Twin Paradox Problem: Do Twins Age Differently?

[Dale]

Then at what point does the clock discrepency occur?

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?

 

[Jeronimus]

It comes from the Relativistic Doppler Factor:

√((1+β)/(1-β)) = √((1+0.9)/(1-0.9)) = √((1.9)/(0.1)) = √19 ≈ 4.359

Have you actually done an analysis of the Twin Paradox by this method? Can you show us some numbers? Have you demonstrated that the traveling twin will see exactly what I described in post #5 or do you think your analysis will come up with something different? I want to see your numbers, please.

This is the most misleading method to get someone to understand what really happens.

But he's not omniscient and neither are we so we have to settle for what we can actually see and not some wishful thinking about remote viewing.

Omniscient is the shortcut for placing an army of observers all along the frame the traveling twin is at rest in. The army of observers has synced clocks and writes down each event happening inside that frame including the time when Earth's clock is passing by them and the time it shows when passing by them (locally).
The twin takes the event data sent to him and composes an x/t diagram.
He then is able to check how long one second measured on the Earth twin's clock takes within his frame.
He will find that ANY clock which is within the Earth's frame at rest, will run at the SAME pace, slowed down by a factor of about 0,4359s.
Otherwise said. The clocks which are at rest in Earth's frame, advance ~0,4359s for every second passed within the traveling twin's frame.



According to you, the pace at which clocks which are at rest in Earth's frame seen from the traveling twin's perspective depends on if the clocks are in front or behind the twin. So someone who is behind the twin or in front, will come to different conclusions on the same clock. This is ridiculous.


As for the calculations:

I describe here,
https://www.physicsforums.com/showthread.php?t=640671

how to get Δt

And yes, i have done it. And yes, Δt is the same NO matter if you passed a clock or it is moving towards you...
Also, because of

"The laws of physics are the same in all inertial frames of reference."

clocks at rest in System A, traveling at vrel to System B will run slower by the same amount as clocks at rest in System B run slower in System A where they move at vrel.

 

[Jeronimus]

Sorry, it's almost the contrary! What they literally see are true observations (measurements), which SR must be able to predict. And what they calculate based on arbitrary assumptions is in conflict with what others calculate based on different arbitrary assumptions; it's therefore erroneous (and leading to paradoxes!) to think that the traveling twin is able to calculate the real pace of the Earth twin's clock. If he knows SR, then he understands that he can not do such a thing. Different inertial reference systems assign different clock rates to that clock.

[ah I see that George was faster than me ]

Not really. 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.(BTW - 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. See above)

 

[ghwellsjr]

You are only half correct. When the traveling twin departs at 90%c, they both do see each others clock equally slow--a factor of 0.2294 times their own. But this is true only for the outbound portion of the trip. Things are different for the inbound portion of the trip. As soon as the traveling twin turns around, he immediately sees the Earth twin's clock going fast--4.359 times his own. Since he spends an equal amount of time going out as coming in, you can easily calculate how much of a difference there will be in the amount the two twins aged by simply taking an average of the two factors. The average of 0.2294 and 4.359 is 2.2942 so however much the traveling twin aged during the trip, his Earth twin will age 2.2942 times as much. Simple, isn't it?

Then at what point does the clock discrepency occur?
1) During acceleration?
If so, then the difference of age is a function of acceleration. If so, then then it's wrong because we know that the longer the twin travels, the less he ages, and it has nothing to do with how long he accelerates.
2) During traveling?
If so, then your statement doesn't make sense to me if you are saying that they both see their clocks run equally fast or slow (it may make sense to someone else). The only way it would make sense is if they would both see their clocks run slower outbound and faster inbound but by different amounts.
Edir: Or are you saying that the Earth twin still sees the other twin's clock run slower during inbound trip?

By the way, I think that I understand relativistic principles, but I can't deal with formula answers.

I didn't make any comment about what the Earth twin sees of the traveling twin's clock beyond the beginning of the trip but since you asked, I'll fill in the details.

The Earth twin will see the traveling twin's clock run slower than his own for much more than half the trip because he has to wait for the image of the distant turn around event to reach him over that long distance and this is (partly) what results in him seeing his twin's clock with less time on it when they finally get back together.

Another way of putting this is that he will see his twin traveling away from him for way more than half the time (while he sees his clock running slow) and then he sees him turn around and come back for a rather short time (while he sees his clock running fast). So I wouldn't say that the Earth twin sees the other twin's clock run slower during the inbound trip, it's that he sees the outbound trip last longer than the inbound trip. Remember, when he sees outbound, he sees a slow clock, when he sees inbound, he sees a fast clock.

How do you like that? No formulas!

 

[Jeronimus]

Just to make sure, i will explain how I would measure the pace of a moving clock within a system i am at rest in...Two observers which are at a distance x. They are both at rest to each other. They both synced their clocks.

A moving clock passes by the front observer, which writes down the time it passed by him AND the time the moving clock shows. To keep it simple, assume it was at t1=0s for the observer's clock and clockt1=0s.

The clock keeps moving at vrel and reaches the second observer at t2=10s. At vrel = 0.5c and a distance of 5 lightseconds, the moving clock at the second observer would display clockt2 ~8.66s.

Are we in agreement with this?
Are we in agreement that any two observers at a distance of 5 lightseconds within the SAME inertial frame of reference would observe the same if a similar clock moved by them at vrel = 0.5c?The two observers would sent each other the data, allowing them to conclude the pace of the moving clock.

 

[kamenjar]

I think that this whole thread and a multitude of others are here on this forum because some of the GR/SR are simply "stubborn" when making claims. Stubborn in a sense that some claims keep repeating all over posts and texts that are not true when all facts are accounted for.

To my understanding (and I may be wrong), there is no twin paradox and there's nothing paradoxical about it. The only paradoxical thing is the lame statement that says something about "both travelers agreeing that other person's CLOCK runs slower". That is just plain absurd. You can not make statement about the other person's clock when you are moving by just observing the rate of the ticks measured by your own clock rate.

The only way to make a statement about the other twin's CLOCK is to account for redshift. Then you can actually make statements about which clock runs slower and I believe that if someone (more capable than myself) that did calculations conclude that the traveling twin's clock is slower during the whole trip AND that both the traveler and the guy on Earth can do those measurements and calculations and conclud the same - the traveling twin's clock runs slower. So as ghwellsjr said, it's probably simple and there's nothing paradoxical or confusing.

 

[ghwellsjr]

This is the most misleading method to get someone to understand what really happens.

Ok, we'll agree to use your definition of what real means:

(BTW - 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. See above)

Omniscient is the shortcut for placing an army of observers all along the frame the traveling twin is at rest in.

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?

The army of observers has synced clocks and writes down each event happening inside that frame including the time when Earth's clock is passing by them and the time it shows when passing by them (locally).
The twin takes the event data sent to him and composes an x/t diagram.
He then is able to check how long one second measured on the Earth twin's clock takes within his frame.
He will find that ANY clock which is within the Earth's frame at rest, will run at the SAME pace, slowed down by a factor of about 0,4359s.
Otherwise said. The clocks which are at rest in Earth's frame, advance ~0,4359s for every second passed within the traveling twin's frame.

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.

According to you, the pace at which clocks which are at rest in Earth's frame seen from the traveling twin's perspective depends on if the clocks are in front or behind the twin. So someone who is behind the twin or in front, will come to different conclusions on the same clock. This is ridiculous.

It would be ridiculous if I ever said that. Can you point to the post where I said that?

As for the calculations:

I describe here,
https://www.physicsforums.com/showthread.php?t=640671

how to get Δt

And yes, i have done it. And yes, Δt is the same NO matter if you passed a clock or it is moving towards you...
Also, because of

"The laws of physics are the same in all inertial frames of reference."

clocks at rest in System A, traveling at vrel to System B will run slower by the same amount as clocks at rest in System B run slower in System A where they move at vrel.

But what I asked you to do was calculate what the traveling twin sees of the Earth's twin's clock and you haven't done that.

 

[marty1]

I definitely do not agree that velocity has any anything to do with this effect. Anyone in the universe can claim they are at rest so long as they are not accelerating. Even with constant acceleration they could claim they are in a gravitational field.

We can measure the difference in the speed of clocks on the surface of the Earth compared to those in orbit solely due to the affects of gravity. This is not disputed.

 

[ghwellsjr]

I definitely do not agree that velocity has any anything to do with this effect. Anyone in the universe can claim they are at rest so long as they are not accelerating. Even with constant acceleration they could claim they are in a gravitational field.

But the twins cannot both claim they continue at rest if they separate and then end up together at rest again.

What are you talking about?

We can measure the difference in the speed of clocks on the surface of the Earth compared to those in orbit solely due to the affects of gravity. This is not disputed.

We can measure the difference in the speed of clocks at different elevations on the surface of the Earth due just to the effects of gravity but if you're going to put one in orbit you have to also take into account its velocity. Your comment is disputed.

 

[ghwellsjr]

Just to make sure, i will explain how I would measure the pace of a moving clock within a system i am at rest in...Two observers which are at a distance x. They are both at rest to each other. They both synced their clocks.

A moving clock passes by the front observer, which writes down the time it passed by him AND the time the moving clock shows. To keep it simple, assume it was at t1=0s for the observer's clock and clockt1=0s.

The clock keeps moving at vrel and reaches the second observer at t2=10s. At vrel = 0.5c and a distance of 5 lightseconds, the moving clock at the second observer would display clockt2 ~8.66s.

Are we in agreement with this?
Are we in agreement that any two observers at a distance of 5 lightseconds within the SAME inertial frame of reference would observe the same if a similar clock moved by them at vrel = 0.5c?The two observers would sent each other the data, allowing them to conclude the pace of the moving clock.

What you are saying is correct.

However, what I want you to consider is that the two observers do not have to overtly send any data to the other observers, they can just look at it. EDIT: I went back and see that you want them to send both their own time and the time they see on the other clock, so yes, they do have to overtly send the time on the other clock.

So what I'm asking you to do is figure out how each of the clocks observes the time on each of the other clocks in your scenario.

 

[marty1]

But the twins cannot both claim they continue at rest if they separate and then end up together at rest again.

What are you talking about?

While of course they can resolve that after the fact.

We can measure the difference in the speed of clocks at different elevations on the surface of the Earth due just to the effects of gravity but if you're going to put one in orbit you have to also take into account its velocity. Your comment is disputed.

But the relative difference in speeds of the 2 clocks is zero for a geosynchronous clocks. They are always the same distance from one another. How fast is a clock in geosynchronous orbit moving away from you? It is not moving away from you. My satellite dish is always pointing in the same direction. The difference is the acceleration. Each of the clocks are accelerating toward the Earth at different rates and in effect accelerating around a different curve.

Unless, of course, you want to say that "absolute" speed matters.

I am just wondering now how fast a clock ticks if it follows the Earth in orbit around the sun.

Another point:

When the twins are moving away from one another they are advancing ahead of the beam of light that is the other person observation of the other clock so it will advance slower because you are always getting a little farther away (I think I said that right). If you could move at the speed of light the other clock would appear to stop. If you could move faster the other clock would move backwards.

 

[ghwellsjr]

While of course they can resolve that after the fact.

Resolve what? I don't know what you're talking about.

But the relative difference in speeds of the 2 clocks is zero for a geosynchronous clocks. They are always the same distance from one another. How fast is a clock in geosynchronous orbit moving away from you? It is not moving away from you. My satellite dish is always pointing in the same direction. The difference is the acceleration. Each of the clocks are accelerating toward the Earth at different rates and in effect accelerating around a different curve.

Unless, of course, you want to say that "absolute" speed matters.

In this sense, what you are calling "absolute" speed does matter. The satellites are traveling much faster than you are even though they appear to be stationary in the sky above you and you can't ignore their speed when calculating how much slower they are compared to you

I am just wondering now how fast a clock ticks if it follows the Earth in orbit around the sun.

I don't really know but again, you can't ignore its speed.

Another point:

When the twins are moving away from one another they are advancing ahead of the beam of light that is the other person observation of the other clock so it will advance slower because you are always getting a little farther away (I think I said that right). If you could move at the speed of light the other clock would appear to stop. If you could move faster the other clock would move backwards.

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.

 

[Dale]

To my understanding (and I may be wrong), there is no twin paradox and there's nothing paradoxical about it.

I agree. It isn't a true paradox. It is more like a very famous homework problem that introductory students often get wrong.

The only paradoxical thing is the lame statement that says something about "both travelers agreeing that other person's CLOCK runs slower". That is just plain absurd.

The statement isn't absurd, but it is a little off. It should read, "both travelers agree that the other person's clock runs slower in their reference frame". How fast a clock runs is a frame dependent quantity, so the frame needs to be specified.

I believe that if someone (more capable than myself) that did calculations conclude that the traveling twin's clock is slower during the whole trip

Again, how fast a clock runs is frame dependent. Your belief is true in some frames, but false in others.

 

[marty1]

Resolve what? I don't know what you're talking about.

In this sense, what you are calling "absolute" speed does matter. The satellites are traveling much faster than you are even though they appear to be stationary in the sky above you and you can't ignore their speed when calculating how much slower they are compared to you

I don't really know but again, you can't ignore its speed.

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.

 

[zonde]

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.

Only if and when they switch the inertial reference frames that they use, as I mentioned. For example the astronauts in the international space-station likely stick to the ECI frame instead of continuously changing distant "now".
I did not introduce any accelerated reference frames, which need additional definitions and the introduction of such is only good for creating more confusion when discussing SR. In the standard twin paradox the turn-around is of negligible duration so that the effect of acceleration on accumulated clock time of the accelerated clock is negligible, even if the so-called "clock hypothesis" were not applicable.

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).

 

[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?