Einstein's Clock Synchronization Convention

[Aether]

Just a friendly reminder, the first postulate of the special theory of relativity, namely that the speed of light c is the same in all inertial frames, only holds true in view of Einstein's clock synchronization convention. There is no experimental basis whatsoever for preferring this convention over absolute clock synchronization. "Thus the much debated question concerning the empirical equivalence of special relativity and an ether theory taking into account time dilation and length contraction but maintaining absolute simultaneity can be answered affirmatively." -- R. Mansouri & R.U. Sexl, A Test Theory of Special Relativity: I. Simultaneity and Clock Synchronization, General Relativity and Gravitation, Vol. 8, No. 7 (1977), pp. 497-513.

This paper by Mansouri & Sexl is the first of a series of three papers, the other two papers are:

R. Mansouri & R.U. Sexl, A Test Theory of Special Relativity: II. First Order Tests, General Relativity and Gravitation, Vol. 8, No. 7 (1977), pp. 515-524.

R. Mansouri & R.U. Sexl, A Test Theory of Special Relativity: III. Second Order Tests, General Relativity and Gravitation, Vol. 8, No. 10 (1977), pp. 809-814.

This series of papers by Mansouri & Sexl is referenced by most, if not all, of the subsequently published experimental tests of Local Lorentz Invariance (LLI).

 

[JesseM]

Just a friendly reminder, the first postulate of the special theory of relativity, namely that the speed of light c is the same in all inertial frames, only holds true in view of Einstein's clock synchronization convention. There is no experimental basis whatsoever for preferring this convention over absolute clock synchronization.

The experimental aspect is that there's no experiment you can do that will lead to one observer's definition of synchronization being preferred over another, because all the most fundamental laws of physics favored by experiment have the property of Lorentz-invariance.

 

[pervect]

Just a friendly reminder, the first postulate of the special theory of relativity, namely that the speed of light c is the same in all inertial frames, only holds true in view of Einstein's clock synchronization convention.

True

There is no experimental basis whatsoever for preferring this convention over absolute clock synchronization. "

False.

There is an extremely good basis for preferring Einstein's convention. This is the conservation and isotropy of momentum.

The primary reason to synchronize clocks is to be able to measure velocities. When we demand that an object of mass m and velocity v moving north have an equal and opposite momentum to an object of mass m and velocity v moving south, we require Einsteinan clock synchronization.

Empirically, this means that we require an two objects of equal masses moving at the same speed in opposite directions to stop when they collide inelastically.

It is indeed *possible* to use non-Einsteinain clock synchronizations, and under some circumstances it is more-or-less forced on us. In such circumstances, one must not remember that momentum is not isotropic.

Note that Newton's laws assume that momentum is isotropic (an isotropic function of velocity). Therfore Newton's laws (with the definition of momentum as p=mv) cannot be used unless Einstein's clock synchronization is used. Some other definition of momentum other than p=mv must be used if it is to remain a conserved quantity when non-standard clock synchronizations are used.

The ability to use Newton's laws at low velocities was what motivated Einstein to define his method of clock synchronization.

 

[Hurkyl]

Just a friendly reminder, the first postulate of the special theory of relativity, namely that the speed of light c is the same in all inertial frames, only holds true in view of Einstein's clock synchronization convention. There is no experimental basis whatsoever for preferring this convention over absolute clock synchronization.

I would like to remind you that this is analogous to reminding people that there is no experimental basis for using an orthogonal coordinate system when doing plane geometry either. (As opposed to a system where, say, the angle between the x and y axes is 45°)

Your choice of how to define coordinates is not a physical choice -- the result of any physical experiment will be the same no matter what coordinate system you opt to use.

Einstein's coordinates are used because they're nice -- Einstein's coordinate systems are precisely the rectilinear coordinate systems whose coordinate axes are orthogonal. (and non-null)

 

[Aether]

Newton's laws (with the definition of momentum as p=mv) cannot be used unless Einstein's clock synchronization is used. Some other definition of momentum other than p=mv must be used if it is to remain a conserved quantity when non-standard clock synchronizations are used.

The ability to use Newton's laws at low velocities was what motivated Einstein to define his method of clock synchronization.

The general linear transformation that Mansouri & Sexl use has three parameters (\epsilon{_x},\ \epsilon{_2y}, and\ \epsilon{_3z}) that are determined by synchronization procedures. I expect that momentum is conserved when these are taken into account and that it is isotropic, albeit with three extra synchronization parameters to keep track of.

Einstein's coordinate systems are precisely the rectilinear coordinate systems whose coordinate axes are orthogonal. (and non-null)

Mansouri & Sexl needed to add three synchronization parameters to transform the time coordinate while maintaining absolute simultaneity with only four coordinates. Six coordinates, three for time, would seem to be a more natural choice.

 

[Hurkyl]

Mansouri & Sexl needed to add three synchronization parameters to transform the time coordinate while maintaining absolute simultaneity with only four coordinates. Six coordinates, three for time, would seem to be a more natural choice.

Except for the fact we do have empirical evidence for a single time dimension.

 

[Aether]

Except for the fact we do have empirical evidence for a single time dimension.

Why is it that the invariant interval ds absolutely positively has to be a scalar?

 

[Hurkyl]

Why is it that the invariant interval ds absolutely positively has to be a scalar?

Because that's what it's defined to be. I suspect this is not the question you meant to ask.


My comment about the empirical evidence for four dimensions has nothing to do with that: the evidence is the fact that, historically, we've been able to describe any point in space-time that we please by specifying 4 coordinates: where, and when.

 

[pervect]

The general linear transformation that Mansouri & Sexl use has three parameters (\epsilon{_x},\ \epsilon{_2y}, and\ \epsilon{_3z}) that are determined by synchronization procedures. I expect that momentum is conserved when these are taken into account and that it is isotropic, albeit with three extra synchronization parameters to keep track of.

Mansouri & Sexl needed to add three synchronization parameters to transform the time coordinate while maintaining absolute simultaneity with only four coordinates. Six coordinates, three for time, would seem to be a more natural choice.

I don't have the papers you cite, and I couldn't find them or arxiv, either. Your description of them isn't really very enlightening, alas.

The point I want to make is that while it is indeed, possible and sometimes even desirable to use non-Einsteinian clock synchronziation, it is *not* possible to do so and to also assume that Newton's laws work with such a synchronization method.

In other words, when one maks the speed of light anisotropic, one also makes the behavior of matter anisotropic, as well. Light may go "faster" in one direction when you play around with clock synchronziations, but so do racecars, and electron beams, and everything else in the world. (The effect of syncronization "twiddling" is most important for objects which move at high velocities, however).

Basically, one is playing "word games" with the definition of velocity. It is a matter of "convention" that one does not measure the velocity of an airplane by looking at the difference of the clocks at which it takes off in the PST timezone, and the clock at which it lands in the CST timezone. One insists that to get the "fair" speed of the airplane, one uses clocks that are synchronized according to a convention, using the same time-zone for both takeoff and landing times.

Abandoning this convention is possible, but it is not possible to use the "speeds" defined in such an unconventional way with Newton's laws.

 

[Aether]

Because that's what it's defined to be. I suspect this is not the question you meant to ask.

I'm asking that question because ds=c_0d \tau, and three time coordinates merely implies that ds has direction; not necessarily that there are two additional "dimensions": one dimension of time + absolute simultaneity.

My comment about the empirical evidence for four dimensions has nothing to do with that: the evidence is the fact that, historically, we've been able to describe any point in space-time that we please by specifying 4 coordinates: where, and when.

Only by abandoning absolute simulaneity, or else as pervect suggests perhaps by sacrificing the form of Newton's laws.

 

[Hurkyl]

I'm asking that question because ds=c_0 \tau, and three time coordinates merely implies that ds has direction; not necessarily that there are two additional "dimensions": one dimension of time + absolute simultaneity.

Yes, but τ is a scalar, and there aren't three time coordinates.

Of course, in an entirely different theory, you could get entirely different answers.

3 spatial coordiantes + 3 temporal coordinates = 6 dimensions.

Only by abandoning absolute simulaneity, or else as pervect suggests perhaps by sacrificing the form of Newton's laws.

No, we did a pretty good job of describing any point in space-time with 4 coordinates, even when we do adopt absolute simultaneity. "Where and when" was not an invention of Einstein.

 

[Aether]

Yes, but τ is a scalar, and there aren't three time coordinates.

OK, then we break c_0 into components.

No, we did a pretty good job of describing any point in space-time with 4 coordinates, even when we do adopt absolute simultaneity. "Where and when" was not an invention of Einstein.

I'm talking about the issue of momentum/Newton claimed by pervect to be a problem with absolute simultaneity.

 

[Perspicacious]

Just a friendly reminder, ... the speed of light c is the same in all inertial frames, only holds true in view of Einstein's clock synchronization convention. There is no experimental basis whatsoever for preferring this convention over absolute clock synchronization.

You are especially correct on this one point: It's impossible to change the laws of physics by merely resetting clocks.

http://arxiv.org/abs/gr-qc/0409105

"Thus the much debated question concerning the empirical equivalence of special relativity and an ether theory taking into account time dilation and length contraction but maintaining absolute simultaneity can be answered affirmatively."

The greatest importance is in realizing the enormous role arbitrariness and conventionality play in modern physics.

http://www.everythingimportant.org/relativity/

 

[Aether]

Thank you Perspicacious, I will read those articles.

Therfore Newton's laws (with the definition of momentum as p=mv) cannot be used unless Einstein's clock synchronization is used. Some other definition of momentum other than p=mv must be used if it is to remain a conserved quantity when non-standard clock synchronizations are used...The point I want to make is that while it is indeed, possible and sometimes even desirable to use non-Einsteinian clock synchronziation, it is *not* possible to do so and to also assume that Newton's laws work with such a synchronization method.

A particle's 4-momentum is p^\mu=mu^\mu, and we could define its 7-momentum as p^a=mu^a where ds^2=c_0^2 d\tau^2=(dx^4)^2+(dx^5)^2+(dx^6)^2, and the form of Newton's laws are retained. By moving ds^2 to the right side of the line element, we have 0=c_0^2dt^2-(dx^1)^2-(dx^2)^2-(dx^3)^2-(dx^4)^2-(dx^5)^2-(dx^6)^2 instead of ds^2=c_0^2dt^2-(dx^1)^2-(dx^2)^2-(dx^3)^2.

Why isn't this preferable to abandoning absolute simultaneity? Shouldn't abandoning absolute simultaneity be used only as a weapon of last resort?

For example, consider P.A.M. Dirac, The Principles of Quantum Mechanics - Fourth Edition, Oxford University Press, 1958. Section 69 The motion of a free electron, p. 262: "...we can concude that a measurement of a component of the velocity of a free electron is certain to lead to the result +or- c...Since electrons are observed in practice to have velocities considerably less than that of light, it would seem that we have here a contradiction with experiment. The contradiction is not real, though, since the theoretical velocity in the above conclusion is the velocity at one instant of time while observed velocities are always average velocities through appreciable time intervals. We shall find upon further examination of the equations of motion that the velocity is not at all constant, but oscillates rapidly about a mean value which agrees with the observed value."

So, to be clear, it is my intention that 0=c_0^2dt^2-(dx^1)^2-(dx^2)^2-(dx^3)^2-(dx^4)^2-(dx^5)^2-(dx^6)^2 should be viewed in this context of rapid oscillation about a mean value which agrees with the observed value.

Of course, in an entirely different theory, you could get entirely different answers.

3 spatial coordiantes + 3 temporal coordinates = 6 dimensions.

Or 1 temporal coordinate + 6 spatial coordinates: where three spatial coordinates are instantaneous, and the other three are observed as averages of the first three (same basis vectors) through appreciable time intervals. Isn't that in fact what is observed in nature?

I submit that ds only appears to be a scalar when observation is averaged over any appreciable time interval, but when observed at one instant of time...it is a vector.

 

[Hurkyl]

Aether: you seem to be confusing reality with the mathematical model.

ds is a scalar because that's what it's defined to be: it's part of the mathematical construct that is Minowski space, which Special Relativity asserts acts as a model of reality.


Even if you are exactly correct about the behavior of reality, that doesn't change the fact that the ds of Special Relativity is a scalar.

where three spatial coordinates are instantaneous

I have no idea what that would mean.

 

[Aether]

Aether: you seem to be confusing reality with the mathematical model.

ds is a scalar because that's what it's defined to be: it's part of the mathematical construct that is Minowski space, which Special Relativity asserts acts as a model of reality.


Even if you are exactly correct about the behavior of reality, that doesn't change the fact that the ds of Special Relativity is a scalar.

OK, thanks. I'll go back and look at it again with that in mind.

I have no idea what that would mean.

I mean three spatial coordinate differentials, evaluated instantaneously since according to Dirac they change rapidly.

 

[DrGreg]

Shouldn't abandoning absolute simultaneity be used only as a weapon of last resort?

I think the vast majority of physicists and mathematicians would say that abandoning symmetry should be used only as a weapon of last resort.

In the end it's a matter of your own personal philosophy.

 

[Aether]

I think the vast majority of physicists and mathematicians would say that abandoning symmetry should be used only as a weapon of last resort.

In the end it's a matter of your own personal philosophy.

Referencing an arbitrarily chosen frame is an application of your own personal philosophy, but referencing a locally preferred frame is not.

Absolute simultaneity merely implies a locally preferred frame of reference, whereas relative simultaneity implies an arbitrary frame of reference. Where is there an abandonment of symmetry?

Suppose that the SI system were to one day adopt a locally preferred frame as a standard reference: SR could still be recovered completely for transformations between any two arbitrary reference frames by first transforming from one arbitrary frame to the SI standard frame, and then transforming from the SI standard frame to the second arbitrary frame.

For example, the money in your pocket references a locally preferred frame (of sorts); without that, we could all be wondering "how many chickens per gallon is gas going to cost me today...assuming that I can even get any??".

 

[DrGreg]

Suppose I measure the position of an object on my desk using xy coordinates aligned to the side of my desk. Then I want to convert the position to XY coordinates aligned to the side of my room.

Which is easier: perform a direct transformation from one to the other, or use an intermediate step of a "locally preferred frame" of eastings/northings?

(In fact, I think you are saying I really ought not to use a rectangular grid but a parallelogram grid -- a Y-Easting grid?)


The symmetry in SR is that anyone is entitled to assume that their own frame is the preferred frame; there's no need to look for any other preferred frame.

 

[Aether]

Which is easier: perform a direct transformation from one to the other, or use an intermediate step of a "locally preferred frame" of eastings/northings?

It is often easier to perform a direct transform.

(In fact, I think you are saying I really ought not to use a rectangular grid but a parallelogram grid -- a Y-Easting grid?)

All I'm saying is that you have a choice. This doesn't matter much until you try to extrapolate your grid to the entire universe.

The symmetry in SR is that anyone is entitled to assume that their own frame is the preferred frame; there's no need to look for any other preferred frame.

That's OK as a local approximation, but it doesn't scale to the universe as a whole. For example, the amount of matter-energy in the universe is finite, and it is distributed in a certain pattern. Let's say that it is evenly distributed throughout a 3-sphere of radius R(t). There is a center to that sphere, the center of mass of the universe, which we can take to be at rest, and that is a locally preferred frame.

 

[JesseM]

That's OK as a local approximation, but it doesn't scale to the universe as a whole. For example, the amount of matter-energy in the universe is finite, and it is distributed in a certain pattern. Let's say that it is evenly distributed throughout a 3-sphere of radius R(t). There is a center to that sphere, the center of mass of the universe, which we can take to be at rest, and that is a locally preferred frame.

Most cosmological models assume that if space is infinite, then the amount of matter/energy in the universe is infinite too; likewise, if the amount of matter/energy is finite, then that's because space is finite too (it 'wraps around' like the surface of a sphere). Either way, matter/energy would be distributed in a fairly equal way throughout space, with no center to the distribution.

 

[Aether]

Most cosmological models assume that if space is infinite, then the amount of matter/energy in the universe is infinite too; likewise, if the amount of matter/energy is finite, then that's because space is finite too (it 'wraps around' like the surface of a sphere). Either way, matter/energy would be distributed in a fairly equal way throughout space, with no center to the distribution.

The WMAP estimate for the age of the universe is 13.4Gyr; plugging that into the Friedmann cosmological model I get a mass density of 1.00E-26 \ kg/m^3, a spatial volume of 8.54E78 \ m^3, and a finite mass of 8.54E52 \ kg. OK so far? Why is there no center to this distribution?

 

[JesseM]

The WMAP estimate for the age of the universe is 13.4Gyr; plugging that into the Friedmann cosmological model I get a mass density of 1.00E-26 \ kg/m^3, a spatial volume of 8.54E78 \ m^3, and a finite mass of 8.54E52 \ kg. OK so far? Why is there no center to this distribution?

Because that's not how the Friedmann cosmological model works--the big bang is not modeled as an explosion of matter from a specific point in a preexisting space, it's modeled as an expansion of space itself, with matter being equally distributed throughout all of space at all finite times (that's what they mean when they say the model assumes a 'homogenous and isotropic universe' on this page).

 

[Aether]

Because that's not how the Friedmann cosmological model works--the big bang is not modeled as an explosion of matter from a specific point in a preexisting space, it's modeled as an expansion of space itself, with matter being equally distributed throughout all of space at all finite times (that's what they mean when they say the model assumes a 'homogenous and isotropic universe' on this page).

Does this cosmological model assume relative simultaneity, or absolute simultaneity?

 

[JesseM]

Does this cosmological model assume relative simultaneity, or absolute simultaneity?

The model is based on general relativity, and general relativity reduces to SR locally, so in any local region you should find that the laws of physics work the same way in all reference frames, and thus relative simultaneity is implied too. At the same time, GR allows you different choices of how to slice up 4D spacetime into a stack of 3D spacelike hypersurfaces, a procedure known as a "foliation", and there is only a single way to foliate the spacetimes used in these cosmological models in such a way that matter is homogeneous throughout each slice, giving what's known as a "hypersurface of homogeneity" (discussed on pages 724-725 of the GR textbook Gravitation by Misner, Thorne and Wheeler). So locally I assume this means there will be only a single frame whose definition of simultaneity coincides with that of the hypersurface of homogeneity, but this isn't a "preferred reference frame" in the sense that the laws of physics themselves work any differently in this frame than in others.

 

[Aether]

The model is based on general relativity, and general relativity reduces to SR locally, so in any local region you should find that the laws of physics work the same way in all reference frames, and thus relative simultaneity is implied too..

"Clearly, the universe cannot look isotropic to all obervers...Only an observer who is moving with the cosmological fluid can possibly see things as isotropic." --MTW, p. 714.

At the same time, GR allows you different choices of how to slice up 4D spacetime into a stack of 3D spacelike hypersurfaces, a procedure known as a "foliation", and there is only a single way to foliate the spacetimes used in these cosmological models in such a way that matter is homogeneous throughout each slice, giving what's known as a "hypersurface of homogeneity" (discussed on pages 724-725 of the GR textbook Gravitation by Misner, Thorne and Wheeler). So locally I assume this means there will be only a single frame whose definition of simultaneity coincides with that of the hypersurface of homogeneity, but this isn't a "preferred reference frame" in the sense that the laws of physics themselves work any differently in this frame than in others.

Isn't a hypersurface of homogeneity frame independent?

"The surfaces of t=constant of such a coordinate system will coincide with the hypersurfaces of homogeneity of the universe." -- MTW, p. 715

 

[JesseM]

"Clearly, the universe cannot look isotropic to all obervers...Only an observer who is moving with the cosmological fluid can possibly see things as isotropic." --MTW, p. 714.

Sure--that's exactly what I was saying when I pointed out there's only one way to foliate spacetime that will result in each 3D hypersurface being homogeneous. All other ways of slicing up spacetime into 3D hypersurfaces result in matter being non-isotropic in each hypersurface.

Isn't a hypersurface of homogeneity frame independent?

It is frame-independent in the sense that all observers will agree on what the hypersurface of simultaneity is, and they'll all agree on which observer has a definition of simultaneity that uniquely coincides with this foliation. But again, this is not a "preferred reference frame" in the sense of being preferred by the laws of physics, it's only preferred in terms of the distribution of matter/energy in space.

 

[Aether]

Sure--that's exactly what I was saying when I pointed out there's only one way to foliate spacetime that will result in each 3D hypersurface being homogeneous. All other ways of slicing up spacetime into 3D hypersurfaces result in matter being non-isotropic in each hypersurface.

It seems that one could hope to find one and only one local frame in which the universe appears homogeneous and isotropic.

It is frame-independent in the sense that all observers will agree on what the hypersurface of simultaneity is, and they'll all agree on which observer has a definition of simultaneity that uniquely coincides with this foliation. But again, this is not a "preferred reference frame" in the sense of being preferred by the laws of physics, it's only preferred in terms of the distribution of matter/energy in space.

OK, but we agreed that the distribution of matter/energy is space is uniform. Doesn't the distribution of matter/energy in space have a center? What might "preferred by the laws of physics" look like if it existed?

 

[JesseM]

It seems that one could hope to find one and only one local frame in which the universe appears homogeneous and isotropic.

That's what I've been saying!

OK, but we agreed that the distribution of matter/energy is space is uniform. Doesn't the distribution of matter/energy in space have a center?

No. The mainstream cosmological models assume that in the hypersurface of homogeneity, matter/energy is uniformly distributed throughout all of space--if space is infinite, that means an infinite amount of matter/energy. And even if space is finite, it would be analogous to the surface of a sphere--a uniform distribution of 2D matter on a spherical 2D space has no center of mass within that 2D space.

What might "preferred by the laws of physics" look like if it existed?

Well, before relativity physicists imagined that Maxwell's laws would only hold exactly in one frame, that in other frames they'd have to be modified by a Galilei transformation (so that in one of these frames, light would travel at c+v in one direction and c-v in the opposite direction, for example). If this were the case, that frame would be preferred by the laws of physics.

In general, for a particular frame to be "preferred by the laws of physics" it should be true that if you're confined to a small windowless box, there's some experiment that you can do that will allow you to find your velocity relative to this frame. Of course, general relativity says that if you're in a small windowless box with no non-gravitational forces on it, there's no experiment you can do that will tell you how you're moving relative to other objects in the universe--that's implied by the equivalence principle.

 

[Aether]

Well, before relativity physicists imagined that Maxwell's laws would only hold exactly in one frame, that in other frames they'd have to be modified by a Galilei transformation (so that in one of these frames, light would travel at c+v in one direction and c-v in the opposite direction, for example). If this were the case, that frame would be preferred by the laws of physics.

Assuming that a Galilei transformation can be used in order to maintain absolute simultaneity is clearly wrong, but Mansouri & Sexl show that an ether theory maintaining absolute simultaneity is empirically equivalent to SR. They use a more general linear transform than the Lorentz transform, and I presume that Maxwell's equations transform properly either way. Three additional parameters are needed to transform the temporal coordinate in this way.

In general, for a particular frame to be "preferred by the laws of physics" it should be true that if you're confined to a small windowless box, there's some experiment that you can do that will allow you to find your velocity relative to this frame. Of course, general relativity says that if you're in a small windowless box with no non-gravitational forces on it, there's no experiment you can do that will tell you how you're moving relative to other objects in the universe--that's implied by the equivalence principle.

Mansouri & Sexl I says that "A theory maintaining the concept of absolute simultaneity... is empirically equivalent to special relativity, at least as far as kinematics is concerned." Do you need a windowed box to do experiments in kinematics, and a windowless box to do experiments in electrodynamics? Are the extra experiments that one can do in a windowed box not governed by the laws of physics?

 

[Aether]

False.

There is an extremely good basis for preferring Einstein's convention. This is the conservation and isotropy of momentum.

The primary reason to synchronize clocks is to be able to measure velocities. When we demand that an object of mass m and velocity v moving north have an equal and opposite momentum to an object of mass m and velocity v moving south, we require Einsteinan clock synchronization.

Empirically, this means that we require an two objects of equal masses moving at the same speed in opposite directions to stop when they collide inelastically.

It is indeed *possible* to use non-Einsteinain clock synchronizations, and under some circumstances it is more-or-less forced on us. In such circumstances, one must not remember that momentum is not isotropic.

Note that Newton's laws assume that momentum is isotropic (an isotropic function of velocity). Therfore Newton's laws (with the definition of momentum as p=mv) cannot be used unless Einstein's clock synchronization is used. Some other definition of momentum other than p=mv must be used if it is to remain a conserved quantity when non-standard clock synchronizations are used.

The ability to use Newton's laws at low velocities was what motivated Einstein to define his method of clock synchronization.

Do you still believe this, pervect? If so, could you please show a simple example using the transformation equations that I posted in message #92 of the "consistency of the speed of light" thread? How is p=mv affected by the choice of one clock synchronization convention over another? We can measure v using round-trip light signals which are isotropic regardless of one's choice of clock synchronization convention.

 

[pervect]

Do you still believe this, pervect? If so, could you please show a simple example using the transformation equations that I posted in message #92 of the "consistency of the speed of light" thread? How is p=mv affected by the choice of one clock synchronization convention over another? We can measure v using round-trip light signals which are isotropic regardless of one's choice of clock synchronization convention.

I haven't changed my mind about what I wrote.

I don't really see how what you suggest is possible.

To measure the velocity of something, we measure the distance it covers in a certain amount of time.

While we could measure the "rapidity" of an object by having the object carry its own clock along with it, this technique would not allow us to measure the velocity of light, which doesn't have a reference frame. It would also be totally inconsistent with your scheme of different clock synchronizations.

If we are self-consistent, we will measure the velocity of physical objects and the velocity of light in the same manner. I have been assuming that the measurment of velocity is consistent, and done in the usual manner, which I will describe below.

This entails setting up two clocks, synchronizing them, and measuring the time on the second clock when the object passes it, subtracting fromt this the time on the first clock when the object passed it, and dividing the distance by this time difference.

It should be obvious from this that the velocity we measure will depend on the clock synchronization. As we twiddle your synchronization parameter, we will change the measured velocity for a given object or light beam.

If we use the Einstein clock synchronization for a reference, we get v=d/t for the Einstein case.

In the alternate synchronization scheme, v' = d/ (t-delta) (going in one direction), and v' = d/(t+delta) going in the other direction.

Just imagine setting up a racetrack, carefully synchronizing your clocks, measure the speed of a car going both ways, then carefully de-synchronizing your clocks by adding time to one of them, and calculating the effect on the measured velocity of the car.

Until we introduce the conservation of momentum, we don't have any justification for chosing the Einstein velocity over any other velocity. But wehn we do interoduce momentum, only one method of synchronizing clocks will give the relation that p(m,v) = p(m,-v), i.e. that the momentum of two objects of the same mass moving at the same velocity in opposite directions is zero. This is why Einstein chose the method he chose.

The effect will be very tiny at low velocities, but measurable for very fast objects. (Relativistic electrons come to mind).

 

[Aether]

I don't really see how what you suggest is possible...I have been assuming that the measurment of velocity is consistent, and done in the usual manner, which I will describe below.

This entails setting up two clocks, synchronizing them, and measuring the time on the second clock when the object passes it, subtracting fromt this the time on the first clock when the object passed it, and dividing the distance by this time difference.

It should be obvious from this that the velocity we measure will depend on the clock synchronization. As we twiddle your synchronization parameter, we will change the measured velocity for a given object or light beam.

If we use the Einstein clock synchronization for a reference, we get v=d/t for the Einstein case.

In the alternate synchronization scheme, v' = d/ (t-delta) (going in one direction), and v' = d/(t+delta) going in the other direction.

Suppose that we're on a ship using radar to monitor the approach of two unidentified aircraft coming in from opposite directions. At time t_1 we transmit a microwave pulse in one direction, and at time t_2 we receive an echo from the first plane; then at time t_3 we transmit a microwave pulse in the opposite direction, and at time t_4 we receive an echo from the second plane: t_2-t_1=t_4-t_3=0.001 000000\ seconds. Then one second later we repeat the process and get t_2-t_1=t_4-t_3=0.000 996 998. Evidently, both aircraft are approaching our ship at v=450m/s from a range of 149.446km and they will converge over our position in 5.535 minutes and fall right on top of us in a heap of flaming scrap metal if someone doesn't warn one or both of them in time. What difference does it make how our clocks are synchronized?

Can you show an actual example of where p=mv holds with Einstein synchronization but doesn't hold with absolute synchronization?

 

[pervect]

If you know that the speed of light is constant, you can use the radar method you describe. However, to say that the velocity of the planes is exactly 450 m/s, you need to know that the speed of light is 'c' in both directions. If the speed of light is not the same in both directions, your calculations will not be correct.

As far as an example goes, I'm not sure why you need one, it's really *very* simple, but here goes.

You have a 100 meter course. You set up clocks at both end. You adjust the clock synchronization so that light takes 333 ns to travel 100 meters (that's 3e8 m/sec) in both directions.

Now you take something moving at half the speed of light. You find that it takes 666 ns for the object to traverse the course in each direction.

You also find that the two objects, moving at .5c with the same mass m (whatever that is) have the same momentum. You measure momentum with a ballistic pendulum, or some other means (it will be tricky at these velocities). The point is that if you had an ideal inelastic collision, both objects would come to a complete stop when they collided.

Now, you adjust the clock synchronization so that the far clock is 10 ns ahead, and you repeat the experiments.

You find that it takes light 343 ns to traverse the course in one direction, and 323 ns in the other.

Hopefully you see why this is the case? Imagine a third clock if you need to, one that is set to read 10 ns later than clock #2.

You also find that it takes the massive object moving at .5c 676 ns to traverse the course in one direction, and 656 ns to traverse the course in the other direction.

(The reasoning is the same as above).

You calculate the velocities v' with the new synch method, and find them to be

147,928 m/s in one direction
152,439 m/s in the other direction

Because of these, we now argue that the massive object is not moving "at the same velocity" in both directions with the new clock synchronization.

However, changing the clock synchrhonization did not change the momentum of the objects. They came to a complete stop when they collided before, and they will still do that, even though we now say that their velocities are different.

Unless you believe that changing the setting of a clock will change the result of a completely separate physical experiment, perhaps, but that is so inconsistent I don't see how you could believe that).

We therefore conclude that the physical quantity momentum is not equal to m*v with the new clock synchronization scheme. Basically p is now a function of m, v, and direction - i.e. it's not an isotropic function of velocity.

 

[Aether]

If you know that the speed of light is constant, you can use the radar method you describe. However, to say that the velocity of the planes is exactly 450 m/s, you need to know that the speed of light is 'c' in both directions. If the speed of light is not the same in both directions, your calculations will not be correct.

Let's say that the CMB rest frame is our ether frame, so that our frame is at v=368km/s with respect to the ether frame. If the speed of light in one direction is c_0^2/(c_0+v) and the distance to the radar reflector is 149.446km then the light-speed travel time to the target is 0.000 499 110 112 159 seconds, and if the speed of light in the opposite direction is c_0^2/(c_0-v) then the light-speed travel time from the target is 0.000 497 886 283 261 seconds. The round-trip light signal travel time is the sum of these two one-way travel times and is equal to 0.000 996 996 395 420 seconds. This is exactly the same round-trip light signal travel time that you get when the speed of light is the same in each direction, c_0=299,792,458 m/s.

As far as an example goes, I'm not sure why you need one, it's really *very* simple, but here goes.

I just think that it's constructive to have a concrete example.

Thanks for the 100 meter course example. Simple is good. I'm still studying it, and will have more to say about it shortly.

 

[yogi]

Aether - are you saying that summing the two velocities leads to a correct average velocity over the total round trip distance

 

[Ich]

yogi - are Aether´s tranformation equations those that you call "Selleri Transformation"?

 

[Aether]

Aether - are you saying that summing the two velocities leads to a correct average velocity over the total round trip distance

Over the total round trip distance, the sum of the reciprocal speeds of light in LET is the same as the sum of the reciprocal speeds of light in SR: (c_0+v)/c_0^2+(c_0-v)/c_0^2=1/c_0+1/c_0.

 

[DrGreg]

Aether

Here's an algebraic proof from your beloved transformation equations. Combine the LET transform

t_{LET} = T / \gamma
x_{LET} = \gamma (X - vT)

with the Lorentz transform

t_{SR} = \gamma (T - vX / c^2)
x_{SR} = \gamma (X - vT)

(where \gamma = 1 / \sqrt{1 - v^2/c^2}, X and T are ether co-ordinates, and c is the absolute speed of light. Quantities with an SR subscript are measured in Special Relativity theory. Quantities with an LET subscript are measured in Lorentz Ether Theory. Quantities without a subscript are measured relative to the ether and therefore are valid in both theories.)

You get

t_{LET} = t_{SR} + v \ x_{SR} / c^2
x_{LET} = x_{SR}

For a particle with velocity u (along the x-axis), substitute x_{SR} = u_{SR} \ t_{SR} and x_{LET} = u_{LET} \ t_{LET} to obtain

\displaystyle{ u_{LET} = \frac{u_{SR}} {1 + v \ u_{SR} / c^2} }

So, two particles of equal mass with equal and opposite values of u_SR do not have equal and opposite values of u_LET. The last four paragraphs of pervect's last post (#34) now apply.

 

[Aether]

Here's an algebraic proof from your beloved transformation equations. Combine the LET transform

t_{LET} = T / \gamma
x_{LET} = \gamma (X - vT)

with the Lorentz transform

t_{SR} = \gamma (T - vX / c^2)
x_{SR} = \gamma (X - vT)

(where \gamma = 1 / \sqrt{1 - v^2/c^2}, X and T are ether co-ordinates, and c is the absolute speed of light. Quantities with an SR subscript are measured in Special Relativity theory. Quantities with an LET subscript are measured in Lorentz Ether Theory. Quantities without a subscript are measured relative to the ether and therefore are valid in both theories.)

DrGreg, this equation is not the one that I gave from M&S: t_{SR} = \gamma (T - vX / c^2), but you have mixed it in with the other three that I did give. I think that yours implies a metric signature of -2, but M&S' imply a metric signature of +2?

You get

t_{LET} = t_{SR} + v \ x_{SR} / c^2
x_{LET} = x_{SR}

For a particle with velocity u (along the x-axis), substitute x_{SR} = u_{SR} \ t_{SR} and x_{LET} = u_{LET} \ t_{LET} to obtain

\displaystyle{ u_{LET} = \frac{u_{SR}} {1 + v \ u_{SR} / c^2} }

So, two particles of equal mass with equal and opposite values of u_SR do not have equal and opposite values of u_LET. The last four paragraphs of pervect's last post (#34) now apply.

Starting out with all of our clocks Einstein synchronized we have this:
x_{SR}=u_{SR}(T/\gamma-vx_{SR}/c_0^2)

and then after we readjust these clocks according to M&S-I p. 502 Eq. (3.5) f(x,v)=-vx_{SR}/c_0^2 we have this:
x_{LET}=u_{LET}(T/\gamma)

u_{SR} and u_{LET} are completely unaffected by this, so I suspect that these issues of the conservation of momentum will go away when this clock synchronization step is properly accounted for. There really isn't any empirical difference between these two theories; the only difference is that in one the speed of light is relative and simultaneity is absolute, but in the other the speed of light is absolute and simultaneity is relative. Thank you for the algebraic proof; this is not my final word on it, only a preliminary one.

 

[pervect]

The problem isn't going away. There is a very simple proof that one, and only one clock synchronization scheme will give isotropy of momentum.

Let us have three clocks

t1......t2,t3

Assume that t1 and t2 are synchronized so that momentum is isotropic. Assume t3 has some synchronization difference delta, i.e. t3 = t2 + delta.

We have an object #1 moving left to right with some fixed momentum p that starts out on the left at t1=0. At the same time, we have an object #2 moving right to left with the same mass and the same fixed momentum p that starts ou on the right at t2=0.

Object #1 will have a velocity of d/t2 according to clock 2, and a velocity of d/t3 according to clock three. Since t3 = t2 + delta, the velocity according to clock 2 will be d/(t2+delta).

Object #2 will have a velocity of d/t1 according to clock 2 (which starts out at 0), and a velocity of d/(t1-delta) according to clock 3 (which starts out at a value of t2+delta, and we've already mentioned that t2 was zero.

So if d/t2 = d/t1, then for any delta other than zero, d/(t2+delta) cannot be equal to d/(t1-delta.

Thus if we assume that the velocities are the same for some particular clock synchronization, any other clock synchronization will make them different.

Now here is the part where experiment steps in.

Relativity makes the claim that the _same_ clock synchronization that makes the speed of light isotropic (the same in both directions) is this unique clock synchronization that makes the velocities of objects with equal momentum isotropic.

 

[DrGreg]

DrGreg, this equation is not the one that I gave from M&S: t_{SR} = \gamma (T - vX / c^2), but you have mixed it in with the other three that I did give. I think that yours implies a metric signature of -2, but M&S' imply a metric signature of +2?

The Lorentz equations I quoted are in the form you will find them in most textbooks (maybe using different letters, etc). You can invert them to get

T = \gamma (t_{SR} + v \ x_{SR} / c^2)
X = \gamma (x_{SR} + v \ t_{SR})

The first of these equations rearranges to give the first equation you quoted in this message: https://www.physicsforums.com/showpost.php?p=755432&postcount=92. I haven't assumed any metric signature.

...so I suspect that these issues of the conservation of momentum will go away when this clock synchronization step is properly accounted for

My algebra already takes account of the clock synchronisation -- it is included within the LET transformation formula. I stand by my argument. The only way you can account for momentum is to give it an anisotropic formula in LET coordinates.

 

[Aether]

Let us have three clocks

t1......t2,t3

Assume that t1 and t2 are synchronized so that momentum is isotropic. Assume t3 has some synchronization difference delta, i.e. t3 = t2 + delta.

We have an object #1 moving left to right with some fixed momentum p that starts out on the left at t1=0. At the same time, we have an object #2 moving right to left with the same mass and equal and opposite fixed momentum p that starts ou on the right at t2=0.

Object #1 will have a velocity of d/t2 according to clock 2, and a velocity of d/t3 according to clock three. Since t3 = t2 + delta, the velocity according to clock 2 will be d/(t2+delta).

Object #2 will have a velocity of d/t1 according to clock 2 (which starts out at 0), and a velocity of d/(t1-delta) according to clock 3 (which starts out at a value of t2+delta, and we've already mentioned that t2 was zero.

So if d/t2 = d/t1, then for any delta other than zero, d/(t2+delta) cannot be equal to d/(t1-delta.

The x-coordinate of each clock determines a unique value of delta=-vx/c_0^2 for that clock so I have labeled each delta according to the clock that it belongs to, and restated the travel times more explicitly to show that the delta terms fall out for any time interval that is measured by a single clock. This always leads to the opposite conclusion from your example when any time interval is measured by a single clock, so I presume that you will want to specify that each of these time intervals is to be measured using two different clocks?


Let us have three clocks

t1......t2,t3

Assume that t1 and t2 are synchronized so that momentum is isotropic. Assume t3 has some synchronization difference delta, i.e. t3 = t2 + delta2.

We have an object #1 moving left to right with some fixed momentum p1=\gamma 1 m(v+u) that starts out on the left at t1_0=0. At the same time, we have an object #2 moving right to left with the same mass and a fixed momentum p2=\gamma 2 m(v-u) that starts out on the right at t2_0=0.

Object #1 will have a relative velocity of u12=d/(t2_1-t2_0) according to clock 2, and a relative velocity of u13=d/(t3_1-t3_0) according to clock three. Since t3 = t2 + delta2, the relative velocity according to clock 2 will be u12=d/(t2_1+delta2-t2_0-delta2).

Object #2 will have a relative velocity of u21=d/(t1_1-t1_0) according to clock 1 (which starts out at 0), and a relative velocity of u23=d/(t1_1+delta1-t1_0-delta1) according to clock 3 (which starts out at a value of t3_0=t2_0+delta2, and we've already mentioned that t2_0=0.

So if d/(t2_1-t2_0) = d/(t1_1-t1_0), then for any delta whatsoever, d/(t2_1+delta2-t2_0-delta2) is also equal to d/(t1_1+delta1-t1_0-delta1).

 

[Aether]

You get

t_{LET} = t_{SR} + v \ x_{SR} / c^2
x_{LET} = x_{SR}

For a particle with velocity u (along the x-axis), substitute x_{SR} = u_{SR} \ t_{SR} and x_{LET} = u_{LET} \ t_{LET} to obtain

\displaystyle{ u_{LET} = \frac{u_{SR}} {1 + v \ u_{SR} / c^2} }

Let’s be more explicit and say that x_{SR1}-x_{SR0}=u_{SR}(t_{SR1}-t_{SR0}), x_{LET1}-x_{LET0}=u_{LET}(t_{LET1}-t_{LET0}), and x_{LET1}-x_{LET0}=x_{SR1}-x_{SR0}.

u_{LET}=u_{SR}(t_{SR1}-t_{SR0})/(t_{LET1}-t_{LET0})=
u_{SR}(t_{SR1}-t_{SR0})/(t_{SR1}+vx/c_0^2-t_{SR0}-vx/c_0^2)=u_{SR}.

So, two particles of equal mass with equal and opposite values of u_SR do not have equal and opposite values of u_LET. The last four paragraphs of pervect's last post (#34) now apply.

False.

 

[pervect]

Combinging two previous posts by Aether

The x value of each clock determines a unique value of delta=-vx/c_0^2 for that clock so I have labeled each delta according to the clock that it belongs to, and restated the travel times more explicitly to show that the delta terms fall out for any time interval that is measured by a single clock. This always leads to the opposite conclusion from your example when any time interval is measured by a single clock, so I presume that you will want to specify that each of these time intervals is to be measured using two different clocks?


Let’s be more explicit and say that x_{SR1}-x_{SR0}=u_{SR}(t_{SR1}-t_{SR0}), x_{LET1}-x_{LET0}=u_{LET}(t_{LET1}-t_{LET0}), and x_{LET1}-x_{LET0}=x_{SR1}-x_{SR0}.

u_{LET}=u_{SR}(t_{SR1}-t_{SR0})/(t_{LET1}-t_{LET0})=
u_{SR}(t_{SR1}-t_{SR0})/(t_{SR1}+vx/c_0^2-t_{SR0}-vx/c_0^2)=u_{SR}.

If we take your argument at face value, you have just proven that all velocities in LET are the same as all velocities in SR for any value of velocity, including 'c', therefore the speed of light is isotropic in LET.

But the speed of light is not supposed to be isotropic in LET, if I have been following your exposition of the theory correctly.

You have written vx/c_0^2, but you have also said that x is unique for each clock. If we assume that the x is different for clock 1 and clock 0, we get Dr Greg's results, and we also get an anisotropic speed of light.

Note that if we use your equations, you are adding the _same_ value of time to _every_ clock regardless of position. It is not terribly surprising that this does not give us any variation in velocity. It also gives us an isotropic speed for light.

But making the value of time added depend on position gives us

<br /> T_{LET1} = T_{SR1} + \frac{v X1}{c_0^2}<br />
<br /> T_{LET0} = T_{SR0} + \frac{v X0}{c_0^2}<br />

here v is a hypothetical "ether velocity", and X0 and X1 are "ether coordinates", which are different for the two different events because they don't happen at the same point in space.

This gives us directly that
<br /> T_{LET1} - T_{LET0} = T_{SR1} - T_{SR0} + \frac{v \left( X1-X0 \right) } {c_0^2}<br />

Substituting this into your first equation (which I think is correct though I haven't double checked it) we get

<br /> \frac {u_{LET}}{u_{SR}} = \frac { T_{SR1} - T_{SR0} }{ T_{SR1} - T_{SR0} + \frac{v \left( X1-X0 \right) }{c^2} }<br />

which gives anisotropy of velocity for v=c and for v<c.

 

[DrGreg]

pervect (post #45) is nearly correct.

The correct equations are

t_{LET1} = t_{SR1} + v \ x_{SR1} / c_0^2
t_{LET0} = t_{SR0} + v \ x_{SR0} / c_0^2

But x_{SR1} \neq x_{SR0} (except if u_SR = 0) so the rest of his argument still applies.

 

[DrGreg]

We have an object #1 moving left to right with some fixed momentum p1=\gamma 1 m(v+u) that starts out on the left at t1_0=0.

When you quote the momentum as p=\gamma_{v+u} m(v+u) there are two things wrong with this.

1. pervect and myself are talking about momentum relative to the observer (who is moving at absolute velocity v). You seem to be talking about absolute momentum, relative to the ether. As behaviour relative to the ether is agreed between SR and LET, absolute momentum isn't an issue, only relative momentum.

2. The formula p=\gamma_{v+u} m(v+u) isn't valid for absolute momentum. v is the velocity of the observer relative to the ether. u is the velocity of the particle relative to the observer. But (v+u) is not the velocity of the particle relative to the ether, neither in SR nor in LET.

(Added)

Incidentally, the reasoning that momentum is p=\gamma_u m u rather than Newtonian p=m u requires an application of the postulates of SR. How would you justify this formula without invoking SR?

 

[Aether]

When you quote the momentum as p=\gamma_{v+u} m(v+u) there are two things wrong with this.

1. pervect and myself are talking about momentum relative to the observer (who is moving at absolute velocity v). You seem to be talking about absolute momentum, relative to the ether. As behaviour relative to the ether is agreed between SR and LET, absolute momentum isn't an issue, only relative momentum.

OK. I don't mean to ignore this thread, but the other one is keeping me occupied at the moment. I think that I proved here that velocities and momentums measured by any single clock are the same for both SR and LET, and I gather that you and pervect want to talk only about time intervals measured by two different clocks? It is not at all clear to me what either of you are saying when you specify a time interval as "t2" when there is a clock labeled t2, etc..

2. The formula p=\gamma_{v+u} m(v+u) isn't valid for absolute momentum. v is the velocity of the observer relative to the ether. u is the velocity of the particle relative to the observer. But (v+u) is not the velocity of the particle relative to the ether, neither in SR nor in LET.

As long as we agree to restrict our discussion to motion along the x-axis then this is OK. Wouldn't it be a needless complication to do otherwise?

Incidentally, the reasoning that momentum is p=\gamma_u m u rather than Newtonian p=m u requires an application of the postulates of SR. How would you justify this formula without invoking SR?

Do you not use the Lorentz transformation to justify this? I would then use the LET transformation.

 

[pervect]

You can't measure the one-way speed of light (or anything else, for that matter) without two clocks. I was not the person who introduced "one-way" speeds into this thread.

Look at the thread title and originating post, for instance:

Einstein's Clock Synchronization Convention

Just a friendly reminder, the first postulate of the special theory of relativity, namely that the speed of light c is the same in all inertial frames, only holds true in view of Einstein's clock synchronization convention. There is no experimental basis whatsoever for preferring this convention over absolute clock synchronization

It seems a bit revisionist to me to start saying

I think that I proved here that velocities and momentums measured by any single clock are the same for both SR and LET

when the entire thread, from its inception, has been about clock synchronization, which of course implies using more than one clock. (You can't synchronize a clock to itself!).'

What I've been doing is attempting to point out the importance of Einstein's clock synchronization convention. It's not a matter of "don't use this convention and nothing will happen". It's a matter of "don't use this convention and Newton's laws will fail and the momentum of a body will depend on it's direction of travel".

 

[Aether]

You can't measure the one-way speed of light (or anything else, for that matter) without two clocks. I was not the person who introduced "one-way" speeds into this thread.

Look at the thread title and originating post, for instance:



It seems a bit revisionist to me to start saying



when the entire thread, from its inception, has been about clock synchronization, which of course implies using more than one clock. (You can't synchronize a clock to itself!).'

What I've been doing is attempting to point out the importance of Einstein's clock synchronization convention. It's not a matter of "don't use this convention and nothing will happen". It's a matter of "don't use this convention and Newton's laws will fail and the momentum of a body will depend on it's direction of travel".

I'm only asking for you to label the coordinates in your example so that I can see which clock they came from at which snapshot in time, and where you specify a time interval to label it explicitly so that I can see exactly how it is supposed to be measured. My reality check has been to measure velocities using a radar pulse and a single clock, so I know that LET must give the same velocity as that method. If we use two clocks, then the synchronizations of those clocks will have to be accounted for, and therefore the two clocks will have to be explicitly represented within any time interval.