I Synchronizing clocks in an inertial frame if light is anisotropic

[Sagittarius A-Star]

I don’t think this is correct.

But I think it is correct: The reason is, that the term $m ({dx \over d\tau}, {dy \over d\tau}, {dz \over d\tau})$ does not contain the coordinate-time and the momentum is conserved.

Additional argument:
The relativistic 3-momentum $\mathbf v = ({dx \over d\tau}, {dy \over d\tau}, {dz \over d\tau})$ is unchanged in Anderson-coordinates because of
$\tilde {\gamma} \tilde {\mathbf v} = \gamma \mathbf v$.

Source (Wikipedia):

https://en.wikipedia.org/wiki/One-w...ansformations_with_anisotropic_one-way_speeds

 

[James Hasty]

The purpose of my post was to demonstrate a way to synchronize two clocks located in the same inertial frame. I thought this would be simple enough. But what you have taught me is that IF light is anisotropic, some kind of convention must be established for the speed of light in one direction vs. another. It doesn't matter what the convention is, for example: "infinite" speed one-way and "c/2" in the other. But the physics is so much simpler for a constant "c" in both directions.
After having carefully studied this. I have concluded : it is not possible to synchronize two clocks, even if both are in the same inertial frame and stationary with respect to each other, without defining a convention for the 1WSOL in each direction.
I have attached my study which includes a spacetime diagram and my conclusion. It is just 2 pages.

 

[Dale]

The purpose of my post was to demonstrate a way to synchronize two clocks located in the same inertial frame. I thought this would be simple enough. But what you have taught me is that IF light is anisotropic, some kind of convention must be established for the speed of light in one direction vs. another. It doesn't matter what the convention is, for example: "infinite" speed one-way and "c/2" in the other. But the physics is so much simpler for a constant "c" in both directions.
After having carefully studied this. I have concluded : it is not possible to synchronize two clocks, even if both are in the same inertial frame and stationary with respect to each other, without defining a convention for the 1WSOL in each direction.
I have attached my study which includes a spacetime diagram and my conclusion. It is just 2 pages.

That is good work. It is much more difficult to do than it appears at first glance.

 

[cianfa72]

That is good work. It is much more difficult to do than it appears at first glance.

Yes, as far as I can tell the key point is that basically, before the exchange of light signals between observer A and B takes place, it isn't known in advance where the event labeled as 0 by the local clock is located along each of the two observer's worldline.

 

[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.

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 $$ \frac {L} {2L/c} = c/2$$

 

[James Hasty]

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 agree this works, but only because you have defined the convention for the 1-way speed of light in both directions.

 

[cianfa72]

I agree this works, but only because you have defined the convention for the 1-way speed of light in both directions.

Yes exactly. For instance this convention does prescribe to reset to 0 the B's clock at event B2 (i.e. label it as T=0 along B's worldline) when B receives back the light signal emitted at event A1 that encodes T=0 within it. Thus this procedure/convention defines OWSOL as infinite in A -> B direction.

 

[pervect]

Indeed, if I am reading my notebook correctly the Lagrangian for a free particle in an Anderson synchronized frame is $$L=-(\dot T+(\kappa-1)\dot X)(\dot T +(\kappa+1)\dot X)+\dot Y^2 + \dot Z^2$$ This leads to a conserved energy $$E=\frac{m(1+V_X \kappa)}{\sqrt{1-V_Y^2-V_Z^2+2V_X\kappa+V_X^2(\kappa^2-1)}}\approx m+\frac{1}{2}m V^2-m\kappa V^3$$ and a conserved momentum $$p=\left( \begin{array}{c}
\frac{m(\kappa+V_X(\kappa^2-1))}{\sqrt{1-V_Y^2-V_Z^2+2 V_X\kappa + V_X^2(\kappa^2-1)}}\\
\frac{m V_Y}{\sqrt{-V_Y^2-V_Z^2+(1+V_X(\kappa-1))(1+V_X(1+\kappa))}}\\
\frac{m V_Z}{\sqrt{-V_Y^2-V_Z^2+(1+V_X(\kappa-1))(1+V_X(1+\kappa))}}
\end{array} \right)\approx -m \kappa + m V$$

Notice that for energy it is not a huge effect, coming in at 3rd order in V. But for momentum it is a 0th order effect in V. Indeed, I think that would make it fair to say that Newton's laws don't work in that frame even though technically you could modify stuff so that some version of it would technically work.

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. 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. Would you happen to know the timelike Killing vector associated with Anderson's metric?

I'm not sure why quoting your post doesn't format the Latex - sorry for the messy quote.

add:

I'm thinking that we just rewrite this as

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

making the underlying time symmetry obvious and setting the Killing vector to $dT + \kappa dX$, unless I've made a sign error (or other error).

More thoughts:

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.

 

[pervect]

Well, it looks like I was overcomplicating the analysis. Unless I've made an error (quite possible), [1,0,0,0] and [0,1,0,0] are both killing vectors. Or if you prefer $\partial/\partial T$ and $\partial / \partial X$.

And all the Christofel symbols vanished, too. I wasn't expecting that, but it makes sense that the equations of motion make the second derivative of X,Y, and Z with respect to proper time for force-free motion vanish.

A series expansion in terms of $\beta$ of the inner product of the metric of the two Killing vectors with a normalized 4-velocity gives an energy/unit mass of $1 + \beta^2 - \kappa \beta^3$ + higher order terms, while the momentum is $-\kappa + \beta -1.5 \kappa \beta^2$+higher order terms in $\beta$

The normalized four velocity was:
$$\frac{1}{ \sqrt{1 + 2\beta\kappa - \beta^2(1-\kappa^2) } } [1, \, \beta, \, 0, \, 0]$$

The full expressions for the X component of a conserved momentum and energy were
$$E_x = \frac{1 + \beta \kappa}{ \sqrt{1 + 2\beta\kappa - \beta^2(1-\kappa^2)}} \quad M_x=\frac{\beta - \kappa - \beta \kappa^2} { \sqrt{1 + 2\beta\kappa - \beta^2(1-\kappa^2) } }$$

The choice of Killing vectors has some degree of freedom up to linear combinations so these results aren't unique. Also, we're free to add a constant to either term, so I think we could get rid of the term that makes the momentum for a stationary particle with $\beta=0$ nonzero.

 

[Sagittarius A-Star]

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

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

 

[Sagittarius A-Star]

The normalized four velocity was:
$$\frac{1}{ \sqrt{1 + 2\beta\kappa - \beta^2(1-\kappa^2) } } [1, \, \beta, \, 0, \, 0]$$

I think this must be multiplied with $m$ to get the four-momentum.

 

[PAllen]

The one-way kinetic energy $m (\tilde \gamma -1)$ depends significantly on the clock-synchronization scheme. It can't be measured without clock synchronization.

Really? How does a calorimeter depend on clock synchronization? Or, simpler, if crude, crater depth in 'standard sandy ground' ?

 

[Sagittarius A-Star]

Really? How does a calorimeter depend on clock synchronization? Or, simpler, if crude, crater depth in 'standard sandy ground' ?

A change in kinetic energy is $\int \vec F \cdot d\vec s$. A force to decelerate the moving body by mechanical contact is in the end an electric force on a moving charge. But the time-component of the 4-current (charge density) depends on the clock synchronization, therefore also the electric force does.

Another aspect may be how the force transforms from the moving frame to the rest-frame under a generalalized Lorentz transformation.

Source (chapter 2.3.3):
https://ui.adsabs.harvard.edu/abs/1998PhR...295...93A/abstract

 

[Dale]

Really? How does a calorimeter depend on clock synchronization? Or, simpler, if crude, crater depth in 'standard sandy ground' ?

A calorimeter doesn’t measure KE.

 

[PAllen]

A calorimeter doesn’t measure KE.

Sure it does. And if particles with identical motion are measured with 2 different calorimeters moving relative to each other, they will measure different KE. Calorimeters are the main way particle accelerator experiments measure KE in the lab frame.

 

[Dale]

Sure it does

No, KE is frame dependent and the measurement of a calorimeter is frame independent.

measure KE in the lab frame

That is a different quantity. It requires a specification of the reference frame. That includes the simultaneity convention.

 

[PAllen]

A change in kinetic energy is $\int \vec F \cdot d\vec s$. A force to decelerate the moving body by mechanical contact is in the end an electric force on a moving charge. But the time-component of the 4-current (charge density) depends on the clock synchronization, therefore also the electric force does.

Another aspect may be how the force transforms from the moving frame to the rest-frame under a generalalized Lorentz transformation.

Source (chapter 2.3.3):
https://ui.adsabs.harvard.edu/abs/1998PhR...295...93A/abstract

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

 

[PAllen]

No, KE is frame dependent and the measurement of a calorimeter is frame independent.

That is a different quantity. It requires a specification of the reference frame. That includes the simultaneity convention.

KE is dependent on relative velocity between detecter and particle. However, it not dependent on synchronization convention. It is, in fact, the inner product of the detector’s 4 velocity and the particles 4 momentum. This is a local invariant.

 

[Dale]

It is, in fact, the inner product of the detector’s 4 velocity and the particles 4 momentum. This is a local invariant.

That is indeed an invariant, but it is not equal to $\frac{1}{2}mv^2$ in a frame with an anisotropic speed of light.

 

[PAllen]

That is indeed an invariant, but it is not equal to $\frac{1}{2}mv^2$ in a frame with an anisotropic speed of light.

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.

 

[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.