Speed of the light and dilation of time

[ghwellsjr]

OK, if you insist then we'll take this in smaller steps:

Do you agree with Einstein's claim that if we had one inertial clock and a second non-inertial clock always equidistant from the first one but traveling at some speed in a circle, then the non-inertial clock will tick "more slowly" than the inertial clock ticks?

 

[GAsahi]

OK, if you insist then we'll take this in smaller steps:

Do you agree with Einstein's claim that if we had one inertial clock and a second non-inertial clock always equidistant from the first one but traveling at some speed in a circle, then the non-inertial clock will tick "more slowly" than the inertial clock ticks?

Nowhere does Einstein make the above claim that you keep mis-attributting. I provided you with the exact quote from his paper.
Both clocks tick at the same rate, one second per second. The clock that traverses the longest path will show the least elapsed time. Nothing to do with any "ticking".

 

[ghwellsjr]

So if I change the word "tick" to "go", will you agree with Einstein:

Do you agree with Einstein's claim that if we had one inertial clock and a second non-inertial clock always equidistant from the first one but traveling at some speed in a circle, then the non-inertial clock will "go more slowly" than the inertial clock goes?

 

[GAsahi]

So if I change the word "tick" to "go", will you agree with Einstein:

Do you agree with Einstein's claim that if we had one inertial clock and a second non-inertial clock always equidistant from the first one but traveling at some speed in a circle, then the non-inertial clock will "go more slowly" than the inertial clock goes?

It means that the non-inertial clock will accumulate less elapsed time. I explained that to you many posts ago. You are back to comingling "clock rate" with "elapsed time".

 

[ghwellsjr]

The non-inertial clock accumulates less elapsed time than what? And over what interval?

 

[GAsahi]

The non-inertial clock accumulates less elapsed time than what? And over what interval?

See https://www.physicsforums.com/showpost.php?p=3966589&postcount=28, first part of the post.

 

[ghwellsjr]

I can understand how your post establishes an interval between the time when those two clocks were united and after the traveling clock makes a round trip and they are reunited but Einstein's non-inertial clock is on the Earth's equator and his inertial clock is at one of the Earth's poles. They were never united nor will they ever reunite. So how do you specify the interval in Einstein's claim?

 

[GAsahi]

I can understand how your post establishes an interval between the time when those two clocks were united and after the traveling clock makes a round trip and they are reunited but Einstein's non-inertial clock is on the Earth's equator and his inertial clock is at one of the Earth's poles. They were never united nor will they ever reunite. So how do you specify the interval in Einstein's claim?

You realize that the clock at rest at the pole and the one at rest on the Equator are in the same (almost) inertial frame, right?

 

[Austin0]

This is further compounding the confusion, the Doppler effect on frequency follows a different set of rules in accelerated frames. The revolving observer is continuously accelerating , so you cannot extrapolate from the Doppler effect in inertial frames. The only thing that you got right is the fact that, quantitatively, the observers notice a mutual blueshift when they approach each other. When they are separating from each other, they are experiencing a mutual redshift.
In both cases the effect is mutual i.e. you cannot have:

I think in this instance ghwellsjr is correct.
The mutual red and blue shift you are referencing here is regarding the relationship of received signal frequency to emitted (proper) frequency. Obviously valid but it does not necessarily imply that the received signals are going to be shifted in that manner relative to their own proper rates.
To put it in context:
If we assume a gamma of 31.4 and a circular path of 314 ls then when the moving clock reaches the opposite side, the inertial clock will have emitted 157 1 second signals.
100 of those signals will still be in transit meaning that 57 signals would have been received by the accelerated system. The accelerated system at this point would have emitted 5 signals and have an elapsed time of 5 sec.
Obviously this is an approximation based on the assumption of angular velocity close to c.
but aside from that is there anything fundamentally wrong with this picture??

 

[ghwellsjr]

You realize that the clock at rest at the pole and the one at rest on the Equator are in the same (almost) inertial frame, right?

Almost, but how does that answer my question?

 

[GAsahi]

Almost, but how does that answer my question?

It tells you that there is no difference , the observer at the pole in the Einstein scenario might as well been located on the Equator. You couldn't figure this out by yourself?

 

[abanks]

It tells you that there is no difference , the observer at the pole in the Einstein scenario might as well been located on the Equator. You couldn't figure this out by yourself?

Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.


http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[GAsahi]

Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.


http://www.fourmilab.ch/etexts/einstein/specrel/www/

You obviously do not understand which are the observers in discussion.

 

[ghwellsjr]

It tells you that there is no difference , the observer at the pole in the Einstein scenario might as well been located on the Equator. You couldn't figure this out by yourself?

Finally, you have answered my earlier question, that you do not agree with Einstein when he said:

Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.

You obviously do not understand which are the observers in discussion.

Einstein said nothing about any observers because all observers agree that there is a difference in how the two clocks "go", the one at the equator going slower than the one at the pole, "under otherwise identical conditions", meaning we are to ignore other influences on the clocks such as gravity, temperature, humidity, etc.

Are you sure you want to continue with this position?

 

[GAsahi]

Finally, you have answered my earlier question, that you do not agree with Einstein when he said:Einstein said nothing about any observers because all observers agree that there is a difference in how the two clocks "go", the one at the equator going slower than the one at the pole,

You are pretty consistent in mistaking elapsed time for clock rate.

"under otherwise identical conditions", meaning we are to ignore other influences on the clocks such as gravity, temperature, humidity, etc.

Are you sure you want to continue with this position?

If the only thing that you are able is misinterpreting Einstein 1905 paper, I do not see much point continuing this discussion.

 

[DaveC426913]

Point of order: GAsahi, on PF, your arguments stand on their merits, not on condescension or sarcasm. ghwellsjr is disagreeing as strongly as you, but is managing to remain polite and address your arguments, not your person.

Carry on.

 

[ghwellsjr]

I think in this instance ghwellsjr is correct.
The mutual red and blue shift you are referencing here is regarding the relationship of received signal frequency to emitted (proper) frequency. Obviously valid but it does not necessarily imply that the received signals are going to be shifted in that manner relative to their own proper rates.
To put it in context:
If we assume a gamma of 31.4 and a circular path of 314 ls then when the moving clock reaches the opposite side, the inertial clock will have emitted 157 1 second signals.
100 of those signals will still be in transit meaning that 57 signals would have been received by the accelerated system. The accelerated system at this point would have emitted 5 signals and have an elapsed time of 5 sec.
Obviously this is an approximation based on the assumption of angular velocity close to c.
but aside from that is there anything fundamentally wrong with this picture??

There's nothing wrong with your picture but I think it needs more explanation. You have covered the first half of the loop from the time the moving clock leaves the inertial clock until it gets half way around during which time it will have received 57 signals and emitted 5 which is an average red Doppler shift of 1/11.4 but now you should continue for the other half of the trip by saying that it will receive the 100 signals that were in transit, plus another 157 signals for a total of 257 while still only emitting 5 for a blue Doppler shift of 51.4. So the ratio per loop of the total signals received to the total signals emitted is 31.4 which is equal to gamma. But what is not obvious is that the maximum Doppler shift will occur at the closest approach of the two clocks where the Doppler shift will equal approximately two times gamma or 62.8 and then flip to the reciprocal as it passes. And the minimum Doppler shift occurs at the other side where it equals zero. These are all Dopplers as observed by the moving clock. The Dopplers as observed by the inertial clock are similar but reciprocals and skewed in time so they are not symmetrical as the moving clock goes around the circle and much more difficult to analyze and describe. Suffice it to say that the average is the reciprocal of gamma.

However, that scenario only serves to show that the red and blue shifts vary tremendously during the course of the loop with only the average equaling gamma for the moving clock and one over gamma for the inertial clock.

I think a better scenario would be similar to the one Einstein ended section 4 with which is to put the inertial clock in the center of the circle so that it is always 50 ls away from the circular path of the moving clock. This means that the number of signals in transit going from the inertial clock to the moving clock is 50 and has no bearing on the Doppler shift, it is merely gamma for the moving clock all the time. The number of signals in transit going from the moving clock to the inertial clock is 50/31.4 or barely more than 1 but again has no bearing on the Doppler and is the reciprocal of gamma for the inertial clock. In other words, the inertial clock sees the moving clock ticking slow all the time while the moving clock sees the inertial clock ticking fast all the time. This is the claim that Einstein made.

 

[GAsahi]

In other words, the inertial clock sees the moving clock ticking slow all the time

Correct.

while the moving clock sees the inertial clock ticking fast all the time. This is the claim that Einstein made.

Incorrect, this is the claim that you mis-attribute to Einstein. Nowhere in his text does he make such a claim. Here is the complete quote:

"From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by $\frac{1}{2}tv^2/c^2$(up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.

It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide.

If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the traveled clock on its arrival at A will be $\frac{1}{2}tv^2/c^2$ second slow. Thence we conclude that a balance-clock7 at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

Besides , you keep changing scenarios, sometimes your inertial observer is in the center of the circle, resulting into a transverse Doppler effect of $\tau=\frac{t}{\gamma}$, other times your inertial observer is located on the Equator, resulting into a Doppler effect due to rotation (you will need to learn how the Doppler effect operates in accelerated frames) of $\nu_{observed}=\nu_{source}\sqrt{\frac{1-\omega R/c}{1+\omega R/c}}$ or
$\nu_{observed}=\nu_{source}\sqrt{\frac{1+\omega R/c}{1-\omega R/c}}$ depending on direction.
To make matters even worse, you also co-mingle the Doppler effects with the calculation of total elapsed time (that is , what Einstein did in his paper) $\tau=\frac{t}{\gamma}$ , hich happens to have the same formula as the one used by the transverse Doppler effect.

 

[ghwellsjr]

In other words, the inertial clock sees the moving clock ticking slow all the time

Correct.

Well, although you finally agree with this statement, it is not what Einstein said. He said the equator clock goes slower than the pole clock.

while the moving clock sees the inertial clock ticking fast all the time. This is the claim that Einstein made.

Incorrect, this is the claim that you mis-attribute to Einstein. Nowhere in his text does he make such a claim.

But to say that the equator clock goes slower than the pole clock is to also say that the pole clock goes faster than the equator clock. So if you allow me to extrapolate what Einstein specifically said about the comparison of the rates at which the clocks tick to saying that one clock can see the other one ticking slower then you shouldn't have any problem allowing me to extrapolate in the other direction and to say that the other clock sees the one clock ticking faster.

Here is the complete quote:

"From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by $\frac{1}{2}tv^2/c^2$(up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.

It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide.

If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the traveled clock on its arrival at A will be $\frac{1}{2}tv^2/c^2$ second slow. Thence we conclude that a balance-clock7 at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

I am well aware of what Einstein said and I am well aware that when you have two inertial clocks with a relative speed, then the time dilation effect is reciprocal and you cannot say that one clock is ticking slower than the other one without specifying the frame in which this is true. However, when you have one inertial clock and a second non-inertial one that moves in a circle and continually returns to the first one, then you can make a frame independent statement about how the clocks, on average, tick at different rates, with the non-inertial one ticking slower (and accumulating less time per lap) than the inertial clock which, of course, means that the inertial clock, on average, is ticking faster (and accumulating more time per lap) than the non-inertial one.

Besides , you keep changing scenarios, sometimes your inertial observer is in the center of the circle, resulting into a transverse Doppler effect of $\tau=\frac{t}{\gamma}$, other times your inertial observer is located on the Equator, resulting into a Doppler effect due to rotation (you will need to learn how the Doppler effect operates in accelerated frames) of $\nu_{observed}=\nu_{source}\sqrt{\frac{1-\omega R/c}{1+\omega R/c}}$ or
$\nu_{observed}=\nu_{source}\sqrt{\frac{1+\omega R/c}{1-\omega R/c}}$ depending on direction.

I'm only talking about these two different scenarios because Einstein talked about them. I agree that being able to define the Doppler in the first of Einstein's scenarios is extremely complicated and I never tried to do that. I said it fluctuates wildly but the average is easy to calculate.

To make matters even worse, you also co-mingle the Doppler effects with the calculation of total elapsed time (that is , what Einstein did in his paper) $\tau=\frac{t}{\gamma}$ , hich happens to have the same formula as the one used by the transverse Doppler effect.

I'm not co-mingling the Doppler effect with the elapsed time, I'm co-mingling it with the "time dilation" and that's because Relativistic Doppler is composed of two parts, the "time dilation" part and the part for the light transit time. For two inertial observers, that second part is continuously changing but specifically for Einstein's second scenario, that second part is unchanging and so has no bearing on the Doppler and so the Doppler and the "time dilation" factors are the same. But you seem unwilling or unable to see that it works both ways so that the equator clock can continuously see the pole clock as ticking faster than itself.

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.

 

[GAsahi]

I am well aware of what Einstein said

Then, you should stop making up claims about what he said. I posted the complete paragraph, nowhere in the original paper does he say what you claim he says. You need to stop mis-attributting your erroneous claims to Einstein.
There is a simple way of putting an end to this, since you never put your claims in a mathematical form, why don't you write the mathematical formalism supporting your claims? For example, you claim:

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.

This is , of course, false. The Doppler effect is a function of speed. Speed is frame-variant, so , the effect cannot be "the same in all frames". Write the math and you'll prove it to yourself.

 

[ghwellsjr]

I am well aware of what Einstein said

Then, you should stop making up claims about what he said. I posted the complete paragraph, nowhere in the original paper does he say what you claim he says. You need to stop mis-attributting your erroneous claims to Einstein.
There is a simple way of putting an end to this, since you never put your claims in a mathematical form, why don't you write the mathematical formalism supporting your claims? For example, you claim:

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.

This is , of course, false. The Doppler effect is a function of speed. Speed is frame-variant, so , the effect cannot be "the same in all frames". Write the math and you'll prove it to yourself.

Isn't that interesting: "I invite you to do the calculation in whatever frame would convince you" but instead you challenge me to do the same thing. I already have done this numerous times on this forum. For example, see A&B moving in opposite directions @ 0.6 c comparison with sound, especially post #4.

Now, can you please show me any example where the relative speed between two inertial observers traveling inline comes out differently in different frames and then can you show me any example where the Doppler that these two observers see of the other one is different in different frames?

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.

 

[Austin0]

Originally Posted by Austin0

I think in this instance ghwellsjr is correct.
The mutual red and blue shift you are referencing here is regarding the relationship of received signal frequency to emitted (proper) frequency. Obviously valid but it does not necessarily imply that the received signals are going to be shifted in that manner relative to their own proper rates.
To put it in context:
If we assume a gamma of 31.4 and a circular path of 314 ls then when the moving clock reaches the opposite side, the inertial clock will have emitted 157 1 second signals.
100 of those signals will still be in transit meaning that 57 signals would have been received by the accelerated system. The accelerated system at this point would have emitted 5 signals and have an elapsed time of 5 sec.

There's nothing wrong with your picture but I think it needs more explanation.

Well of course it is your picture as per the conditions of the original scenario. I did not provide more explanation as I assumed the extrapolations were self evident.

GAsahi was claiming that your claim was incorrect because Doppler shift was reciprocal so I provided an example with your boundary conditions where this was clearly controverted.
The first half of the trip was the interval where the signals would be red shifted.
If in that interval the accelerating system received more signals than proper time it is clear that those signals were NOT red shifted relative to that systems proper time.

It is also clear that although the instantaneous relative velocity/gamma would vary with angle of motion wrt the inertial clock the same basic relationship would necessarily pertain. I.e. The accelerating system would receive more signals than sent. If those signals were video footage then as you say the motion of the inertial actors would appear sped up.





You have covered the first half of the loop from the time the moving clock leaves the inertial clock until it gets half way around during which time it will have received 57 signals and emitted 5 which is an average red Doppler shift of 1/11.4 but now you should continue for the other half of the trip by saying that it will receive the 100 signals that were in transit, plus another 157 signals for a total of 257 while still only emitting 5 for a blue Doppler shift of 51.4. So the ratio per loop of the total signals received to the total signals emitted is 31.4 which is equal to gamma. But what is not obvious is that the maximum Doppler shift will occur at the closest approach of the two clocks where the Doppler shift will equal approximately two times gamma or 62.8 and then flip to the reciprocal as it passes. And the minimum Doppler shift occurs at the other side where it equals zero. These are all Dopplers as observed by the moving clock. The Dopplers as observed by the inertial clock are similar but reciprocals and skewed in time so they are not symmetrical as the moving clock goes around the circle and much more difficult to analyze and describe. Suffice it to say that the average is the reciprocal of gamma.

However, that scenario only serves to show that the red and blue shifts vary tremendously during the course of the loop with only the average equaling gamma for the moving clock and one over gamma for the inertial clock.

I think a better scenario would be similar to the one Einstein ended section 4 with which is to put the inertial clock in the center of the circle so that it is always 50 ls away from the circular path of the moving clock. This means that the number of signals in transit going from the inertial clock to the moving clock is 50 and has no bearing on the Doppler shift, it is merely gamma for the moving clock all the time. The number of signals in transit going from the moving clock to the inertial clock is 50/31.4 or barely more than 1 but again has no bearing on the Doppler and is the reciprocal of gamma for the inertial clock. In other words, the inertial clock sees the moving clock ticking slow all the time while the moving clock sees the inertial clock ticking fast all the time. This is the claim that Einstein made.

 

[GAsahi]

Isn't that interesting: "I invite you to do the calculation in whatever frame would convince you" but instead you challenge me to do the same thing. I already have done this numerous times on this forum. For example, see A&B moving in opposite directions @ 0.6 c comparison with sound, especially post #4.

You must mean this post. I have news for you, it does not prove that the RD (Relativistic Doppler effect) is frame invariant, quite the opposite, it is frame - variant. The fact that you doctored some elementary calculation, does not give creed to your point, quite the opposite.

Now, can you please show me any example where the relative speed between two inertial observers traveling inline comes out differently in different frames and then can you show me any example where the Doppler that these two observers see of the other one is different in different frames?

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.

 

[Austin0]

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 start. What you did, is you missed in your calculations that $\beta=v/c$ is frame - VARIANT. Now, that I showed your mistake, redo your post 4 correctly, please.

What do you think w represents?
Taken at face value from F' if u = velocity of F wrt F' then v would represent the velocity of the source wrt F' in the additions equation.
In which case w would represent the velocity of the source as measured in F =v

On the other hand if you are running the formula from F then w would be the velocity of the source as measured in F'
Which is it?

 

[GAsahi]

What do you think w represents?
Taken at face value from F' if u = velocity of F wrt F' then v would represent the velocity of the source wrt F' in the additions equation.
In which case w would represent the velocity of the source as measured in F =v

On the other hand if you are running the formula from F then w would be the velocity of the source as measured in F'
Which is it?

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.

 

[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.

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.