Speed of the light and dilation of time

[ghwellsjr]

Is v frame variant?

Is u frame variant?

Is w frame variant?

Or is it just β that is frame variant?

 

[GAsahi]

Is v frame variant?

Is u frame variant?

Is w frame variant?

Or is it just β that is frame variant?

All are frame variant.

 

[ghwellsjr]

Easy, in frame F, the realtive speed between source and receiver is v. In frame F', moving with speed u wrt F, the relative speed is w=\frac{u \pm v}{1 \pm uv/c^2}. So, your "proof" falls appart right from the start. You missed in your calculations that \beta=v/c is frame - VARIANT. Now, that I showed your mistake, redo your post 4 correctly, please.

You are not applying the formula correctly. Here's what you should be saying:

In frame F, the source has speed u and the receiver has speed v. The relative speed between them can then be calculated using the formula for w.

If you want, you can transform the scenario into frame F' in which the source now has a new speed u' and the receiver has a new speed v' and you can calculate the relative speed between them using the formula for w but you will get the same answer as before.

I did use the formula for w in my post on the other thread and everything there is correct so you would please delete your post that it is flawed?

And please answer my request:

I invite you to do the calculation in whatever frame would convince you rather than to just continually claim that the equator clock won't see the pole clock running faster than itself.

 

[GAsahi]

If you want, you can transform the scenario into frame F' in which the source now has a new speed u' and the receiver has a new speed v' and you can calculate the relative speed between them using the formula for w but you will get the same answer as before.[/color]

Prove it. Use math. This is the best way to find your mistakes.

 

[Austin0]

Please stop butting in. If you do not know that speed is a frame variant quantity (this answers your question about the meaning of w), you should not interfere in this thread. Thank you.

Yes I understand the meaning of w and the additive velocities equation. I was unsure of which frame you were applying it from and what significance you were attaching to the result.

The velocity of anyone frame is obviously a relative quantity.
But as far as I know the internally measured relative velocity between any two frames S and S' is frame invariant.
Do you think this is incorrect?

 

[ghwellsjr]

Prove it. Use math. This is the best way to find your mistakes.

OK, I'll use the same scenario that I described in the other thread where speeds are expressed as fractions of c:

In frame F, the speed of the source is u=0.6 and the speed of the receiver is v=-0.6. The relative speed between them is:

w = (u-v)/(1-uv) = (0.6+0.6)/(1+0.6*0.6) = 1.2/1.36 = 0.882

Now let's transform the speeds into frame F' moving at j=0.365854 wrt frame F.

u' = (u-j)/(1-uj) = (0.6-0.365854)/(1-0.6*0.365854) = 0.234146/0.7804876 = 0.3

v' = (v-j)/(1-vj) = (-0.6-0.365854)/(1+0.6*0.365854) = -0.965854/1.2195124 = -0.792

Now we plug these two speeds into the formula for w to get w':

w' = (u'-v')/(1-u'v') = (0.3+0.792)/(1+0.3*0.792) = 1.092/1.2376 = 0.882

In both frames the relative speed between the source and the receive is 0.882.

Now I've done everything you've asked, will you please do what I requested:

I invite you to do the calculation in whatever frame would convince you rather than to just continually claim that the equator clock won't see the pole clock running faster than itself.

 

[GAsahi]

OK, I'll use the same scenario that I described in the other thread where speeds are expressed as fractions of c:

In frame F, the speed of the source is u=0.6 and the speed of the receiver is v=-0.6. The relative speed between them is:

w = (u-v)/(1-uv) = (0.6+0.6)/(1+0.6*0.6) = 1.2/1.36 = 0.882

Let's make things simpler, if the relative speed between source and receiver in frame F is v, so the Doppler effect in frame F is:

\frac{\nu_r}{\nu_s}=\sqrt{\frac{1-v/c}{1+v/c}}

The relative speed between source and receiver in frame F' is w=\frac{u-v}{1-uv/c^2} so the Doppler effect in frame F' is:

\frac{\nu'_r}{\nu'_s}=\sqrt{\frac{1-w/c}{1+w/c}}=\sqrt{\frac{1-u/c}{1+u/c}}\sqrt{\frac{1-v/c}{1+v/c}}Contrary to your repeated erroneous claims, the Doppler effect is not frame invariant.
As an aside, any quantity that depends on speed , such as total energy, kinetic energy, momentum, Lorentz force, relative speed between two objects is NOT frame invariant. This is textbook SR, if you get a result that shows frame invariance, you have done something wrong. I'll let you figure what you did wrong in your "proof".

 

[GAsahi]

Do you think this is incorrect?

Yes. Please stop.

 

[PAllen]

Let's make things simpler, if the relative speed between source and receiver in frame F is v, so the Doppler effect in frame F is:

\frac{\nu_r}{\nu_s}=\sqrt{\frac{1-v/c}{1+v/c}}

The relative speed between source and receiver in frame F' is w=\frac{u-v}{1-uv/c^2} so the Doppler effect in frame F' is:

\frac{\nu'_r}{\nu'_s}=\sqrt{\frac{1-w/c}{1+w/c}}=\sqrt{\frac{1-u/c}{1+u/c}}\sqrt{\frac{1-v/c}{1+v/c}}Contrary to your repeated erroneous claims, the Doppler effect is not frame invariant.
As an aside, any quantity that depends on speed , such as total energy, kinetic energy, momentum, Lorentz force, relative speed between two objects is NOT frame invariant. This is textbook SR, if you get a result that shows frame invariance, you have done something wrong. I'll let you figure what you did wrong in your "proof".

The doppler effect is completely determined by the relative velocity of source and target. Both of these velocities are frame dependent, but the relative velocity between emitter at event of emission and receiver at event of detection is frame invariant. Thus all observers agree on the doppler measured by a given detector from a given source.

Mathematically, relative speed is defined by parallel transport of 4-velocity from one event to another (in SR, this is path independent, thus unique), then dot product of transported source 4-velocity with unit 4-vector 4-orthogonal to target 4-velocity. Dot products are invariant - period. (In a standard inertial frame in SR, parallel transport leaves a vector unchanged).

[Note: ghwellsjr's calculation is correct, and the above argument simply provides the general principles which guarantee it must come out this way].

 

[PAllen]

Let's make things simpler, if the relative speed between source and receiver in frame F is v, so the Doppler effect in frame F is:

\frac{\nu_r}{\nu_s}=\sqrt{\frac{1-v/c}{1+v/c}}

The relative speed between source and receiver in frame F' is w=\frac{u-v}{1-uv/c^2} so the Doppler effect in frame F' is:

\frac{\nu'_r}{\nu'_s}=\sqrt{\frac{1-w/c}{1+w/c}}=\sqrt{\frac{1-u/c}{1+u/c}}\sqrt{\frac{1-v/c}{1+v/c}}


Contrary to your repeated erroneous claims, the Doppler effect is not frame invariant.
As an aside, any quantity that depends on speed , such as total energy, kinetic energy, momentum, Lorentz force, relative speed between two objects is NOT frame invariant. This is textbook SR, if you get a result that shows frame invariance, you have done something wrong. I'll let you figure what you did wrong in your "proof".

I gave a general argument that this is all wrong. The specific errors in the above math are as follows:

Given (for example) stationary (in unprimed frame) source, and target at speed v, you have relative velocity v.

In primed frame moving at u in the same direction as v, you now have:

target moving: (u-v)/(1-uv/c^2) = v' [as given by GAsahi ]

source moving at: -u

relative speed of source and target in primed frame is then:

(v'+u)/(1+uv'/c^2)

If one simply does the algebra, the result is v, validating gwellsjr's calculation.

 

[DrGreg]

...the Doppler effect is not frame invariant...

To show that the Doppler effect is frame invariant you don't have to do any maths at all -- just look at the definition \nu_r / \nu_s.

\nu_s is measured by the source using the source's own proper-time clock.

\nu_r is measured by the receiver using the receiver's own proper-time clock.

All observers, whatever coordinates they are using, agree on what the source and receiver's proper times are (associated with a 2\pi phase change), therefore agree on the two frequencies, therefore agree on the Doppler factor.

 

[GAsahi]

The doppler effect is completely determined by the relative velocity of source and target. Both of these velocities are frame dependent, but the relative velocity between emitter at event of emission and receiver at event of detection is frame invariant. Thus all observers agree on the doppler measured by a given detector from a given source.

Even if the relative velocity between source and receiver were frame invariant, the Doppler effect is not frame invariant by virtue of the fact that:

\frac{\nu_o}{\nu_r}=\frac{1-\frac{v}{c} cos (\theta)}{\sqrt{1-(v/c)^2}}

\theta, the angle between the relative velocity and the wave vector is known not to be frame invariant (see the aberration formula).

Mathematically, relative speed is defined by parallel transport of 4-velocity from one event to another (in SR, this is path independent, thus unique), then dot product of transported source 4-velocity with unit 4-vector 4-orthogonal to target 4-velocity. Dot products are invariant - period. (In a standard inertial frame in SR, parallel transport leaves a vector unchanged).

Dot product of 4-vectors is indeed frame-invariant. Unfortunately, the vectors involved in the Doppler effect are THREE - vectors, not 4-vectors:

\frac{\nu_o}{\nu_r}=\frac{1-\frac{\vec{v}\vec{e}}{c}}{\sqrt{1-(v/c)^2}}

whre \vec{e} is the wave vector. The dot product of two 3-vectors is not frame invariant.

 

[jartsa]

Let us consider observer A circling around observer B, and one instantaneous moment of this situation. It looks like A is just passing B at high speed, so we can guess A and B see a mutual time dilation.

 

[PAllen]

Even if the relative velocity between source and receiver were frame invariant, the Doppler effect is not frame invariant by virtue of the fact that:

\frac{\nu_o}{\nu_r}=\frac{1-\frac{v}{c} cos (\theta)}{\sqrt{1-(v/c)^2}}

\theta, the angle between the relative velocity and the wave vector is known not to be frame invariant (see the aberration formula).

This is wrong too. The angle here is the angle as measured in the comoving inertial frame of the detector. This is frame invariant.

Let me ask this: Do you really think the measurement on a given instrument depends on who looks at the instrument?

E
Dot product of 4-vectors is indeed frame-invariant. Unfortunately, the vectors involved in the Doppler effect are THREE - vectors, not 4-vectors:

\frac{\nu_o}{\nu_r}=\frac{1-\frac{\vec{v}\vec{e}}{c}}{\sqrt{1-(v/c)^2}}

whre \vec{e} is the wave vector. The dot product of two 3-vectors is not frame invariant.

This is incorrect. The doppler is determined by the source 4-velocity (at emission event) represented in the basis of the target frame at event of detection. This is determined by dot products of the source 4-velocity with the target basis unit vectors, as I previously explained. This result is manifestly frame invariant.

 

[DeepSpace9]

I totally understand how moving at the speed of light can make you travel into the future, but how does it make you go back in time?

 

[GAsahi]

This is wrong too. The angle here is the angle as measured in the comoving inertial frame of the detector. This is frame invariant.

...measured from an arbitrary frame in motion wrt the frame of the detector this angle changes (via the aberration formula).

Let me ask this: Do you really think the measurement on a given instrument depends on who looks at the instrument?

Wrongly formulated question, not "who looks at the instrument" but "from what frame is the measurement made". This is the essence of judging frame invariance.

This is incorrect.

Look up p.62, formula (90), "The theory of relativity", C. Moller.
Look up p.114, formula (311a) , Theory of relativity", W. Pauli,
Look up...

 

[PAllen]

...measured from an arbitrary frame in motion wrt the frame of the detector this angle changes (via the aberration formula).

The only measurement that matters for Doppler is the angle measured from the detector's frame.

.

Wrongly formulated question, not "who looks at the instrument" but "from what frame is the measurement made". This is the essence of judging frame invariance.

The measurement is made by a detector. Any measurement by any detector is frame invariant.

.Look up p.62, formula (90), "The theory of relativity", C. Moller.
Look up p.114, formula (311a) , Theory of relativity", W. Pauli,
Look up...

I read Pauli 40 years ago. I think I understand what he wrote perfectly well.

 

[GAsahi]

The only measurement that matters for Doppler is the angle measured from the detector's frame.

From the perspective of another frame, moving wrt. the detector, the angle changes.

I read Pauli 40 years ago. I think I understand what he wrote perfectly well.

This is not a valid answer, the formula I cited can be found in both Moller and Pauli, and in any standard textbook, perhaps it is time for you to read Pauli again.

 

[PAllen]

From the perspective of another frame, moving wrt. the detector, the angle changes.

That is not relevant. You misinterpret the physics of the Doppler formula. The angle in the formula is the angle as measured from the detector frame.

This is not a valid answer, the formula I cited can be found in both Moller and Pauli, and in any standard textbook, perhaps it is time for you to read Pauli again.

The formula for Doppler is not the issue. The issue is your misinterpretation of the variables in the formula. I certainly don't need to re-read Pauli. I have a copy and enjoy it regularly. The issue, again, is physical understanding of the formulas in it.

 

[GAsahi]

That is not relevant. You misinterpret the physics of the Doppler formula. The angle in the formula is the angle as measured from the detector frame.

No one denies that. How does the formula transform from the perspective of a moving frame , this is the issue being discussed.

This is not a valid answer, the formula I cited can be found in both Moller and Pauli, and in any standard textbook, perhaps it is time for you to read Pauli again.

The formula for Doppler is not the issue. The issue is your misinterpretation of the variables in the formula. I certainly don't need to re-read Pauli. I have a copy and enjoy it regularly. The issue, again, is physical understanding of the formulas in it.

The "variable" in the formula is , contrary to your claim, a 3-vector, not a 4-vector. While the dot product of two 4-vectors is invariant, the dot product of 3-vectors is not. I suggest that you re-read Pauli. While you are at it, read (or get) Moller. His book is better than Pauli's.

 

[PAllen]

No one denies that. How does the formula transform from the perspective of a moving frame , this is the issue being discussed.






The "variable" in the formula is , contrary to your claim, a 3-vector, not a 4-vector. I suggest that you re-read Pauli. While you are at it, read (or get) Moller. His book is better than Pauli's.

The 3-vector in the formula is constructed by taking a 4-vector's dot products with the basis vectors of the detector frame. These dot products are frame invariant.

Let's focus on physics. If you say two detectors in different states of motion measure different Doppler, of course, this is true. It is also true that a given detector's measurement of Doppler is invariant. This is about all there is to the physics of the situation. Do you dispute any of this? If so, you are simply wrong.

 

[GAsahi]

The 3-vector in the formula is constructed by taking a 4-vector's dot products with the basis vectors of the detector frame. These dot products are frame invariant.

There is no 4-vectors in any of the two derivations I cited. You can check your copy of the Pauli book. Even better, get the Moller book, it is a better book.

Let's focus on physics. If you say two detectors in different states of motion measure different Doppler, of course, this is true.

So, are we done here?

 

[PAllen]

There is no 4-vectors in any of the two derivations I cited. You can check your copy of the Pauli book. Even better, get the Moller book, it is a better book.
So, are we done here?

I don't know. Gwellsjr was simply stating that the relative velocity and doppler between a given source and detector was frame invariant, and calculated a specific example showing this. Everything about his calculation and interpretation is correct. You disputed this. So I don't know if we are done. Gwellsjr said nothing about two different detectors in different states of motion measuring the same thing.

[as for 3-vectors versus 4-vectors: an event at a world line has an associated tangent 4-vector. If you express this in a particular basis, you get a 3-vector (3-velocity). The way to get this expression is to form dot products of the 4-velocity with a frame's basis vectors. The basis vector's of a detector's frame are unit 4-vectors, and the dot products are invariant. I would readily state that Pauli does not use this more modern language, but (unlike you) he makes no mistakes at all about the physics of the situation under discussion. He understands that the variables in the Doppler formula relate to measurements in the detector frame, and thus are determined purely by the motion of detector and source, and not by the motion of some other observer.]

 

[ghwellsjr]

Post #24:

Yes, on average, your son would see you in slow motion and you would see all the people in fast motion, but I wouldn't say "thousands of times actual Earth speed", just thousands of times your speed.

This is false, time dilation is symmetrical, each observer sees the other in slow motion.

Post #54:

One thing about Doppler, since it is a local effect, you can calculate it using any frame and it will be the same in all other frames. I realize that I have not calculated it in a non-inertial frame in which the equator clock is at rest but I already know the answer from the frame in which the pole clock is at rest so I don't have to do all that extra work. But if you don't believe it, then I invite you to do the calculation in whatever frame would convince you rather than to just continually claim that the equator clock won't see the pole clock running faster than itself.

And since I have complied with your request, can you please comply with mine:

I invite you to do the calculation in whatever frame would convince you rather than to just continually claim that the equator clock won't see the pole clock running faster than itself.

And please answer my request:

I invite you to do the calculation in whatever frame would convince you rather than to just continually claim that the equator clock won't see the pole clock running faster than itself.

Now I've done everything you've asked, will you please do what I requested:

I invite you to do the calculation in whatever frame would convince you rather than to just continually claim that the equator clock won't see the pole clock running faster than itself.

We're not done until GAsahi performs my simple request to show that Einstein's claim that the equator clock is going more slowly than the pole clock also applies to the pole clock that it is going more slowly than the equator clock and that

time dilation is symmetrical, each observer sees the other in slow motion.

.

 

[GAsahi]

[as for 3-vectors versus 4-vectors: an event at a world line has an associated tangent 4-vector. If you express this in a particular basis, you get a 3-vector (3-velocity). The way to get this expression is to form dot products of the 4-velocity with a frame's basis vectors. The basis vector's of a detector's frame are unit 4-vectors, and the dot products are invariant. I would readily state that Pauli does not use this more modern language, but (unlike you) he makes no mistakes at all about the physics of the situation under discussion.

You are trying to move the goalposts, I posted a formula for the Doppler effect in terms of 3-vectors (post 72), you claimed that it was wrong (post 74). I gave you two references that show the formula to be correct (post 76).

He understands that the variables in the Doppler formula relate to measurements in the detector frame, and thus are determined purely by the motion of detector and source, and not by the motion of some other observer.]

There is no reference to the invariance of the Doppler formula in the Pauli text. Nor is it any reference in the Moller text.

 

[PAllen]

You are trying to move the goalposts, I posted a formula for the Doppler effect in terms of 3-vectors (post 72), you claimed that it was wrong (post 74). I gave you two references that show the formula to be correct (post 76).

No, I said your interpretation of the variables in it was wrong. In a different post, I showed where you made an error computing relative velocity between source and target in different frames, and that after correcting the error, the result is that it is the same.

 

[GAsahi]

We're not done until GAsahi performs my simple request to show that Einstein's claim that the equator clock is going more slowly than the pole clock also applies to the pole clock that it is going more slowly than the equator clock and that.

It is very simple really: Have the observer traveling around the Equator unroll a fiberoptic cable connecting him with the stationary observer. From time to time, each observer sends a pulse towards the other observer (not at the same time). Each observer will measure the other's frequency redshifted by : \sqrt{\frac{1-\omega R/c}{1+\omega R/c}}. The Doppler effect is mutual, the roles of the source and the receiver are interchangeable. You are mixing the Doppler effect with Einstein's calculation of total elapsed time. In that experiment, the roles of the inertial and the non-inertial observer are not interchangeable.

 

[GAsahi]

No, I said your interpretation of the variables in it was wrong..

Here is exactly what you said:

This is incorrect. The doppler is determined by the source 4-velocity (at emission event) represented in the basis of the target frame at event of detection. This is determined by dot products of the source 4-velocity with the target basis unit vectors, as I previously explained. This result is manifestly frame invariant.

 

[PAllen]

Here is exactly what you said:

My explanation is referring to how the 3 vectors in the formula are actually frame invariant because they are referred to the basis determined by the 4-velocity of the detector. I may not have been clear as I should have been, but the thrust of my statement was this: the definition of the variables in the formula make the result frame invariant.

 

[PAllen]

It is very simple really: Have the observer traveling around the Equator unroll a fiberoptic cable connecting him with the stationary observer. From time to time, each observer sends a pulse towards the other observer (not at the same time). Each observer will measure the other's frequency redshifted by : \sqrt{\frac{1-\omega R/c}{1+\omega R/c}}. The Doppler effect is mutual, the roles of the source and the receiver are interchangeable. You are mixing the Doppler effect with Einstein's calculation of total elapsed time. In that experiment, the roles of the inertial and the non-inertial observer are not interchangeable.

The Doppler effect is mutual between two inertial world lines (each acting as source and target relative to the other). Between an inertial world line and a non-inertial world line it is not mutual. A light source in forced circular motion around an inertial target is always red shifted. A target in forced circular motion about an inertial light source will see its light always blue shifted. The simplest way to derive this latter fact is to compute, in the inertial frame, the proper time interval for the circular target between arrival of successive circular wave fronts emitted by the central source.