Clock postulate and differential aging

[ThomasT]

In a recent thread about differential aging in the archetypal twin scenario, I suggested that the periods of oscillators are affected by accelerations, or in other words that a clock's tick rate is affected by changes in its speed.

This statement was disagreed with by some, who said that it was contradicted by the clock hypothesis or postulate.

However, the clock postulate just says that the rate of the earthbound clock is always related to the rate of the traveling clock by (1-v²)^-1/2 .

From the above mentioned thread, when asked for an equation relating tick rate and acceleration, I replied:

What about this:
The tick rate is r(1-(v/c)2)0.5, where v is the relative velocity of the clock and the observer determining a tick rate, and r is the tick rate of the clock at rest.

When the velocity of the traveling clock changes (when it's accelerated), then its tick rate (wrt the stationary earthbound observer) changes.

To which DaleSpam replied:

That is the same as the equation I gave, \frac {d\tau} {dt} = \sqrt {1-v^2/c^2} . So if that is what you mean, then we are obviously in agreement. I am glad we straightened that out.

Note that v is speed.

Are we in agreement then? Has it been straightened out?

 

[Dmitry67]

No.

I had already provided an example:
2 twin scenatio with identical acceleration, with the same distance where second twin is accelerated, but with 2 different "arms" - total travel distance where the accelerated twin is moving without acceleration

Based on your hypotesis, age difference will be the same in both cases (as acceleration is identical), while it is not (in a case with a longer arm there is more difference)

 

[Dale]

Are we in agreement then?

I believe so, yes. We use the same formula for making predictions about clock rates and elapsed time. Everything else is either a philosophical argument about alternative interpretations of the same equations or a semantic argument about the best way to translate the math into English.

 

[Frame Dragger]

I believe so, yes. We use the same formula for making predictions about clock rates and elapsed time. Everything else is either a philosophical argument about alternative interpretations of the same equations or a semantic argument about the best way to translate the math into English.

"To translate the math into English..." Let me finish that for yah... "Which is basically impossible, and totally impossible right now for SQM."

:)

It's true, but it does make life a little difficult sometimes.

 

[bcrowell]

For a couple of good discussions of this, see p. 9 of Dieks http://www.phys.uu.nl/igg/dieks/rotation.pdf and this page by Baez http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html . I find Dieks' argument very persuasive and physically enlightening; you don't need a separate postulate, because it follows directly from SR. Both Baez and Dieks agree that this statement of clocks holds iff the clocks are small enough (infinitesimally small).

 

[Jamneutron]

I have seen in many places and accept that the traveling twin clock slows with respect to the Earth clock caused by relativity.
Can anyone explain or point to an explanation for how this mechanism works.

 

[ThomasT]

I believe so, yes. We use the same formula for making predictions about clock rates and elapsed time. Everything else is either a philosophical argument about alternative interpretations of the same equations or a semantic argument about the best way to translate the math into English.

Thanks DaleSpam. I thought that my original statement that precipitated the confusion was really pretty innocuous and not very informative -- and, I'm glad you've helped confirm that.

And I remain absolutely amazed and fascinated by the physical phenomenon of differential aging.

 

[ThomasT]

No.

I had already provided an example:
2 twin scenatio with identical acceleration, with the same distance where second twin is accelerated, but with 2 different "arms" - total travel distance where the accelerated twin is moving without acceleration

Based on your hypotesis, age difference will be the same in both cases (as acceleration is identical), while it is not (in a case with a longer arm there is more difference)

I agree that if you run one clock longer at the higher speed, even though the acceleration histories are otherwise identical, then that clock will accumulate less time.

Neither your scenario nor the clock postulate-hypothesis contradict the statement that changes in speed (accelerations) affect the periods of oscillators (the tick rates of clocks).

 

[Dale]

Neither your scenario nor the clock postulate-hypothesis contradict the statement that changes in speed (accelerations) affect the periods of oscillators (the tick rates of clocks).

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.

 

[Dale]

I have seen in many places and accept that the traveling twin clock slows with respect to the Earth clock caused by relativity.
Can anyone explain or point to an explanation for how this mechanism works.

Here is my favorite:

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_spacetime.html

 

[bcrowell]

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.

The only thing that makes me want to point people to Dieks in addition to Baez is that Dieks explains how it's not necessary to make it a separate postulate. It follows logically from the usual postulates of SR.

 

[Dale]

I am still reading that one, but it looks good so far.

 

[russ_watters]

I agree that if you run one clock longer at the higher speed, even though the acceleration histories are otherwise identical, then that clock will accumulate less time.

Neither your scenario nor the clock postulate-hypothesis contradict the statement that changes in speed (accelerations) affect the periods of oscillators (the tick rates of clocks).

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.

 

[Al68]

When the velocity of the traveling clock changes (when it's accelerated), then its tick rate (wrt the stationary earthbound observer) changes.

Sure that's a true statement, but incomplete, since the following statement is just as true:

When the velocity (relative to the observer) of the traveling clock changes (when the clock is not accelerated), then its tick rate (wrt the observer) changes.

So while your statement is perfectly true, it would be just as true if the clock doesn't accelerate, and for the same reason.

So a more precise statement would be: If the relative velocity of a clock changes (whether it accelerates or not), then its tick rate (wrt the observer) changes.

As an example, in the start of the twins paradox, the ship accelerates away from earth. The tick rate of Earth's clock (the periods of oscillations of Earth's clock) changed relative to the ship observer due to the change in the relative velocity of Earth's clock relative to the ship. Would you say that the tick rate of Earth's clock was affected by the ship's acceleration?

 

[Frame Dragger]

Sure that's a true statement, but incomplete, since the following statement is just as true:

When the velocity (relative to the observer) of the traveling clock changes (when the clock is not accelerated), then its tick rate (wrt the observer) changes.

So while your statement is perfectly true, it would be just as true if the clock doesn't accelerate, and for the same reason.

So a more precise statement would be: If the relative velocity of a clock changes (whether it accelerates or not), then its tick rate (wrt the observer) changes.

As an example, in the start of the twins paradox, the ship accelerates away from earth. The tick rate of Earth's clock (the periods of oscillations of Earth's clock) changed relative to the ship observer due to the change in the relative velocity of Earth's clock relative to the ship. Would you say that the tick rate of Earth's clock was affected by the ship's acceleration?

Hmmm... I suppose that would depend on which frame of reference you looked at. If you were born on that ship during its accelerating period you'd percieve the tr on Earth to be faster relative to your own clock. From the point of view of a third observer at rest relative to the ship and Earth, what would they percieve? Wouldn't the change in tick rate be purely a matter of which Intertial Frame you happen to be in? A third observer is still going to agree on the order of events and the way they unfold will be consistant, regardless of which tr you decide is valid. They are, after all... relative, aren't they?

 

[MikeLizzi]

Here is my favorite:

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_spacetime.html

Hi DaleSpam,
Good reference. I am quite familiar with the different versions of the diagram shown in that reference.

But I have a problem.
When the earthbound twin is the observer (presumably inertial) I plot events for the earthbound twin using the orthagonal axes and for the astrounat twin plot events using rotated axes. That's what I see shown in the diagram.

So far so good.

But the skeptic wants to see corroboration. The skeptic asks "What about when the astronaut is the observer?" So I draw the diagram with the astonaut as the observer. So then the astronaut gets events ploted using the orthagonal axes and the earthbound twin gets events ploted using the rotated axes. The diagram looks the same but the people and agings are reversed.

I get a parodox.

 

[atyy]

The only thing that makes me want to point people to Dieks in addition to Baez is that Dieks explains how it's not necessary to make it a separate postulate. It follows logically from the usual postulates of SR.

But the time read by an ideal accelerated clock surely it requires an additional definition, as Dieks's limit, ie. the proper time. As I understand it, the clock hypothesis is a definition of an ideal clock.

 

[Frame Dragger]

Hi DaleSpam,
Good reference. I am quite familiar with the different versions of the diagram shown in that reference.

But I have a problem.
When the earthbound twin is the observer (presumably inertial) I plot events for the earthbound twin using the orthagonal axes and for the astrounat twin plot events using rotated axes. That's what I see shown in the diagram.

So far so good.

But the skeptic wants to see corroboration. The skeptic asks "What about when the astronaut is the observer?" So I draw the diagram with the astonaut as the observer. So then the astronaut gets events ploted using the orthagonal axes and the earthbound twin gets events ploted using the rotated axes. The diagram looks the same but the people and agings are reversed.

I get a parodox.

You're missing something. In the case of the astronaut both observers on Earth and "the rocket ship" agree that the rocket is accellerating in relation to a (relatively) stationary Earth. If you're going to invert the experiment the way you did, then you have to take into account that both Earthbound observers and the Astronaut need to "agree" that the EARTH to be accelerating away from the astronaut. It's one of those "lost in translation" issues between the math and the thought experiment.

What matters is what is under acceleration relative to the Intertial Frames you're comparing it to (returning to in the case of the astronaut, to Earth). The body under acceleration experiences the effect, and in this thought experiment we can't equivocate "velocity and acceleration", because an astronaut MUST undergo a period of acceleration and then braking in relation to his initial Frame.

 

[MikeLizzi]

You're missing something. In the case of the astronaut both observers on Earth and "the rocket ship" agree that the rocket is accellerating in relation to a (relatively) stationary Earth. If you're going to invert the experiment the way you did, then you have to take into account that both Earthbound observers and the Astronaut need to "agree" that the EARTH to be accelerating away from the astronaut. It's one of those "lost in translation" issues between the math and the thought experiment.

What matters is what is under acceleration relative to the Intertial Frames you're comparing it to (returning to in the case of the astronaut, to Earth). The body under acceleration experiences the effect, and in this thought experiment we can't equivocate "velocity and acceleration", because an astronaut MUST undergo a period of acceleration and then braking in relation to his initial Frame.

Thanks for your reply Frame Dragger. I am familiar with the arguments you present. But I was talking about using a Minkowski Diagram to describe the the action when the astronaut is the observer. What do your arguments have to do with that? I'm asking how one draws a Minkowski Diagram for the Twins problem when the astronaut is the observer.

 

[Frame Dragger]

Thanks for your reply Frame Dragger. I am familiar with the arguments you present. But I was talking about using a Minkowski Diagram to describe the the action when the astronaut is the observer. What do your arguments have to do with that? I'm asking how one draws a Minkowski Diagram for the Twins problem when the astronaut is the observer.

That's easy... http://en.wikipedia.org/wiki/Twin_paradox Halfway down this issue is addressed. Here is the relevant portion...

Difference in elapsed times: how to calculate it from the ship
Note that the above formula


(here τ is another customary notation for the time of the non-inertial observer K') gives us the dependence of the proper time τ on the time elapsed in the inertial frame K, assumed the velocity v(t) is known. Actually, this equation answers the problem of the dependence between proper time and inertial time for the inertial observer at rest in K but not for the moving observer. Indeed, the data v(t) is not always easily accessible from the ship. The observer in the non-inertial frame K' may find it difficult to measure its velocity with respect to K, not to mention that this velocity has to be parametrized with respect to the inertial time of K, a step which needs the dependence between proper and inertial time to be known in advance. Can the non-inertial observer predict what will be the time elapsed for the twin which remained at home using only easily accessible data? The answer is affirmative and at least for a unidirectional motion has a simple answer given by[10]


where a(τ) is the acceleration of the non-inertial observer as measured by himself (for instance with an accelerometer) in the whole round-trip and should not be confused with the non-relativistic acceleration d2x / dt2. It can also be shown that the inequality Δt > Δτ follows from the previous expression by using the Cauchy-Schwarz inequality:

Check the page for the equations, I can't copy-paste them, and I'm not writing them out here when I can link lol.

 

[MikeLizzi]

That's easy... http://en.wikipedia.org/wiki/Twin_paradox Halfway down this issue is addressed. Here is the relevant portion...



Check the page for the equations, I can't copy-paste them, and I'm not writing them out here when I can link lol.

Frame Dragger:
Your response and the wikipedia reference have nothing to do with the issue I presented.

 

[Frame Dragger]

Frame Dragger:
Your response and the wikipedia reference have nothing to do with the issue I presented.

Both address the issue that no paradox arises from a formulation of diagram from the "traveler's" Intertial Frame.

 

[MikeLizzi]

Both address the issue that no paradox arises from a formulation of diagram from the "traveler's" Intertial Frame.

Frame Dragger:
I sorry but that's not what I was asking about. I guess I didn't explain myself very well. I was asking how you draw a Minkowski Diagram for the twins problem with the astronaut as the observer. The diagram in the wikipedia article is with the earthboud twin as the orserver.

I should add a little more or I will appear impolite. 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.

 

[Frame Dragger]

Frame Dragger:
I sorry but that's not what I was asking about. I guess I didn't explain myself very well. I was asking how you draw a Minkowski Diagram for the twins problem with the astronaut as the observer.

Ah... I thought your problem was the apparent paradox. I see. I don't know how to draw a Minkowski Diagram from the TP from that point of view. I see why you're having a problem... sorry about the confusion on my end.

Here is an excerpt that might help: http://www.springerlink.com/content/j603l05128p27727/fulltext.pdf?page=1

 

[MikeLizzi]

Ah... I thought your problem was the apparent paradox. I see. I don't know how to draw a Minkowski Diagram from the TP from that point of view. I see why you're having a problem... sorry about the confusion on my end.

Here is an excerpt that might help: http://www.springerlink.com/content/j603l05128p27727/fulltext.pdf?page=1

Hi again,
Yep. I scanned the article that you reference. I've read so many of these I can't go through them anymore. That reference is just repeating, using slightly different language, what the wikipedia article said. They all require the astronaut to add a time jump to the earthbound twins clock. Now somebody's clock can go slower or faster than yours but it can't jump. It would lead to all kinds of paradoxes. Solving a problem by including an artificial jump to somebody's clock is not doing physics. Its playing games with a calculation to make sure the numbers come out right. 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.

 

[Mentz114]

Calculating proper-time

For those who want to do a proper-time calculation that doesn't involve jumps, and have access to Maxima, the script below does this. It calculates the position, velocity etc for a space-ship fillowing a worldline defined by

<br /> x = t^2(t^2-2)d<br />
d is a parameter ( actually the furthest point reached). This worldline describes a comfortable trip along the x-axis and back again to base. The journey will last 2 time units on the base clock.

The script plots the variables and works out the time on the ships clock.
( Warning : this script may be wrong. I cannot be held responsible for any problems it may cause)

/*-------------------------------------------------------
This script calculates the velocity and acceleration
required to make a 'comfortable' return journey to a point
d units away in 2 time units. The proper length of the 
trip is found.
---------------------------------------------------------*/

/* plot the position function xt = d*(t^2-1)^4 */
/* d is the furthest point on the trip. The Romberg integration fails around d=1/2  */
d:1/100; /* change this to do different trips */

xt:d*t^4*(t-2)^4;
wxplot2d([xt], [t,0,2]);

/* Get velocity */
vt:diff(xt,t,1);

/* plot velocity */
wxplot2d([vt],[t,0,2]);

/* get acceleration */
diff(vt,t,1);

/* plot acceleration */
wxplot2d([%],[t,0,2]);

/* Proper time by numerical integration */
romberg(sqrt(1-(vt)^2), t, 0, 2);


/* try some different trips */
d:1/50; 
xt:d*t^4*(t-2)^4;
vt:diff(xt,t,1);

/* Proper time by numerical integration */
romberg(sqrt(1-(vt)^2), t, 0, 2);

d:1/20; 
xt:d*t^4*(t-2)^4;
vt:diff(xt,t,1);

/* Proper time by numerical integration */
romberg(sqrt(1-(vt)^2), t, 0, 2);

d:1/10; 
xt:d*t^4*(t-2)^4;
vt:diff(xt,t,1);

/* Proper time by numerical integration */
romberg(sqrt(1-(vt)^2), t, 0, 2);

d:1/5; 
xt:d*t^4*(t-2)^4;
vt:diff(xt,t,1);

/* Proper time by numerical integration */
romberg(sqrt(1-(vt)^2), t, 0, 2);

 

[Frame Dragger]

This is all very interesting... but what's the value in it? I understand the philosophical objection to cutting corners, but there is a reason the corner is often cut; it can afford to be.

Mentz has kindly and clearly showed the "how", but I'm still a bit confused as to the "why", except as a mental exercise.

 

[Dale]

I'm asking how one draws a Minkowski Diagram for the Twins problem when the astronaut is the observer.

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.

 

[Frame Dragger]

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.

I still reiterate my question about the purpose of this. A skeptic already has better direct observations which confirm SR/GR than a thought experiment. Is this just spinning the mental wheels?

 

[MikeLizzi]

I still reiterate my question about the purpose of this. A skeptic already has better direct observations which confirm SR/GR than a thought experiment. Is this just spinning the mental wheels?

Hi Frame Dragger:

The skeptic is asking for corroboration. We solve the problem from the point of view of the earthbound twin because its the easier task. Since we believe SR is internally consistent, we are satisfied. But the skeptic asks "How can you get a corroborating answer from the point of view of the astronaut?" 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.

 

[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