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The Twin Paradox & Frame of Reference


dr_mabeuse

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look at it from the point of view of the stationary clocks

at first rocket is stationary and all the clocks are in synch with it clock onboard the rocket.

the rocket accelerates to velocity v and is time dilated by gamma

At this point, the clocks are not in sync, and more to the point, each stationary clock disagrees with the rocket clock by a different amount

 

but the clock onboard the rocket is adjusted to run at gamma times faster than normal.

from the point of view of the stationary clocks the rocket is now in synch.

You have to do more than assert this. I gave you an equation that shows that this is not true; the discrepancy between t' and t depends on distance. You have acknowledged as much. Now you insist that it's not true.

 

from the rockets point of view all the clocks are out of synch

but when the rocket and a given clock are at the same place at the same time then that given clock must agree with the clock onboard the rocket

No. Why would they agree if they originally disagreed and the clocks are running at the same apparent rate?

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each stationary clock disagrees with the rocket clock by a different amount?

 

not from the point of view of the stationary clocks

How can the rocket clock see each stationary clock as showing a different time, if each stationary clock disagrees by the same amount? How do you reconcile that with the Lorentz transform equation? x is different for each clock. The transform has to work in both directions.

 

 

Look at it this way. The Andromeda clock sees the rocket as being a day off, right?

 

But what of the clock partway along that the rocket sees as being a half a day off? Does it see the rocket as being a day off, too? Does the clock next to the rocket suddenly see it as being a day off? How does that work? How does T = 0 suddenly become T = 1 day?

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From the point of view of the stationary clocks all of the clocks are in sync and agree with the Rockets clock provided the rocket has sped up his Clock to keep up with the normal time

 

 

From the Rockets point of view things look differently

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From the point of view of the stationary clocks all of the clocks are in sync and agree with the Rockets clock provided the rocket has sped up his Clock to keep up with the normal time

Repeating yourself doesn't make this right all of the sudden. Show this with the equations of relativity.

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If you have two rockets traveling at the same speed. Their clocks are synched with each other. One is at the beginning of the trek, one at the end. According to your analysis, their clock must disagree with the stationary clocks by the same amount, because they are synched.

 

But you also state that the discrepancy differs with distance. The clock at the beginning of the travel shows equal time, while the clock at the end is off by a day.

 

You can't have both be true. This is a contradiction. You have a bad assumption in play.

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Then they will be a day apart from the point of view of the stationary clocks

 

 

Draw A Spacetime diagram. Let the x-axis a space and the y-axis be time. Draw clocks on the x-axis at time equals zero.

 

Then draw a diagonal line representing 1 simultaneous moment from the point of view of the rocket.

 

 

 

How to draw a spacetime diagram:

https://www.physicsforums.com/threads/how-to-draw-a-spacetime-diagram.314080/

Edited by granpa
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The Andromeda paradox is dependent upon a calculation that you make of when you expect to receive information about events that are happening simultaneous to "now" at a distant location. That expectation is frame dependent, and thus changes what information you will determine to be simultaneous when you receive it if you change frames.

 

At no point are you ever observing any clocks actually running backwards, or jumping backwards or otherwise moving in a direction other than forward. Once the light has reached you, or has had time to reach you regardless of whether any light actually made the trip, there is no going backwards.

 

The only thing that can go backwards is when you calculate that you will receive information that is simultaneous to something going on in your frame, but that does not mean you will see any clocks run backwards.

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At no point do you see the clocks being out of synch?

 

If you see the clocks being in sync with your clock and then you accelerate and see the clocks being out of sync and then you decelerate and see them being in sync again then at some point some of the clocks must run backwards (if they are far enough away)

Edited by granpa
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The clocks will be out of sync because some will run faster and some will run slower. But they will never run backwards.

 

At no point do you see the clocks being out of synch?

 

If you see the clocks being in sync with your clock and then you accelerate and see the clocks being out of sync and then you decelerate and see them being in sync again then at some point some of the clocks must run backwards (if they are far enough away)

Clock A and B say that it is currently 2:30 and are in sync. I slow clock B down to half speed. After two minutes, Clock A days that it is 2:32 and Clock B says it is 2:31. They are out of sync.

 

Now I return Clock B to running at it's original rate. After 2 minutes, Clock A days it is 2:34 and Clock B says that it is 2:33. They are now ticking in sync but the times are out of sync.

 

Now I slow Clock A down to half speed. After 2 minutes, Clocm A days that it is 2:35 and Clock B says it is 2:35.

 

I return Clock A back to its original tick rate. Both clocks now say that it is the same time and are moving at the same rate.

 

The clocks were in sync, then they went out of sync, then they were returned to being in sync again.

 

At no point did either clock run backwards.

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You will never see the clock at infinity or minus infinity because the light will never reach you. You are causally completely disconnected from any clock at infinity and you will never be connected to it in any way.

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At no point do you see the clocks being out of synch?

 

 

That's the second time you've asked this when nobody has made a claim remotely resembling it. "Our of sync" means the rate and the phase differ. That's all. A clock running backwards is a special case of clocks being out of sync.

At no point do you see the clocks being out of synch?

 

If you see the clocks being in sync with your clock and then you accelerate and see the clocks being out of sync and then you decelerate and see them being in sync again then at some point some of the clocks must run backwards (if they are far enough away)

 

They never get in sync again. That's one of the problems with this scenario. You need to show this rather than assert it.

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But what would you calculate it's time as?

I don't see how you could. You calculate simultaneity by figuring out when you will receive the signal that was released at the moment that you are trying to determine simultaneity with. You can check the rate at which time is passing at the distant location based on how it is moving with respect to you, and then figure out how much time will have passed at the location between what you are seeing presently and what you will see at the time you determine you will receive the signal that would have been released at the present time.

 

But for an infinitely distant clock, you don't see any time on it right now because the light from that clock hasn't had time to reach you, and you will never receive a signal that is released "right now" because the light cannot cover the infinite distance in any finite time.

 

There is literally no way I can see to establish simultaneity in any frame with an infinitely distant clock.

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I don't see how you could. You calculate simultaneity by figuring out when you will receive the signal that was released at the moment that you are trying to determine simultaneity with. You can check the rate at which time is passing at the distant location based on how it is moving with respect to you, and then figure out how much time will have passed at the location between what you are seeing presently and what you will see at the time you determine you will receive the signal that would have been released at the present time.

 

But for an infinitely distant clock, you don't see any time on it right now because the light from that clock hasn't had time to reach you, and you will never receive a signal that is released "right now" because the light cannot cover the infinite distance in any finite time.

 

There is literally no way I can see to establish simultaneity in any frame with an infinitely distant clock.

 

 

The short answer is that the calculated result would be infinite, and as such, meaningless.

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Let's consider rocket on earth, between planet P and Andromeda. Before the launch of the rocket, the clock time in the rocket on earth is synch with Andromeda clock, Earth clock, and planet P clock: 12:00 o'clock.

When the rocket accelerates, the rocket frame of simultaneity events rotates counter clockwise. This means Andromeda clock time event beyond 12:00 shows up in rocket frame of simultaneous events. And Planet P clock time event BEFORE 12:00 become part rocket frame of simultaneous events. Per rocket frame clock P runs backward from 12:00 to 11:00 during acceleration.

 

During the voyage the Andomeda clock rate runs slow relative to rocket time rate (hence f.ex. when rocket time is 13:45, Andromeda time is 14:30. When rocket ticked from 12:00 to 13:45, Andromeda clock ticked from 13:00 to 14:30.)

When rocket decelerates, the rocket frame of simultaneous events rotates clockwise making the Andromeda clock time run backwards per rocket frame. (f.ex from 14:30 to 14:00)

post-114399-0-74288000-1468353081_thumb.jpg

Edited by VandD
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"frame of simultaneous events" is an unusual phrase, but I don't think it works the way you are describing. I think it means that an even that happened at 1100 at P and at 1300 at Andromeda will be seen to be simultaneous.

 

But to see a clock run backward means the signal that left P at 1100 must overtake the one that left at 1200, and that's just ridiculous. The signals travel at c. There's no way for that to happen. It gets especially ludicrous if we work up an example where I've already gotten that signal. How would I get it again?

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"They never being in sync"......

Hmmmmm,

it influences decay rate,

which is measured in quantities per second, Bq https://en.wikipedia.org/wiki/Becquerel ,

if one second is different in different frames,

amount of decaying stuff (whatever it is, but in the case of exploding nova/supernova, it's plentiful) is completely different..

Edited by Sensei
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Let's consider rocket on earth, between planet P and Andromeda. Before the launch of the rocket, the clock time in the rocket on earth is synch with Andromeda clock, Earth clock, and planet P clock: 12:00 o'clock.

When the rocket accelerates, the rocket frame of simultaneity events rotates counter clockwise. This means Andromeda clock time event beyond 12:00 shows up in rocket frame of simultaneous events. And Planet P clock time event BEFORE 12:00 become part rocket frame of simultaneous events. Per rocket frame clock P runs backward from 12:00 to 11:00 during acceleration.

 

During the voyage the Andomeda clock rate runs slow relative to rocket time rate (hence f.ex. when rocket time is 13:45, Andromeda time is 14:30. When rocket ticked from 12:00 to 13:45, Andromeda clock ticked from 13:00 to 14:30.)

When rocket decelerates, the rocket frame of simultaneous events rotates clockwise making the Andromeda clock time run backwards per rocket frame. (f.ex from 14:30 to 14:00)

attachicon.gifclock backwards.jpg

You need to distinguish clock indication and time coordinates of simultaneity. :P

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"frame of simultaneous events" is an unusual phrase,

 

Stop nitpicking.

A frame determines which events an observer considers happening simultaneously. They all have same time coordinate. Didn't you know that?

In my spacetime diagram the event clock P showing 11:00, and Andromeda clock showing 13:00 happen simultaneously per purple frame.

 

 

I think it means that an even that happened at 1100 at P and at 1300 at Andromeda will be seen to be simultaneous.

What you think is wrong. In fact you tell me you cannot read spacetime diagrams nor what they stand for.

 

Will be seen to be? I'm not talking about when an event is literally SEEN by an observer. I talk about when an event happens for an observer. You should know the difference, don't you?

 

 

 

 

But to see a clock run backward

Once more: I'm not talking about SEEING a clock run backwards. Granpa deals with a clock running backwards per observer's frame. Sure you know the difference. Don't you?

You need to distinguish clock indication and time coordinates of simultaneity. :P

Same time coordinate means happening simultaneously. Whatever the time indication on the clock is. Clock P showing 11:00, and Andromeda clock showing 13:00 happen simultaneously per purple frame. Both events have same time coordinate per purple frame.

Edited by VandD
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