What Causes Time Dilation in the Twin Paradox?

[matheinste]

Hello neopolitan.

Quote:-

---One thing that is a little confusing about all of the bent path diagrams I have seen is that they are all vertical, implying that "A" is absolutely stationary. This just isn't the case, "A" is relatively stationary, in that "A"'s inertial frame does not change.----

The fact that A's path is vertical i.e. along the time axis means that A is stationary with respect to the coordinates of the spacetime diagram. In other words the analysis is taking place from the point of view of A. This is basic.

Matheinste.

 

[Fredrik]

"B" knows that by not accelerating, "A" will take a route through spacetime which has a higher time component than anyone who is initially stationary wrt to "A", accelerates to travel away and then accelerates back towards "A".

"B" can therefore work out the correct result from first principles, without having rely on simultaneity at all.

Yes, that's the point of the approach that DaleSpam is advocating. The correct final ages follow immediately from the postulate that what a physical clock measures is the integral of \sqrt{-g_{\mu\nu} dx^\mu dx^\nu} along the curve in Minkowski space that represents the clock's motion.

One thing that is a little confusing about all of the bent path diagrams I have seen is that they are all vertical, implying that "A" is absolutely stationary. This just isn't the case, "A" is relatively stationary, in that "A"'s inertial frame does not change.

It doesn't imply that A is absolutely stationary. Any object that moves with a constant velocity is stationary in some inertial frame, and in this case it's simply convenient to draw the diagram using the inertial frame in which A is stationary.

What you can do, however, is swing "A"'s path through spacetime around in any direction,

I didn't read all the details of your argument, but a quick glance was enough for me to see that it's pointless. It's obvious that a third frame isn't going to tell you anything that you can't see by considering A's rest frame, and if you're not going to talk about simultaneity, your best option is to use DaleSpam's approach anyway.

 

[Fredrik]

This is wrong, you did, in fact, consider B's rest frame to be an inertial frame.

No, I didn't. I just used one frame for the first half and another for the second half.

you don't have to "work backwards" from the answer. I can work the example quantitatively in all three inertial frames if you wish.

You still seem to be missing the point. Maybe I'm missing your point too, but you're definitely missing mine. I can do those calculations too, obviously, but our task isn't to find multiple ways to calculate the correct final ages. It's to explain what's wrong with the incorrect solution. Your approach can only tell us that there's something wrong with it (by telling us the correct final answer), but it doesn't tell us why it's wrong. OK, it points out that there's a bend in B's path, but it doesn't explain why that matters. And you are going to have to work backwards from the answer to find the magnitude of the error (i.e. 25.6 years), unless you do what I did, which is to consider the simultaneity lines of the two inertial frames associated with B's motion.

 

[RandallB]

No, I didn't. I just used one frame for the first half and another for the second half.

But nothing is SR grants you permission to do that.
SR requires that you solve any problem like this wrt just one reference frame. Not two of your choice.
Pick the right two and you can make almost any weird thing appear in that approch including Backwards Causality.

That includes the “two straight pieces of B's path” if by that you mean the outbound reference frame and returning reference frame—those are two different frames.

SR requires you pick one of those frames or the A frame or any other fourth frame your care to make up. But your must work the problem all the way though using just one frame and translate all time and locations for other frames based on the 0 starting position and Lorentz transforms from the one reference frame. No matter what frame you pick you will get the same result on B returning to A.

 

[Dale]

No, I didn't. I just used one frame for the first half and another for the second half.

That is practically the definition of a non-inertial reference frame!

our task isn't to find multiple ways to calculate the correct final ages. It's to explain what's wrong with the incorrect solution. Your approach can only tell us that there's something wrong with it (by telling us the correct final answer), but it doesn't tell us why it's wrong. OK, it points out that there's a bend in B's path, but it doesn't explain why that matters.

Don't you see? The fact that there is a bend in B's path is the reason why the other approach is wrong. In Minkowski geometry the longest timelike interval between two points is a straight line, so a path with a bend in it is always shorter than a straight path. So the existence of a bend in a path is an important geometrical feature. The other approach is wrong geometrically because it straightens out a bent path and bends a straight path. In physical terms it is a non-inertial reference frame.

 

[neopolitan]

An early morning postscript to my post from last night.

The third observer will not work out that "A" and "B" have cashed in 40 units of spacetime, because according this observer, "A" will have converted a certain amount of spacetime into spatial translation. Therefore the amount of spacetime cashed in, according to the third observer, will be a lower value given by

sqrt ('time elapsed according to "A"' squared plus 'distance traveled according to the third observer')

and this value will be the same as the time elapsed according to the third observer.

However, the third observer will still work out that "A" and "B" cash in the same amount of spacetime overall, and "A" will have cashed in more as time, due to the fact that "A"'s single inertial path is the shortest between the start and finish points.

And, yes, this is the same solution as DaleSpam's but just worded differently.

As for Randall and asking a simultaneity based question about the relationship between "A" Proper time and "B" Proper time at the turnaround point ... is that really the crux of the twin paradox as generally stated?

I thought it was that one twin ends up older than the other twin. If the crux of the twin paradox is that you run into problems with working out Proper times at the turnaround, I would take this as a reason to avoid using simultaneity altogether when it comes to solving these sorts of problems and use the approach advocated by DaleSpam (and, to a more clumsy extent, myself). After all, Dale and I don't need to know Proper time to show that there really isn't a problem. You just shift the problem.

cheers,

neopolitan

 

[atyy]

If the crux of the twin paradox is that you run into problems with working out Proper times at the turnaround

Everyone ages according to the integral of his own proper time, under the assumption that he is an ideal clock whose time dilation depends on velocity and not acceleration. We know this is true for individual particles, and even for atomic clocks.

In an inertial frame, the observer A at spatial coordinate x=0 has proper time s_A which is the global time coordinate t of the frame ds_A²=dt². For any other observer B, accelerating or not, ds_B²=dt²-dx².

If we take the "point of view" of B, we get new global space and time coordinates p,q. By definition of taking his point of view, ds_B²=dq². Since this is not an inertial frame, the proper time of any other observer A is generally not ds_A²=dq²-dp², unless B is an inertial observer. So we can get the twin paradox even by integrating the proper time simply by using the wrong proper time formula.

In the derivation of the time dilation formula, there is a critical part where the proper time of one observer is identified with the global time axis of an inertial frame, and the proper time of the other observer is identified with the global time axis of another inertial frame. The final part of the derivation comes by noting that two inertial frames are related by a Lorentz transformation. So the presentation of the twin paradox using the time dilation formula is about proper times anyway.

It is tempting for me to want something to happen to B's clock at the bend, since surely it is his acceleration that is absolute. If B uses an ideal clock or an atomic clock, it will not be affected by acceleration. The most common non-ideal clock is the pendulum, but that will not work in outer space with no gravity (SR has trouble with gravity anyway). I guess I should make B use a quantum gravity clock?

 

[Antenna Guy]

It is tempting for me to want something to happen to B's clock at the bend, since surely it is his acceleration that is absolute.

Surely contracted meters per dilated seconds squared cannot be considered "absolute".

Regards,

Bill

 

[atyy]

Surely contracted meters per dilated seconds squared cannot be considered "absolute".

I just mean B will know that he is not an inertial observer.

 

[Al68]

Everyone ages according to the integral of his own proper time, under the assumption that he is an ideal clock whose time dilation depends on velocity and not acceleration. We know this is true for individual particles, and even for atomic clocks.

Do we? Or are you just talking about SR time dilation, not GR time dilation? It seems to me that if we're talking about an accelerated observer, we have to consider GR time dilation as well (equivalence principle).

Of course this would lead to crazy conclusions like Earth's clock running faster than the ship's clock (from the ship's POV) during the acceleration. Oh, wait...

 

[Fredrik]

Do we? Or are you just talking about SR time dilation, not GR time dilation? It seems to me that if we're talking about an accelerated observer, we have to consider GR time dilation as well (equivalence principle).

Many of us (including me) consider that a part of SR, not GR. The special theory of relativity consists of a mathematical model (Minkowski space) and a set of postulates about identifications between the real world and the model (e.g. "what a clock measures is the proper time along the curve that represents the clock's motion"). It is possible to come up with a list of such postulates that make inertial frames the only valid coordinate systems, and you could (and apparently do) call that theory SR. I think of that theory as "partial SR", "1905 SR" or "crippled SR", and the theory you get with a more complete set of identifications as "SR".

 

[Fredrik]

But nothing is SR grants you permission to do that.
SR requires that you solve any problem like this wrt just one reference frame. Not two of your choice.

I know. If you had read enough of the discussion to see what it was about, you would have known that the result I was trying to obtain was the "paradox" part of the twin paradox. That's an incorrect result, and it obviously isn't possible to obtain an incorrect result without doing something wrong.

 

[Fredrik]

Don't you see? The fact that there is a bend in B's path is the reason why the other approach is wrong. In Minkowski geometry the longest timelike interval between two points is a straight line, so a path with a bend in it is always shorter than a straight path.

No, I don't see. That last sentence tells us that B must be younger when he meets A again because his world line is bent. No disagreement there. What I don't see is how that tells us what's wrong with the naive use of the time dilation formula that concludes that A is younger.

This is what's wrong with it: It neglects what happens when you switch from one inertial frame to another in the middle of a calculation. My approach explains what happens when you do. Your approach is coordinate independent, so it can't explain it.

Your approach is the best way to quickly find out which one of the twins is right, but I still don't consider it a complete resolution of the paradox. The paradox is a mistake in a coordinate dependent calculation, so no coordinate independent approach can tell us what the mistake is.

 

[neopolitan]

Fredrik,

I note that you never answered the question about a clock on "B"'s spaceship which is designed to give "B" the real universe ageing of "A".

Do you agree that the discontinuity from 7.2 to 32.8 years is part of the incorrect way of looking at the twin scenario which leads to it appearing to be a paradox, or are you giving it as the solution?

If you agree that there is no real need to advance such a clock from 7.2 to 32.8 years at the turnaround, and that the 7.2 and 32.8 year figures are representative of incorrectly using two inertial frames as if they were one then we are not in disagreement.

cheers,

neopolitan

BTW When I say real universe ageing of "A", I mean taking into account that "A" undergoes no acceleration - it is not meant to imply that there is some absolute ageing of "A".

 

[granpa]

the point of the twins paradox isn't that one twin ages less. we know why that is. it follows from time dilation. the question is why is it that the moving twin ages less yet from his point of view it is the other twin that is aging more slowly. the answer is that there is a shift that occurs when the moving twin changes speed.

the answer you are giving is simply the answer you would get if the moving twin were to completely stop. so the 2 really agree completely.

 

[matheinste]

Hello neopolitan.

The real time and ageing for B, and the real time and ageing for A are seen differently by each other.They are both equally real but not numerically equal.

Matheinste.

 

[neopolitan]

the point of the twins paradox isn't that one twin ages less. we know why that is. it follows from time dilation. the question is why is it that the moving twin ages less yet from his point of view it is the other twin that is aging more slowly. the answer is that there is a shift that occurs when the moving twin changes speed.

the answer you are giving is simply the answer you would get if the moving twin were to completely stop. so the 2 really agree completely.

No, granpa.

The reason the moving twin ages less is because the moving twin starts off stationary in "A"'s frame, changes inertial frame to one which is not stationary wrt to "A", changes to another inertial frame which is also not stationary wrt to "A" and ends up stationary again in "A"'s frame.

"A" remains in "A"'s frame the whole time.

There is no shift when the moving twin changes speed. If there were, can you at least answer why there is no corresponding shift during the accelerations that get "B" on the way and, at the end, bring "B" to an inertial frame which is stationary wrt "A". Of course there is a shift, from "being in one inertial frame to being in another inertial frame" obviously, but there is no jumping forward or backwards of the ageing clock. Why?

(BTW I think I can come up with the staff answer, I just want to check if you have it to hand. But, having the staff answer doesn't mean that I agree with the shift at turnaround.)

cheers,

neopolitan

 

[Dale]

Surely contracted meters per dilated seconds squared cannot be considered "absolute".

The term "absolute" means "frame invariant". Proper acceleration is the acceleration measured by an accelerometer, it's norm is the norm of the four-acceleration. Therefore proper acceleration is frame invariant aka absolute.

 

[Dale]

The paradox is a mistake in a coordinate dependent calculation, so no coordinate independent approach can tell us what the mistake is.

I disagree. The paradox is a mistake in the treatment of a non-inertial reference frame as an inertial reference frame. Students need to learn to identify and avoid non-inertial reference frames much more than they need to learn how to calculate simultaneity in an accelerating reference frame.

 

[RandallB]

I disagree. The paradox is a mistake in the treatment of a non-inertial reference frame as an inertial reference frame. Students need to learn to identify and avoid non-inertial reference frames much more than they need to learn how to calculate simultaneity in an accelerating reference frame.

I think (I hope) you meant to say;
“to learn how THEY CANNOT calculate simultaneity in an accelerating reference frame”.
Or Maybe
“to learn how to establish a framework of synchronized clocks in an accelerating reference frame”.

The rules drawn from simultaneity do not allow identifying spatially separated events as simultaneous in an accelerating reference frame (even if they are defined as carrying clocks synchronized in that frame), it merits no special or preferred consideration over an inertial frame in that respect.

If it did you could redefine the twin problem as B departing A at a high speed with a low fixed acceleration that will turn B around at some distant point and return to A with the same high speed vectored in the opposite direction.
It would still allow B time to run faster than A at the beginning of the trip, with the time rate of B at the return being much slower than both A and what B was at the beginning.

The crux of fully completing the Twin Problem analysis is when you confirm you cannot declare if time is passing the same for B in both frames it uses, or if one of those frames it may be running fast – even faster than A. Upon confirming that concept you have essentially established the essence of what is meant by Einstein Simultaneity. And using an accelerating reference frame does not avoid Simultaneity.

 

[Fredrik]

Fredrik,

I note that you never answered the question about a clock on "B"'s spaceship which is designed to give "B" the real universe ageing of "A".

If you agree that there is no real need to advance such a clock from 7.2 to 32.8 years at the turnaround, and that the 7.2 and 32.8 year figures are representative of incorrectly using two inertial frames as if they were one then we are not in disagreement.

There's no need to have a clock at all, and if we do put a clock on the ship, we can program it to display whatever we want. You can't say that there's one correct way to program the clock without first specifying exactly what it's supposed to display.

If my job is to program a clock on the ship to show A's age "now", in the inertial frame that's co-moving with B, I would make it get the ship's speed once a second or so, and use SR to calculate what time to display. This would make it jump from 7.2 to 32.8 years at the turnaround event. I would also make the clock display the ship's speed relative to Earth at all times, because the concept of "now" doesn't make sense without a coordinate system, and the speed is what you need to know to understand what coordinate system the clock is using.

BTW When I say real universe ageing of "A", I mean taking into account that "A" undergoes no acceleration - it is not meant to imply that there is some absolute ageing of "A".

Then what are you implying? Are you just saying that if we only consider A's point of view, there's no way to get a result that looks like a paradox? This is true, but it isn't a resolution of the paradox. It's just a way to ignore it.

Do you agree that the discontinuity from 7.2 to 32.8 years is part of the incorrect way of looking at the twin scenario which leads to it appearing to be a paradox, or are you giving it as the solution?

It's the resolution of the apparent paradox. It's what's missing in the naive and incorrect calculation that says that A will only age 14.4 years. It's what you have to add to the incorrect calculation to make it correct.

 

[Fredrik]

The rules drawn from simultaneity do not allow identifying spatially separated events as simultaneous in an accelerating reference frame

I had a discussion with DrGreg about that https://www.physicsforums.com/showthread.php?t=249815", and we agreed that the standard definition of simultaneity is sufficient to define a coordinate system on a region that contains the world line. It may not be well-defined very far from the world line, but at least it's well-defined in the immediate vicinity.

 

[RandallB]

I had a discussion with DrGreg about that https://www.physicsforums.com/showthread.php?t=249815", and we agreed that the standard definition of simultaneity is sufficient to define a coordinate system on a region that contains the world line. It may not be well-defined very far from the world line, but at least it's well-defined in the immediate vicinity.

I'm thinking NO.

Simultaneity does not IMO allow for a preferred observer and their world line any more than a preferred frame. That would be an assumption of simultaneous-ness inconsistent with simultaneity.

 

[Fredrik]

Huh? Who said anything about a preferred observer? I'm saying that given any world line, you can use the standard definition of simultaneity to associate a local coordinate system with it. Now, I could be wrong about that, but you'd need a mathematical argument to prove that. If you have one, post it in the other thread.

 

[Dale]

The rules drawn from simultaneity do not allow identifying spatially separated events as simultaneous in an accelerating reference frame (even if they are defined as carrying clocks synchronized in that frame)

I don't understand your comments. If you have a pair of synchronized clocks then two events are simultaneous if and only if they occur when their local clocks read the same. That's what it means to synchronize a pair of clocks.

 

[Al68]

There is no shift when the moving twin changes speed. If there were, can you at least answer why there is no corresponding shift during the accelerations that get "B" on the way and, at the end, bring "B" to an inertial frame which is stationary wrt "A". Of course there is a shift, from "being in one inertial frame to being in another inertial frame" obviously, but there is no jumping forward or backwards of the ageing clock. Why?

There is, but since it depends on the distance between the ship and earth, it's negligible when the distance is small, and goes away if we consider the distance to be zero.

 

[RandallB]

I don't understand your comments. If you have a pair of synchronized clocks then two events are simultaneous if and only if they occur when their local clocks read the same. That's what it means to synchronize a pair of clocks.

NO
Being synchronized does not mean the same times separated by distance are considered simultaneous. SR shows this with other frames all defining the frame you are using as having all its clocks out of synchronization. Unless you can demonstrate how a frame or item can declare it own system of synchronization correct over other frames you cannot assume it is the correct judge of when spatially separated events actually simultaneous.

Simultaneity does not mean you can pick any frame or world line and declare it preferred just because if you do use it as the only reference and ignore all others will give good results.
Simultaneity says that any other frame can do just as well using a different standard for synchronization.

That is the whole point made by Simultaneity no matter how much you love the frame you live in you can only know that events happening collocated in space and time are simultaneous. And that is why B can only know what time it is in the A frame at the turnaround and only at that place in the A frame. Because even the A’ observer waiting at the turnaround point cannot KNOW what is simultaneously being displayed on the clock back at A.

Does that not make commonsense? Well yah hey – that the point! the rules of SR boxes you into that result and it will take something outside of SR to get outside that box. A priori assumptions, like picking some timeline, won’t do.

 

[Dale]

Simultaneity says that any other frame can do just as well using a different standard for synchronization.

Yes, this is what is meant by the relativity of simultaneity.

Simultaneity is a frame-dependent concept. "Simultaneous" simply means that two events share the same time coordinate in the specified reference frame. The specified reference frame doesn't have to be inertial, but obviously the Einstein synchronization convention cannot be applied in non-inertial frames.

 

[Fredrik]

...obviously the Einstein synchronization convention cannot be applied in non-inertial frames.

You can apply it to the co-moving inertial frame at every point on the world line. This gives you a local definition of simultaneity. This defines the spatial coordinates of a coordinate system, and the proper time along the world line defines the time coordinate. So the result is a local coordinate system that's defined in a region that contains the world line.

Edit: I should mention that I'm not 100% sure that this procedure actually works. If the world line is curved, the simultaneity lines at two nearby points on the world line will intersect at some distance from the world line. If this distance goes to zero when we let the distance between the nearby points go to zero, the procedure doesn't work.

 

[granpa]

furthermore what you are wanting to do with the Earth frame is exactly what the people onboard the rocket will want to do with the rocket frame. they will consider the rocket to be stationary. to them it is the most convenient frame.