Clock postulate and differential aging

[atyy]

The rule that I understand for building a minkowski diagram is that the observer gets the orthagonal axis. There is no way the person with the orthagonal axis can come out younger. The wikipedia article says that the observer must adjust the earthboud twins time during the turn around. That's llike saying "Well, we don't get the right answer if we follow the rules so let's add another rule." Time doesn't jump, not for anybody.

Not a Minkowski diagram, in which the coordinate time axis is an inertial worldline, with coordinate time on that axis the proper time along that worldline. We can choose coordinates in which the coordinate time axis is a noninertial worldline, and coordinate time on that axis the proper time along that worldline. From one of George Jones's posts: http://arxiv.org/abs/gr-qc/0104077 (see Figs. 8, 9).

 

[MikeLizzi]

The astronaut is a non-inertial observer. There is no Minkowski diagram in which his worldline is straight.

In Minkowski diagrams inertial observers have straight worldlines, and two Minkowski diagrams are related to one another via a Lorentz transform, which is a linear transform. Because it is a linear transform if a line is bent in one Minkowski diagram it will be bent in all of them.

Hi DaleSpam:
I agree. My understanding of how to draw a Minkowski diagram is that the observer gets the orthagonal axis, always. Then rotated axis are calculated and drawn based on the relative velocity of the observed to the observer. Your comment would seem to support my belief that I have not yet see a correct Minkowski diagram with the astronaut as the observer. (In fact I believe it can't be done) The references given by Frame Dragger give the astronaut the orthagonal axis and then offer a kind of compensation for the ensuing contradiction. Yet those references are considered canon.

 

[MikeLizzi]

Not a Minkowski diagram, in which the coordinate time axis is an inertial worldline, with coordinate time on that axis the proper time along that worldline. We can choose coordinates in which the coordinate time axis is a noninertial worldline, and coordinate time on that axis the proper time along that worldline. From one of George Jones's posts: http://arxiv.org/abs/gr-qc/0104077 (see Figs. 8, 9).

Hi atyy,
Just reading the summary of the reference you provided tells me it has nothing to do with the issue I am trying to raise. A lot of posting have been put up and I wonder if the issue I am trying to raise is getting lost.

 

[Frame Dragger]

Did you look at figure 9?

 

[MikeLizzi]

Did you look at figure 9?

Just took a quick look. Is that a Minkowski diagram?

 

[atyy]

Hi atyy,
Just reading the summary of the reference you provided tells me it has nothing to do with the issue I am trying to raise. A lot of posting have been put up and I wonder if the issue I am trying to raise is getting lost.

Yes, the issue you are trying to raise is getting lost. Although in the twin paradox, we can do what you are asking, more generally, about the point of view of various observers in terms of coordinates is irrelevant. The elapsed proper time along a worldline is a geometric invariant, and that is all there is to it.

 

[ThomasT]

If the clocks accumulate less time for longer periods at high speed, then the acceleration has to cause a permanent change in speed, not just a change in speed while the clock is accelerating. Is that what you mean, because your wording doesn't imply that to me.

I don't know what you mean by a permanent change in speed. Do you mean constant or uniform speed for a certain interval? If the speed is changing, then the clock is accelerating, right?

What I meant was that the interval that you run one of the clocks at the higher (constant) speed is longer, so, even though the acceleration histories are identical, the clock that you ran longer at the higher (constant) speed will accumulate less time for the two intervals associated with the two clocks (including the accelerations) being compared.

Did I just confuse myself again?

 

[MikeLizzi]

Yes, the issue you are trying to raise is getting lost. Although in the twin paradox, we can do what you are asking, more generally, about the point of view of various observers in terms of coordinates is irrelevant. The elapsed proper time along a worldline is a geometric invariant, and that is all there is to it.

Hi atyy,

Exactly. So if someone draws a diagram in which the elasped proper time of the astronaut is greater than that of the earthbound twin you have a paradox. Right?.

 

[atyy]

So if someone draws a diagram in which the elasped proper time of the astronaut is greater than that of the earthbound twin you have a paradox. Right?.

What do you mean by "paradox"?

 

[Mentz114]

MikeLizzi:

To the skeptic, trying to work the problem on a Minkowski diagram, it looks like you draw the exact same diagram, only with the twins switch places. And that would be a paradoxical answer.

Yes, that would be wrong.

The accelerating observer carries his own orthogonal frame along his worldline. At any point you can construct a set of axes tangential to the worldline. If you then rotate the space-time so the relevant tangent is vertical, you have a spacetime diagram from the point of view of an inertial observer instantaneously at rest wrt to the accelerating observer.

The accelerating world-line is still curved, and the other observers worldline is tilted.

It's not exactly what you wanted but it's the best you can do.

 

[ThomasT]

I think part of the original problem is the wording of statements like this. All changes in speed involve acceleration, but not all accelerations involve changes in speed. They are not always equivalent.

I think Baez said it well in the link bcrowell provided above, and I would highly recommend a careful reading of it. But in the end, if we use the same equation then in my book we agree on everything important.

Thanks. I skimmed over Baez, but will give it a more careful reading.

 

[Frame Dragger]

MikeLizzi:

Yes, that would be wrong.

The accelerating observer carries his own orthogonal frame along his worldline. At any point you can construct a set of axes tangential to the worldline. If you then rotate the space-time so the relevant tangent is vertical, you have a spacetime diagram from the point of view of an inertial observer instantaneously at rest wrt to the accelerating observer.

The accelerating world-line is still curved, and the other observers worldline is tilted.

It's not exactly what you wanted but it's the best you can do.

Thank you, but I kept saying this and it doesn't seem to make an impact. I think he's reaching for a metaphysical interpretation that just isn't there, or he's seeing a problem that likewise, is not there.

 

[MikeLizzi]

What do you mean by "paradox"?

Hi again,

The canonical analysis of the twins example puts the worldline of the earthbound twin on the vertical ct axis. The astronaut gets two world lines one with positive slope and one negative eventually joining up with the worldline of the earthbound twin. Spacetime intervals must be equal. The astronaut's spacetime interval has a longer space component so he must have a shorter time component. He comes back younger.

So if someone draws the worldline of the astronaut along the ct axis and givess the earthbound twin the two legs of the triangle, the above conclusion would have to be reversed. One would have to declare that the astronaut came back older. And that would be a praradox.

Can we agree on that?

 

[MikeLizzi]

Thank you, but I kept saying this and it doesn't seem to make an impact. I think he's reaching for a metaphysical interpretation that just isn't there, or he's seeing a problem that likewise, is not there.

Frame Dragger.

?? I hate metaphysics. I'm just tying to draw a Minkowski diagram.

 

[Frame Dragger]

Hi again,

The canonical analysis of the twins example puts the worldline of the earthbound twin on the vertical ct axis. The astronaut gets two world lines one with positive slope and one negative eventually joining up with the worldline of the earthbound twin. Spacetime intervals must be equal. The astronaut's spacetime interval has a longer space component so he must have a shorter time component. He comes back younger.

So if someone draws the worldline of the astronaut along the ct axis and givess the earthbound twin the two legs of the triangle, the above conclusion would have to be reversed. One would have to declare that the astronaut came back older. And that would be a praradox.

Can we agree on that?

No, because that isn't the reverse of the twin problem. What you're doing is a bit like expecting binoculars to work regardless of which end you look through, ignoring relative orientation. That would be a metaphorical, metaphysical interpreation of the very sound answer that Mentz gave you.

 

[MikeLizzi]

MikeLizzi:

Yes, that would be wrong.

The accelerating observer carries his own orthogonal frame along his worldline. At any point you can construct a set of axes tangential to the worldline. If you then rotate the space-time so the relevant tangent is vertical, you have a spacetime diagram from the point of view of an inertial observer instantaneously at rest wrt to the accelerating observer.

The accelerating world-line is still curved, and the other observers worldline is tilted.

It's not exactly what you wanted but it's the best you can do.

Hi Mentz114,

I understand exactly what you are saying. May I conclude that you agree with my belief that one cannot draw a Minkowski with the astronaut as an observer?

 

[atyy]

Hi again,

The canonical analysis of the twins example puts the worldline of the earthbound twin on the vertical ct axis. The astronaut gets two world lines one with positive slope and one negative eventually joining up with the worldline of the earthbound twin. Spacetime intervals must be equal. The astronaut's spacetime interval has a longer space component so he must have a shorter time component. He comes back younger.

So if someone draws the worldline of the astronaut along the ct axis and givess the earthbound twin the two legs of the triangle, the above conclusion would have to be reversed. One would have to declare that the astronaut came back older. And that would be a praradox.

Can we agree on that?

The canonical analysis says 4>3, but if I write 3=6, and since 6>4, then 3>4, which would be a paradox?

 

[MikeLizzi]

No, because that isn't the reverse of the twin problem. What you're doing is a bit like expecting binoculars to work regardless of which end you look through, ignoring relative orientation. That would be a metaphorical, metaphysical interpreation of the very sound answer that Mentz gave you.

Frame Dragger,

I didn't say it was the reverse of the Twins problem. I know it is not. I'm asking how to draw a Minkowski diagram with the astronaut as the observer.

 

[MikeLizzi]

The canonical analysis says 4>3, but if I write 3=6, and since 6>4, then 3>4, which would be a paradox?

Hi aty,

I agree. I didn't intend to include my second paragraph as canonical analysis, only the first paragraph. Sorry if I didn't make that clear.

 

[Dmitry67]

Which one?
In twin paradox astronaut has 2 different diagrams: on the way away and when he returns. You can not have just one, as astronaut changes the frame

 

[MikeLizzi]

Which one?
In twin paradox astronaut has 2 different diagrams: on the way away and when he returns. You can not have just one, as astronaut changes the frame

Hi Dmitry67,

Thank you for your post. If I may restate what you wrote... (I think english is not your native language)

In the twins paradox (with the astronaut as the observer) one must use 2 different diagrams. One for the case where the Earth is moving away and one where the Earth returning.

OK?

 

[Dmitry67]

yes, correct.

 

[MikeLizzi]

yes, correct.

Hi Dmitry67,

Thank you. I put down some numbers for the first diagram. I assume gamma is 2. The first column is the time on the astronaut's clock. The second column is the calculated time on an Earth clock. OK?

First Diagram:

Astronaut Time____Earth Time
(observer)
0 _______________ 0
1 _______________ .5
2 _______________ 1.0
3 _______________ 1.5
4 _______________ 2.0
5 _______________ 2.5
Turn around

 

[Mentz114]

No single space-time diagram can represent the view-point of the accelerated observer, you need a different one for every instant along the worldline.

But what is the importance of that ? We know that elapsed time is equated with the proper length, so I don't really see that it matters.

 

[Frame Dragger]

No single space-time diagram can represent the view-point of the accelerated observer, you need a different one for every instant along the worldline.

But what is the importance of that ? We know that elapsed time is equated with the proper length, so I don't really see that it matters.

Again, I've been asking this question several times without any meaningful effect. *shrug*

EDIT: Al68: I pointed out 2 pages ago that there is an implied period of accelleration and braking, but it didn't seem to faze him. :/ I'm sticking with my "looking the wrong way through binoculars" metaphor. This is the use of a tool in the manner it wasn't intended, then drawing conclusions that a paradox exists, or a skeptic requires satisfaction in this particular form.

 

[Al68]

Now somebody's clock can go slower or faster than yours but it can't jump...What is really happening is that the astronaut determines the earthbound twin's clock is running FASTER than his during the spaceship turnaround. Skipping that calculation is why this artificial jump needs to be added.

Yep, exactly. Of course time doesn't jump, it only appears to jump as an artifact of the simplifying assumption of instantaneous turnaround, which equally can't happen. In the accelerated frame of the ship, Earth's clock runs fast. The greater the acceleration, the faster Earth's clock runs in the ship's frame. It's only modeled as a jump associated with instantaneous turnaround to simplify the math.

In reality there must be a finite interval for the turnaround acceleration and for Earth's clock to advance. They are both only treated as instantaneous to simplify the math, not to claim it's physically possible.

I've never seen a Minkowski diagram for the ship twin either, but it couldn't be in the standard form. It would have to be modified to allow for gravitational time dilation of clocks in an accelerated reference frame, but I doubt it could then be called a Minkowski diagram.

Einstein's own 1918 Twins paradox resolution is unique in that it does consider the non-inertial reference frame in which the ship is "stationary" the entire trip. It skips the math, presumably because his intended audience wouldn't need it to be shown. You can find it here: http://en.wikisource.org/wiki/Dialog_about_objections_against_the_theory_of_relativity

 

[Dale]

Your comment would seem to support my belief that I have not yet see a correct Minkowski diagram with the astronaut as the observer. (In fact I believe it can't be done) The references given by Frame Dragger give the astronaut the orthagonal axis and then offer a kind of compensation for the ensuing contradiction. Yet those references are considered canon.

It is certainly ok to use non-inertial coordinates in which a non-inertial observer is given a constant coordinate position, but then those diagrams are not Minkowski diagrams because the Minkowski metric would not be valid in such coordinates. That would be a more general class of coordinate charts where general metrics are permitted rather than limiting it to the Minkowski metric. You certainly can use such coordinate systems, provided you are careful with your math and if you do so you always get the same result for the twins.

 

[Frame Dragger]

It is certainly ok to use non-inertial coordinates in which a non-inertial observer is given a constant coordinate position, but then those diagrams are not Minkowski diagrams because the Minkowski metric would not be valid in such coordinates. That would be a more general class of coordinate charts where general metrics are permitted rather than limiting it to the Minkowski metric. You certainly can use such coordinate systems, provided you are careful with your math and if you do so you always get the same result for the twins.

...Which brings the whole thing back to... why?! The original statement that a "skeptic" would somehow require a valid reversal of the diagram is still baffling.

 

[MikeLizzi]

Yep, exactly. Of course time doesn't jump, it only appears to jump as an artifact of the simplifying assumption of instantaneous turnaround, which equally can't happen. In the accelerated frame of the ship, Earth's clock runs fast. The greater the acceleration, the faster Earth's clock runs in the ship's frame. It's only modeled as a jump associated with instantaneous turnaround to simplify the math.

In reality there must be a finite interval for the turnaround acceleration and for Earth's clock to advance. They are both only treated as instantaneous to simplify the math, not to claim it's physically possible.

I've never seen a Minkowski diagram for the ship twin either, but it couldn't be in the standard form. It would have to be modified to allow for gravitational time dilation of clocks in an accelerated reference frame, but I doubt it could then be called a Minkowski diagram.

Einstein's own 1918 Twins paradox resolution is unique in that it does consider the non-inertial reference frame in which the ship is "stationary" the entire trip. It skips the math, presumably because his intended audience wouldn't need it to be shown. You can find it here: http://en.wikisource.org/wiki/Dialog_about_objections_against_the_theory_of_relativity

Hi Al68,

Yours is the point I was trying to make. Thank you very much for stating it.

To those who think I was being unscientific:

Are you now going to declare Al68 unscientific too?

Just adding the following: I am thu with this thread.

 

[Dmitry67]

Hi Dmitry67,

Thank you. I put down some numbers for the first diagram. I assume gamma is 2. The first column is the time on the astronaut's clock. The second column is the calculated time on an Earth clock. OK?

First Diagram:

Astronaut Time____Earth Time
(observer)
0 _______________ 0
1 _______________ .5
2 _______________ 1.0
3 _______________ 1.5
4 _______________ 2.0
5 _______________ 2.5
Turn around

Yes, then

5 (after turnaround) __ 17.5
6 _______________ 18
7 _______________ 18.5
8 _______________ 19
9 _______________ 19.5
10 ______________ 20

You are probably surprised by the gap in "Earth time"? It is not something really physical. It is just a a position of Earth at 4D miskovski spacetime based on the calculations of the spaceship. When you turn around, you rotate that diagram, and positions of the points change.