual, and bad for relativistic computations per the usual argument of why we keep the rest energy as part of the total energy in relativistic physics, but not necessarily bad for someone wanting to use these coordinates for Newtonian physics.

 

[Dale]

Also, interestingly enough, I found that all the Christoffel symbols vanished, which was initially surprising to me but probably also shouldn't have been.

It was surprising to me also. I assumed that there would be fictitious forces that would come into play when making a U turn, for example.

 

[pervect]

But that directly measured local quantity is the only thing that can sensibly be called KE. So perhaps you should say the Newtonian formula only equals KE with conventional synchronization. That would support @pervect ’s point of view that Newtonian physics (e.g. the Newtonian KE formula matching measured KE) selects for isotropic synchronization.

I like the definition of measuring the energy with a calorimeter - or by the ability to penetrate an armor plate if one has a militaristic bent. Momentum measuring is less common, but one could measure the velocity of a large stationary sandbag after an inelastic collision to find the momentum of a bullet. For an isotrorpic clock synchronization, the measured kinetic energy (or momentum) should not depend on the direction of motion. For a non-isotorpic clock synchronization, I expect it will. Exact errors and dependency of these measurements are still TBD, I think. Mostly I'd have to be a LOT more careful before I wanted to give figures to someone who wasn't in a position to check my work due to a difference in background - I'm way too rusty to dash of a quick accurate analysis.