Twin Paradox Problem: Do Twins Age Differently?

[harrylin]

Probably I was not clear enough about what I was talking.
The effect applies to remote (Earth) clock, not local clock. So this has nothing to do with local clock and accumulated clock time.
When rocket has finished it's turn-around then Earth clock has made a "jump" ahead. This "jump" ahead is a coordinate effect.

And my example was intended to show that we can avoid acceleration part but the effect is still there. Imagine that the twin that is heading away from Earth instead of turning back pass another astronaut that is heading toward Earth. And when they pass each other they simply exchange their clock readings. But they will disagree what time it is on Earth now and that is the same effect as the one of changing frames (accelerating).

Certainly I was not clear enough in posts #8 and #27, but what was not clear?? Perhaps George's explanation now solved that issue, but just in case:

At the moment that you make a turn-around, you:
1. can not influence what happens on earth
2. have only one inertial reference system at your disposal, which is the one of the Earth (ECI frame).
Next, after the turn-around you can decide to still indirectly use the ECI frame (just as astronauts always have done until now in real life), or set up a new inertial reference system by re-synchronizing your clocks. That system maps a different distant time as the other ones.

When that is understood, it is immediately clear that it's just a matter of switching reference frames, so that alternative scenario's with fly-by at the same velocities cannot have a different effect. There is no problem with that illustration, but it should not be presented as spooky action at a distance.

 

[harrylin]

I'm not talking about the accumulated difference of age, the exact amount of that difference is not the important part of the "resolution of the apparent paradox", the relevant part is the breaking of the "false symmetry", and for that it is vital that one frame is inertial and the other is not, do you really not agree about this?

No, that is still a misunderstanding. This is how it works:
- Pick any inertial frame you want (most handy is the solar system rest frame)
- Calculate the retardation due to motion of the traveler
- Calculate the retardation due to motion of the Earth (however, you will see that that is negligible for this case)
- Subtract the two retardations, and you find the difference; that is the main observed phenomenon.
- The LT guarantee that if you choose to use any different inertial frame, or if you like to switch inertial frames mid-way or at any other time, the same observations will be predicted.
Mathematicians such as Poincare understood this from the start (they "form a group").

 

[marty1]

If what you are saying is true, the traveling twin would see the Earth twin's clock going at (1-0.9)/1 = 0.1 of his own clock on the outbound half of the trip and he would see it at (1+0.9)/1 = 1.9 of his own on the inbound half of the trip. The average of these two numbers is (0.1+1.9)/2 = 1 which means that their clocks accumulate the same amount of time during the trip which is not the case in reality.

The case in reality is that you cannot ignore your acelleration nor the resulting ambient gravitation field you are in as a result of your instantaneous velocity against the background of the entire universe at any moment.

 

[stevendaryl]

Would you say the inertiality of the Earth twin is an irrelevant factor in the "twin paradox"?

Yes, in the sense that the gravitational field of the Earth would make very little difference to the results.

A still idealized situation, but one which takes gravity into account would be this: you have the Earth as the only source of gravity in the universe. A rocket ship launches from Earth, travels several light years away at nearly speed c, and returns. A clock aboard the rocket ship is compared with a clock on the Earth the whole time. The discrepancy between the elapsed times on the two clocks will be almost the same as the pure SR prediction. The effect of gravity only makes much of a difference while the rocket is near the Earth.

 

[stevendaryl]

My point is that seeing the clocks moving slower is NOT due to time dilation but the fact that you are not seeing the light pass you from a fixed point but always seeing it later and later with the extreme of going toward stopping and even reversing... all the while there was no time warping or dilation involved.

You are confusing two different effects: time dilation and Doppler shift. Doppler shift is as you describe: if a source of light signals is moving away from you, then every signal will have a slightly longer transit time than the previous. So this will give the visual appearance of the source running slow. But that is not time dilation. Time dilation is the difference in "clock rates" *AFTER* Doppler shift has been taken into account.

In the case of a satellite in geosynchronous orbit, there is no Doppler shift, since the distance between the satellite and the point under it on the surface of the Earth remains constant. But there is velocity-dependent time dilation.

 

[TrickyDicky]

No, that is still a misunderstanding.

What misunderstanding?, you are also picking at least one inertial frame (it doesn't matter if it is the Earth or the solar system).

This is how it works:
- Pick any inertial frame you want (most handy is the solar system rest frame)
- Calculate the retardation due to motion of the traveler
- Calculate the retardation due to motion of the Earth (however, you will see that that is negligible for this case)
- Subtract the two retardations, and you find the difference; that is the main observed phenomenon.
- The LT guarantee that if you choose to use any different inertial frame, or if you like to switch inertial frames mid-way or at any other time, the same observations will be predicted.

This is fine with me. How does this show any misunderstanding in what I said?

Mathematicians such as Poincare understood this from the start (they "form a group").

What group are you referring to?, there are several.

 

[TrickyDicky]

Yes, in the sense that the gravitational field of the Earth would make very little difference to the results.

Where did I introduce gravity? I'm trying to stick to SR here.
My point was that in SR you need some inertial rest frame you can compare the noninertial traveller twin with. If as harrylin said one can consider the acceleration of the traveller twin as absolute, that makes the inertial frame that you are using as reference to be absolute wrt that acceleration.

 

[robinpike]

The problem with Relativity's explanation for the Twin paradox, is that, once back on earth, for the traveling twin's clock to have a lesser time than the stay at home twin's clock, it can be deduced that the rate of time on the traveling twin's clock must have slowed down at some point during the journey.

Once a clock has had its rate of time slowed down by acceleration, Relativity has no mechanism to return the rate of time back to 'normal' - since any further acceleration can only cause the clock's rate of time to slow down even more...

 

[harrylin]

[..] I don't know how the Earth twin can be considered inertial, so it all seems like a purely imaginary exercise, of course we all know SR is not the theory suited for the real situation in which no pure inertial frames exist.

No. I mean what I said, the real situation where the Earth twin is not inertial. [..]the important thing was that the traveling one was noninertial and the other observer was inertial. Would you say the inertiality of the Earth twin is an irrelevant factor in the "twin paradox"?

No, that is still a misunderstanding. [..]

What misunderstanding?, you are also picking at least one inertial frame (it doesn't matter if it is the Earth or the solar system). [..] How does this show any misunderstanding in what I said?

See the citations above. Once more: in post #62 I did not assume that the Earth is an inertial frame. Thus there is no need to think that it is important that the Earth twin ("the other observer") was inertial, contrary to what you continued to say. I picked the solar system rest frame, which is practically inertial over the course of two hundred years; and in that protocol neither twin was at any time "inertial".

What group are you referring to?, there are several.

I'm not a mathematician. Here it is:
https://en.wikisource.org/wiki/On_the_Dynamics_of_the_Electron_%28June%29

 

[stevendaryl]

Where did I introduce gravity? I'm trying to stick to SR here.
My point was that in SR you need some inertial rest frame you can compare the noninertial traveller twin with. If as harrylin said one can consider the acceleration of the traveller twin as absolute, that makes the inertial frame that you are using as reference to be absolute wrt that acceleration.

Sorry for misunderstanding you. When you said "the real situation where the Earth twin is not inertial", in what sense were you saying that the Earth twin is not inertial?

 

[TrickyDicky]

Sorry for misunderstanding you. When you said "the real situation where the Earth twin is not inertial", in what sense were you saying that the Earth twin is not inertial?

I'm just separating the "formal plane" of the paradox solution in SR versus the real universe where unlike in flat Minkowski space, there are no pure inertial frames but they are nevertheless used in idealized situations where its use is irrelevant to the problem at hand.
When there is no actual inertial reference, I at least wouldn't know how to compute the time dilation that accumulates between the twins.

 

[stevendaryl]

The problem with Relativity's explanation for the Twin paradox, is that, once back on earth, for the traveling twin's clock to have a lesser time than the stay at home twin's clock, it can be deduced that the rate of time on the traveling twin's clock must have slowed down at some point during the journey.

No, that's not true. Not in any objective sense.

Here's an analogy. Suppose you have a system of roads, and every road has markers on it every 100 meters. You have two roads that meet at a point, diverge, and then come back together at a second point. You can compare the number of markers along the two different roads between the first meeting and the second meeting. You might find that one road has a greater number of markers between the two points than the other road. Does that mean that the road with the greater number of markers must have its markers closer together, or that the road with the small number of markers must have its markers farther apart? No, it's just that the distance between two points depends on the path taken. The number of markers is an accurate measure of these two distances.

In SR, the analogy of "number of markers along a road" is "number of ticks of a clock along a spacetime path". Two spacetime paths meet at some point, diverge, and then come back together later. The number of clock ticks between the two points is different for the two different paths. Does this mean that one clock's ticks come closer together, or that the other clock's ticks come farther apart? No, it's just that the proper time between two points depends on the spacetime path taken.

 

[TrickyDicky]

See the citations above. Once more: in post #62 I did not assume that the Earth is an inertial frame. Thus there is no need to think that it is important that the Earth twin ("the other observer") was inertial, contrary to what you continued to say. I picked the solar system rest frame, which is practically inertial over the course of two hundred years; and in that protocol neither twin was at any time "inertial".

Because you picked the solar system as inertial rest frame to refer and compute the accumulated time dilations of both twins, it makes no difference, I was stressing you need at least one inertial reference.

I'm not a mathematician. Here it is:
https://en.wikisource.org/wiki/On_the_Dynamics_of_the_Electron_%28June%29

You don't have to be to know that, none of us(surely I'm not) may be physicists and we are talking about SR right? ;-)

That must be the Lorentz group, but since you mentioned Poincare I thought you might be talking about the Poincare group which in this case is not needed because the origin is fixed in this problem.

 

[stevendaryl]

I'm just separating the "formal plane" of the paradox solution in SR versus the real universe where unlike in flat Minkowski space, there are no pure inertial frames but they are nevertheless used in idealized situations where its use is irrelevant to the problem at hand.
When there is no actual inertial reference, I at least wouldn't know how to compute the time dilation that accumulates between the twins.

I still don't understand what you're talking about. The reason that there are no inertial frames in the real universe is because of gravity. In the absence of gravity, Earth would be at rest in an inertial frame. But you seem to want to say that Earth is not inertial, and ALSO you want to ignore gravity.

There is no way to accurately compute elapsed times on clocks near the Earth without taking gravity into account. The elapsed time on a clock as it travels between events A and B is given by:

\tau = \int_A^B \sqrt{g_{ij} dx^i dx^j}

where g_{ij} is the metric tensor coefficients.

For the Earth, a good approximation to \tau for slow velocities is given by:

\tau = \int_A^B (1 - \dfrac{G M}{c^2 r} - \dfrac{1}{2} \dfrac{v^2}{c^2}) dt

 

[TrickyDicky]

I still don't understand what you're talking about. The reason that there are no inertial frames in the real universe is because of gravity. In the absence of gravity, Earth would be at rest in an inertial frame. But you seem to want to say that Earth is not inertial, and ALSO you want to ignore gravity.

There is no way to accurately compute elapsed times on clocks near the Earth without taking gravity into account. The elapsed time on a clock as it travels between events A and B is given by:

\tau = \int_A^B \sqrt{g_{ij} dx^i dx^j}

where g_{ij} is the metric tensor coefficients.

For the Earth, a good approximation to \tau for slow velocities is given by:

\tau = \int_A^B (1 - \dfrac{G M}{c^2 r} - \dfrac{1}{2} \dfrac{v^2}{c^2}) dt

I want to ignore gravity when talking about the SR solution, not in general.

 

[harrylin]

No, that's not true. Not in any objective sense. [..]

For me, "objective" in the context of SR is similar to "absolute": if all inertial frames agree that a statement is true. Evidently you mean something else with "objective sense", but what?
Your example didn't clarify that, as you merely explained a certain sense of interpreting the statement that you claim to be "not true" (and probably not corresponding to the way it was meant).

 

[ghwellsjr]

You mean the real situation where the traveling twin instantly accelerates to 90%c?

No. I mean what I said, the real situation where the Earth twin is not inertial. Besides, harrylin already gave the pertinent quote from Einstein himself stating wrt the results of the time dilation it doesn't matter whether the traveling twin moves in a curve or in a polygon line (which we all know is an unphysical way of accelerating), the important thing was that the traveling one was noninertial and the other observer was inertial.

Of course it's a purely imaginary exercise intended like all exercises to ignore all irrelevant factors.

Would you say the inertiality of the Earth twin is an irrelevant factor in the "twin paradox"?

Yes, it is, if you are referring to your comment, "the important thing was that the traveling one was noninertial and the other observer was inertial". It is neither important, relevant, significant, nor is it a requirement for the Twin Paradox.

In the simplest presentation of the Twin Paradox, we talk about the Earth twin as if the Earth had no gravity and no acceleration, which are of course not true and so the Earth and the Earth twin are considered to be inertial for the purpose of discussing the scenario. In this simplest presentation the other twin is non-inertial and so without knowing anything else, we can always say that the traveling twin is the one who ages less because he experienced acceleration whereas the Earth twin did not. This leads many people to falsely jump to the conclusion that it is the acceleration that causes the difference in aging between the twins and they look for explanations (a non-inertial frame or jumping between two frames) that support that incorrect idea. These people tend to reject the simple explanation that I offered in post #42 and quoted in post #52.

But we can complicate the Twin Paradox by having both twins follow exactly the same accelerations. They can both take off in identical spaceships and achieve 90%c. Then one of them immediately turns around and lands back on Earth for the rest of the time while the other twin continues on for a long time before repeating what the first twin did. So now both twins are non-inertial in exactly the same way so we can't jump to the false conclusion that the acceleration is what caused the difference in aging. But we can still analyze this more complicated Twin Paradox using the same inertial Earth reference frame as before. It's the simplest one to use, mainly because it is the one that is used to describe their motions.

In fact, it doesn't matter how the two twins accelerate or what speeds they achieve or what directions they travel in (polygons or circles or some of each) or where they end up together. We can always analyze their individual aging during the entire process from the standpoint of the inertial frame that we use to describe their activities.

Furthermore, with a little more work, we can determine what each twin sees of the other ones clock during the entire scenario, not just at the beginning and the end, and any analysis we do (whatever frame(s) we use) will all provide the same answers. We can make this as complicated as we want. But the more complicated analyses will not provide any more insight or information into what is happening.

 

[harrylin]

[..]I was stressing you need at least one inertial reference. [..]

Sorry that I understood what you said, and not what you meant.

So yes, SR refers to inertial frames, just like classical physics. And you now understand my explanation of how to calculate it accounting for the non-inertial Earth observer. Good.

[EDIT:] However, why then do you state in-between:

When there is no actual inertial reference, I at least wouldn't know how to compute the time dilation that accumulates between the twins.

That was what I thought to have explained you in post #62, and it appeared that you understood it...

But perhaps you mean that if we lack some necessary data (such as specifications that can be converted to an inertial frame), then we can't compute the outcome? That's very true!

 

[TrickyDicky]

ghwell, You completely missed my point, see my previous posts.
I was referring to the fact that there must be some inertial frame in the problem, not that it must be one of the twins.

 

[TrickyDicky]

Sorry that I understood what you said, and not what you meant.

So yes, SR refers to inertial frames, just like classical physics. And you now understand that the non-inertiality of the Earth twin is not a problem. Good.

Great, but I understood that before.

 

[stevendaryl]

For me, "objective" in the context of SR is similar to "absolute": if all inertial frames agree that a statement is true.

I don't consider that good enough for a statement to be objective. I think that it must also be the case that the terms mentioned in that statement have a meaning that is independent of observers. A statement involving "clock rate" can't be an objective statement, because there is no such thing as a clock rate. There is only a clock rate relative to a coordinate system.

It is objectively true that an inertial path connecting two spacetime points has a longer proper time than an accelerated path connecting the same two points. It is not objectively true that a clock following the accelerated path has a lower clock rate.

 

[stevendaryl]

I want to ignore gravity when talking about the SR solution, not in general.

I really don't understand what you are saying, then. For the twin paradox, you can either include gravity or not. Whether you do or not, you can calculate the elapsed times for the two twins.

 

[TrickyDicky]

I don't consider that good enough for a statement to be objective. I think that it must also be the case that the terms mentioned in that statement have a meaning that is independent of observers. A statement involving "clock rate" can't be an objective statement, because there is no such thing as a clock rate. There is only a clock rate relative to a coordinate system.

It is objectively true that an inertial path connecting two spacetime points has a longer proper time than an accelerated path connecting the same two points. It is not objectively true that a clock following the accelerated path has a lower clock rate.

I agree with this and in a way summarizes what I was saying in my previous posts.

 

[TrickyDicky]

I really don't understand what you are saying, then. For the twin paradox, you can either include gravity or not. Whether you do or not, you can calculate the elapsed times for the two twins.

Let's drop it, I don't have the slightest idea what you are talking about, sorry.

 

[harrylin]

[..] I was referring to the fact that there must be some inertial frame in the problem, not that it must be one of the twins.

Yes, I think that I understand you now. Your remark is true in general for physics. Without any idea of the state of motion of the participating objects, little can be predicted with certainty about the physical effects. In that sense is motion not just "relative", and Newton illustrated that famously with the bucket experiment. Einstein tried to solve that with GR, but if I correctly understand his later clarifications, he did not manage to make it truly Machian.

 

[stevendaryl]

Let's drop it, I don't have the slightest idea what you are talking about, sorry.

What I'm talking about is that I don't have the slightest idea what you are talking about.

 

[Nugatory]

When there is no actual inertial reference, I at least wouldn't know how to compute the time dilation that accumulates between the twins.

I want to ignore gravity when talking about the SR solution, not in general.

When we're talking about the SR solution, either we're conducting a thought experiment in an idealized flat space or we're conducting a real experiment in an environment where the gravitational effects are negligible. Either way, we use the Minkowski metric. (That's pretty much the definition of special relativity as a special case of the general theory).

We calculate the time elapsed for each twin as \tau = \int_A^B \sqrt{g_{ij} dx^i dx^j} where the g_ij are the Minkowski metric components... and that calculation requires no actual inertial reference.

[Thanks to stevndaryl for the plagiarized latex]

 

[stevendaryl]

Because you picked the solar system as inertial rest frame to refer and compute the accumulated time dilations of both twins, it makes no difference, I was stressing you need at least one inertial reference.

What you need to compute proper times is a coordinate system with a known metric tensor.

 

[TrickyDicky]

Yes, I think that I understand you now. Your remark is true in general for physics. Without any idea of the state of motion of the participating objects, little can be predicted with certainty about the physical effects. In that sense is motion not just "relative", and Newton illustrated that famously with the bucket experiment. Einstein tried to solve that with GR, but if I correctly understand his later clarifications, he did not manage to make it truly Machian.

Right.

 

[TrickyDicky]

What I'm talking about is that I don't have the slightest idea what you are talking about.

It's nice when people reaches this level of agreement.