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

[Dale]

This means that from Speedo's reference frame, he can still apply Time Dilation to Goslo.

First, you have to define Speedo's reference frame. As I mentioned in post 11 there is no standard way of doing so, which means that you have to be explicit.

What exactly do you mean when you say "Speedo's reference frame".

 

[A.T.]

Therefore, the only explanation as to how Goslo got older is that as Speedo changed his direction, he also switched to another world line, and he observes a time gap, i.e. Goslo suddenly becomes older.

Comparing distant clocks depends on certain conventions. According to the usual one, time flows at different rates at different positions in an accelerating frame (just like in a gravitational field, see Equivalence principle). So yes, in the accelerating frame of Speedo time is running very fast at the very distant position of Goslo, allowing Goslo to age quickly.

This observed non-uniform aging of Goslo is just a consequence of the non-inertial frame of Speedo. Describing the world from non-inertial frames is always "strange", even in classical mechanics where suddenly inertial forces appear. In Relativity non-inertial become a bit more "strange".

 

[ghwellsjr]

I don't say that. I say that as Speedo undergoes acceleration at the half-way point of his trip, he immediately sees an increase in the rate at which Goslo's clock ticks but this is only because he is moving toward the images of Goslo's clock that were already in transit toward him. It's just a Doppler shift. But Goslo doesn't see any such change until way later when the images of Speedo undergoing acceleration finally reach him. It's this imbalance in the time that each one sees the other ones clock ticking slower then faster that allows them to see a difference in the accumulated times on each others clock.

So Goslo's clock ticking faster as seen by Speedo is a result of the Doppler effect.

This means that from Speedo's reference frame, he can still apply Time Dilation to Goslo.

Therefore, the only explanation as to how Goslo got older is that as Speedo changed his direction, he also switched to another world line, and he observes a time gap, i.e. Goslo suddenly becomes older.

Amirite?

Again, please note that I have been talking about what each twin actually sees, not how you can resolve what happens according to an arbitrary non-inertial reference which to me is a pointless exercise and shouldn't be pursued until you thoroughly understand how you can resolve what happens from an arbitrary inertial reference frame. Do you understand how to analyze the situation from an inertial reference frame, for example, one in which Goslo remains at rest throughout the scenario?

 

[AntonL]

And which twin aged?

Suppose the two twins are each in their respective spaceships traveling at some or other high speed as measured in reference system C, so A and B stay young while C ages.

A & B do not know about C so let's forget about C and just look at A & B in their respective reference frames

A & B sync their clocks:
1) A sees B depart at some high speed and return, A aged and B remained young
but also at the same time
2) B sees A depart at some high speed and return, B aged and A remained young

So whoever ages depends into which spaceship you place the reference frame.

Now if above is all gibberish then SR is to blame. Sorry I rephrase: The science must be correct it is just that I do not understand SR. Maybe someone can give me a nice logical explanation and proof to explain to me which clock actually slowed down when A and B meet again.

PS:Now let's go back to reference system C and relativity says A & B clocks tick slower than C's clock.

3) Now A sees B depart but actually B slows to C so B clock is faster than A, but in reference system A, B clock is slower than A

As I said, I am totally but I am sure their must be an explanation forthcomming

 

[Dale]

A & B sync their clocks:
1) A sees B depart at some high speed and return, A aged and B remained young
but also at the same time
2) B sees A depart at some high speed and return, B aged and A remained young

So whoever ages depends into which spaceship you place the reference frame.

This is not true. All inertial frames will agree which is younger when they return. Once you have completely specified the problem in anyone inertial frame you can determine what happens in any other.

The science must be correct it is just that I do not understand SR. Maybe someone can give me a nice logical explanation and proof to explain to me which clock actually slowed down when A and B meet again.

Just use the following formula to calculate the time accumulated for A and for B. The formula is valid in any inertial frame:
http://en.wikipedia.org/wiki/Proper_time#Mathematical_formalism

PS:Now let's go back to reference system C and relativity says A & B clocks tick slower than C's clock.

3) Now A sees B depart but actually B slows to C so B clock is faster than A, but in reference system A, B clock is slower than A

As I said, I am totally but I am sure their must be an explanation forthcomming

Just pick any inertial frame, use the formula above, and look at the result. Then, Lorentz transform to any other inertial frame, and do the same. Compare the results.

 

[ghwellsjr]

A & B sync their clocks:
1) A sees B depart at some high speed and return, A aged and B remained young
but also at the same time
2) B sees A depart at some high speed and return, B aged and A remained young

So whoever ages depends into which spaceship you place the reference frame.

Now if above is all gibberish then SR is to blame. Sorry I rephrase: The science must be correct it is just that I do not understand SR. Maybe someone can give me a nice logical explanation and proof to explain to me which clock actually slowed down when A and B meet again.

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

Please note that 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. You can't blame SR or your lack of knowledge of SR for the fact that we can determine the age difference between the two twins in either of your two scenarios.

You do realize that you have described two different scenarios even though you said "but at the same time", don't you? What matters is which twin is the one that "returns", that is, which one fires his rockets and turns around. If B is the one that does the returning as you said in scenario 1, then B ages less. If A is the one that does the returning as you said in scenario 2, then A is the one that ages less. As I pointed out in post #4, whichever twin is the one that returns is the one that immediately sees the image of the other ones clock go from ticking slow to ticking fast while the other twin does not see that until much later.

Like I said in post #4, it's really very simple.

 

[stevendaryl]

According to Lorentz, time is always dilated.


So think of two twins, Speedo and Goslo.

- Goslo stays on Earth and drinks tea.
- Speedo gets into a rocket, zooms off into outer space and zooms back.


Imagine YOU are Speedo. You zoom off, and when you return to Earth, you find that Goslo is 20 years older than you.

Here's a related "paradox" in good old planar Euclidean geometry.

Suppose you are walking along a road and you come to a second road that intersects the first at an angle theta (for definiteness, assume that it is 8.05°). Suppose that each road has distance markers, one every meter.

While you are traveling on your road, you see that when you pass marker number 100 on your road, you look straight to your left and see marker number 101 on the other road. When you pass marker number 200, you see marker number 202 on the other. Etc. When you get to marker number 500, the other road makes a turn toward your road. You continue to compare markers: when you pass marker number 600, the corresponding marker on the other road is 606, etc. Finally, when you reach marker number 1000, the two roads come together again, and the second road shows distance marker 1010.

Letting N be the marker number on your road, and N' be the marker number on the other road, we can show geometrically that the formula relating the two distances is:

ΔN' = ΔN √(1+m²)
where m is the slope (tangent of 8.05°), which is .1414. For our choice of m, this turns out to be:
ΔN' = 1.01 ΔN

This is not too surprising: a straight line is the shortest distance between two points, so it makes sense that the "bent" road is slightly longer than the "straight" road. But now let's look at things from the point of view of a traveler on the "bent" road. When he comes to marker number 101, he looks straight to his right and sees that the corresponding marker on your road is not 100, but 102. When he comes to maker number 202, he looks straight to his right and he sees, not marker number 200, but marker number 204. The relationship that he notices is just the opposite of the pattern you noticed. Instead of

ΔN' = 1.01 ΔN

he finds

ΔN = 1.01 ΔN'

When the traveler on the "bent" road comes to the bend, which is at marker number 505, he looks straight to his right, and sees marker number 510 on your road. So far, from the point of view of the other traveler, it looks like your road is going to be the longer one, not his. But then he turns the corner...

As he is turning, he is looking to his right at your road, and he sees distance markers on your road moving past. Before the turn, he sees marker number 510, but then as he turns, he sees to his right distance markers number 509, 508, ..., 500, 499, ... 490. Immediately after making the turn, when he looks to his right, he sees marker number 490 on your road. From then on, he sees your numbers increase by the same formula as before:

ΔN = 1.01 ΔN'

So when he travels another 505 meters (for a total of 1010 meters counting both legs of the trip), he sees your road's markers advance by 510 meters, for a total of 490 + 510 = 1000.

So, the moral of the story is that only the "straight-line" traveler can use the formula
ΔN' = ∫dN √(1+m²)
to compute the length of another road. A traveler along a "bent" road will get the wrong answer using that formula, unless he takes into account the effects of turning a corner.

The twin paradox is very similar. Any inertialtraveler can use the formula
tau = ∫√(1-(v/c)²) dt
to compute the amount of aging for another traveler. But a noninertial traveler cannot use that formula, unless he takes into account the effects of turning around.

Proper time in SR is analogous to distance along a path in Euclidean geometry.
Relative speed in SR is analogous to relative slope in Euclidean geometry.
Changing velocity in SR is analogous to turning a corner in Euclidean geometry.

 

[Dale]

Here's a related "paradox" in good old planar Euclidean geometry...

Excellent analogy!

 

[ghwellsjr]

Stevendaryl, isn't your explanation identical to the one Jonathan Scott presented in post #2? And I have the same complaint about yours as I did about his: it doesn't address the question that Michio Cuckoo asked:

Discuss this but only consider Speedo's point of view. Because if you are Speedo, what would you see?

You talk about what your two travelers would see in your analogy but it isn't at all like anything that Speedo (and Goslo) can see, is it? What they see is what I described in post #4 and is all you need to show that Speedo ages less than Goslo, don't you agree?

 

[stevendaryl]

Stevendaryl, isn't your explanation identical to the one Jonathan Scott presented in post #2? And I have the same complaint about yours as I did about his: it doesn't address the question that Michio Cuckoo asked:

You talk about what your two travelers would see in your analogy but it isn't at all like anything that Speedo (and Goslo) can see, is it? What they see is what I described in post #4 and is all you need to show that Speedo ages less than Goslo, don't you agree?

I agree that the Doppler shift explanation explains the difference in the two ages, but I think it leaves a bit of a puzzle as to why can the inertial twin use the formula:

τ= ∫√(1-(v/c)² dt

to compute the age τ of the noninertial twin, but the noninertial twin cannot use that formula to compute the age of the inertial twin. Why not? After all, the noninertial traveler is only noninertial for a tiny moment; how can that tiny moment make such a big difference?

That formula for proper time is exactly analogous to the formula for path length in planar Euclidean geometry:

L = ∫√(1+m² dx

Trying to apply the time dilation formula to compare the ages of the twins is very much like trying to apply the length formula to compare the lengths of two roads.

 

[Nugatory]

Stevendaryl, isn't your explanation identical to the one Jonathan Scott presented in post #2? And I have the same complaint about yours as I did about his: it doesn't address the question that Michio Cuckoo asked:

You talk about what your two travelers would see in your analogy but it isn't at all like anything that Speedo (and Goslo) can see, is it? What they see is what I described in post #4 and is all you need to show that Speedo ages less than Goslo, don't you agree?

IMHO: That's a fair criticism, but there's something else going on that's also worth noting:

Online discussions of the twin paradox tend to veer back and forth between "What do we see?" and "How can that be?". The former is Michio Cuckoo's original question, best answered by your post #4 and Dopplerish explanations; the latter is often better answered with a spacetime diagram analysis, and both Jonathan Scott's and Stevendaryl's answers are of that flavor.

 

[Mentz114]

I agree that the Doppler shift explanation explains the difference in the two ages, but I think it leaves a bit of a puzzle as to why can the inertial twin use the formula:

τ= ∫√(1-(v/c)² dt

to compute the age τ of the noninertial twin, but the noninertial twin cannot use that formula to compute the age of the inertial twin. Why not?

Anyone can calculate the proper length of any worldline if they know it.

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

Yes, it kinks the worldline and reduces its proper length.

That formula for proper time is exactly analogous to the formula for path length in planar Euclidean geometry:

...

Trying to apply the time dilation formula to compare the ages of the twins is very much like trying to apply the length formula to compare the lengths of two roads.

The proper length uses Lorentzian distance, but otherwise the two calculations are the same.

See https://www.physicsforums.com/showthread.php?t=587949&page=8&post=126

(the link isn't right so you need to scroll down the page)

 

[ghwellsjr]

I agree that the Doppler shift explanation explains the difference in the two ages, but I think it leaves a bit of a puzzle as to why can the inertial twin use the formula:

τ= ∫√(1-(v/c)² dt

to compute the age τ of the noninertial twin, but the noninertial twin cannot use that formula to compute the age of the inertial twin. Why not? After all, the noninertial traveler is only noninertial for a tiny moment; how can that tiny moment make such a big difference?

Of course the noninertial twin can use that formula, just like you can use the formula or anybody else can, we just have to use it from any arbitrary inertial Frame of Reference. The only reason why you say that the inertial twin can use the formula is because you have bought into the false notion that the twins can only use a frame in which they are at rest which leads to another false notion that the noninertial twin must use a noninertial reference frame.

As I pointed out in post #13, we can use a frame in which Goslo remains at rest throughout the scenario and all the time dilation falls on Speedo making the explanation trivially simple. But we could also use a frame in which Speedo is at rest during the first half of the trip in which all the time dilation falls on Goslo. But during the second half of the trip, Speedo will have even more time dilation while Goslo's continues the same. However, this inertial frame is more complicated to analyze because we have to calculate both twins' time dilation to see which one ends up with more.

Please don't get me wrong, I'm not saying that noninertial frames are invalid, I'm just saying that they are very complicated to analyze and don't offer anything more than any inertial frame or no frame at all (Doppler analysis) and most significantly, they don't provide the twins with any more insight into what is happening to the other twin--none whatsoever--there is no advantage to using a noninertial frame except to demonstrate the mathematical prowess of the presenter.

 

[Samshorn]

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.

 

[stevendaryl]

Of course the noninertial twin can use that formula, just like you can use the formula or anybody else can, we just have to use it from any arbitrary inertial Frame of Reference.

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).

 

[AntonL]

This is not true. All inertial frames will agree which is younger when they return. Once you have completely specified the problem in anyone inertial frame you can determine what happens in any other ... Compare the results.

really?


and

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.

 

[Mentz114]

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.

You are making it difficult for yourself. Firstly, the time on any clock, irrespective of what is going on elsewhere is given by a simple formula which relies *only* on information about the worldline of the clock in question. So if you want to compare the elapsed time for different clocks, just calculate the proper time of the worldline between the events for each worldline, then compare them. As easy as measuring shoes.

 

[stevendaryl]

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.

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

 

[Dale]

really?

Really, really. It is a pretty typical homework problem. I recommend working through it for your own instruction.

Suppose in some inertial frame that A leaves Earth at v=.6c and reaches a planet 3 ly away in a time of 5 y and then immediately turns around at v=.6c returning 10 y later, while B stayed at Earth the whole time. Calculate the age of the twins at the reunion:
1) in the frame where the Earth is at rest the whole time
2) in a frame where the Earth is moving at .333 c the whole time

 

[Nugatory]

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...

Unfortunately, that reformulated problem is waaaay harder than the flat spacetime problem in the original formulation of the twin paradox. Your best bet is to work through post 4 and the FAQ at http://www.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/TwinParadox/twin_paradox.html until you have nailed the flat spacetime problem, then think about the problem of two traveling twins and one stay-at-home twin, still in flat spacetime with no gravity, no circular orbits, travel in a straight line, ...

 

[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.