Solving the twin paradox with special relativity

[alan123hk]

I saw a book that uses special relativity to solve the twin paradox, the inference process is roughly as follows.

Suppose a spacecraft sets off from the Earth to travel to a distance black hole and then return to the earth. We divide this process into three stages, that is, the process of navigating to this black hole (a→ b), the short moment of reverse turning (b→c) and the process of homing (c→d).

Vc = speed of the spacecraft , in light year
Lc = distance between the Earth and the black hole, in light year

The following is from the perspective of an observer in the spacecraft .

When he reached the black hole (a→ b), he found that the time of C2 was $\left( \frac {L_c} {V_c} \right)$, but he doesn't think this is the time elapsed by the observer on earth, because he thinks that the time of C2 is already $\left( \frac {V_c L_c} {c^2} \right)$ faster than the C1, so the actual elapsed time of the Earth should be $\left( \frac {L_c} {V_c} -\frac {V_c L_c}{c^2} \right)$

The moment the spacecraft turns to go in opposite direction (b→c) , he found that time of C1 suddenly changed from $\left( \frac {V_c L_c} {c^2} \right)$ behind C2 to $\left( \frac {V_c L_c} {c^2} \right)$ ahead of C2, so the time on Earth suddenly increased by $\left( 2\frac {V_c L_c} {c^2} \right)$

When the spacecraft returns (c→d), the Earth also passed time $\left( \frac {L_c} {V_c} -\frac {V_c L_c}{c^2} \right)$

So in the end, the observer in the spacecraft thinks that the total elapsed time of the observer on the Earth is $\left[ 2 \left( \frac {L_c} {V_c} -\frac {V_c L_c}{c^2}\right) + 2\frac {L_c V_c} {c^2} = 2\frac {L_c} {V_c} \right]$, which is elapsed time by the observer on the earth.

On the other hand, the time elapsed by the observer in the spacecraft can be calculated from the observer on the Earth according to the time dilation, which is $\left[\left( 2\frac {L_c}{V_c} \right) \sqrt {1-\frac {V_c^2} {C^2}} \right]$

The final conclusion is that the ratio of time elapsed on Earth to the time elapsed in the spacecraft is $\frac {1} { \sqrt {1-\frac {V_c^2} {C^2}}}$ , so for this twin brother, the brother who came back from the spacecraft is younger than the brother on earth.

I admire the above derivation process described in this book, but I still have a question in my heart, because the uniform linear motion is relative, why the observer on the Earth can’t think that the spacecraft is not in motion, but it is actually he carries the clock C1 Move in the opposite direction with the earth? So in this case, will observers on Earth come to the completely opposite conclusion?

 

[FactChecker]

I admire the above derivation process described in this book, but I still have a question in my heart, because the uniform linear motion is relative, why the observer on the Earth can’t think that the spacecraft is not in motion, but it is actually he carries the clock C1 Move in the opposite direction with the earth? So in this case, will observers on Earth come to the completely opposite conclusion?

The situation is completely symmetric while the motion is uniform, but this is not such a case.
For accelerated motion, the twin who turns around is different from the twin who stays on Earth. There are a few ways to accept this:
From a purely physical, real-world, point of view, the traveling twin feels the acceleration when he turns around. The stationary twin never feels acceleration.
From a purely mathematical point of view, the traveling twin knows that he does not remain on a non-accelerating path in a single space-time reference frame. He changes his space-time reference frame.

 

[PeroK]

I admire the above derivation process described in this book, but I still have a question in my heart, because the uniform linear motion is relative, why the observer on the Earth can’t think that the spacecraft is not in motion, but it is actually he carries the clock C1 Move in the opposite direction with the earth? So in this case, will observers on Earth come to the completely opposite conclusion?

The Lorentz Transformation and the rules on time dilation and length contraction apply when measuring in an IRF (Inertial Reference Frame). In this scenario, the Earth remains approximately at rest in the same IRF throughout. The rocket, on the other hand, is at rest in one IRF on the outbound journey and at rest in a different IRF on the inbound journey.

When you analyse the experiment from the rocket's frame of reference you must (by some means) take this change of IRF into account. You can't simply pretend that the rocket has been moving inertially throughout.

 

[PeroK]

PS to see this imagine a series of rockets sent out at intervals to repeat the experiment. The rockets on the outbound journey would be at rest relative to each other; and the rockets on the inbound journey would be at rest relative to each other. But, there would be no doubt that the rockets on the inbound journey would be moving relative to the rockets on the outbound journey.

If each rocket has a corresponding "twin" on Earth, then all these observers on Earth would remain at rest relative to each other.

 

[Orodruin]

This may or may not help
https://www.physicsforums.com/insights/geometrical-view-time-dilation-twin-paradox/

 

[pervect]

Suppose you were flying from Las Vegas to Denver. I imagine you wouldn't be at all surprised that if you flew straight from Las Vegas to Denver, that your total air mileage was shorter than if you took a detour to Albuquerque, i.e. if you flew from Las Vegas, to Albuquerque to Denver, you'd have more air miles. And you'd say the reason is simple, that the shortest distance between two points is a straight line.

If you were to draw a space-time diagram, you should come to a very similar conclusion, with one important distance. While the shortest distance between two points in a Euclidean plane is a straight line, for timelike curves on a space-time diagram, the "straight line" path the longest path, called in some references the path of "maximal aging", rather than the shortest path.

Because of one's knowledge of Euclidean geometry, it's obvious to most people that the shortest distance is a straight line. It's not really intuitive that on a space-time diagram this is the path of maximal aging, which is the opposite of what one would expect from the analogy. But from the spatial analogy, it's clear that in general there is no reason to expect that taking a detour will not change the total path length, and that "straight line" paths are different from "bent" paths.

The "triangle paradox" corresponding to the "twin paradox" is obviously not a paradox at all, in fact it's usually called the triangle inequality, which says that the hypotenuse is shorter than the sum of the other sides of the triangle. There is just no reason to expect that the total path length will remain the same when one takes a detour. The interesting difference between the geometry of Euclid and the Lorentzian geometry of space-time is that in the former, one uses the pythogerean theorem where $ds^2 = dx^2 + dy^2 + dz^2$, while in the space-time case, one uses the space-time line element $d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$

 

[Mister T]

I admire the above derivation process described in this book, but I still have a question in my heart, because the uniform linear motion is relative, why the observer on the Earth can’t think that the spacecraft is not in motion, but it is actually he carries the clock C1 Move in the opposite direction with the earth? So in this case, will observers on Earth come to the completely opposite conclusion?

No. The motion is not linear for the traveling twin, but it is for the staying twin.

 

[Grampa Dee]

I saw a book that uses special relativity to solve the twin paradox, the inference process is roughly as follows.

Suppose a spacecraft sets off from the Earth to travel to a distance black hole and then return to the earth. We divide this process into three stages, that is, the process of navigating to this black hole (a→ b), the short moment of reverse turning (b→c) and the process of homing (c→d).

View attachment 293674​

Vc = speed of the spacecraft , in light year
Lc = distance between the Earth and the black hole, in light year

The following is from the perspective of an observer in the spacecraft .

When he reached the black hole (a→ b), he found that the time of C2 was $\left( \frac {L_c} {V_c} \right)$, but he doesn't think this is the time elapsed by the observer on earth, because he thinks that the time of C2 is already $\left( \frac {V_c L_c} {c^2} \right)$ faster than the C1, so the actual elapsed time of the Earth should be $\left( \frac {L_c} {V_c} -\frac {V_c L_c}{c^2} \right)$

The moment the spacecraft turns to go in opposite direction (b→c) , he found that time of C1 suddenly changed from $\left( \frac {V_c L_c} {c^2} \right)$ behind C2 to $\left( \frac {V_c L_c} {c^2} \right)$ ahead of C2, so the time on Earth suddenly increased by $\left( 2\frac {V_c L_c} {c^2} \right)$

When the spacecraft returns (c→d), the Earth also passed time $\left( \frac {L_c} {V_c} -\frac {V_c L_c}{c^2} \right)$

So in the end, the observer in the spacecraft thinks that the total elapsed time of the observer on the Earth is $\left[ 2 \left( \frac {L_c} {V_c} -\frac {V_c L_c}{c^2}\right) + 2\frac {L_c V_c} {c^2} = 2\frac {L_c} {V_c} \right]$, which is elapsed time by the observer on the earth.

On the other hand, the time elapsed by the observer in the spacecraft can be calculated from the observer on the Earth according to the time dilation, which is $\left[\left( 2\frac {L_c}{V_c} \right) \sqrt {1-\frac {V_c^2} {C^2}} \right]$

The final conclusion is that the ratio of time elapsed on Earth to the time elapsed in the spacecraft is $\frac {1} { \sqrt {1-\frac {V_c^2} {C^2}}}$ , so for this twin brother, the brother who came back from the spacecraft is younger than the brother on earth.

I admire the above derivation process described in this book, but I still have a question in my heart, because the uniform linear motion is relative, why the observer on the Earth can’t think that the spacecraft is not in motion, but it is actually he carries the clock C1 Move in the opposite direction with the earth? So in this case, will observers on Earth come to the completely opposite conclusion?

I believe that I also will need some help with this one, as I view the scenario in the same light as Allan, due to the source of gravitation being the cause for the rocket to "change" it's inertial reference frame.
How can the observers in the spacecraft prove to themselves that they have changed their inertial frame of reference if no force due to acceleration was ever detected?

 

[FactChecker]

I believe that I also will need some help with this one, as I view the scenario in the same light as Allan, due to the source of gravitation being the cause for the rocket to "change" it's inertial reference frame.
How can the observers in the spacecraft prove to themselves that they have changed their inertial frame of reference if no force due to acceleration was ever detected?

Does it really matter to the physics that they are convinced? Mathematically, if the traveling twin is not following an unaccelerated spacetime path in a single inertial reference frame, isn't that enough?

 

[PeterDonis]

the source of gravitation being the cause for the rocket to "change" it's inertial reference frame.

If gravitation due to massive bodies is present, there is no such thing as a global inertial frame and you cannot use the forms of analysis used in SR that assume the existence of such a frame. You need more general tools.

The most general tool is simply geometry: look at the length of each twin's path in spacetime and compare them. This can be done for any scenario in any spacetime.

 

[PeterDonis]

Suppose a spacecraft sets off from the Earth to travel to a distance black hole and then return to the earth.

Is the spacecraft supposed to turn around using its own rockets, or use the black hole's gravity to do it?

If the latter, this scenario cannot be analyzed using special relativity.

 

[Grampa Dee]

Is the spacecraft supposed to turn around using its own rockets, or use the black hole's gravity to do it?

If the latter, this scenario cannot be analyzed using special relativity.

Thank you Peter for both posts. Yes, I see. I will then try to explain it in a different way, but it will still have some element of GR within it. I don't know if this will be accepted.

Let’s say that, in a similar situation, we have two twins, each in separate rockets, side by side, sharing the same inertial frame of reference. Twin #2 decides to accelerate for a while and then becomes inertial again and in doing so,would have changed it’s inertial frame of reference relative to twin # 2, who remains on the same inertial frame as before. Now, it’s important to note that it is known who changed frames because only one of the twins detected a force of acceleration , that is, twin # 2.

Now, up ahead in the direction of travel of twin #2, there is a black hole (this is known only to you and me, and not the twins ), which gives rise to a gravitational acceleration for twin # 2. Both observers will detect a continuous change in relative velocities between each other, but who can tell who’s the one that is actually in the process of acceleration?

 

[PeterDonis]

Let’s say that, in a similar situation, we have two twins, each in separate rockets, side by side, sharing the same inertial frame of reference.

"Sharing the same inertial frame of reference" is meaningless. What I assume you actually mean is "at rest relative to each other". But if that's what you mean, you should say that.

A general rule: since no actual physics depends on any particular choice of a frame of reference, if you find yourself phrasing your problem description in terms of frames of reference, stop and rephrase it. You should be able to describe any scenario without using frames of reference at all.

Twin #2 decides to accelerate for a while

How? Does he fire rockets? If so, you should say that.

and then becomes inertial again

Meaning what? That he stops firing his rockets? If so, you should say that.

in doing so,would have changed it’s inertial frame of reference relative to twin # 2, who remains on the same inertial frame as before.

See my note above about frames. Can you rephrase this so it does not use any frames of reference?

it’s important to note that it is known who changed frames because only one of the twins detected a force of acceleration , that is, twin # 2.

No, feeling a force does not mean you know that you "changed frames". It just means you felt a force. The actual physics here is that twin #1 never feels a force and twin #2 does. That's it.

up ahead in the direction of travel of twin #2, there is a black hole (this is known only to you and me, and not the twins ), which gives rise to a gravitational acceleration for twin # 2.

"Gravitational acceleration" is a frame-dependent concept. You should rephrase this to eliminate it.

Also, is this supposed to be a continuation of the same scenario, or a different scenario?

Both observers will detect a continuous change in relative velocities between each other

How will they detect this? Be specific.

who can tell who’s the one that is actually in the process of acceleration?

Does either one feel any force? If not, then neither one is "actually in the process of acceleration".

If neither one feels a force, but they detect that their relative velocity is changing (note that, as above, you should be specific about how they detect this, to make sure they are detecting something that doesn't depend on any choice of reference frame), then they know they are in a curved spacetime and Special Relativity cannot be used to analyze their situation.

 

[alan123hk]

First of all, I want to say that even the special theory of relativity, which is simpler than the general theory of relativity, is a very difficult subject for me. I am just a junior learner and I am very interested in this topic. Therefore, my views expressed here may be wrong, and of course I have no intention of challenging any experts here. I just want to stimulate my thinking in the discussion and help me understand easier and faster.

Is the spacecraft supposed to turn around using its own rockets, or use the black hole's gravity to do it?

In fact, the book I read only said that one of the twin brothers traveled far in a spacecraft at a speed close to the speed of light, and then returned to Earth. The book does not mention black holes, nor does it mention what kind of energy the spacecraft uses. I use the name black hole because I think it will make this space travel story more interesting.

If the latter, this scenario cannot be analyzed using special relativity.

For a beginner, who does not understand the difficult mathematics of relativity, I cannot understand this. My own imagination is that no matter what energy the spacecraft uses to propel it, the steering of the spacecraft is to use black hole gravity or nuclear fusion force, etc. Shouldn't the observer inside the spacecraft only feel an equivalent force? So what is the difference in the impact on his time-space measurement?

Moreover, the analysis method mentioned in this book seems to assume that all the acceleration and deceleration times of the spacecraft are extremely short, so the influence of these processes is ignored, and the special theory of relativity is used reasonably for analysis. So whether this space uses the gravitational force of the black hole to change the navigation direction does not seem to have much influence on this matter.

 

[PeterDonis]

the book I read

Which book?

I use the name black hole because I think it will make this space travel story more interesting.

If you're trying to learn physics, you're not telling a story, and certainly should not be adding irrelevant details to make the "story" more interesting. You should stick to the bare essentials of the physics you want to understand.

My own imagination is that no matter what energy the spacecraft uses to propel it, the steering of the spacecraft is to use black hole gravity or nuclear fusion force, etc.

If by "steering" you mean "changing direction relative to the original starting point" (which is Earth in your scenario, but really should be imagined as the stay-at-home twin floating freely in space, since Earth itself is a gravitating massive body and would make the scenario not analyzable using special relativity), then you are quite wrong, it makes a huge difference whether the spacecraft changes direction by firing its rockets--which means it feels the force of the rockets pushing it--or by a gravitational "slingshot" maneuver around a black hole--which means it is in free fall the whole time and feels zero force. The first case-- spacecraft fires rockets to turn around--can be analyzed using special relativity (and that is probably the case the book you read had in mind). The second cannot.

 

[alan123hk]

Which book?

This book is written in Chinese. I recently found it in the storage box. I bought it more than 20 years ago and it is now worn out, so I did not mention its details on this English discussion platform.

 

[alan123hk]

it makes a huge difference whether the spacecraft changes direction by firing its rockets--which means it feels the force of the rockets pushing it--or by a gravitational "slingshot" maneuver around a black hole--which means it is in free fall the whole time and feels zero force. The first case-- spacecraft fires rockets to turn around--can be analyzed using special relativity (and that is probably the case the book you read had in mind). The second cannot.

Okay, I see what you mean. I didn't realize this difference before.

I believe this may be a higher level question involving general relativity, so can I ask a simple question first.

If the gravitational "slingshot" maneuvering around the black hole is used to reverse the direction and return to the earth, will the age of the twin returning to the Earth be younger or greater than the age at which the rocket thrust changed direction?

 

[Ibix]

If the gravitational "slingshot" maneuvering around the black hole is used to reverse the direction and return to the earth, will the age of the twin returning to the Earth be younger or greater than the age at which the rocket thrust changed direction?

I think you have a stay at home and two travellers, one who slingshots around a black hole and one who reverses course using rockets, but who both set off at the same speed relative to Earth and who turn around the same distance from Earth. The short answer is that the travellers return at the same age, younger than the stay at home.

The more detailed answer is that the exact result depends on details of the slingshot orbit and thrust profile, but by making the distance traveled very large and the turnaround period very small you can make any differences between the travellers' ages negligibly small compared to their ages.

 

[FactChecker]

Being closer to the gravitation of a large mass makes time pass slower. So this would make both the stationary and traveling twin calculate that the stationary twin is older.
The effect of the GR space curvature can be larger than the SR velocity effect. In the GPS system, the speed of the GPS satellites clocks would run slower by 7 ms/day in SR but they would run 45ms faster due to GR space curvature. The net result is that GPS satellite clocks run 38ms/day faster.

by making the distance traveled very large and the turnaround period very small you can make any differences between the travellers' ages negligibly small compared to their ages.

Doesn't a long, far, trip keep them at different gravitational potentials longer and with greater differences? It seems more complicated than just the effect of the turnaround.

 

[PeroK]

Okay, I see what you mean. I didn't realize this difference before.

I believe this may be a higher level question involving general relativity, so can I ask a simple question first.

If the gravitational "slingshot" maneuvering around the black hole is used to reverse the direction and return to the earth, will the age of the twin returning to the Earth be younger or greater than the age at which the rocket thrust changed direction?

The general idea in the twin paradox is that the turnaround, however it is achieved, has a very short duration compared to the outbound and inbound journeys. The major factor in terms of differental ageing between the traveling twin (younger) and stay-at-home twin (older) is the overall journey time - this is easy to calculate. A minor factor is the difference caused by the precise physical nature of the turnaround.

To describe the precise details of a gravitational turnaround requires GR - as GR is the relativistic theory of gravity. But, this will give you only a minor correction to the overall differential ageing. Note that the gravitational time dilation of the black hole will be negligible for 99% of the journey, so we can treat the outbound and inbound journeys as effectively inertial motion in approximately flat spacetime - which is the domain of SR. This is what I did in the analysis above.

If you are learning SR, then you should treat the gravitational slingshot as just another turnaround mechanism. An understanding of that in itself will have to wait until you have learned enough SR to progress to GR.

 

[PeroK]

Doesn't a long, far, trip keep them at different gravitational potentials longer and with greater differences? It seems more complicated than just the effect of the turnaround.

Not really. For an interstellar space flight at relativistic speed, the dominant factor is the relative velocity over a period of years. If you simply slingshot round a black hole then you have a short period (a few days perhaps according to Earth) where there is significant gravitational time dilation.

For the typical scenario you have a differential ageing of, say, 10 years plus or minus a few days perhaps depending on whether you calculate the effect of the turnaround itself.

Note that including up to four realistic acceleration and deceleration phases at a proper acceleration of $1g$ would be significant, as each of these phases would take a year or two. That would change the calculations somewhat, but the not overall conclusion that the traveller would be younger on return.

 

[Ibix]

Doesn't a long, far, trip keep them at different gravitational potentials longer and with greater differences?

The gravitational effect of the black hole is negligible for the vast majority of the journey, certainly if the traveling is done at large fractions of $c$.

 

[PeroK]

Moreover, the analysis method mentioned in this book seems to assume that all the acceleration and deceleration times of the spacecraft are extremely short, so the influence of these processes is ignored, and the special theory of relativity is used reasonably for analysis. So whether this space uses the gravitational force of the black hole to change the navigation direction does not seem to have much influence on this matter.

Precisely.

 

[FactChecker]

Note that including up to four realistic acceleration and deceleration phases at a proper acceleration of $1g$ would be significant, as each of these phases would take a year or two. That would change the calculations somewhat, but the not overall conclusion that the traveller would be younger on return.

Wouldn't the traveling twin's being closer to the black hole increase how much younger he would be? I think that, in fact, it helps the traveling twin to agree that the stationary twin is older.

 

[Grampa Dee]

"Sharing the same inertial frame of reference" is meaningless. What I assume you actually mean is "at rest relative to each other". But if that's what you mean, you should say that.

A general rule: since no actual physics depends on any particular choice of a frame of reference, if you find yourself phrasing your problem description in terms of frames of reference, stop and rephrase it. You should be able to describe any scenario without using frames of reference at all.How? Does he fire rockets? If so, you should say that.Meaning what? That he stops firing his rockets? If so, you should say that.See my note above about frames. Can you rephrase this so it does not use any frames of reference?No, feeling a force does not mean you know that you "changed frames". It just means you felt a force. The actual physics here is that twin #1 never feels a force and twin #2 does. That's it."Gravitational acceleration" is a frame-dependent concept. You should rephrase this to eliminate it.

Also, is this supposed to be a continuation of the same scenario, or a different scenario?How will they detect this? Be specific.Does either one feel any force? If not, then neither one is "actually in the process of acceleration".

If neither one feels a force, but they detect that their relative velocity is changing (note that, as above, you should be specific about how they detect this, to make sure they are detecting something that doesn't depend on any choice of reference frame), then they know they are in a curved spacetime and Special Relativity cannot be used to analyze their situation.

I apologize for not being clear,Peter, I will try to answer your questions...
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Peter wrote:
"Sharing the same inertial frame of reference" is meaningless. What I assume you actually mean is "at rest relative to each other". But if that's what you mean, you should say that.
---------
yes; they were at rest relative to each other.

--------
Peter wrote:
A general rule: since no actual physics depends on any particular choice of a frame of reference, if you find yourself phrasing your problem description in terms of frames of reference, stop and rephrase it. You should be able to describe any scenario without using frames of reference at all.
-------
I'm not sure what you mean; instead of saying that the rockets have a velocity relative to each other, isn't it better to say that they are within a different inertial frame of reference when speaking of Relativity?

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Peter wrote:
How? Does he fire rockets? If so, you should say that.
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yes.

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Peter wrote:
Meaning what? That he stops firing his rockets? If so, you should say that.
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yes.

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Grampa Dee wrote:
in doing so,would have changed it’s inertial frame of reference relative to twin # 2, who remains on the same inertial frame as before.
-----------
I'm sorry; I meant that twin #1 not #2 remained as it was before...did not accelerate.

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Peter wrote:
No, feeling a force does not mean you know that you "changed frames". It just means you felt a force. The actual physics here is that twin #1 never feels a force and twin #2 does. That's it.
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...but before, I wrote that twin #2 accelerated for awhile ( stepped on the gas), which in itself must claim that it experienced a force of acceleration or is this wrong as well?

---------------
Peter wrote:
"Gravitational acceleration" is a frame-dependent concept. You should rephrase this to eliminate it.
Also, is this supposed to be a continuation of the same scenario, or a different scenario?
---------------

I understand that both twins will indeed experience an acceleration. I'm assuming that twin#2 which has already a velocity pointing towards the black hole, and is itself much closer to it therefore will experience a greater acceleration than twin #1, something that they will be able to detect by light signals.
Yes, it is a continuation of the same scenario in order to identify two types of accelerations; the first being caused by the firing of the engines of twin #2,and so, there will be a change of reference between the two twins ( I'm sorry for using the term "frame of reference" again, but it is
what I am questioning...two frames where acceleration is concerned...one is non inertial (the firing of engines)and the other one caused by gravity, what I believe to be an inertial frame, which I know I can be wrong ...being the reason why I am asking about this particular case.)

----------------
Peter wrote:
Does either one feel any force? If not, then neither one is "actually in the process of acceleration".

If neither one feels a force, but they detect that their relative velocity is changing (note that, as above, you should be specific about how they detect this, to make sure they are detecting something that doesn't depend on any choice of reference frame), then they know they are in a curved spacetime and Special Relativity cannot be used to analyze their situation.
------------------Yes, I do understand that we're not speaking of special relativity due to the gravitational potential involved, and so, I'm not really in focus with this actual thread. The reason why I responded to this thread was because of the "blackhole" mentioned in the scenario that Allen gave in his first post...I understand that I'm off track when it concerns SR...I'm sorry.

I was thinking that the detection of acceleration between the two twins, were done by light signals...red shifts.
While I agree that the detection of acceleration will be identified as being caused by a curved spacetime,since no one is experiencing a force of acceleration, is there a way that they can detect where the black hole is located? Whether it is located in the direction closer to Twin #2 ,or in the opposite direction, beyond twin #1?

 

[PeroK]

Wouldn't the traveling twin's being closer to the black hole increase how much younger he would be? I think that, in fact, it helps the traveling twin to agree that the stationary twin is older.

Yes, but not by a significant amount. If you are traveling at $0.9c$, say, relative to Earth, you'd need to get very close to a black hole in order to execute an unpowered slingshot and that would be a fast turnaround relative to Earth time. Hence, very little differential ageing.

The deceleration and acceleration option, on the other hand, would make a significant practical difference to the final ages, as there would be 4 times 2 years, say, at less than cruising speed. Assuming it takes 2 years to reach 0.9c at a guess (in the Earth frame).

But, neither of those adjustments is fundamental to the spacetime characteristic being illustrated by the twin paradox.

 

[PeterDonis]

If the gravitational "slingshot" maneuvering around the black hole is used to reverse the direction and return to the earth, will the age of the twin returning to the Earth be younger or greater than the age at which the rocket thrust changed direction?

In terms of the relative aging of the twins, this scenario is basically the same as the standard twin paradox where the traveling twin fires rockets to turn around: the traveling twin is younger when they meet again. The only difference is that the traveling twin does not feel any force while turning around if the gravitational "slingshot" is used to do it. But the relative aging of the twins is not determined by the forces they feel or don't feel; it's determined by the lengths of their two paths through spacetime, i.e., by geometry. The geometry of the two scenarios is basiaclly the same; there is a difference in the neighborhood of the black hole, but that difference occupies only a small part of the total path through spacetime of the traveling twin so it doesn't significantly affect the relative aging.

 

[PeterDonis]

Being closer to the gravitation of a large mass makes time pass slower.

True, but irrelevant to the case under discussion, since the traveling twin only spends a very small portion of their total path through spacetime close to the black hole.

The effect of the GR space curvature can be larger than the SR velocity effect. In the GPS system, the speed of the GPS satellites clocks would run slower by 7 ms/day in SR but they would run 45ms faster due to GR space curvature. The net result is that GPS satellite clocks run 38ms/day faster.

The GR effect in the GPS system is not due to "space curvature". It's due to altitude above the Earth. At the GPS satellite speeds, which are much slower than the speed of light, any correction due to "space curvature" is negligible compared to the altitude effect.

Doesn't a long, far, trip keep them at different gravitational potentials longer and with greater differences?

First, the "gravitational potential" in question is only present in a (drastically non-inertial) frame in which the traveling twin is always at rest.

Second, a longer, farther trip does have the effect you describe compared to a shorter trip, and the result is that the difference in aging between the twins is greater for a longer, farther trip. But the difference between the two types of turnaround, which is what @PeroK was talking about, becomes smaller and smaller as a fraction of the total difference in aging as the trip gets longer and farther.

 

[PeterDonis]

I think that, in fact, it helps the traveling twin to agree that the stationary twin is older.

If the traveling twin has a proper understanding of the physics and spacetime geometry involved, he will agree on the differential aging regardless of the specifics of how the turnaround is done.

 

[PeterDonis]

instead of saying that the rockets have a velocity relative to each other, isn't it better to say that they are within a different inertial frame of reference when speaking of Relativity?

Not at all. Relative velocity can be directly measured (for example, by timing round-trip light signals). Inertial frames are abstractions (and global inertial frames don't even exist if gravitating masses are present) and the physics is independent of any choice of frame, so focusing on them just obfuscates the physics.

before, I wrote that twin #2 accelerated for awhile ( stepped on the gas), which in itself must claim that it experienced a force of acceleration

It means twin #2 felt a force. "Force of acceleration" is either redundant (since a felt force is identical to proper acceleration, what is measured by an accelerometer) or frame-dependent (since coordinate acceleration depends on your choice of frame and is therefore irrelevant to the physics).

I understand that both twins will indeed experience an acceleration.

Not in your scenario, no. You say twin #1 never fires his rockets; that means he never feels a force at all.

If twin #2 fires his rockets, he feels a force. But if he does not fire his rockets, but just let's the gravity of the black hole determine his motion, he feels no force either.

I'm assuming that twin#2 which has already a velocity pointing towards the black hole, and is itself much closer to it therefore will experience a greater acceleration than twin #1, something that they will be able to detect by light signals.

Neither twin will "experience" any acceleration--they both feel no force. They can measure that their relative velocity is changing by exchanging round-trip light signals, yes; that was the method I had in mind. But that measured change in relative velocity, when they both feel no force, does not mean they are "experiencing acceleration" even though they feel no force. It means the spacetime they are in is curved, i.e., there is a gravitating mass present.

.two frames where acceleration is concerned...one is non inertial (the firing of engines)and the other one caused by gravity, what I believe to be an inertial frame,

Physics has nothing to do with frames. It has to do with actual observables. Feeling a force (for example, when twin #2 fires his rockets) is a direct observable. So is spacetime curvature (as measured by round-trip light signals between the twins when they are at different distances from the black hole and neither one is firing rockets). The latter measurement is sometimes referred to as "relative acceleration" in the GR literature (and I'll refer to it that way below), but it has nothing to do with any choice of frame.

I was thinking that the detection of acceleration between the two twins, were done by light signals...red shifts.

Yes, as noted above.

is there a way that they can detect where the black hole is located? Whether it is located in the direction closer to Twin #2 ,or in the opposite direction, beyond twin #1?

From just the relative acceleration between them, measured as described above, I don't think so. They would need a third "twin" separated from them tangentially (i.e., in a direction perpendicular to the direction of the black hole), so that each pair of twins could measure the relative acceleration between them. Those multiple relative acceleration measurements could determine the direction to the hole.

 

[Grampa Dee]

Not at all. Relative velocity can be directly measured (for example, by timing round-trip light signals). Inertial frames are abstractions (and global inertial frames don't even exist if gravitating masses are present) and the physics is independent of any choice of frame, so focusing on them just obfuscates the physics.It means twin #2 felt a force. "Force of acceleration" is either redundant (since a felt force is identical to proper acceleration, what is measured by an accelerometer) or frame-dependent (since coordinate acceleration depends on your choice of frame and is therefore irrelevant to the physics).Not in your scenario, no. You say twin #1 never fires his rockets; that means he never feels a force at all.

If twin #2 fires his rockets, he feels a force. But if he does not fire his rockets, but just let's the gravity of the black hole determine his motion, he feels no force either.Neither twin will "experience" any acceleration--they both feel no force. They can measure that their relative velocity is changing by exchanging round-trip light signals, yes; that was the method I had in mind. But that measured change in relative velocity, when they both feel no force, does not mean they are "experiencing acceleration" even though they feel no force. It means the spacetime they are in is curved, i.e., there is a gravitating mass present.Physics has nothing to do with frames. It has to do with actual observables. Feeling a force (for example, when twin #2 fires his rockets) is a direct observable. So is spacetime curvature (as measured by round-trip light signals between the twins when they are at different distances from the black hole and neither one is firing rockets). The latter measurement is sometimes referred to as "relative acceleration" in the GR literature (and I'll refer to it that way below), but it has nothing to do with any choice of frame.Yes, as noted above.From just the relative acceleration between them, measured as described above, I don't think so. They would need a third "twin" separated from them tangentially (i.e., in a direction perpendicular to the direction of the black hole), so that each pair of twins could measure the relative acceleration between them. Those multiple relative acceleration measurements could determine the direction to the hole.

Thank you Peter; you've answered pretty much what I was asking... I'm sorry for my poor choice of words...it's not easy to express myself very clearly when it concerns scientific experiments.

I just wanted to add...

Grampa Dee said:
I understand that both twins will indeed experience an acceleration.

I wanted to say that both twins can measure (light signals) each others relative increase of velocities.

 

[PeterDonis]

both twins can measure (light signals) each others relative increase of velocities.

Yes. However, that is not the same as either twin feeling any force, which is what "experience acceleration" would normally mean.

 

[Grampa Dee]

Yes. However, that is not the same as either twin feeling any force, which is what "experience acceleration" would normally mean.

I understand; however, acceleration, for me, is first and foremost a change in velocity, which is usually measured with measuring rods (or light signals) as opposed to forces, and so I visualized more the experience of acceleration with measuring velocities...again, it's not easy for me to express myself scientifically due to the exactitude in speech that is needed.

 

[Sagittarius A-Star]

I understand; however, acceleration, for me, is first and foremost a change in velocity, which is usually measured with measuring rods (or light signals) as opposed to forces

You should distinguish between "coordinate acceleration" (what you describe) and "proper acceleration" (can be measured with an accelerometer).

 

[PeterDonis]

it's not easy for me to express myself scientifically due to the exactitude in speech that is needed.

When in doubt, it's best to not use words that can be ambiguous, but just to directly describe what you're measuring and how you're measuring it. Saying you're timing round-trip light signals, or saying you're using an accelerometer, makes it clear what you're measuring without having to worry about the ambiguity of terms like "experience" or "acceleration".