What Causes Time Dilation in the Twin Paradox?

[granpa]

The "SR simultaneity rule" is not a law of physics, it's just true by convention.

it may not be a law of physics but if you use something different then you will have to rewrite the laws of physics for that frame.

after all the whole point of relativity is that the laws of physics are the same in every frame. that is where the Einstein simultaneity rule comes from.

 

[Al68]

it may not be a law of physics but if you use something different then you will have to rewrite the laws of physics for that frame.

after all the whole point of relativity is that the laws of physics are the same in every frame. that is where the Einstein simultaneity rule comes from.

Of course, but we're never required to even consider what Earth's clock reads simultaneous with any event in the ship's frame. I didn't say the rule was wrong, just that we don't need to use it.

Al

 

[Al68]

why not consider 3 grids of synchronized clocks. one for Earth frame. one for outbound rocket frame. and one for inbound rocket frame.

I like this one. No need to worry about what a distant clock in a different frame reads, nothing "weird" happens at the turnaround, and the ship can just look out the window and see what time it is in Earth's frame (local to the ship) at any time. And that clock is the time that the Earth clock would read "now" if the ship decides to return to Earth's inertial frame (stop) near that clock.

Al

 

[Fredrik]

Of course, but we're never required to even consider what Earth's clock reads simultaneous with any event in the ship's frame.

We don't have to think about simultaneity to prove that there's no paradox, or to find the correct final ages of the twins, but we have to do it if we want to explain what's wrong with the calculation that says A is younger.

why not consider 3 grids of synchronized clocks. one for Earth frame. one for outbound rocket frame. and one for inbound rocket frame.

That's what my spacetime diagram does. The jump from 7.2 years to 32.8 years is the correction that's needed for the error that's introduced by simply switching from the first of B's inertial frames to the second.

No need to worry about what a distant clock in a different frame reads, nothing "weird" happens at the turnaround, and the ship can just look out the window and see what time it is in Earth's frame (local to the ship) at any time. And that clock is the time that the Earth clock would read "now" if the ship decides to return to Earth's inertial frame (stop) near that clock.

You only need one of those "grids" (inertial frames) for that, so I assumed that he had something else in mind when he started talking about three of them.

 

[Al68]

Although the sudden jump simultaneity convention works in this particular version of the twin paradox, it's kinda weird. George Jones posted a paper with a nice simultaneity convention that's a lot smoother for B (surprisingly simple too): https://www.physicsforums.com/showthread.php?p=1893032&highlight=accelerated#post1893032

It's only weird because looking at the coordinate distance of the ship from Earth (or equivalently, clock time on earth) in the ship's frame before, during, and after the turnaround is, IMO, just a weird way to look at it. It's like the only reason to look at it this way is to have an exercise in the simultaneity rule for its own sake.

I'll have to look at that other thread.

Al

 

[granpa]

It's only weird because looking at the coordinate distance of the ship from Earth (or equivalently, clock time on earth) in the ship's frame before, during, and after the turnaround is, IMO, just a weird way to look at it. It's like the only reason to look at it this way is to have an exercise in the simultaneity rule for its own sake.

I'll have to look at that other thread.

Al

as has already been pointed out several times, we don't need simultaneity to figure out how much less the twin will age. we need it to explain why the twin that ages less observes the other twin aging more slowly on the outbound and inbound parts of the trip. that is the whole point of the paradox.

 

[Al68]

We don't have to think about simultaneity to prove that there's no paradox, or to find the correct final ages of the twins, but we have to do it if we want to explain what's wrong with the calculation that says A is younger.

Sure, if someone makes the erroneous calculation.

The jump from 7.2 years to 32.8 years is the correction that's needed for the error that's introduced by simply switching from the first of B's inertial frames to the second.

Sure, if we care about Earth's clock reading simultaneous with the turnaround in the ship's frame.

You only need one of those "grids" (inertial frames) for that, so I assumed that he had something else in mind when he started talking about three of them.

Well, the other two grids could be used to look at the ship's frame(s) from Earth's pov.

Fredrik, do you believe that the "jump" in the coordinate position of the Earth (and Earth's clock, since it's directly dependent) in the ship's frame during the acceleration is more than just a result of changing the pov/reference frame?

Al

 

[Al68]

as has already been pointed out several times, we don't need simultaneity to figure out how much less the twin will age. we need it to explain why the twin that ages less observes the other twin aging more slowly on the outbound and inbound parts of the trip. that is the whole point of the paradox.

Only if the ship's twin looks at it in this "weird" way. If he ignores what time on Earth is simultaneous with local time, he could just look out his window at a clock in the grid. That won't tell him what time it is on Earth "now", but that's OK if he's not asking. It will tell him how much time will have elapsed on Earth if he chooses to return to Earth's frame (stop) near any of the clocks. That should make it clear to him that, since each clock shows shows more elapsed time than his own (by the gamma factor), that his twin on Earth will be older whenever and wherever he returns to Earth's frame.

Al

 

[granpa]

Only if the ship's twin looks at it in this "weird" way. If he ignores what time on Earth is simultaneous with local time, he could just look out his window at a clock in the grid. That won't tell him what time it is on Earth "now", but that's OK if he's not asking. It will tell him how much time will have elapsed on Earth if he chooses to return to Earth's frame (stop) near any of the clocks. That should make it clear to him that, since each clock shows shows more elapsed time than his own (by the gamma factor), that his twin on Earth will be older whenever and wherever he returns to Earth's frame.

Al

it wouldn't seem weird to the people on the rocket. thear time would seem perfectly natural to them.

if the Earth twin accelerates and catches up with the moving twin and they spent the rest of their lives traveling at relativistic speed then Earth time would weird. its all relative.

 

[neopolitan]

It's only weird because looking at the coordinate distance of the ship from Earth (or equivalently, clock time on earth) in the ship's frame before, during, and after the turnaround is, IMO, just a weird way to look at it. It's like the only reason to look at it this way is to have an exercise in the simultaneity rule for its own sake.

I'll have to look at that other thread.

Al

Woot, that is a far more complex version of what I was thinking. Although I never thought about in terms of that prism-like effect in Figures 5 and 6, I do think I understand how it works.

Thanks for that link, Al.

cheers,

neopolitan

 

[Al68]

Woot, that is a far more complex version of what I was thinking. Although I never thought about in terms of that prism-like effect in Figures 5 and 6, I do think I understand how it works.

Thanks for that link, Al.

cheers,

neopolitan

Well, atyy provided the link, I only quoted his post.

Al

 

[Fredrik]

Fredrik, do you believe that the "jump" in the coordinate position of the Earth (and Earth's clock, since it's directly dependent) in the ship's frame during the acceleration is more than just a result of changing the pov/reference frame?

No.

(What else would it be? I can't even think of a wrong answer to that question.)

 

[Al68]

No.

(What else would it be? I can't even think of a wrong answer to that question.)

OK, something in another post made me wonder if I was missing something. My bad.

Al

 

[Dale]

We don't have to think about simultaneity to prove that there's no paradox, or to find the correct final ages of the twins, but we have to do it if we want to explain what's wrong with the calculation that says A is younger.

Q: What's wrong with the calculation that says A is younger?

A: It treats a non-inertial frame as an inertial frame.

No simultaneity needed.

 

[Fredrik]

Q: What's wrong with the calculation that says A is younger?

A: It treats a non-inertial frame as an inertial frame.

No, it doesnt. It doesn't even try to associate a coordinate system with B's world line. All it does is to combine the result of two calculations performed in two different coordinate systems. It must seem quite plausible to someone less experienced with relativity calculations that that should work, since the two frames agree that A's aging rate is 60% of B's. You clearly don't have to believe that B's path is a geodesic to think that the incorrect calculation looks correct.

Usually when someone makes the mistake of switching coordinate systems in the middle of a calculation, the reason why that doesn't work is that the two coordinate systems disagree about the specific thing you're calculating. How long did your plane trip take? Calculate it as local arrival time minus local departure time and you get the wrong result if the destination is in another time zone. The answer is wrong because the coordinate systems (time zones) disagree about the time. But in the twin paradox, the coordinate systems don't disagree about the aging rate. The only relevant thing they disagree about is simultaneity.

 

[granpa]

Q: What's wrong with the calculation that says A is younger?

A: It treats a non-inertial frame as an inertial frame.

No simultaneity needed.

but it is a fact that the moving twin perceives the stationary twin to be aging more slowly both on the outbound and the inbound parts of the trip. how do you explain this to a student?

 

[JesseM]

but it is a fact that the moving twin perceives the stationary twin to be aging more slowly both on the outbound and the inbound parts of the trip. how do you explain this to a student?

But he doesn't really "perceive" this in any direct sense (that certainly isn't what he sees), it's just that during each part of the trip, if he uses an inertial coordinate system where he is at rest during that phase, then in that coordinate system the other twin will be aging more slowly than himself. I would just explain to the student that you can't combine the elapsed ages for the inertial twin in the two coordinate systems for each leg of the trip, because the definition of simultaneity used in the first coordinate system is different from the definition in the second, so the inertial twin's age at the moment of the turnaround in the first one is very different from the inertial twin's age at the moment of the turnaround in the second one.

 

[granpa]

But he doesn't really "perceive" this in any direct sense (that certainly isn't what he sees),

it is exactly what he 'sees' as long as he takes light travel time into account. where are you getting the idea that it isnt. this is simple relativity.

 

[Fredrik]

I would just explain to the student that you can't combine the elapsed ages for the inertial twin in the two coordinate systems for each leg of the trip, because the definition of simultaneity used in the first coordinate system is different from the definition in the second,

What we're really discussing here is if the second half of the quoted text above needs to be included at all. DaleSpam is of the opinion that all we need to say is that there's no inertial frame in which B is stationary during the whole trip. "The End". My opinion is that this doesn't really explain why it's wrong to just use the time dilation formula on the two straight parts of B's world line separately. I think the only thing that can explain that is what you just said.

 

[Dale]

No, it doesnt. It doesn't even try to associate a coordinate system with B's world line.

Yes it does, specifically it tries to associate a coordinate system where B is at rest the whole time (B's world line is straight and vertical at all points). That is a non-inertial coordinate system.

All it does is to combine the result of two calculations performed in two different coordinate systems.

If you want to combine results performed in two different coordinate systems you must always properly transform your results from one into the other. Since you don't explicitly perform a coordinate transform you are implicitly working in a single coordinate system and that coordinate system is non-inertial.

 

[granpa]

so you accept that the moving twin sees the stationary twin aging more slowly during the outbound part of his trip and also during the inbound part of his trip. and you accept that it is the fact that he accelerates during the turn around that causes him to actually age less. so what exactly are you arguing?

 

[Dale]

but it is a fact that the moving twin perceives the stationary twin to be aging more slowly both on the outbound and the inbound parts of the trip. how do you explain this to a student?

Again, this is the same point I have been making all along. While the twins are not together they cannot make local comparisons of their clocks, so any "relative aging" claims are actually statements about the time coordinate in a given reference frame. Any reference frame where the stationary twin is aging more slowly on both legs is a non-inertial reference frame.

so you accept that the moving twin sees the stationary twin aging more slowly during the outbound part of his trip and also during the inbound part of his trip.

Only in a non-inertial reference frame.

and you accept that it is the fact that he accelerates during the turn around that causes him to actually age less.

The fact that he accelerates during the turn around is what indicates that his rest frame is non-inertial.

so what exactly are you arguing?

I am arguing that the spacetime geometric approach is completely sufficient for resolving the paradox because the metric is different in non-inertial reference frames and you cannot get a twin paradox in flat spacetime without using a non-inertial reference frame.

IMO, it is more important to teach a student to identify and avoid non-inertial reference frames than to teach them about confusing and arbitrary simultaneity conventions that can arise in non-inertial reference frames. Most students struggle with the relativity of simultaneity more than any other concept even in inertial reference frames. The extra confusion of simultaneity conventions in non-inertial frames is not necessary to the resolution and therefore should be avoided.

 

[granpa]

Again, this is the same point I have been making all along. While the twins are not together they cannot make local comparisons of their clocks, so any "relative aging" claims are actually statements about the time coordinate in a given reference frame.

they are statements about what each will actually see. while they are not accelerating

Any reference frame where the stationary twin is aging more slowly on both legs is a non-inertial reference frame.

that is exactly what we are saying. why do you say that as though it disagrees with what we are saying.
non-inertial=accelerating. acceleration=change in simultaneity.

 

[granpa]

if we assume that the laws of physics are the same in every frame then we must conclude that simultaneity is lost and that it follows the Einstein convention. so do you not believe that the laws of physics are the same in every frame?

 

[JesseM]

What we're really discussing here is if the second half of the quoted text above needs to be included at all. DaleSpam is of the opinion that all we need to say is that there's no inertial frame in which B is stationary during the whole trip. "The End". My opinion is that this doesn't really explain why it's wrong to just use the time dilation formula on the two straight parts of B's world line separately. I think the only thing that can explain that is what you just said.

If this is an accurate summary of DaleSpam's view I think I'd agree with you and disagree with DaleSpam--after all, if we want to integrate along a curve to find its length from point A to C in Euclidean space, we're free to pick some point B along the curve between A and B, and use one cartesian coordinate system to do an integral which gives us the length from A to B, and a different cartesian coordinate system to do an integral which gives us the length from B to C, and then add these two lengths. Here we are not using a single non-cartesian coordinate system, but rather adding results from two different cartesian coordinate systems, which is perfectly valid in this situation. If different inertial frames in SR agreed on simultaneity you could do something similar in the twin paradox, using one inertial frame to find the time elapsed on the inertial twin's clock between the moment the other twin departed and the moment the other twin turned around, and using a different inertial frame to find the time elapsed on the inertial twin's clock between the moment the other twin turned around and the moment they reunited. So knowing about the relativity of simultaneity is key to understanding why you can't just add partial elapsed times from two different frames in this way to get the total elapsed time.

 

[JesseM]

it is exactly what he 'sees' as long as he takes light travel time into account. where are you getting the idea that it isnt. this is simple relativity.

Light travel time depends on knowing the distance that the signal was when it was emitted, and this depends on your choice of coordinate system as well. I object to your use of the word "perceive", which makes it sound like it's just a straightforward observation that doesn't depend on choosing a particular coordinate system in which to define measurements. For example, if the twin that turns around uses an inertial coordinate system where he is at rest during the outbound phase, and then continues to use that same coordinate system during the inbound phase (rather than switching to a new coordinate system where he is at rest during the inbound phase), then he will find that the inertial twin's clock is ticking faster than his own during the inbound phase, not slower. This will be true even if the only thing he uses the coordinate system for is to calculate the distance of the inertial twin from himself when the light from each clock tick was emitted, in order to figure out how long ago each tick "really" happened by subtracting the light travel time.

 

[granpa]

the phrase 'what he sees' implies that he is using his current frame in which he is at rest.

 

[JesseM]

the phrase 'what he sees' implies that he is using his current frame in which he is at rest.

Most physicists seem to use the word "observes" for what is happens in a given observer's rest frame, while "sees" refers to actual visual appearances.

 

[granpa]

lol.

 

[JesseM]

Well, whatever you think of the terminology, my original point stands: "perceives" makes it sound too physical, when in fact your statement depends on the non-inertial observer first picking one coordinate system during the outbound leg and another during the inbound leg, any time you have a non-inertial observer the choice of what coordinate system(s) represent his "perceptions" is pretty arbitrary. I could equally well invent a coordinate system where the non-inertial observer is at rest and in which the inertial twin's clock alternates between ticking faster and slower, and then say based on this that the non-inertial twin "percieves" the inertial twin's alternates between fast and slow ticking throughout the journey--this statement would be no more or less physical than your own.