Twin paradox (Pete and MacM)

[MacM]

You need to read the post carefully.

I have shown that your sequence only proves that the signals will be received simultaneously in a particular reference frame (frame MV) - and that is not the reference frame of either clock.

Frame MV (for Mean Velocity) is the reference frame in which clock A and clock B have equal and opposite velocities.

OK. I would have to agree that in your MV they would indeed have equal relative velocity but I point out that the velocity from your MV is only 1/2 that of the relative velocity as viewed from the respective clocks.

Further that they also have equal and relative velocity to each other considered at rest. That is the equal relative velocity that counts.

You cannot argue that I see you going 100 Mph on the highway but that you see me going only 50 Mph. Our relative velocity is either 50 Mph or it is 100 Mph for both observers. To be relative in the first instance the views must be recipocal or reversable 100%.

 

[Pete]

There's no problem with B assuming that A is still transmitting at 1MHz locally.

But, to state that that frequency is the same in B's frame (not the received frequency, mind you... we're talking the actual time between transmitted wave peaks) relies on a premise that times in A's frame are equal to times in B's frame, which is what you set out to prove!

 

[MacM]

Then you have a fundamental misconception about what relativity is.

Blatantly false.

Relativity is all about the differences between measurements made in different frames of reference.

Absolutely.

If you make all measurements in one frame of reference only, then you will never need relativity. It is only if you ever want to compare measurements between two different frames that relativity is needed.

True. But that is if you want to record the percieved view of the event and not speak of the reality of the event.

It is your mistake not mine to be calling perception reality. Nobody is disagreeing that motion does not distort the perception of reality according to your mathematics. The disagreement is in what you are calling reality.

The problem is that your clock example tries to compare local times on clock A and B using two different reference frames at once. It can't be done.

False. I know that has been your agruement but you are clearly wrong. I can, I have synchronized two reference frames in such a manner that the distorted perception is exposed.

The clocks don't tick at the same rate, as seen from the reference frame of either clock. There is one reference frame where they do tick at the same rate, but you haven't mentioned that frame yet.

No need to try and divert here. Lets address the issue.

Your failure to understand your own topic is clear by your choice of words.

Clocks do not tick in any other frame but their own. They are only observed ticking in other frames.

Show where in my sequence any stipulation is false, without also falsifying the premises of Relativity.

 

[Pete]

OK. I would have to agree that in your MV they would indeed have equal relative velocity but I point out that the velocity from your MV is only 1/2 that of the relative velocity as viewed from the respective clocks.

Further that they also have equal and relative velocity to each other considered at rest. That is the equal relative velocity that counts.

You cannot argue that I see you going 100 Mph on the highway but that you see me going only 50 Mph. Our relative velocity is either 50 Mph or it is 100 Mph for both observers. To be relative in the first instance the views must be recipocal or reversable 100%.

You are trying to work in two frames at once.
Pick one, and only one:
you are at 100Mph, and I am at 0Mph
you are at 0Mph, and I am at -100Mph
you are at 50Mph, and I am at -50Mph
you are at 200Mph, and I am at 100Mph
you are at x Mph, and I am at (x-100)Mph


If we say:
"you are at 100Mph and I am at 100Mph"
then we'll get the wrong results, because we're trying to work in two frames at once.

The relativistic view is that it doesn't matter in which frame we work... but it obviously does matter that we're consistent!

 

[Pete]

Clocks do not tick in any other frame but their own.

Disagree. Do clocks stop ticking when they are moving?

A clock tick is an event. Events are separated by time and/or distance.
In a clock's own frame, each tick is separated by time, but not distance.
In another frame, each tick is separated by both time and distance.

The time separation can be quantified, and is referred to as the tick rate.

They are only observed ticking in other frames.

Through observation of ticks and adjusting for signal delays, the tick rate of a clock can be determined in another frame.

Let me find an old post to show you...

 

[MacM]

You are trying to work in two frames at once.
Pick one, and only one:
you are at 100Mph, and I am at 0Mph
you are at 0Mph, and I am at -100Mph
you are at 50Mph, and I am at -50Mph
you are at 200Mph, and I am at 100Mph
you are at x Mph, and I am at (x-100)Mph


If we say:
"you are at 100Mph and I am at 100Mph"
then we'll get the wrong results, because we're trying to work in two frames at once.

The relativistic view is that it doesn't matter in which frame we work... but it obviously does matter that we're consistent!

You apparently misunderstood my post. You have just confirmed my point. "Relative Velocity" between two observers is always the same value.

My point was if you now add a third car whare the state trooper is parked at rest and I am doing the speed limit of 65 Mph and you are going 130 Mph.

While my view (your MV example) is that you have a relative velocity of 65 Mph to me and I have a 65 Mph velcotity to the trooper.

Yours and the troopers view is that you each have a 130 Mph relative velocity and assuming his car can exceed 130 Mph you just got one hell of a speeding ticket.

 

[MacM]

Disagree. Do clocks stop ticking when they are moving?

Just how do you propose a clock moves unless it is relative to some other observer? The clock sees itself always at rest in in an inertial frame.

A clock tick is an event. Events are separated by time and/or distance.

True.

In a clock's own frame, each tick is separated by time, but not distance.

True.

In another frame, each tick is separated by both time and distance.

True.

The time separation can be quantified, and is referred to as the tick rate.

False. It can be defined as the observed tick rate. Not the event tick rate in reality of the clock. The clock ticks. You only observe. It is your observation that changes not the clocks tick rate.

Through observation of ticks and adjusting for signal delays, the tick rate of a clock can be determined in another frame.

Certainly. You can make appropriate adjustments by knowing the distortion made to your observation and calculate the actual physical tick rate of clock events. That is what I have said all along. Thanks for agreeing.

Let me find an old post to show you...

Go ahead it will not change the result here.

 

[Pete]

Here we go - this is an example of how to measure a clock's tick rate in a different frame. In this case the "clock" is an EMF oscillator.

Doppler shift, signal delay, and time dilation.

A signal that was transmitted at from a source separating from us at V = 0.9c = 2.7x10^8m/s is received at 229kHz.

Let's figure out how the signal delay has affected the transmitted signal, and determine the original transmission frequency.

We receive one wave every 1/229000 seconds.
We know that the source is transmitting one wave every T seconds - but we don't know the value of T. We need to find out.

We know that the source is moves some distance (D) away between each wave.
So, each wave has to go D distance further than the last.
That means that each wave takes D/c longer to reach us.
That means that our time between receiving waves = T + D/c

Since the source is moving away at V, and transmitting one wave every T seconds, we know that D = VT

So, our time between receiving waves = T + VT/c

We know our time between receiving waves (1/229000 s).
We know V (0.9c) and c.

So, we can figure out T

1/229000 = T + 0.9T = 1.9T
T = 1/(229000x1.9) seconds = 1/435100 seconds

Great!
We know how much time between wave transmissions, so we turn it upside down to get the transmission frequency:

f = 435kHz

 

[MacM]

Ack!
So B is assuming that A is still transmitting at 1MHz?

Why did you say differently before?
Remember?

I was showing the affect of the Relavistic view. Now the reality. The reality is according to Relativity (which is a direct conflict in premises that they saty the same but then are physically different in reality).

The "thinks" part means the observed doppler shifted carrier beam without knowledge of relative motion. But in this case we know the actual operating frequency of A and hence know the doppler shifted frequency is not its true frequency and hnce anc caldulate our relative velocity.

If that were not the case that A and B are still transmitting at a common frequency then doppler shift would be meaningless and you could not calculate velocity.

So for Relativity to actually work my position is affirmed. A and B still are operating at the common frequency of 1 MHz.

 

[Pete]

You apparently misunderstood my post. You have just confirmed my point. "Relative Velocity" between two observers is always the same value.

You sequence in which you synchronized the clocks relies on two velocities measured in the same frame. Not relative velocities between two observers.

To mix analogies, your sequence shows that the doppler shifts are received simultaneously according to your car (in which myself and the tropper have equal and opposite velocities), but not in my car (in which we don't) nor in the trooper's car.

 

[MacM]

Wrong.
The claim of SR is that A's tick rate is 1/sec in A's frame, and 0.436/sec in B's frame.
B's view (0.229/sec) is not suggested to be the reality.

If you are now willing to admit that B's view is not reality then we have no diagreement. But that has not been James R's position on this. He claims B's view is reality. I am saying it is perception.

 

[Pete]

A and B still are operating at the common frequency of 1 MHz.

There is no disagreement (and never has been) that A and B are operating at the same frequency in their own frames.
You continue to assume that if A is at 1MHz in A's frame, then A must be at 1MHz in B's frame. This is the point of contention.
You're assuming that times and frequencies can be transferred unchanged from one frame to another - which is what you set out to prove!


If you work out the signal delay (as I have just done) you find differently.
If B recieves A's signal at 229kHz, and A is receding from B at 09.c, then A must be operating at 435kHz in B's frame.

(I think there are rounding errors in that 435kHz figure, since the received frequency isn't exactly 229kHz.)

 

[MacM]

Here we go - this is an example of how to measure a clock's tick rate in a different frame. In this case the "clock" is an EMF oscillator.

Doppler shift, signal delay, and time dilation.

A signal that was transmitted at from a source separating from us at V = 0.9c = 2.7x10^8m/s is received at 229kHz.

Let's figure out how the signal delay has affected the transmitted signal, and determine the original transmission frequency.

We receive one wave every 1/229000 seconds.
We know that the source is transmitting one wave every T seconds - but we don't know the value of T. We need to find out.

We know that the source is moves some distance (D) away between each wave.
So, each wave has to go D distance further than the last.
That means that each wave takes D/c longer to reach us.
That means that our time between receiving waves = T + D/c

Since the source is moving away at V, and transmitting one wave every T seconds, we know that D = VT

So, our time between receiving waves = T + VT/c

We know our time between receiving waves (1/229000 s).
We know V (0.9c) and c.

So, we can figure out T

1/229000 = T + 0.9T = 1.9T
T = 1/(229000x1.9) seconds = 1/435100 seconds

Great!
We know how much time between wave transmissions, so we turn it upside down to get the transmission frequency:

f = 435kHz

I can't remember now but I'm not sure we were still at 0.9c. The 229KHz signal was James R's number for the doppler. But if that has thrown you off sorry.

The point would be to run your calculation backwards and determine the relative velocity which reduces the 1 MHz to 229KHz. James r may have made an error but I don't recall it. I recall agreeing with his conversion so the relative velocity may be different or you have converted not using the correct doppler hift formula. VT seems Newtonian.

 

[Pete]

If you are now willing to admit that B's view is not reality then we have no diagreement. But that has not been James R's position on this. He claims B's view is reality. I am saying it is perception.

Neither James nor I have ever claimed that Doppler shift equates to time dilation.

If you bothered to read our posts, you'd find the repeated statement that Doppler shift is the result of signal delay and time dilation.

The reality is that in B's frame 1/436000 seconds pass between the transmission of each wave from A.

Your assertion that Relativity states that the doppler shifted signal equates to A's tick rate in B's frame is in ignorance of Relativity. I really wish that you would learn what Relativity is before trying to argue against it, so I wouldn't have to correct these silly mistakes. I'm embarrassed for you.

 

[Pete]

I can't remember now but I'm not sure we were still at 0.9c.

I can remember. We are at 0.9c

 

[Pete]

VT seems Newtonian.

D = VT is from the definition of velocity.

 

[MacM]

D = VT is from the definition of velocity.

Actually I just checked and indeed you are making an error.

f (Relativity) = f (Newton)/gamma.

f = f0 [(1-v/c)/(1+v/c)]^.5

For 0.9c that will give you 0.2294157339

Your error is likely forgetting that your "D" has changed in relativity.

 

[MacM]

You sequence in which you synchronized the clocks relies on two velocities measured in the same frame. Not relative velocities between two observers.

To mix analogies, your sequence shows that the doppler shifts are received simultaneously according to your car (in which myself and the tropper have equal and opposite velocities), but not in my car (in which we don't) nor in the trooper's car.

The third car is your MV and I have said it is incorrect to assume that the only equal relative velocity is from the MV perspective. It would be equal but only 1/2 have the correct value of the equal relative velocity between you and the trooper.

That ws my very point about your claim regarding the MV view.

 

[Pete]

D = VT is from the definition of velocity.

Actually I just checked and indeed you are making an error.

f (Relativity) = f (Newton)/gamma.

f = f0 [(1-v/c)/(1+v/c)]^.5

For 0.9c that will give you 0.2294157339

Your error is likely forgetting that your "D" has changed in relativity.

Mac,
Your claim to understand relativity is laughable.
Does that 0.229 figure look familiar?
It's the doppler shift. The doppler shift does not reflect A's transmitted frequency in B's frame.
A's transmitted frequency in B's frame can be determined from the doppler shifted signal, as shown earlier - it is 435kHz.

What is the time dilation factor at 0.9c, I wonder?
If relativity is consistent, it should be 0.435 (allowing for rounding error)...

 

[Pete]

The third car is your MV and I have said it is incorrect to assume that the only equal relative velocity is from the MV perspective. It would be equal but only 1/2 have the correct value of the equal relative velocity between you and the trooper.

That ws my very point about your claim regarding the MV view.

Mac,
From the point of view of your car, do your car and my car have the same speed?
From the point of view of my car, do your car and my car have the same speed?
From exactly how many points of view do your car and my car have the same speed?