Is the Twin Paradox Truly Unresolvable from Speedo's Perspective?

[Mentz114]

If your problem involves orbits, then you can't solve it using Special Relativity.

This is true, but there is an invariant proper interval in GR, so the calculation is similar, but using a different metric from the Minkowski.

 

[ghwellsjr]

Well, that's the issue: why must it be an inertial frame of reference? Why can't it be a "piecewise inertial" frame of reference? That is, why can't you use

τ= ∫√(1-(v/c)2 dt

where t is proper time for the "traveling" twin, and v is the speed of the "stay-at-home" twin relative to the instantaneous rest frame of the traveling twin? You can't because there are pieces of the worldline of the stay-at-home twin that are left unaccounted for. (In the Euclidean analog, the problem is that there are pieces of the path of the "straight" road that are counted twice by the corresponding formula for length).

That formula is defined for an inertial Frame of Reference. Its definition is not applicable to a noninertial frame. If you want to come up with your own definition of a noninertial frame, then you have to be precise about what you mean and you have to come up with your own formula for calculating whatever you want to calculate.

But again, my complaint about your analogy is that when you say that one traveler can look over at the other one and see how far he has progressed, it gives the impression that Speedo can look over at Goslo and see how far he has progressed, meaning, of course, where he is at any particular time, but he can't do that. No explanation provided by SR, either with inertial or noninertial frames can do that. All that these "explanations" are doing is echoing back the definitions that the frames employed in the first place. They don't provide any new information to the twins. It's nothing more than saying, "I will call the time on Goslo's clock just before Speedo turns around t_t- and the time just after Speedo turns around t_t+ and then subtracting the times and saying that Goslo's clock did something weird when Speedo turned around" (for a noninertial frame). Or saying, "the time on Goslo's clock when Speedo turned around was t_t" (for an inertial frame). They are both based on arbitrary definitions and none is more right or wrong or more enlightining to either twin.

We already dissuaded Michio Cuckoo from using insta-beams (see his post #7) and now you have just reintroduced them in your analogy. That's my complaint.

 

[ghwellsjr]

My explanation is not based on SR, it is based only on the Principle of Relativity, Einstein's first postulate, but not on his second postulate. His second postulate is also needed in order to construct a Frame of Reference. So even without resorting to a reference frame, you can still prove unambiguously which twin ages less and which twin ages more.

Your explanation tacitly invokes the light-speed postulate, and also a frame of reference. Remember, Galilean relativity also satisfies the principle of relativity, but it doesn't imply unequal aging of the twins. So the unequal aging can't be explainable in terms of ONLY the principle of relativity. When you talk about Doppler effects that each twin will see, you are tacitly assuming things about the propagation of light being unaffected by the motion of the source, and so on, things that are tantamount to the light-speed postulate, in addition to the principle of inertia (i.e., the principle of relativity), and these things collectively are sufficient to construct inertial coordinate systems. You can't get any of the unique effects of special relativity (as distinct from Galilean relativity) purely from the relativity postulate alone. The light-speed postulate (or something equivalent to it) is necessary.

Normal Doppler, where there is a medium such as air for sound, is not relativistic, meaning that two observers don't hear the same thing coming from the other one because we have to take into account their relative speed in the medium.

If we assume the Principle of Relativity for light, we are assuming that what each twin sees of the other one is symmetrical and not dependent on their relative speed in any medium. It can be easily demonstrated that two inertial observer with a relative motion between them, traveling along the same line will see a Doppler factor while approaching that is the reciprocal of the Doppler factor while receding. This by itself is all that is necessary to show that if one of those twins remains inertial while the other one travels away at some constant speed creating a constant Doppler factor less than one and then turns around and approaches at that same constant speed, he will observe a Doppler factor that is the inverse of the first one, a number greater than one. Let's say that the return Doppler factor is N, a number greater than one and the departing Doppler factor is 1/N. We take the average of these two numbers to get the total ratio of their accumulated ages. This number will always be greater than one and in fact is equal to gamma. Speedo will watch Goslo's clock constantly advancing during his entire trip, first slower then his own and then at turn-around faster than his own and when they meet, Goslo's clock will have advance gamma times the amount his own clock has advanced. This is exactly addressing the question that Michio Cuckoo asked in his first post.

Remember, Einstein's second postulate is that the propagation of light is c, meaning that the unmeasurable one-way time is equal to one-half of the round-trip time and the Doppler analysis does not require that or depend on that in any way. In fact it works the same in any frame even in those for which the two one-way times for a round trip propagation of light are not equal. In other words, it is making no statement about the synchronization of the clocks of the two twins while they are separated, only the final outcome of the time difference when they reunite.

 

[ghwellsjr]

I already did that in post #4. Did you read it? Do you understand it?

No I don't understand post 4

Nor have you seemed to understood the problem I posed. Let's reformulate the "Which twin aged?" problem.

Let each twin be in a separate spaceship A and B orbiting a planet C in circular orbits. We assume the planet to be non rotating. A orbits clockwise and B orbits anticlockwise. Both A and B clocks as seen from C are ticking at the same rate but at a reduced rate to that of C due to the orbit velocity and some increased rate due to gravitational effects as compared to if they where stationary on planet C. each time A and B pass (twice per orbit) the clocks have the same time. The GPS satellites pride themselves as proving SR as the clocks are corrected for both gravitational and velocity aspects as described.

Lets go into spaceship A and set our reference frame in this spaceship A. That is A is now experienced as stationary an B is moving relative to A and planet C is orbiting around A. Please describe what we expect from clock B now relative to A.

As I said I am totally confused as I cannot find a logical pleasing answer.

Usually, when people discuss the Twin Paradox, one of the twins does all the traveling while the other one stays home. Since you didn't explicitly state in your scenario in post #34 that they were in symmetrical orbits around a planet (you didn't even say there was a planet) and since you called them twins and since the subject of this thread is the Twin Paradox, I assumed that you were presenting Twin Paradox scenarios. But now that you have provided further details, I see that you are not asking about the Twin Paradox but something for which there is not even an apparent paradox.

Twin A will observe that twin B's clock advances more slowly than his own during most of twin B's orbit around the planet, assuming a transparent planet, and then he will see twin B's clock advancing faster than his own so that when twin B gets back to twin A, their clocks display exactly the same time. Why is this confusing to you?

 

[DrGreg]

After all, the noninertial traveler is only noninertial for a tiny moment; how can that tiny moment make such a big difference?

Analogy: suppost you are looking directly at the Moon and you then turn your head sideways through 90° in one second. That brief moment when your head is non-inertial causes the moon to move over one third of a million miles in one second relative to your head's "frame of reference". Sometimes a small change can have a big effect.

 

[ghwellsjr]

And as everyone knows, the sun travels many millions of miles per hour in its orbit around the Earth relative to the Earth's noninertial rest frame.

 

[Whovian]

And as everyone knows, the sun travels many millions of miles per hour in its orbit around the Earth relative to the Earth's noninertial rest frame.

But this is a thought experiment assuming the Earth is stationary, I think. Plus, as predicted by General Relativity, the Earth's orbit is actually an inertial FoR.

 

[Austin0]

And as everyone knows, the sun travels many millions of miles per hour in its orbit around the Earth relative to the Earth's noninertial rest frame.

Yes but that's due to rotation, yeah. If the rotational and orbital periods were the same (like the OP's rocket around his planet) it would be the whole cosmos revolving around a stationary earth-sol systems at many times c, even better huh?.
;-)

 

[ghwellsjr]

But this is a thought experiment assuming the Earth is stationary, I think. Plus, as predicted by General Relativity, the Earth's orbit is actually an inertial FoR.

I was talking about earth, not its orbit.

 

[Whovian]

I was talking about earth, not its orbit.

What do you mean? I meant the Earth itself, which is an inertial reference frame, but an observer actually on the Earth, due to gravity and rotation, is in a non-inertial reference frame.

 

[Samshorn]

If we assume the Principle of Relativity for light, we are assuming that what each twin sees of the other one is symmetrical and not dependent on their relative speed in any medium.

Not true. There are other ways that light could propagate, such as a classical (c+v) ballistic theory, which is perfectly relativistic, not reliant on any speed relative to a supposed medium, and yet such a theory wouldn't imply the time dilation effects and asymmetrical aging of special relativity, because the speed of light would not be independent of the speed of the source. (Try working out your prediction for the twins' ages based on a Newtonian ballistic theory of light, in which "what each twin sees of the other one is symmetrical and not dependent on their relative speed in any medium". You won't get any asymmetrical aging.) Only by combining BOTH of those conditions, i.e., the principle of relativity (classically associated with ballistic theories) AND the principle of source-independence (classically associated with wave theories) do you get the effects of special relativity. The latter is the light-speed postulate, which you are tacitly assuming.

 

[Whovian]

Why are we "tactically assuming" this? Experiments confirm it.

 

[A.T.]

After all, the noninertial traveler is only noninertial for a tiny moment; how can that tiny moment make such a big difference?

Because of the big distance. In an accelerated frame time flows at different rates along the acceleration direction. The further away along that line, the bigger the difference in clock rates.

See:
https://www.physicsforums.com/showpost.php?p=3920630&postcount=32

 

[omg!]

i think this is a much simpler thought experiment:

imagine you have two clocks C1, C2 in an inertial frame and showing the same time. a third clock C3 moves uniformly on a trajectory parallel to the line made by C1, C2. at time 0, C3 crosses the point of shortest distance to C1 on its trajectory. all the clocks read 0 at this point. After some time t2, an observer at C2 sees C3 crossing and reads off the time t3 on C3 and he finds t3<t2.

now consider the inertial frame associated with C3. would an observer traveling with C3 in that frame see that t3>t2 when C3 crosses C2?

 

[ghwellsjr]

I was talking about earth, not its orbit.

What do you mean? I meant the Earth itself, which is an inertial reference frame, but an observer actually on the Earth, due to gravity and rotation, is in a non-inertial reference frame.

I was talking about the approximately 24-hour rotation of the earth, not the annual orbital period of the Earth which would produce a speed for the sun of only a very small fraction of a million miles per hour. And just like an observer on the surface of the earth, the Earth itself can define a noninertial reference frame. It would be noninertial even without gravity. Rotation is all that I'm considering. In that noninertial frame, the sun travels at millions of miles per hour.

And as Austin0 pointed out, if we include stars beyond our sun, they are traveling at many times c.

 

[ghwellsjr]

If we assume the Principle of Relativity for light, we are assuming that what each twin sees of the other one is symmetrical and not dependent on their relative speed in any medium.

Not true. There are other ways that light could propagate, such as a classical (c+v) ballistic theory, which is perfectly relativistic, not reliant on any speed relative to a supposed medium, and yet such a theory wouldn't imply the time dilation effects and asymmetrical aging of special relativity, because the speed of light would not be independent of the speed of the source. (Try working out your prediction for the twins' ages based on a Newtonian ballistic theory of light, in which "what each twin sees of the other one is symmetrical and not dependent on their relative speed in any medium". You won't get any asymmetrical aging.) Only by combining BOTH of those conditions, i.e., the principle of relativity (classically associated with ballistic theories) AND the principle of source-independence (classically associated with wave theories) do you get the effects of special relativity. The latter is the light-speed postulate, which you are tacitly assuming.

Well, we don't live in a world where the ballistic theory of light is true and you are correct that Einstein did include the principle of source-independence in his second postulate but you have ignored the salient point of his second postulate which I specifically addressed at the end of the post you quoted:

Remember, Einstein's second postulate is that the propagation of light is c, meaning that the unmeasurable one-way time is equal to one-half of the round-trip time and the Doppler analysis does not require that or depend on that in any way. In fact it works the same in any frame even in those for which the two one-way times for a round trip propagation of light are not equal. In other words, it is making no statement about the synchronization of the clocks of the two twins while they are separated, only the final outcome of the time difference when they reunite.

So, yes, I am taking for granted the source-independence of light because that part of Einstein's second postulate is measurable and proven to be true.

So now would you agree that based solely on Einstein's first postulate of the Principle of Relativity and the principle of the source-independence of light but ignoring the rest of Einstein's second postulate and without establishing any frame of reference, we can determine that the Doppler factors for approaching and retreating at the same relative speed are reciprocals of one another and from this alone we can determine that Goslo ages more than Speedo by the factor gamma? And would you agree that this analysis directly answers Michio Cuckoo's question in post #1?

 

[GeorgeDishman]

i think this is a much simpler thought experiment:

imagine you have two clocks C1, C2 in an inertial frame and showing the same time. a third clock C3 moves uniformly on a trajectory parallel to the line made by C1, C2. at time 0, C3 crosses the point of shortest distance to C1 on its trajectory. all the clocks read 0 at this point. After some time t2, an observer at C2 sees C3 crossing and reads off the time t3 on C3 and he finds t3<t2.

now consider the inertial frame associated with C3. would an observer traveling with C3 in that frame see that t3>t2 when C3 crosses C2?

Yes, he considers that clocks C1 and C2 are not synchronised by the amount t3-t2.

He also considers C1 and C2 are both ticking too slowly (equally) compared to C3.

He sees both C1 and C3 saying t=0 when they pass because the difference in their tick rates integrates to t3-t2 over the journey from C1 to C2.

 

[stevendaryl]

That formula is defined for an inertial Frame of Reference. Its definition is not applicable to a noninertial frame..

Uh, that's what I said. The point of my post was to explain what goes
wrong if you try to use it in a noninertial frame, to give a feel for why it doesn't work in noninertial frames.

But again, my complaint about your analogy is that when you say that one traveler can look over at the other one and see how far he has progressed, it gives the impression that Speedo can look over at Goslo and see how far he has progressed, meaning, of course, where he is at any particular time, but he can't do that.

It's almost never the case that we understand what's going on in the world
by just looking. Looking gives us information which we then have to piece
together to get the complete story. You can certainly, after the fact,
compute the quantity:

T_1(t) = the age of twin 1 at the time twin 2 has age t, as measured in the
frame in which twin 2 is instantaneously at rest.

and

T_2(s) = the age of twin 2 at the time twin 1 has age s, as measured in
the frame in which twin 1 is instantaneously at rest.

Then these quantities will satisfy

dT_1/dt = square-root(1-(v/c)²)
where v is the speed of twin 1 as measured in the frame in which twin 2 is at rest, and

dT_2/dt = square-root(1-(v/c)²)
where v is the speed of twin 2 as measured in the frame in which twin 1 is at rest,

except during times of acceleration.

 

[stevendaryl]

Analogy: suppost you are looking directly at the Moon and you then turn your head sideways through 90° in one second. That brief moment when your head is non-inertial causes the moon to move over one third of a million miles in one second relative to your head's "frame of reference". Sometimes a small change can have a big effect.

That's exactly the point of the Euclidean analogy that I (and others) gave.

 

[Whovian]

I was talking about the approximately 24-hour rotation of the earth,

Fair enough. The point of the analogy still holds.

not the annual orbital period of the Earth which would produce a speed for the sun of only a very small fraction of a million miles per hour.

Still please note that if the Earth were replaced by a point (which won't happen, but for the sake of example let's assume it did,) that point would be a completely inertial reference frame, despite the fact that it's orbiting the Sun, one of the primary postulates of General Relativity.

And just like an observer on the surface of the earth, the Earth itself can define a noninertial reference frame. It would be noninertial even without gravity. Rotation is all that I'm considering. In that noninertial frame, the sun travels at millions of miles per hour.

Also true. The point of the analogy also holds.

And as Austin0 pointed out, if we include stars beyond our sun, they are traveling at many times c.

I feel like doubting this cos velocities don't add in the same sense as classic relativity tells us it does. I'd have to claim that you'd get near c.

 

[stevendaryl]

Still please note that if the Earth were replaced by a point (which won't happen, but for the sake of example let's assume it did,) that point would be a completely inertial reference frame, despite the fact that it's orbiting the Sun, one of the primary postulates of General Relativity.

No, that's not completely correct. Freefall is only equivalent to inertial motion in a limiting sense: An experiment conducted in freefall will give the same result as the same experiment conducted in an inertial frame, provided that tidal effects can be ignored (that is, the variation of gravity with location). That's probably true for most experiments that you might perform in a falling elevator, but is definitely not true if the experiment involves orbits.

 

[Samshorn]

Yes, I am taking for granted the source-independence of light [speed] because that part of Einstein's second postulate is measurable and proven to be true.

You're taking more than that for granted. Your Doppler analysis tacitly assumes not only that the speed of light is independent of the source, but that it is the same in all directions in terms of coordinates in which the laws of Newtonian mechanics hold good (to the first approximation). If this were not true, none of the relativistic effects would occur. It is indispensable in the derivation of all relativistic effects. Try giving your Doppler explanation assuming light propagates at a speed independent of the source but has a different speed in different directions in terms of coordinates such that the laws of mechanics hold good. You will not get any asymmetric aging.

Of course, it would be possible to derive the Doppler explanation in such terms, but only by introducing the requisite compensatory effects due to the use of those coordinates, so this really doesn't save you from needing to account for the full premises and effects of special relativity, one way or another. It is exactly like claiming that classical mechanics can dispense with the principle of inertia because we can use non-inertial coordinates to describe phenomena. The fallacy of the argument is that, when such non-inertial coordinates are used, it is necessary to include compensatory terms (e.g., fictitious forces) which are simply defined based on the deviation of the non-inertial coordinates from the inertial coordinates. So it doesn't really avoid the reliance on the principle of inertia, it just disguises the reliance.

So now would you agree that based solely on Einstein's first postulate of the Principle of Relativity and the principle of the source-independence of light but ignoring the rest of Einstein's second postulate and without establishing any frame of reference, we can determine that the Doppler factors for approaching and retreating at the same relative speed are reciprocals of one another and from this alone we can determine that Goslo ages more than Speedo by the factor gamma?

No, you've only gotten part-way there. You've now accepted the need for the light-speed postulate to the extent that you must invoke source-independence (good!), but you still need to understand that you are also assuming directional independence in terms of coordinates in which the laws of mechanics hold good, and moreover you are assigning the specific value of c to that source-independent and direction-independent quantity (to give actual numerical predictions). In other words, you are invoking Einstein's second postulate (along with the relativity postulate). This shouldn't surprise you. It's obvious that relativistic effects (such as asymmetric aging of the twins) do not follow simply from the principle of relativity.

 

[Mentz114]

Well, that's the issue: why must it be an inertial frame of reference? Why can't it be a "piecewise inertial" frame of reference? That is, why can't you use

τ= ∫√(1-(v/c)2 dt

where t is proper time for the "traveling" twin, and v is the speed of the "stay-at-home" twin relative to the instantaneous rest frame of the traveling twin? You can't because there are pieces of the worldline of the stay-at-home twin that are left unaccounted for. (In the Euclidean analog, the problem is that there are pieces of the path of the "straight" road that are counted twice by the corresponding formula for length).

That formula is defined for an inertial Frame of Reference. Its definition is not applicable to a noninertial frame.

The proper length is frame independent. The formula above can be used to calculate the proper time of any worldline, inertial or not.

 

[Samshorn]

Originally Posted by stevendaryl: "Well, that's the issue: why must it be an inertial frame of reference? Why can't it be a "piecewise inertial" frame of reference? That is, why can't you use

τ= ∫√(1-(v/c)2 dt

where t is proper time for the "traveling" twin, and v is the speed of the "stay-at-home" twin relative to the instantaneous rest frame of the traveling twin?"

Originally Posted by ghwellsjnr: "That formula is defined for an inertial Frame of Reference. Its definition is not applicable to a noninertial frame."

The proper length is frame independent. The formula above can be used to calculate the proper time of any worldline, inertial or not.

You're both right. The formula can be applied to any worldline, but only if v and t are defined in terms of a single standard inertial system of space and time coordinates.

 

[stevendaryl]

The proper length is frame independent. The formula above can be used to calculate the proper time of any worldline, inertial or not.

The issue is not whether the worldline is inertial, but whether an inertial frame is used to compute the t and v appearing in the integral:

τ= ∫√(1-(v/c)2 dt

What doesn't work is the following:
Let t be the proper time of one observer O₁.
Let v(t) be the speed of a second observer O₂, as measured in a frame in which O₁ is instantaneously at rest.

Then the above integral does not give you the age of O₂, except in the case where O₁ is inertial.

 

[Mentz114]

The issue is not whether the worldline is inertial, but whether an inertial frame is used to compute the t and v appearing in the integral:

τ= ∫√(1-(v/c)²) dt

What doesn't work is the following:
Let t be the proper time of one observer O₁.
Let v(t) be the speed of a second observer O₂, as measured in a frame in which O₁ is instantaneously at rest.

Then the above integral does not give you the age of O₂, except in the case where O₁ is inertial.

OK, because worldlines have to be parameterized by a clock time, the formula requires that t is the time on a clock and in SR, t and proper time are the same in the rest frame of an inertial observer. None of this detracts from the proposition that the invariant proper length of a worldline can be found.

(Thanks to Samshorn also)

I think it's probably time for me to shut-up and calculate. I might try the calculation you outline.

 

[ghwellsjr]

That formula is defined for an inertial Frame of Reference. Its definition is not applicable to a noninertial frame..

Uh, that's what I said. The point of my post was to explain what goes
wrong if you try to use it in a noninertial frame, to give a feel for why it doesn't work in noninertial frames.

In which post did you mention anything about a definition?

But again, my complaint about your analogy is that when you say that one traveler can look over at the other one and see how far he has progressed, it gives the impression that Speedo can look over at Goslo and see how far he has progressed, meaning, of course, where he is at any particular time, but he can't do that.

It's almost never the case that we understand what's going on in the world
by just looking. Looking gives us information which we then have to piece
together to get the complete story. You can certainly, after the fact,
compute the quantity:

T_1(t) = the age of twin 1 at the time twin 2 has age t, as measured in the
frame in which twin 2 is instantaneously at rest.

and

T_2(s) = the age of twin 2 at the time twin 1 has age s, as measured in
the frame in which twin 1 is instantaneously at rest.

Then these quantities will satisfy

dT_1/dt = square-root(1-(v/c)²)
where v is the speed of twin 1 as measured in the frame in which twin 2 is at rest, and

dT_2/dt = square-root(1-(v/c)²)
where v is the speed of twin 2 as measured in the frame in which twin 1 is at rest,

except during times of acceleration.

Instead of pointing out that an arbitrary definition is required to analyze, compare, contrast or otherwise make any statements about remotely located clocks, you (and Jonathan Scott) talk only about making measurements, as if we can learn the truth about such matters if only we are smart enough to apply the correct interpretation. I have repeatedly said that you will make whatever measurements follow from your arbitrary definition of how the measurements should be made and that is a valid exercise but you should not conclude that it is any more correct or meaningful than the measurements made from any other arbitrary definition.

By the way, I don't understand your calculation procedure quoted above or why you put in the acceleration exception, but it doesn't matter, I continue to affirm that a well-defined process for dealing with remote times is perfectly valid. I'm not arguing that, simply that it's just one of an infinite number of arbitrary definitions. But every one of them will agree with what Speedo sees of Goslo's clock and that is the one thread of consistency between them all and it is most easily analyzed using Relativistic Doppler. No other analysis provides any additional insight into what is actually happening.

 

[A.T.]

The issue is not whether the worldline is inertial, but whether an inertial frame is used to compute the t and v appearing in the integral:

τ= ∫√(1-(v/c)2 dt

What doesn't work is the following:
Let t be the proper time of one observer O₁.
Let v(t) be the speed of a second observer O₂, as measured in a frame in which O₁ is instantaneously at rest.

Then the above integral does not give you the age of O₂, except in the case where O₁ is inertial.

OK, because worldlines have to be parameterized by a clock time, the formula requires that t is the time on a clock and in SR, t and proper time are the same in the rest frame of an inertial observer.

I would say this formula doesn't work for in noninertial frames because it assumes the standard pseudo-Euclidean metric. It should work, if you use the appropriate metric for the non-inertial frame. For example the Rindler line element for a frame with uniform proper acceleration g:

 

[stevendaryl]

In which post did you mention anything about a definition?

I just meant that the formula cannot be used (without modifications) in a noninertial frame.

By the way, I don't understand your calculation procedure quoted above or why you put in the acceleration exception, but it doesn't matter, I continue to affirm that a well-defined process for dealing with remote times is perfectly valid. I'm not arguing that, simply that it's just one of an infinite number of arbitrary definitions. But every one of them will agree with what Speedo sees of Goslo's clock and that is the one thread of consistency between them all and it is most easily analyzed using Relativistic Doppler. No other analysis provides any additional insight into what is actually happening.

I disagree whole-heartedly. Whether something provides insight or not is a matter of opinion, of course, but I believe that it can be very helpful to look at things from a wide variety of viewpoints.

 

[ghwellsjr]

But every one of them will agree with what Speedo sees of Goslo's clock and that is the one thread of consistency between them all and it is most easily analyzed using Relativistic Doppler. No other analysis provides any additional insight into what is actually happening.

I disagree whole-heartedly. Whether something provides insight or not is a matter of opinion, of course, but I believe that it can be very helpful to look at things from a wide variety of viewpoints.

I agree that additional analyses from a wide variety of viewpoints can be very helpful, if by that you mean they all agree on what is actually happening--they all agree on what Speedo sees of Goslo's clock and vice-versa--but not if you mean they all provide additional (and divergent) information of what is actually happening. That's the whole point of Special Relativity, you can analyze the same scenario from different reference frames and they all agree on what is actually happening--they don't purport to make any other claims on what is actually happening.

 

[ghwellsjr]

So now would you agree that based solely on Einstein's first postulate of the Principle of Relativity and the principle of the source-independence of light but ignoring the rest of Einstein's second postulate and without establishing any frame of reference, we can determine that the Doppler factors for approaching and retreating at the same relative speed are reciprocals of one another and from this alone we can determine that Goslo ages more than Speedo by the factor gamma? And would you agree that this analysis directly answers Michio Cuckoo's question in post #1?

No, you've only gotten part-way there. You've now accepted the need for the light-speed postulate to the extent that you must invoke source-independence (good!), but you still need to understand that you are also assuming directional independence in terms of coordinates in which the laws of mechanics hold good, and moreover you are assigning the specific value of c to that source-independent and direction-independent quantity (to give actual numerical predictions). In other words, you are invoking Einstein's second postulate (along with the relativity postulate). This shouldn't surprise you. It's obvious that relativistic effects (such as asymmetric aging of the twins) do not follow simply from the principle of relativity.

I'm talking about the physics of the world we live in. It is not necessary to postulate the directional independence of light to show that the Doppler factors for approaching and retreating are reciprocals of each other. In fact, prior to Einstein, with the belief in an absolute but unidentifiable ether in which light propagates at c, so that Speedo and Goslo are traveling through the ether at a constant speed so that light does not take the same time to go between them in different directions, they still will measure their Doppler shifts to be equal when retreating and equal when approaching and they can prove that these two factors will be the inverse from each other. A science does not need to establish a frame of reference with previously synchronized clocks based on the postulate that light takes the same time to get from A to B as it does to get from B to A to determine that these facts would be true. In other words, it works just as well with the postulate that there exists an absolute ether.

 

[SinghRP]

I copy below a file, but the figure may not show up.
Relativistic aging
Very briefly I will present my understanding of relativistic aging.

Special relativity (SR): (“Moving” below means motion at a constant speed along a straight line.)

(1) If we observe an object moving past us with a certain velocity u, it will appear contracted in the direction of its motion by a factor √(1 – u2/c2), where c is the sped of light.

(2) If we observe a clock moving past us with a certain velocity u, it will appear to be losing time, and its rate will be slowed down by a factor 1/√(1 – u2/c2), where c is the sped of light.

(3) If we move with the object and the clock, we measure no length contraction and no time dilatation. Meson example: if we ride with cosmic mesons coming down, we will detect no changes in their lifespan; if we stand on the ground and observe the mesons coming down, we will detect prolonged lifespan.

(4) If Observer A is moving relative to Observer B with speed u, then Observer B is moving relative to Observer A with -u. Because of u2, the direction of u is irrelevant. To each observer, other’s clock appears to be losing time. No relative aging here. (Same logic applies to lengths.)

General relativity (GR): Gravitation is due to the curvature which matter (or antimatter) creates in the field of space-time geometry. The field of space-time geometry is the gravitational field. Particles follow space-time geodesic lines of the field. Why and how matter warps space and time are left unexplained! Remember: there is one and only true theory that can explain and describe gravitational wave/pulse and its structure, speed, and polarization; this is the ultimate test of any gravity theory. Astronomic collisions and interactions among celestial bodies notwithstanding, so far there is no evidence of waves/pulses (or gravitons) in the field of space-time geometry.

(1) A particle vibrates at lower frequency closer to a mass (or in a stronger gravitational field).

(1 a) If the frequency is that of emitted light, spectral lines produced at a mass are redshifted compared to those produced at infinity. That is, emitted spectral wavelengths are longer closer to the mass.

(1 b) If the vibrating particle serves as an atomic clock, its time period is longer at the mass than at infinity. That is, time runs slower closer to the mass.

(1 c) At a black hole, it can be easily deduced from the above that light travels at c but with a virtually flat waveform and the atomic clock’s time run virtually stops.

(1 d) When no mass is present anywhere (absurd but assume it anyway), time is meaningless. At the center of a mass, time is meaningless.

(2) A thin material rod is longer closer to a mass (or in a stronger gravitational field). The rod is deformed in a non-uniform gravitational field.

(2 a) At a black hole, the rod flattens to the point where it disintegrates.

(3) The effects of other fundamental fields (the strong, the weak, and electromagnetic) on time are not known.

Principle of equivalence (PE). Specific thought experiments reveal that there is a pseudogravity in the direction opposite to an applied acceleration. This is a just a pseudo (as if) gravity, not a fundamental-force gravity! I am not aware of any experimental evidence of such a pseudogravity affecting time and length. (Thought experiments are good to have but they can’t be substitutes for real experiments. Sadly, some of the thought experiments have become reality for most physicists. Under the Pseudogravity thread, I made statements on acceleration, weightlessness, and centrifugal force, et al. Please review them.)

Relativistic aging.

(1) A physicist on the first floor of a tall building has a twin sister manager on the top floor. Per GR, the physicist ages more slowly than the manager. (Now you know why I sit at the ground floor.)

(2) Two brothers decide to determine whether SR and GR could help guide them age slowly. Older brother Resto stays at Earth; younger brother Speedo goes on a journey into deep space. They have devised a plan for traveling from Earth to Planet P, which is comparatively quite massive. See the figure below.

[Unable to copy the Figure here.]

Speedo departs Earth at acceleration aE from point E to point A and achieves velocity u with which he continues to Planet P. After 10 years, he reaches close to Planet P. He accelerates at aP from B to P to gain sufficient speed to orbit the planet safely. He orbits for 5 years. He decelerates at aP to speed u from P to B, continues toward the earth, decelerates at aE from A to E to touchdown. He meets his brother Resto. Does Resto find Speedo younger or older?

E to A: No changes in Speedo’s clock due to the applied acceleration (no experimental proof). His atomic/biological clock speeds up as Earth’s gravity gets weaker.

A to B: The speed is uniform; so, according to SR, the brothers age at the same rate relative to each other. Per GR, the clock speeds up as Earth’s gravity gets weaker but slows down as P’s gravity gets stronger.

B to P: No changes in Speedo’s clock due to the applied acceleration (no experimental proof). His atomic/biological clock slows down as P’s gravity gets stronger.

Orbiting P: Speedo’s clock is slower compared to what it was at Earth due to stronger gravity. For 5 years he ages at a slower rate.

P to B: No changes in Speedo’s clock due to the applied deceleration (no experimental proof). His atomic/biological clock speeds up as the P’s gravity gets weaker.

B to A: The speed is uniform; so, according to SR, the brothers age at the same rate relative to each other. Per GR, the clock speeds up as P’s gravity gets weaker but slows down as Earth’s gravity gets stronger.

A to E: No changes in Speedo’s clock due to the applied deceleration (no experimental proof). His atomic/biological clock slows down as Earth’s gravity gets stronger.

The journey’s summary. Any gain/loss during the outward journey from E to P is canceled by the loss/gain during the return journey from P to E. So, ultimately Speedo gets younger but only during the time he orbited the heavier Planet P. Only the difference between Planet P’s gravity and Earth’s gravity contributes to Speedo’s age.

 

[stevendaryl]

I have a whole list of slight disagreements with your writeup. My general complaint is that whether one clock ticks at a faster or slower rate than another clock is dependent on a choice of a coordinate system. It's not relative to an observer, it's relative to a coordinate system. In Special Relativity, we can (often) ignore the difference, because we can associate an (inertial) coordinate system with each observer. But in General Relativity, no coordinate system is inertial over an extended region, and there is no unique way to pick a coordinate system for an observer. This is relevant to the question of whether one observer's clock is "ticking faster" than another clock, because to make such a claim requires figuring out "What time t_A does clock A show when clock B shows time t_B?" To answer such a question requires a coordinate system (at least, a way of determining which events are simultaneous).

Anyway, your thought experiment involves both rockets and gravity, so it's a question for General Relativity. Assuming that the rocket itself is small enough that we don't have to worry about its effect on spacetime curvature, GR gives the following answer for how old a person will be after taking a rocket trip:

τ = ∫√(g_uv dx^udx^v)

where g_uv is the metric tensor and dx^u is the change in coordinate x^u along the trip. Assuming that all speeds are pretty small compared with the speed of light, and that the only relevant mass is the mass of the Earth (whose gravity is pretty mild), then we can use the following approximation (using Schwarzschild coordinates)

τ = ∫ (1 - GM/(c²r) - 1/2 (v/c)²) dt

For Resto, r is constant, and is equal to R, the radius of the Earth, and v=0 (let's ignore the rotation of the Earth). So
τ_RESTO = (1-GM/(c² R)) t

For Speedo, we can simplify things by assuming that he very quickly gets far enough away that the Earth's gravity is irrelevant, and that he spends most of his time traveling at some constant speed v. In that case,

τ_SPEEDO = (1-1/2 (v/c)²) t

This is the slow-speed approximation to the SR result τ = √(1- (v/c)²) t

So which twin ages the most depends on the relative sizes of -GM/R and -1/2 (v/c)²

 

[SinghRP]

Well, I read the posts after mine on Relativistic aging. Too many situations have been mixed up for this issue to be resolved. Thanks anyway.

 

[Samshorn]

It is not necessary to postulate the directional independence of light to show that the Doppler factors for approaching and retreating are reciprocals of each other.

You didn't read what I wrote. Special relativity (with relativistic Doppler, asymmetric aging of twins, etc) isn't based on "directional independence of light speed" (a meaningless phrase, like one hand clapping), it is based on directional independence (indeed, the complete invariance) of light speed in terms of coordinates in which the homogeneous and isotropic laws of mechanics hold good. Do you see the difference? This proposition, from which all the uniquely relativistic phenomena follow, has been abundantly confirmed by experiment.

 

[GeorgeDishman]

Very briefly I will present my understanding of relativistic aging.

Special relativity (SR): (“Moving” below means motion at a constant speed along a straight line.)

(1) If we observe an object moving past us with a certain velocity u, it will appear contracted in the direction of its motion by a factor √(1 – u2/c2), where c is the sped of light.

(2) If we observe a clock moving past us with a certain velocity u, it will appear to be losing time, and its rate will be slowed down by a factor 1/√(1 – u2/c2), where c is the sped of light.

(3) If we move with the object and the clock, we measure no length contraction and no time dilatation. Meson example: if we ride with cosmic mesons coming down, we will detect no changes in their lifespan; if we stand on the ground and observe the mesons coming down, we will detect prolonged lifespan.

(4) If Observer A is moving relative to Observer B with speed u, then Observer B is moving relative to Observer A with -u. Because of u2, the direction of u is irrelevant. To each observer, other’s clock appears to be losing time.

Up to there is fine.

No relative aging here.

That is wrong. You are right that the fact that the speed enters a the square means the sign doesn't matter but that doesn't imply "no relative aging", it means each sees the other as aging more slowly, as you said, each considers that the other's clock is running too slowly.

In order to compare the overall effect, you need to bring them back together so one will have to accelerate and that breaks the symmetry. The one that accelerates has aged least when the meet.