How Does the Twin Paradox Affect Aging?

[JulianM]

I don't understand the question, I'm afraid. Are you asking what happens to a long object that moves with the traveling twin outbound, but doesn't turn round? Nothing happens to it - as I explained above, all of this simultaneity stuff is one or other twin interpreting their sensor readings. If you want a completely consistent interpretation for the traveling twin at turnaround you need to use a non-inertial frame. My favourite is radar coordinates - see https://arxiv.org/abs/gr-qc/0104077

No, I was trying to clarify the point about the clocks going slower as the object moves away and then faster when it comes back.

Imagine that as it returns it passes the stay-at-home twin and the traveling twin is on a reasonably long object. At the point that it comes past Iwithout stopping) the front of the object is moving away from the stay-at-home and the reasr is coming towards him.
This would seem to result in a starnge situation, which I don't understand, of the clocks, on a presumably rigid object, appearing (or in actuality?) running at different speeds according to the stay-at-home (though presumably not for the traveller?)

The purpose of my question was to better understand post #45 which describes traveling away, changing direction and then coming back. I am not clear on whether just velocity, or direction causes this.

 

[bhobba]

But what do we mean when we say that the Frame-A clocks aren't synchronized in Frames B and C?

Frame A, B and C, - forget it.

We have two frames - some inertial frame the rocket takes off from and the rocket frame. We sync the clocks before the rocket takes off - for definiteness we use slow transport ie bring the rocket ship clock next to the inertial frame clock, sync them, then slow transport it to the rocket ship. They are now synced and in the same frame. So acceleration does not affect the clocks we will use atomic clocks. Now the rocket takes off. It's accelerated, so by the definition of an inertial frame as I previously explained is not inertial - since you can't use SR in non inertial frames you must do the analysis from the inertial frame. But you are then faced with the issue the rocket not being inertial how do you apply SR to it. Simple - you approximate the non inertial rocket by a series of constant velocities of short duration. You can use SR during those short durations since its going at constant velocity. Now to the actual analysis. You use world lines as the link I gave details. To get the time in each frame you simply integrate along the world line. In the accelerated world line in the rocket you have this dτ = √1-v^2 dt where dτ is the time change in the rocket clock and dt is the time change in the inertial frame clock (you can take the very small time change as the time increments you divide the acceleration into to apply SR). In the inertial frame dτ is the time change on its clock. You integrate both to get the change in time in each frame. dτ is prety easy in the inertial frame to get the difference τ. But in the rocket ship you have √1-v^2 in front of it which lies between 0 and 1. It can't always be 1 or no acceleration will occur. So when integrated you get a time difference less than the inertial frame ie less than τ.

Its not hard. Some here are way over complicating it.

Thanks
Bill

 

[Ibix]

No, I was trying to clarify the point about the clocks going slower as the object moves away and then faster when it comes back.

Right. The clocks run at the same speed, which is not the same speed as the stay-at-home's clocks. However, they are not synchronised correctly according to the stay-at-home although they are according to the traveling twin. "Leading clocks lag" doesn’t mean that leading clocks run more slowly, just that they were set to a time earlier than clocks behind them, according to a frame where they are moving.

What you actually see depends on the Doppler shift. Clocks coming towards you appear to run fast because each successive tick happens closer to you, so there's less lag and the ticks arrive closer together. And the reverse when they move away. As Sienna pointed out, people do not use "see" consistently and sometimes mean what you actually see (Doppler shift) or how you interpret it (time dilation and a leading-clocks-lag simultaneity change).

 

[bhobba]

No, I was trying to clarify the point about the clocks going slower as the object moves away and then faster when it comes back.

During an infinitesimal time interval the rocket clock is always running either slower or the same rate. It can't run at the same rate for the entire journey otherwise it would not move at all - little alone accelerate. When you add up the infinitesimal rates in each frame (called integration) since at some instants during the journey the rate of the rocket clock is slower it must be less than the stay at home clock. See that v^2 in the equation I wrote - that means direction is irrelevant.

Again its not hard - you just need to look at it correctly.

This - what will it look like if you observe it is a distraction. Understand what I said before thinking about it. IBIX is correct - but please don't try to understand it until you understand what I said or you will likely get confused.

Thanks
Bill

 

[SiennaTheGr8]

Frame A, B and C, - forget it.

We have two frames - some inertial frame the rocket takes off from and the rocket frame. We sync the clocks before the rocket takes off - for definiteness we use slow transport ie bring the rocket ship clock next to the inertial frame clock, sync them, then slow transport it to the rocket ship. They are now synced and in the same frame. So acceleration does not affect the clocks we will use atomic clocks. Now the rocket takes off. It's accelerated, so by the definition of an inertial frame as I previously explained is not inertial - since you can't use SR in non inertial frames you must do the analysis from the inertial frame. But you are then faced with the issue the rocket not being inertial how do you apply SR to it. Simple - you approximate the non inertial rocket by a series of constant velocities of short duration. You can use SR during those short durations since its going at constant velocity. Now to the actual analysis. You use world lines as the link I gave details. To get the time in each frame you simply integrate along the world line. In the accelerated world line in the rocket you have this dτ = √1-v^2 dt where dτ is the time change in the rocket clock and dt is the time change in the inertial frame clock (you can take the very small time change as the time increments you divide the acceleration into to apply SR). In the inertial frame dτ is the time change on its clock. You integrate both to get the change in time in each frame. dτ is prety easy in the inertial frame to get the difference τ. But in the rocket ship you have √1-v^2 in front of it which lies between 0 and 1. It can't always be 1 or no acceleration will occur. So when integrated you get a time difference less than the inertial frame ie less than τ.

Its not hard. Some here are way over complicating it.

Thanks
Bill

Overcomplicated? I don't think three inertial frames is even complicated. If anything it's an oversimplification, since the rocket twin switches directions without undergoing acceleration (physically impossible). As I tried to show in my post above, the "rear clock ahead" effect is key to understanding this three-frame approach. The math is straightforward.

Of course, your "incremental acceleration" approach works too, and if I remember correctly Taylor and Wheeler use it in Spacetime Physics.

Also, SR can handle accelerated motion (and accelerated frames) just fine.

 

[bhobba]

If anything it's an oversimplification

Two frames vs three frames? To each their own I suppose.

And didn't I just explain it handles accelerated frames just fine? - but you must do it the way I described, or something similar. SR works only in inertial frames. You have to use various 'tricks' for other frames. They are not hard - but are required. Its in the very derivation of the Lorentz transformations eg:
http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

Thanks
Bill

 

[SiennaTheGr8]

Two frames vs three frames? To each their own I suppose.

And didn't I just explain it handles accelerated frames just fine? - but you must do it the way I described, or something similar. SR works only in inertial frames. You have to use various 'tricks' for other frames. They are not hard - but are required. Its in the very derivation of the Lorentz transformations eg:
http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

Thanks
Bill

Well, you said:

... since you can't use SR in non inertial frames you must do the analysis from the inertial frame. But you are then faced with the issue the rocket not being inertial how do you apply SR to it. Simple - you approximate the non inertial rocket by a series of constant velocities of short duration. You can use SR during those short durations since its going at constant velocity. ...

(which did seem somewhat contradictory to what you said next about integrating!).

 

[Janus]

When you say LOOKS do you mean it's not real?

First you have to define what you mean by "LOOKS".

In your post above, you ask about a clock that runs fast when approaching and slow when receding. In this case, LOOKS means "what you see visually" and involves Doppler shift. You get a Doppler shift whether you include Relativistic effects or not. You see a clock coming towards you run fast, because the distance between you and it is constantly decreasing and light take less and less time to cross the distance. This causes the light and information about events it carries to "scrunch up" If the clock is receding the distance is constantly increasing, the light takes longer and longer the cross the distance, and things are stretched out.

If we account for this time lag, we can get a better idea of what that clock is doing compared to our own. When we do this, then we find that whether the clock is receding or approaching, it runs slow. A lot of the time you will hear it said that the clock "appears" to slow. This is because from that clock's perspective, it is your clock that it the one that runs slow and neither clock runs slower than the other in an absolute sense. Here, "appears" does not mean "what we see with our eyes", but what we determine as to what the clock is doing.

If we go back to the post by SiennaTheGr8 you quoted, they were talking about "appears" rather than "LOOK"S. In this case, when the Rocket twin changes from traveling away from to traveling towards, he does visually see the Earth clock go from running slow to running fast, but he will determine that it was running slow in both cases.
But what SiennaTheGr8 was referring to is what happens to the Earth clock at that instant he changes direction. If we assume that he changes direction instantaneously, then the light he is getting from the Earth is the same both before and after the change, and he visually sees no change in the Earth's clock's reading. What does change is what time the Earth Clock "appears" to have (the time on the Earth clock "Now" after he accounts for the time it took the light to reach him. What happens is that the Earth clock goes from being lagging behind his clock to leading it by some time. the time he considers it to be on Earth "right now" changes even though the time he "sees" on the Earth clock doesn't change.

 

[robphy]

Can you help with the other part of my question - what happens to a long object as it passes the return point and it has one clock running fast and the other more slowly as it passes?

I don't understand the question, I'm afraid. Are you asking what happens to a long object that moves with the traveling twin outbound, but doesn't turn round? Nothing happens to it - as I explained above, all of this simultaneity stuff is one or other twin interpreting their sensor readings. If you want a completely consistent interpretation for the traveling twin at turnaround you need to use a non-inertial frame. My favourite is radar coordinates - see https://arxiv.org/abs/gr-qc/0104077

No, I was trying to clarify the point about the clocks going slower as the object moves away and then faster when it comes back.

Imagine that as it returns it passes the stay-at-home twin and the traveling twin is on a reasonably long object. At the point that it comes past without stopping) the front of the object is moving away from the stay-at-home and the rear is coming towards him.
This would seem to result in a strange situation, which I don't understand, of the clocks, on a presumably rigid object, appearing (or in actuality?) running at different speeds according to the stay-at-home (though presumably not for the traveller?)

The purpose of my question was to better understand post #45 which describes traveling away, changing direction and then coming back. I am not clear on whether just velocity, or direction causes this.

From my reading of your original question and this bolded part of your reply to Ibix,
I think this situation (of an extended object passing the home frame) will address your question.
I've drawn a spacetime diagram on rotated graph paper.

Alice is at rest.
Bob, moving with velocity (6/10)c=(3/5)c. is traveling with a "co-moving inertial" ruler whose proper length is 10.
The Back of the ruler is in blue.
The ruler can be thought of as a fleet of inertial clocks traveling with Bob... just focus on the front (where Bob is) and the Back.
The clocks on the ruler are synchronized in Bob's frame.
At event O, Alice and Bob meet. For simplicity, they have pre-arranged to have their clocks set to read "0" at this event.DOPPLER:

• Here are Alice's visual observations.
  At event O (when Alice and Bob meet), Alice sees the image of "-10" on the Back clock and the image of "0" on Bob's clock.
  
  When Alice's clock reads 6, she sees 2 on the Back clock and 3 on Bob's clock.
  • The Back clock approaching Alice had an elapsed time of 12 between her readings...
    that is, Alice watched 12 hours of Back's tv-programming in 6 hours...
    this factor of 12/6 is the Doppler factor relating frequencies ("blueshift for approaching")
  • Bob's clock receding from Alice had an elapsed time of 3 between her readings...
    that is, Alice watched 3 hours of Bob's tv-programming in 6 hours...
    this factor of 3/6 is the Doppler factor relating frequencies ("redshift for receding")
  • This factor of 2 and its reciprocal values correspond to v=(3/5)c.
• Between Alice's clock readings at 6 and 12, we highlight the 3-tick intervals from the Back clock and from Bob's clock.
  
• You can draw your own lightlike lines to see how each clock sees the images of each other's clock.
  For instance, how does Bob see the Back clock? How does the Back clock see Bob's clock?

TIME-DILATION:

• In Alice's frame of reference, her time marches on with her lines-of-simultaneity (Minkowski-perpendicular to her worldline).
  • Alice says "the event when her clock reads -5" is simultaneous with both
    • "the event when Bob's clock reads -4"
    • "the event when the Back clock reads 2"
  • Alice says "the event when her clock reads 0" is simultaneous with both
    • "the event when Bob's clock reads 0"
    • "the event when the Back clock reads 6"

• Alice says "the event when her clock reads 5" is simultaneous with both
  • "the event when Bob's clock reads 4"
  • "the event when the Back clock reads 10"

• Alice says when Bob's clock elapses 4 ticks, her clock has elapsed 5 ticks.
  Alice says when the Back clock elapses 4 ticks, her clock has elapsed 5 ticks.
  And this persists before and after the ruler passes Alice.
  This ratio 5/4 is the time-dilation factor corresponding to v=(3/5)c.
  Note that the time-dilation factor $\gamma=\frac{1}{\sqrt{1-v^2}}$ is an even function of velocity...
  so it depends on the relative-speed, not the direction (approaching vs receding).

If you analyze the situation from Bob's frame of reference with his lines-of-simultaneity (Minkowski-perpendicular to his worldline),
he will come to the same conclusion:
Bob says when Alice's clock elapses 4 ticks, his clock has elapsed 5 ticks.
And this persists before and after the ruler passes Alice (or Alice passes the ruler).
This ratio 5/4 is the same time-dilation factor corresponding to v=(3/5)c.

​

 

[PeterDonis]

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Edit: A number of off topic posts have been deleted. The original question has been sufficiently addressed, and the thread will remain closed.