Einstein's Clock Synchronization Convention

[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.

 

[JesseM]

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.

How does this method work, exactly--are you bouncing multiple pulses off an object and seeing the difference between the time interval that the pulses are emitted and the time interval that the pulses are received? If so, what equation do you solve to find the speed of the object? If you assume the radar signals travel at c in both directions as in SR, that the object's velocity is v and its distance at t=0 is d, and that there is a time interval of t_0 between when two pulses are emitted, then the time the first pulse catches up to the object and bounces back can be found by solving this equation for t:

ct = vt + d

and the time the second pulse bounces back can be found by solving this equation:

c (t - t_0 ) = vt + d

Solving the first equation gives t = d/(c-v), solving the second gives t = (d + ct_0 )/(c-v). So, the difference between these times is ct_0 / (c - v), during which time the object will have moved further away by a distance of vct_0 / (c - v), so the second pulse has that much further to get back, and since it also travels back at c according to SR, this will add another vt_0 / (c - v) to the time it takes to return. So, the total time interval between the return of the two pulses will be (ct_0 / (c - v)) + (vt_0 / (c - v)) = t_0 (c + v)/(c - v). So if you measure the time interval between the pulses being emitted as t_0, and the time interval between them returning as t_1, then you can solve t_1 = t_0 (c + v)/(c - v) for v to get an equation that tells you the object's velocity in terms of these two time intervals, giving v = c (t_1 - t_0) / (t_1 + t_0 ). So that's how I think you could use radar signals to measure velocities in SR, tell me if you see any mistakes in my reasoning or my math. (edit: I did make a conceptual mistake, but it's fixed now)

But if you can't assume the radar signals moved at c in both directions, it's not so obvious to me how you'd use radar signals to measure an object's velocity in the LET, or why the fact that the round-trip velocity of light is still c would imply that the answer you'd get for a given object's one-way speed would be the same as the SR answer. In fact, based on the numerical example I did on the "relativity without aether" thread, I'm pretty sure the coordinate velocity of an object would not be the same in a given observer's frame if he was using the LET transform equations as it would if he was using the Lorentz transform equations. If you think it would be, what's your reasoning?

 

[Aether]

How does this method work, exactly--are you bouncing multiple pulses off an object and seeing the difference between the time interval that the pulses are emitted and the time interval that the pulses are received? If so, what equation do you solve to find the speed of the object? If you assume the radar signals travel at c in both directions as in SR, and there is a time interval of t_0 between when two pulses are emitted, then the object's distance will have increased by v t_0 between the time the first and second pulse are emitted (assume the initial distance was d), so the time the first pulse catches up to the object and bounces back can be found by solving this equation for t:

ct = vt + d

and the time the second pulse bounces back can be found by solving this equation:

c (t - t_0 ) = vt + (d + v t_0 )

Solving the first equation gives t = d/(c-v), solving the second gives t = (d + t_0 + v t_0 )/(c-v). So, the difference between these times is t_0 (1 + v) / (c - v), and since the pulses take the same amount of time to return in SR, the time interval between the return of the two pulses to where they were emitted should be 2(1+v) t_0 / (c-v). So if you measure the time interval between the pulses being emitted as t_0, and the time interval between them returning as t_1, then you can solve t_1 = 2(1+v) t_0 / (c-v) for v to get an equation that tells you the object's velocity in terms of these two time intervals, giving v = c (t_1 - 2t_0 ) / (t_1 + 2t_0 ). So that's how I think you could use radar signals to measure velocities in SR, tell me if you see any mistakes in my reasoning or my math.

But if you can't assume the radar signals moved at c in both directions, it's not so obvious to me how you'd use radar signals to measure an object's velocity in the LET, or why the fact that the round-trip velocity of light is still c would imply that the answer you'd get for a given object's one-way speed would be the same as the SR answer. In fact, based on the numerical example I did on the "relativity without aether" thread, I'm pretty sure the coordinate velocity of an object would not be the same in a given observer's frame if he was using the LET transform equations as it would if he was using the Lorentz transform equations. If you think it would be, what's your reasoning?

Please see posts #33 & 35. I'm not saying that I'm sure that the coordinate velocity would be the same in LET as in SR, I'm saying that here's an experiment (like Michelson-Morley) that gives the same result for the velocity of an object using either LET or SR because there is no question of clock synchronization (e.g., only one clock is involved). So, if we can prove that SR gives this same velocity using two clocks but LET doesn't, then that's an example where they are not empirically equivalent. That's not supposed to happen. Unless of course the difference turns out to be something entirely trivial like (v+u) in one direction, and (v-u) in the other.

 

[Hans de Vries]

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 once confronted him with the problem of a wheel and anisotropic speed.
Imagine a rigid wheel with different angular velocities at different places..

The wheel would also have a non-zero momentum into some direction while
rotating in place...

Actually. One could easily create a situation where the rotating wheel moves
in one direction while the momentum points in the other direction. A situation
which corresponds with negative mass..


Regards, Hans

 

[pervect]

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.

Sorry to muddle the waters, you're right. If we invert the standard textbook Lorentz transforms (given in your post), we can solve for T and X, getting the standard result.

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

Substituting in the equation t_{let} = T / \gamma, which defines t_let, (and I think everyone has agreed to this as being the proper equation to define t_let), we get

t_{let} = t_{sr} + v x_{sr}/c^2

from which follows your results directly, and the rest of my original argument.

 

[JesseM]

Please see posts #33 & 35.

OK, in post #33 you provide an example:

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

But what calculation did you do to get this approach speed? This works if I plug in t_0 = 1 and t_1 = 1.000996998 - 0.001000000 = 0.999996998 into my equation v = c (t_1 - t_0) / (t_1 + t_0 ) (the original equation I had in my last post was wrong, but I edited it), but this equation was derived using the assumption that the signal moved at c in both directions, I don't see how you could get an approach speed of 450 m/s without making this assumption.

I'm not saying that I'm sure that the coordinate velocity would be the same in LET as in SR, I'm saying that here's an experiment (like Michelson-Morley) that gives the same result for the velocity of an object using either LET or SR

Every experiment gives the same result using either LET or SR, that's what is meant when it's said they are "empirically equivalent". But the interpretation of the experiment is different--the timing of the radar pulses can no longer tell you the "velocity" of the object in your reference frame in LET, because "velocity" means something different.

So, if we can prove that SR gives this same velocity using two clocks but LET doesn't, then that's an example where they are not empirically equivalent.

You're still confusing physical facts with coordinate-dependent statements. All the physical facts involving the reading on your clock when the radar pulses will be the same, but you can no longer use the timing of these pulses to determine the "velocity" of the object in the same way (in particular, you can't plug the time intervals t_0 and t_1 into the equation v = c (t_1 - t_0) / (t_1 + t_0 )) because the notion of "velocity" itself is coordinate-dependent.

 

[Aether]

OK, in post #33 you provide an example: But what calculation did you do to get this approach speed? This works if I plug in t_0 = 1 and t_1 = 1.000996998 - 0.001000000 = 0.999996998 into my equation v = c (t_1 - t_0) / (t_1 + t_0 ) (the original equation I had in my last post was wrong, but I edited it), but this equation was derived using the assumption that the signal moved at c in both directions, I don't see how you could get an approach speed of 450 m/s without making this assumption.

Since we know (see post #35) that the average round trip speed of light is equal to c_0 then I did use that as a simplifying assumption when I calculated this velocity.

Every experiment gives the same result using either LET or SR, that's what is meant when it's said they are "empirically equivalent". But the interpretation of the experiment is different--the timing of the radar pulses can no longer tell you the "velocity" of the object in your reference frame in LET, because "velocity" means something different. You're still confusing physical facts with coordinate-dependent statements. All the physical facts involving the reading on your clock when the radar pulses will be the same, but you can no longer use the timing of these pulses to determine the "velocity" of the object in the same way (in particular, you can't plug the time intervals t_0 and t_1 into the equation v = c (t_1 - t_0) / (t_1 + t_0 )) because the notion of "velocity" itself is coordinate-dependent.

Is what you're saying equivalent to what DrGreg and pervect are saying?

 

[JesseM]

Since we know (see post #35) that the average round trip speed of light is equal to c_0 then I did use that as a simplifying assumption when I calculated this velocity.

Yeah, but in my derivation of the equation v = c (t_1 - t_0) / (t_1 + t_0 ) I assumed that the one-way velocity of light in each direction was c, not just that the average round-trip velocity was c. I'm pretty sure that in the non-preferred coordinate systems of the LET transform, the round-trip velocity of light would still be c but the velocity of slower-than-light objects would not necessarily be the same as what my formula says, therefore in your example it would no longer be true that the velocity of both aircraft must be 450 m/s in such a coordinate system. I think that in calculating that number, you must have implicitly or explicitly assumed that the one-way velocity of light was c...but if you think I'm wrong, could you show me the steps of how you got that number for the velocity of each airplane?

Is what you're saying equivalent to what DrGreg and pervect are saying?

I dunno, I haven't been following this thread as carefully as I have the "relativity without aether" thread...I'll look over their comments later on and see if I agree with their arguments.

 

[Aether]

Yeah, but in my derivation of the equation v = c (t_1 - t_0) / (t_1 + t_0 ) I assumed that the one-way velocity of light in each direction was c, not just that the average round-trip velocity was c. I'm pretty sure that in the non-preferred coordinate systems of the LET transform, the round-trip velocity of light would still be c but the velocity of slower-than-light objects would not necessarily be the same as what my formula says, therefore in your example it would no longer be true that the velocity of both aircraft must be 450 m/s in such a coordinate system. I think that in calculating that number, you must have implicitly or explicitly assumed that the one-way velocity of light was c...but if you think I'm wrong, could you show me the steps of how you got that number for the velocity of each airplane?

OK. It may be tomorrow before I am able to put all the steps in here, but the jist is that the round-trip speed of light is isotropic (rotation invariance) in LET as well as SR. So, I ping an airplane twice to measure \Delta x/\Delta t.

I didn't calculate this example using LET; I calculated it to show that I could measure a velocity with one clock.

I dunno, I haven't been following this thread as carefully as I have the "relativity without aether" thread...I'll look over their comments later on and see if I agree with their arguments.

OK. Thanks.

 

[pervect]

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.

I hadn't realized that you were using different methods to measure velocities for light and for material objects.

I'm not totally sure I understand all the details of your radar method, because I didn't look at it very closely - it seemed like just another complication to me. Why introduce yet another means of measuring velocities, one that is different for light and material objects? Why not just use the same definition of velocity for both? If you have a method that can measure the speed of light (one-way), it should also (if it is any good!) be able to measure the speed of non-light (one-way). If it is not any good, maybe it is better not to use it at all.

As far as subscripts go, there are two events

Event 0 is when the moving object crosses the position of clock 0. The time at which this occurs will be measured by clock 0 and is called t_sr0 or t_let0, depending on the clock synchronization method used.

Event 1 is when the moving object crosses the position of clock 1. The time will be measured by clock 1. The time at which this occurs will be measured by clock 1 and is called t_sr1 or t_let1, depending again on the clock synchronization method.

The velocity of the moving object is given by v_sr = d_sr/(t_sr1-t_sr0) or v_let = d_let/(t_let1 - t_let0). These numbers will in general be different.

 

[Aether]

I hadn't realized that you were using different methods to measure velocities for light and for material objects.

I'm not totally sure I understand all the details of your radar method, because I didn't look at it very closely - it seemed like just another complication to me. Why introduce yet another means of measuring velocities, one that is different for light and material objects? Why not just use the same definition of velocity for both? If you have a method that can measure the speed of light (one-way), it should also (if it is any good!) be able to measure the speed of non-light (one-way). If it is not any good, maybe it is better not to use it at all.

My apologies for not making that clear. One property of LET is that the round-trip speed of light is isotropic, so I presume that I can use a radar to measure a velocity with one clock just as in SR. I don't expect for two clocks to give me a different answer than the radar, so if they do then I would be worried. That's the only point of that.

As far as subscripts go, there are two events

Event 0 is when the moving object crosses the position of clock 0. The time at which this occurs will be measured by clock 0 and is called t_sr0 or t_let0, depending on the clock synchronization method used.

Event 1 is when the moving object crosses the position of clock 1. The time will be measured by clock 1. The time at which this occurs will be measured by clock 1 and is called t_sr1 or t_let1, depending again on the clock synchronization method.

The velocity of the moving object is given by v_sr = d_sr/(t_sr1-t_sr0) or v_let = d_let/(t_let1 - t_let0). These numbers will in general be different.

OK. At first I was looking for something like (t_let11-t_let10)=light signal travel-time interval on clock 1.