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Twin clocks-is it acceleration?

by phyti
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DaleSpam
#37
Feb26-13, 10:46 AM
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Quote Quote by stevendaryl View Post
I think that the answer "different acceleration profiles" for why one twin ages more than the other is misleading. It certainly is true that in flat Minkowsky spacetime, there is no way for two twins to depart and reunite unless one of them accelerates. But using acceleration as the cause of the difference doesn't generalize to other kinds of topologies of spacetime, and doesn't generalize to General Relativity.

In a cylindrical spacetime, you can have twins start off together, depart and reunite later without either twin accelerating. I guess you could say that such an example goes beyond SR, but it doesn't involve any gravity, so it's not very GR-ish, either. It's just geometry.
I don't think that it is misleading at all.

The explanation must be in terms of some asymmetry between the twins. If there is no asymmetry then their ages must be identical. The reason students are confused by the twin scenario is that they have just been introduced to this idea of symmetry under boosts so they believe that the twins are symmetrical.

Saying that "it's just geometry" is not a good answer, after all, you can have symmetrical geometric figures. Even geometrically, you must identify the asymmetry. In the standard twin paradox, the geometrical asymmetry is the bend in the worldline, i.e. the acceleration. So that is an accurate and not misleading answer. In other scenarios there are other asymmetries, the topology or the curvature or whatever.
A.T.
#38
Feb26-13, 10:57 AM
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Quote Quote by DaleSpam View Post
Geometrical explanations are frame independent, they are not done in any frame.
When you draw a space time diagram or integrate the path element, you are assuming a frame. My point is that to convince people that the result is indeed frame independent, you have to do it for both frames.
PAllen
#39
Feb26-13, 12:22 PM
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Quote Quote by A.T. View Post
When you draw a space time diagram or integrate the path element, you are assuming a frame. My point is that to convince people that the result is indeed frame independent, you have to do it for both frames.
I don't think so. Analogy with plane geometry:

Drawing curves, comparing angles, sliding figures, measuring lengths with a string, etc. [In SR, measuring time with a clock is equivalent to measuring length with a string in plane geometry].

do not involve coordinates. You can then draw on the plane many different cartesian coordinates, or polar coordinates, then do calculations in terms of those.

In SR you have a particular type of flat geometry. You can draw on it without coordinates. It has, objectively, straight lines. If you draw curvilinear coordinates on it, your curves are still curves - you haven't made them straight in a geometric sense.

As for frames, I think a key thing that is hard to convince people is that there is actually no such thing as a global non-inertial frame in SR (just as there is no such thing as a global frame at all in GR - inertial or otherwise). There are many possible accelerated coordinates (equivalent to curvilineary coordinates on a plane), each of which will represent the intrinsic flat geometry superficially differently.
ghwellsjr
#40
Feb26-13, 12:39 PM
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Quote Quote by A.T. View Post
Quote Quote by ghwellsjr View Post
Is your point for the scenario in this thread that it doesn't apply to the Twin Paradox because neither twin is inertial?
I think that it is an interesting variation, which demonstrates that not only the total amount of proper acceleration matters, but its timing. It leads to a more general explanation of the asymmetry, which also applies to the classic version.
Ok, good.
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Quote Quote by ghwellsjr View Post
So the only kind of answer that you will accept is one that shows an inertial explanation for the inertial twin and simply asserts that no such explanation applies to the non-inertial twin (simply because he is non-inertial)?
It is okay to assert that an "inertial explanation" doesn't apply to the rest frame of the non-inertial twin, if that is what you mean. But ideally you would also provide a non-inertial explanation for the non-inertial twin's rest frame, which leads to the same result as the rest-frame of the other twin.
Isn't that what I did in these threads:

http://physicsforums.com/showthread....670653&page=10 post #135
http://physicsforums.com/showthread.php?t=644948&page=6 posts #92 and #93
http://physicsforums.com/showthread.php?t=671398&page=3 posts #35 and #36

In each of these three cases I provided an inertial spacetime diagram for the stay-at-home twin's inertial rest frame and a non-inertial spacetime diagram for the traveling twin's non-inertial rest frame and yet none of the OP's for which I provided these diagrams responded with any statement to the effect that they understood the diagrams. I just don't know if they are effective.
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Quote Quote by ghwellsjr View Post
So when the questioner correctly points out that in the inertial twin's rest frame, there is a constant Time Dilation for the non-inertial twin (assuming instant turn-around), but the rest frame for the non-inertial twin is non-inertial and so it is an incorrect comparison and we simply stop right there. We just call foul?
This is what most "explainers" do. And I think it is not satisfying to most questioners. I would like to see it explained from both frames.
Didn't I explain it from both frames and didn't I show that both rest frames lead to the same results?
Nugatory
#41
Feb26-13, 12:40 PM
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Quote Quote by A.T.
My point is that to convince people that the result is indeed frame independent, you have to do it for both frames.
Quote Quote by PAllen View Post
I don't think so. Analogy with plane geometry....
In SR you have a particular type of flat geometry. You can draw on it without coordinates. It has, objectively, straight lines. If you draw curvilinear coordinates on it, your curves are still curves - you haven't made them straight in a geometric sense.
It depends on who you're trying to convince. You ought to be right, but my experience is that for most audiences A.T. is. It may just be that you get to deal with a better class of audience.
PAllen
#42
Feb26-13, 12:54 PM
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Quote Quote by Nugatory View Post
It depends on who you're trying to convince. You ought to be right, but my experience is that for most audiences A.T. is. It may just be that you get to deal with a better class of audience.
Actually my phrase "I don't think so" was referring to specifically A.T.'s statement:

"When you draw a space time diagram or integrate the path element, you are assuming a frame"

for which I agree computing an integral requires coordinates, but drawing a spacetime diagram does not.

I was not making any statement about pedagogy. On that score, I find different arguments work for different people.
A.T.
#43
Feb26-13, 02:10 PM
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Quote Quote by PAllen View Post
for which I agree computing an integral requires coordinates, but drawing a spacetime diagram does not.
When you draw a diagram with twin A at a constant space coordinate, you have assumed twin's A rest frame.
PAllen
#44
Feb26-13, 02:20 PM
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Quote Quote by A.T. View Post
When you draw a diagram with twin A at a constant space coordinate, you have assumed twin's A rest frame.
I disagree, any more than saying drawing a straight line on a blank paper assumes a coodinate system. Maybe I'll fill in the coordinates at a angle to the line if I choose to add them.
A.T.
#45
Feb26-13, 02:54 PM
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Quote Quote by PAllen View Post
I disagree, any more than saying drawing a straight line on a blank paper assumes a coodinate system. Maybe I'll fill in the coordinates at a angle to the line if I choose to add them.
Sorry, no idea what you mean.
Nugatory
#46
Feb26-13, 03:11 PM
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Quote Quote by A.T. View Post
When you draw a diagram with twin A at a constant space coordinate, you have assumed twin's A rest frame.
You have, but that didn't happen when you drew the twins' worldlines, it happened when you added the axes to the diagram.

One interesting pedagogical trick: Draw up a spacetime diagram without the axes, then draw up a bunch of transparencies, each with their own set of t and x axes - and then drop the various transparencies onto the diagram to show how coordinates work. This has the nice property of making it clear that the proper intervals belong to the coordinate-free picture, and suggests why "frames" are called that.

(There still is one downside - some people fall into the trap of believing that one particular frame, namely the one in which the x and t axes are at a 90-degree angle, is somehow special and distinguished).
A.T.
#47
Feb26-13, 03:17 PM
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Quote Quote by ghwellsjr View Post
In each of these three cases I provided an inertial spacetime diagram for the stay-at-home twin's inertial rest frame and a non-inertial spacetime diagram for the traveling twin's non-inertial rest frame and yet none of the OP's for which I provided these diagrams responded with any statement to the effect that they understood the diagrams. I just don't know if they are effective.
I don't understand them either.



I don't understand the phase where they have constant separation. In reality the blue signals send during that phase should arrive blueshifted, but there is no movement of the source in the diagram. If the signals are simply a picture seen through telescope, the angular size of blue twin, as seen by black twin would be increasing after turnaround. But in the diagram the viewing distance is constant for a few years.

I think it is very difficult to make a smooth and correct diagram for the sharp turnaround version. The simpler case might be the one with constant proper acceleration, then you have just one coordinates chart (Rindler) for the rest frame of the accelerating twin, not two that you have to merge somehow.

See Fig 7:
http://cds.cern.ch/record/497203/files/0104077.pdf
PAllen
#48
Feb26-13, 03:43 PM
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Quote Quote by A.T. View Post
I think it is very difficult to make a smooth and correct diagram for the sharp turnaround version. The simpler case might be the one with constant proper acceleration, then you have just one coordinates chart (Rindler) for the rest frame of the accelerating twin, not two that you have to merge somehow.
Note: Rindler coordinates are radar coordinates for the uniformly accelerating observer; Fermi-normal coordinates (ones modeling a rigid grid of rulers) have extra terms in the metric, and the coordinate axes are different. For an inertial frame in SR, these two coordinate building approaches produce the same result. The is really why (IMO) you can reasonably talk about (global) inertial frames in SR - different reasonable methods of building coordinates agree. It is also why I insist there is actually no such thing as an a (global) accelerated frame even in SR: even for the the simplest acceleration (uniform), the two simplest physical methods of building coordinates produce a different result.
PAllen
#49
Feb26-13, 04:05 PM
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Quote Quote by A.T. View Post
I don't understand the phase where they have constant separation. In reality the blue signals send during that phase should arrive blueshifted, but there is no movement of the source in the diagram. If the signals are simply a picture seen through telescope, the angular size of blue twin, as seen by black twin would be increasing after turnaround. But in the diagram the viewing distance is constant for a few years.
In this version of radar coordinates the horizontal does not represent proper distance along a simultaneity surface. Instead, it represents 1/2 round trip travel time for a signal (from traveler to home and back) as measured by sudden turnaround traveler, times c. For classic sudden turnaround, there is, indeed, a period of time when such signals have constant round trip time. Proper distance along the same simultaneity lines would produce a different diagram. And, of course, since the diagram is drawn from traveler perspective, the simultaneity lines are straight.
A.T.
#50
Feb26-13, 04:18 PM
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Quote Quote by PAllen View Post
Note: Rindler coordinates are radar coordinates for the uniformly accelerating observer; Fermi-normal coordinates (ones modeling a rigid grid of rulers) have extra terms in the metric, and the coordinate axes are different. For an inertial frame in SR, these two coordinate building approaches produce the same result. The is really why (IMO) you can reasonably talk about (global) inertial frames in SR - different reasonable methods of building coordinates agree. It is also why I insist there is actually no such thing as an a (global) accelerated frame even in SR: even for the the simplest acceleration (uniform), the two simplest physical methods of building coordinates produce a different result.
Thanks. Do you have some links on Fermi-normal coordinates in uniformly accelerated frames and Minkowski space-time? The wiki says it is just a generalization of Rindler for curved space-times, so I assumed that for accelerated frames in flat space time they yield the same results.
PAllen
#51
Feb26-13, 05:31 PM
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Quote Quote by A.T. View Post
Thanks. Do you have some links on Fermi-normal coordinates in uniformly accelerated frames and Minkowski space-time? The wiki says it is just a generalization of Rindler for curved space-times, so I assumed that for accelerated frames in flat space time they yield the same results.
For a beautiful derivation that Rindler coordinates are radar coordinates, our very own George Jones recently produced:

http://physicsforums.com/showpost.ph...80&postcount=5

From the wikipedia article, the metric for Rindler coordinates is:

ds^2 = - g^2 x^2 dt^2 + dx^2 ...

I don't know of a good online reference for Fermi-Normal coordinates of a uniformly accelerated observer in SR. However, section 6.6 of MTW derives this (without calling it that). This section is all flat spacetime. The metric becomes:

ds^2 = -(1+ g x)^2 dt^2 + dx^2 ...

Note how this form approaches Minkowski metric for x=0. For Rindler coordinates x=0 represents the horizon. Also note that if you just transform Rindler to have its x=1 line become new x=0, you still don't get Fermi Normal; you get:

ds^2 = -(g+gx)^2 dt^2 + dx^2 ...

What you need is x' = x - 1/g to get from Rindler to Fermi-Normal.

[Edit: On further thought, it appears that Fermi-Normal coordinates as above, for a uniformly accelerating observer would still have the same simultaneity surfaces as Rindler and Radar - just labeled differently. Thus, it seems, you need change in acceleration - e.g. the classic twin with slightly rounded turnaround - to expose the difference between Radar simultaneity and Born Rigid simultaneity (which is what Fermi-Normal uses).]
DaleSpam
#52
Feb26-13, 06:53 PM
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Quote Quote by A.T. View Post
When you draw a space time diagram or integrate the path element, you are assuming a frame.
I think we are talking about different kinds of explanations. What I would call a geometric explanation wouldn't draw any coordinates.
ghwellsjr
#53
Feb26-13, 07:20 PM
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Quote Quote by A.T. View Post
Quote Quote by ghwellsjr View Post
In each of these three cases I provided an inertial spacetime diagram for the stay-at-home twin's inertial rest frame and a non-inertial spacetime diagram for the traveling twin's non-inertial rest frame and yet none of the OP's for which I provided these diagrams responded with any statement to the effect that they understood the diagrams. I just don't know if they are effective.
I don't understand them either.



I don't understand the phase where they have constant separation. In reality the blue signals send during that phase should arrive blueshifted, but there is no movement of the source in the diagram. If the signals are simply a picture seen through telescope, the angular size of blue twin, as seen by black twin would be increasing after turnaround. But in the diagram the viewing distance is constant for a few years.

I think it is very difficult to make a smooth and correct diagram for the sharp turnaround version. The simpler case might be the one with constant proper acceleration, then you have just one coordinates chart (Rindler) for the rest frame of the accelerating twin, not two that you have to merge somehow.

See Fig 7:
http://cds.cern.ch/record/497203/files/0104077.pdf
See Fig 9 of your above link. Barbara is the "Immediate Turn-around" traveler, Alex stays at home. The diagram is showing the non-inertial rest frame of Barbara.
DaleSpam
#54
Feb26-13, 07:33 PM
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Quote Quote by A.T. View Post
I don't understand the phase where they have constant separation. In reality the blue signals send during that phase should arrive blueshifted, but there is no movement of the source in the diagram.
The diagram uses radar coordinates, the same approach as Einstein's convention, but applied to a non inertial observer. Since the resulting coordinates are non inertial you can get red and blue shifting without relative motion. I.e. The standard inertial frame formulas don't apply.


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