Speed of the light and dilation of time

[ghwellsjr]

Same difference, you are mixing up mutual time dilation (the way the two observers see each other moving) with the calculation of total elapsed time (the circular variant of the twins paradox).

You are mixing up time dilation (which is frame dependent and arbitrary) with what observers see (which is relativistic Doppler and not dependent on any arbitrarily selected frame).

Einstein went on to describe:

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.

Einstein was talking about the slow transport of a clock but in this thread it is a fast transport so the only difference is that the clock zipping around the equator will be ticking much more slowly than one at the equator.

Now, if each clock could actually see the other one (without the curvature of the Earth getting in the way), then the zipping one would see the one at the pole as ticking much more quickly all the time and the one at the pole would see the one zipping around as ticking much more slowly all the time. In the Earth's inertial frame, all the time dilation occurs for the zipping clock and it is constant. You can pick a non-inertial frame in which the zipping clock is at rest and the clock at the pole is running faster, not slower, and it is constant. Since the distance between the pole clock and the zipping clock is constant, the relativistic Doppler and the "time dilation" can be made the same.

But in this thread, the stationary clock is not at the pole but at the equator which complicates things. For one, each clock will only be able to see the other one during a small portion of the time when they are close together. During this brief period of time, you can approximate the relative motion as mutual and they each see the other ones clock as ticking faster while approaching then slower while retreating but the time dilation, based on two different approximately inertial frames for each clock will determine that the other clock is ticking much more slowly (I presume this is what HallsofIvy meant in post #21). And if they could see through the earth, they would each continue to see the other ones clock fluctuating in its tick rate, but on average, the zipping clock would see the station clock as going faster than its own and the station clock would see the zipping clock as going slower than its own. That's why I said in post #20 "on average".

I believe uniqueland wanted to avoid all these complications, especially of not being able to see the other one during the entire orbit and so he introduced a couple webcams. Now it will depend on where the mutual antenna is located as to how much fluctuation would be seen by each observer. I mentally put this antenna above the Earth's pole to eliminate any fluctuation but to be more general, I allowed for the antenna or antennas to be located anywhere and so I included "on average".

I'm really sorry that you had to make me go into all these gory details as they have nothing to do with what uniqueland is asking about and I hope it doesn't undo all the work I have been doing in trying to help him understand the answers to his questions.

 

[GAsahi]

You are mixing up time dilation (which is frame dependent and arbitrary) with what observers see (which is relativistic Doppler and not dependent on any arbitrarily selected frame).

1.The two observers are in motion wrt each other, therefore, the observe mutual time dilation (i.e. a slowdown in measured clock rate) when, according to you, they "look at each other's webcam".
2. It is when they get reunited that they notice the discrepancy on total elapsed time. The observer that had the longest spacetime trip has the lowest elapsed time.
3. You are freely mixing the two different effects.

 

[GAsahi]

But in this thread, the stationary clock is not at the pole but at the equator which complicates things. For one, each clock will only be able to see the other one during a small portion of the time when they are close together. During this brief period of time, you can approximate the relative motion as mutual and they each see the other ones clock as ticking faster while approaching then slower while retreating but the time dilation, based on two different approximately inertial frames for each clock will determine that the other clock is ticking much more slowly (I presume this is what HallsofIvy meant in post #21). And if they could see through the earth, they would each continue to see the other ones clock fluctuating in its tick rate, but on average, the zipping clock would see the station clock as going faster than its own and the station clock would see the zipping clock as going slower than its own. That's why I said in post #20 "on average".

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:

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

 

[ghwellsjr]

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 inertial clock will see the non-inertial clock ticking slower than it is ticking and the non-inertial clock will see the inertial clock ticking faster than it is ticking?

 

[GAsahi]

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 inertial clock will see the non-inertial clock ticking slower than it is ticking and the non-inertial clock will see the inertial clock ticking faster than it is ticking?[/color]

The above is your incorrect interpretation of Einstein's claim. Here is the exact quote from his paper:

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 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.

Do you understand the difference between what he's saying and what you are claiming?

 

[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.