Twin Paradox Problem: Do Twins Age Differently?

[ghwellsjr]

You can't move at the speed of light but you can get as close as you want and the clock will be running as slow as you want (but not stopped). And you can't go faster so it's pointless to say what it would do.

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're treating light like it was sound where you can go faster than its speed of propagation and where its speed of propagation is relative to a fixed medium like air. If this were the case, then we could figure out the stationary state of the medium by analyzing how the Doppler shifts are not dependent just on the relative velocity between the source and the receiver and they wouldn't be symmetrical and the coming and going Doppler shifts wouldn't be the inverse of each other and they wouldn't calculate that the traveling twin was younger at the reunion.

EDIT: Let me put some numbers on the above Doppler shifts. 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.

During the outbound portion of the trip, the Earth twin would see the traveling twin's clock going at 1/(1+0.9) = 0.5363 of his own which is no where near 0.1 so the effect is not reciprocal. And at the end of the inbound portion of the trip, he would see 1/(1-0.9) = 1/0.1 = 10 which is no where near the 1.9 so again, it's not reciprocal.

Neither are the outbound and inbound Doppler factors for each twin the inverse of each other. So normal Doppler, such as for sound in air does not apply to light. It requires Relativistic Doppler.

According to the explanation of Special Relativity, you see the clocks moving slower due both to time dilation and to the fact that you are always seeing the light later and later.

 

[ghwellsjr]

Acceleration have coordinate effect on clock reading. That's because acceleration changes the hyperplane of now.
Imagine two observers at the same spot moving at the same speed in respect to Earth but in opposite directions (toward/away). While they see exactly the same picture of Earth they will assign different readings to the "now" of Earth clock.

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).

This is just another of the many ways to analyze the Twin Paradox and they all agree, as you pointed out, concerning their picture of Earth when they meet. And they all agree with the final outcome. And they all agree with everything else in between that is observable. They don't agree on what you are calling remote "now" which is another way of saying "coordinate time" but that is consistent with the calculation of the Proper Time on both clocks. The coordinate times can vary all over the place between these different frames but when you apply the time dilation you get the same Proper Time at each event no matter what frame you use.

You have proposed three inertial observers. You could have proposed analyzing what happens according to each of their rest frames and there would be no frame jumping and no acceleration. I hope you're not suggesting that these three inertial frames are not all equally valid and I hope you're not suggesting that an analysis based on jumping between two of those frames is somehow more valid or better suited to explaining what is "really" happening in the Twin Paradox, are you?

Here, I already explained all this back in post #34:

If the traveling twin actually knows physics, he would be aware that there is no such thing as the "real pace of the Earth twin's clock while traveling towards him". He would know that he can analyze the pace of both of their clocks from any inertial frame of reference and each one can assign different paces to their two clocks, none of which can be considered "real". What's real is the visual data that you call an illusion. Furthermore, each one of these inertial reference frames will agree on exactly what each twin sees throughout the entire trip. You can also analyze the scenario from non-inertial frames or jumping inertial frames and they can assign completely different paces to the two clocks but they will all agree on what each twin really sees.

Do you completely agree with everything I said in the above quote?

If you do, then please read this quote from post #42:

You just said that real meant within a given inertial frame of reference and now you want to talk about a frame that the traveling twin is a rest in. But it cannot be an inertial frame for the entire trip so how does that work?

I don't know why you want to make this so complicated. Let's do what you said and pick as our given inertial frame of reference the one in which the Earth twin is at rest and in which the traveling twin starts out and ends up at rest. In this frame the Earth twin's clock runs normally.

Now the traveling twin accelerates instantly to a speed of 90%c. Gamma at this speed is 2.294 (not 0.4359 as you claim in your linked diagram). That means that a clock traveling at 90%c will run slower by a factor of 2.294. The traveling twin's clock will run slower than the Earth twin's clock by this amount during his entire trip so when he gets back the Earth clock has advanced by 2.294 times whatever his clock advanced. This is what really happens according to your definition of real and it's exactly what I said would happen in post #5 and so I don't know why you say it's misleading.

Do you completely agree with everything I said in the above quote?

If you do, then don't you think it is important to point out that whatever frame provides us with the simplest way to determine what will happen is just as valid as any other frame(s) and no other analysis based on any other frame(s) will provide us with any additional insight or information into what is happening or what any observer observes and so there is no point in discussing other frame(s) except to show that they all agree on what each observer observes throughout the entire scenario?

 

[TrickyDicky]

As I said all this is fine, but in the actual case 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.

 

[ghwellsjr]

As I said all this is fine, but in the actual case 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.

You mean the real situation where the traveling twin instantly accelerates to 90%c? Of course it's a purely imaginary exercise intended like all exercises to ignore all irrelevant factors.

 

[TrickyDicky]

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"?

 

[TrickyDicky]

If you have a triangle then the sum of the lengths of two sides is greater than the length of the third side. At what point does this discrepancy occur?

Actually in the case of Minkowski space the inequality is reversed, but it doesn't affect much the solution of the paradox except to conclude that the older one at reunion is the Earth twin instead of the traveling twin as it would be using the Euclidean triangle inequality you mention.

 

[harrylin]

[..] What george says is that the pace of a specific clock moving at vrel relative to an army of observers at rest within a given inertial reference system, depends on which of the observers you ask. [..] If that sounds good to you, so be it. But that is not really what is the case. [..] real was meant as WITHIN a given inertial frame of reference. Of course the pace is different for each different frame, but i never objected that. [..]

(emphasis mine). Certainly you misunderstood what he meant; and I find your use of "real" peculiar (it just means a real calculation??), but never mind!

 

[harrylin]

No. I mean what I said, the real situation where the Earth twin is not inertial. [..] Would you say the inertiality of the Earth twin is an irrelevant factor in the "twin paradox"?

Yes indeed: for the accumulated difference in age, the speed of the Earth surface in its trajectory around the sun is negligible compared to the speed of the capsule.

BTW, SR also neglects the effect from the Earth's gravitational field.

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.

That is the difference between using a "light" clock hypothesis or "the" clock hypothesis. He obviously assumed any clock to be shock resistant. However, some kinds of clocks may be sensitive to acceleration, in which case a continuous acceleration would result in an additional effect on clock rate that is not accounted for.

 

[TrickyDicky]

Yes indeed: for the accumulated difference in age, the speed of the Earth surface in its trajectory around the sun is negligible compared to the speed of the capsule.

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?

 

[Dale]

Actually in the case of Minkowski space the inequality is reversed, but it doesn't affect much the solution of the paradox except to conclude that the older one at reunion is the Earth twin instead of the traveling twin as it would be using the Euclidean triangle inequality you mention.

Yes, I know. The purpose of the question however is to get the reader to realize that even in Euclidean geometry you cannot identify a single point where the discrepancy occurs. So it doesn't make sense to try to determine it for Minkowski geometry either.

 

[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.

 

[TrickyDicky]

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]

Of course you can substitute the inertial reference in SR with Minkowski metric tensor, that is known since 1907.

 

[harrylin]

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. [..]

Perhaps for you an "objective statement" may only relate to invariants? I'm not that demanding... Check,http://dictionary.reference.com/browse/objective?s=t :
5. not influenced by personal feelings, interpretations, or prejudice; based on facts; unbiased

Based on the dictionary, I conclude that definitely also the sense in which robinpike seems to have meant the statement of post #68, is "objective". It's merely a different kind of "being objective" than yours.

Even more, with your definition, SR-type "time dilation" isn't even possibly part of an "objective statement", or am I mistaken?

 

[ghwellsjr]

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.

I can only go by what you say, not by what you are thinking. But even the statement that there must be some inertial frame in the problem is not correct. There doesn't have to be any frame. Consider this:

Two observers are traveling toward each other at a relative speed of 90%c. They each observe the other ones clock running at 4.359 times their own. When they pass, they reset their clocks to zero. Now they each observe the other ones clock running at 0.2294 times their own. After a while, one of them turns around such that they are now approaching each other at 90%c and, like initially, the one that turned around immediatly sees the other ones clock running at 4.359 times his own. When they pass again, they compare the accumulated times on their respective clocks. The one that turned around sees the time on the other ones clock as 2.2942 times his own. You will note that this is exactly a restatement of Rishavutkarsh's problem in which he did not specify an inertial frame and you will note that I did not use any frame in my analysis in post #5, inertial or non-inertial.

If you want you can use any frame to analyze the problem that Rishavutkarsh stated but it will not provide any more insight or information into what is happening. For example, you could use a frame in which the inertial observer remains at rest. Or you can use a frame in which the non-inertial observer is at rest during the first part of the scenario. Or you can use a frame in which the non-inertial observer is at rest during the last part of the scenario. Or you can use a frame in which during the first part of the problem both observers are traveling in opposite directions at the same speed or in which this is true for the last part of the problem. And note that I always say "a" frame for each of these case because there are an infinite number of frames for each one that you can choose from. Frames are arbitrary and they don't change or influence what the observers see, measure, or observe. No frame is preferred over any other, even the rest frame(s) of the observers.

 

[ghwellsjr]

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

No you don't, I just did it in post #5 and again in post #93.

 

[stevendaryl]

Perhaps for you an "objective statement" may only relate to invariants?

I consider other things besides invariants to be objective, namely vectors and tensors. They have a meaning that is independent of observer (although their components are relative to a coordinate system).

I'm not that demanding... Check,http://dictionary.reference.com/browse/objective?s=t :
5. not influenced by personal feelings, interpretations, or prejudice; based on facts; unbiased

Well, it seems to me that the term "clock rate" is a subjective notion, and so IS based on interpretations.

Based on the dictionary, I conclude that definitely also the sense in which robinpike seems to have meant the statement of post #68, is "objective". It's merely a different kind of "being objective" than yours.

Well, it seems to me that you can always paraphrase a statement that is objective in your sense into a statement that is objective in my sense. You can say "For any coordinate system, there is a time at which the clock rate of the traveling twin is less than the clock rate of the stay-at-home twin, according to that coordinate system." That's objectively true.

Even more, with your definition, SR-type "time dilation" isn't even possibly part of an "objective statement", or am I mistaken?

There is a corresponding objective statement, which is that an inertial path connecting two spacetime events has a greater proper time than an accelerated path connecting the same two points.

 

[stevendaryl]

I wrote:

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

No you don't, I just did it in post #5 and again in post #93.

No, you didn't. You wrote:

Two observers are traveling toward each other at a relative speed of 90%c. They each observe the other ones clock running at 4.359 times their own. When they pass, they reset their clocks to zero. Now they each observe the other ones clock running at 0.2294 times their own.

How did you compute those two numbers, 4.359 and 0.2294, if not by using a metric?

Maybe you are just saying that you can observe the effects of time dilation without using any metric. I wouldn't say that you were computing it.

Maybe I should rephrase what I was saying. If you want to PREDICT how much elapsed time will occur on a moving clock, then you need to know a coordinate system for describing the clock's motion, and you need to know a metric for that coordinate system.

 

[TrickyDicky]

...even the statement that there must be some inertial frame in the problem is not correct. There doesn't have to be any frame. you will note that I did not use any frame in my analysis in post #5, inertial or non-inertial.
...

No frame is preferred over any other, even the rest frame(s) of the observers.

These statements don't make any sense within SR.
As soon as you are talking about percentages of c you are using inertial frames.
The last statement is true as long as there are only inertial frames, all equally valid of course.

 

[D H]

How did you compute those two numbers, 4.359 and 0.2294, if not by using a metric?

No metric is needed. For example, suppose the two ships regularly send time tagged messages to one another. Each message is time tagged with a time of transmission by the sending ship per that ship's clock. A time of receipt is added to the message by the receiving ship per that ship's clock. Computing those rates is simple; it's just a matter of looking over the message log files and comparing transmission times versus reception times. When the ships are approaching one another, each ship will say the other ship's clock is running fast; when moving away from one another, each ship will say the other ship's clock is running slow.

 

[ghwellsjr]

These statements don't make any sense within SR.
As soon as you are talking about percentages of c you are using inertial frames.
The last statement is true as long as there are only inertial frames, all equally valid of course.

Yes, within SR, you need to use frames. But you don't need to use SR to present or solve every relativistic problem which was the case in this thread. Of course, you can use SR if you want.

I disagree that talking about percentages of c means that I am using either SR or frames. Are you saying that prior to Einstein, nobody could talk about percentages of c? Never heard of such a thing. And if what you say is true, then what frame was I talking about in my response to you in post #93?

I wish your point that SR permits only inertial frames so that we could assume an inertial frame whenever in the context of SR we are talking about frames but unfortunately, I have learned that non-inertial frames are also just as valid within the context of SR.

 

[stevendaryl]

No metric is needed. For example, suppose the two ships regularly send time tagged messages to one another. Each message is time tagged with a time of transmission by the sending ship per that ship's clock. A time of receipt is added to the message by the receiving ship per that ship's clock. Computing those rates is simple; it's just a matter of looking over the message log files and comparing transmission times versus reception times. When the ships are approaching one another, each ship will say the other ship's clock is running fast; when moving away from one another, each ship will say the other ship's clock is running slow.

Yes, if you have a detailed record of transmission and reception times on the two ships, you can use that to demonstrate time dilation. What I mean is that there is no way to compute those transmission and reception times without using a coordinate system and a metric for that coordinate system. If I just tell you that I have two ships that start together, move apart, and then come back together, can you tell me what the elapsed times on the two ships will be? No, not without more information.