Synchronizing clocks in an inertial frame if light is anisotropic

[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.

 

[cianfa72]

Just to clarify the point, I drew the following diagram. The base grid $(x,t)$ represents of course the standard inertial coordinate system, in gray the relevant light cones.

View attachment 366014

At event B1, observer B sends a light signal to observer A. A receives it at A1, resets its own clock to read T=0 and sends back the signal to B encoding T=0 within it. B, upon receiving it, resets its own clock to read T=0 as well. From now on their clocks are synchronized and basically define the coordinate time T (uppercase T).

Using this procedure, since observer A and B are rest each other -- as they can ascertain by using round-trip travel time as measured by their own clocks -- the 2WSOL is $c$ (since the light round-trip-time is $2L/c$ for each of them), however what is the OWSOL ?

In A -> B direction OWSOL is infinite while in B -> A direction is $c/2$. Indeed in the latter case B sent at B2 a light signal when the coordinate time was T=0 which reaches observer A when its (now synchronized) clock reads the coordinate time $T = 2L/c$ (event A2). By dividing we get $$ \frac {L} {2L/c} = c/2$$

I was thinking about this again. The synchronization procedure/convention described above assigns the same coordinate time "label" to events that are light-like separated (i.e. connected by light-like geodesics).

Does exist actually a requirement/restriction for a valid synchronization convention that events declared to have the same coordinate time T (i.e. simultaneous) must be spacelike separated ?

 

[Dale]

That is a matter of definitions. Some authors make that requirement, other authors do not.