Notices
Results 1 to 19 of 19
Like Tree2Likes
  • 1 Post By Janus
  • 1 Post By AndrewC

Thread: Variation on the 'twin paradox' with no change of frame?

  1. #1 Variation on the 'twin paradox' with no change of frame? 
    Member
    Join Date
    Jul 2013
    Location
    Scotland
    Posts
    40
    Consider three points in space A,B,F where F is midway between A and B, all in the same inertial frame. Because A,B,F are in the same frame, their clocks can all be synchronised in advance.

    A race is to be held. Two identical spaceships set out from F, at leisure (non-relativistic speeds), and make their way to A and B respectively. Once both ships are in position, a 'start' signal is sent from F, which reaches A and B at the same time - each ship then starts its clock and sets off. Or alternatively, the ships can just agree in advance at what time they will set off, using the already synchronised clocks at A and B as their reference.

    The two ships each race at the same near-light speed to the finish point F, arriving together, and all compare clocks.

    I think that

    • F will consider each ship to be the same age as the other, but having experienced less time than the observer at F.
    • Each ship will consider F to have experienced less time than itself, by the same amount - so disagreeing with F about who is older but agreeing between themselves as to the age of F.
    • Each ship will consider the other to have experienced less time than itself, disagreeing with each other about who is older.

    So - disagreements between the ships and between the ships and F, as to who is older than whom.


    Here however, there is no frame change to reconcile the apparent paradox. The journeys really are symmetrical, and so far as I can see, plotting spacetime diagrams doesn't help resolve this either.

    Is there something wrong with my understanding that all clocks in a single inertial frame can be synchronised (by bouncing light signals between each other, and using the fact that the time taken for a light signal to reach a remote clock is half that of a round trip back to the start point).

    I must be missing the obvious! But what?
    Reply With Quote  
     

  2. #2  
    Senior Member
    Join Date
    Sep 2017
    Posts
    278
    deleted
    Reply With Quote  
     

  3. #3  
    Senior Member
    Join Date
    Sep 2017
    Posts
    278
    Quote Originally Posted by lesaid View Post
    Consider three points in space A,B,F where F is midway between A and B, all in the same inertial frame. Because A,B,F are in the same frame, their clocks can all be synchronised in advance.

    A race is to be held. Two identical spaceships set out from F, at leisure (non-relativistic speeds), and make their way to A and B respectively. Once both ships are in position, a 'start' signal is sent from F, which reaches A and B at the same time - each ship then starts its clock and sets off. Or alternatively, the ships can just agree in advance at what time they will set off, using the already synchronised clocks at A and B as their reference.

    The two ships each race at the same near-light speed to the finish point F, arriving together, and all compare clocks.

    I think that


    F will consider each ship to be the same age as the other, but having experienced less time than the observer at F.
    1.

    More correctly:



    Each ship will consider F to have experienced less time than itself, by the same amount - so disagreeing with F about who is older but agreeing between themselves as to the age of F.
    2. False, see 1. is a more complicated function of



    [*]Each ship will consider the other to have experienced less time than itself, disagreeing with each other about who is older.
    3. False, see 1.

    So - disagreements between the ships and between the ships and F, as to who is older than whom.
    4. False, see 1.



    I must be missing the obvious! But what?
    You are missing the fact that while A and B are symmetrical to each other , the same thing is not true between A and F and B and F.
    Both A and B need to ACCELERATE in order to get to F.
    Reply With Quote  
     

  4. #4  
    Member
    Join Date
    Jul 2013
    Location
    Scotland
    Posts
    40
    Thank you for your reply, and the subsequent 'edits'. I'm sorry about the slow response - thinking about this has to be a 'background task' for me!

    I am afraid I am still struggling to understand the influence of the acceleration on this particular scenario!

    Quote Originally Posted by AndrewC View Post
    You are missing the fact that while A and B are symmetrical to each other , the same thing is not true between A and F and B and F.
    Both A and B need to ACCELERATE in order to get to F
    As far as I can see, the acceleration has two effects. It changes the time dilation calculation, and it results in a change of frame.

    I have done some calculations with a formula for that models an initial acceleration period, and have more or less convinced myself of my intuition; that as the duration of the acceleration tends to zero, the calculated of the whole journey tends to that of the constant speed formula. So by choosing an infinitesimal acceleration period, the difference between the constant speed and accelerating models should be insignificant?

    The frame change would then happen over an infinitesimally short time at the moment of starting with a vanishingly small distance travelled in the accelerating frames. Synchronising clocks after the acceleration, while the spaceship was still infinitesimally close to the start point would effectively remove the frame change from the scenario?

    Similarly, as the acceleration period tended to zero, the observation of would tend to that of a constant speed scenario.

    Alternatively, one could avoid the acceleration period altogether by having the spaceship take a 'run' at the start point (A), passing it at the agreed start time already at constant velocity, and synchronising clocks at the moment of passing. A deceleration at the end of the journey could similarly be avoided.

    With these considerations in mind, I am still struggling to understand how, in this scenario, the acceleration is significant.
    Reply With Quote  
     

  5. #5  
    Senior Member
    Join Date
    Sep 2017
    Posts
    278
    Quote Originally Posted by lesaid View Post

    I have done some calculations with a formula for that models an initial acceleration period, and have more or less convinced myself of my intuition; that as the duration of the acceleration tends to zero, the calculated of the whole journey tends to that of the constant speed formula. So by choosing an infinitesimal acceleration period, the difference between the constant speed and accelerating models should be insignificant?
    But the acceleration DOESN'T "tend to zero". In a realistic scenario is there, and you cannot use hacks to do away with it. (More hacks are at the end of your post).



    The frame change would then happen over an infinitesimally short time at the moment of starting with a vanishingly small distance travelled in the accelerating frames. Synchronising clocks after the acceleration, while the spaceship was still infinitesimally close to the start point would effectively remove the frame change from the scenario?
    No, it wouldn't. The principle of equivalence (GR) teaches you that. You CANNOT do away with the acceleration.
    Look at it this way: your scenario is equivalent to the one that explains the acceleration-induced clock de-synchronization of the GPS satellite clocks from the ground clocks at launch.

    Similarly, as the acceleration period tended to zero, the observation of would tend to that of a constant speed scenario.
    No, it would not, see my first formula. It is that would tend to that of a constant speed scenario.

    Alternatively, one could avoid the acceleration period altogether by having the spaceship take a 'run' at the start point (A), passing it at the agreed start time already at constant velocity, and synchronising clocks at the moment of passing. A deceleration at the end of the journey could similarly be avoided.

    With these considerations in mind, I am still struggling to understand how, in this scenario, the acceleration is significant.
    Err, no. You simply moved the goalposts, the asymmetry between A and F is still there. The clock de-synchronization is DUE to THAT asymmetry, just like in the original twin paradox.
    Reply With Quote  
     

  6. #6  
    Member Janus's Avatar
    Join Date
    Jan 2013
    Location
    Portland, OR
    Posts
    96
    Quote Originally Posted by lesaid View Post
    Thank you for your reply, and the subsequent 'edits'. I'm sorry about the slow response - thinking about this has to be a 'background task' for me!

    I am afraid I am still struggling to understand the influence of the acceleration on this particular scenario!


    As far as I can see, the acceleration has two effects. It changes the time dilation calculation, and it results in a change of frame.

    I have done some calculations with a formula for that models an initial acceleration period, and have more or less convinced myself of my intuition; that as the duration of the acceleration tends to zero, the calculated of the whole journey tends to that of the constant speed formula. So by choosing an infinitesimal acceleration period, the difference between the constant speed and accelerating models should be insignificant?
    The problem is that you are not taking into account how the acceleration of a ship effects what it measures. According to an accelerating clock, clocks in the direction of the acceleration run fast by a factor that is determined by both the distance to the clock and the magnitude of the acceleration. Decreasing the time interval of the acceleration requires that you increase the magnitude of the acceleration to reach the same velocity. As you try to decrease the time interval to zero, the magnitude will approach infinity. In other words, as each ship accelerates from 0 to v, it would measure the other ship's clock run fast, and then run slow once they stopped accelerating, the end result is that if both ship's clocks read zero when they started at A or B, they both will predict and measure that their respective clocks will read the same as they pass each other. it doesn't matter if they were under high acceleration for a short time or low acceleration for a long time (as long as both ship's do the same thing).

    The frame change would then happen over an infinitesimally short time at the moment of starting with a vanishingly small distance travelled in the accelerating frames. Synchronising clocks after the acceleration, while the spaceship was still infinitesimally close to the start point would effectively remove the frame change from the scenario?
    You can't synchronize the clocks on the ships after the acceleration, so that they both agree that their clocks are synchronized (Relativity of Simultaneity)

    Similarly, as the acceleration period tended to zero, the observation of would tend to that of a constant speed scenario.

    Alternatively, one could avoid the acceleration period altogether by having the spaceship take a 'run' at the start point (A), passing it at the agreed start time already at constant velocity, and synchronising clocks at the moment of passing. A deceleration at the end of the journey could similarly be avoided.
    The same problem with the relativity of simultaneity as above. You can have the clocks bot set to zero as they pass A or B, and Both ships will agree upon this, however, they won't agree that they both passed A and B at the same time. Each ship will say that the other ship passed its point and started it clock ticking from zero before they did, and even though it tick slower, it ticked longer. The final result will again be that both ships will agree that their clocks had identical readings when they passed each other.

    With these considerations in mind, I am still struggling to understand how, in this scenario, the acceleration is significant.
    Reply With Quote  
     

  7. #7  
    Member
    Join Date
    Jul 2013
    Location
    Scotland
    Posts
    40
    Thank you for your reply. Your points, and those from AndrewC have raised a number of things for me to think about, and to relate to what I think I already understand. I can only spend limited time on this so my replies will be a day or two in coming - please be patient!

    Rather than try to respond everything at once, I'd like just now to take forward one particular point that you and I think, AndrewC both raised, in different ways. And then will follow up on one or two other points over the next few days.

    Quote Originally Posted by Janus View Post
    The problem is that you are not taking into account how the acceleration of a ship effects what it measures .... In other words, as each ship accelerates from 0 to v, it would measure the other ship's clock run fast, and then run slow once they stopped accelerating, the end result is that if both ship's clocks read zero when they started at A or B, they both will predict and measure that their respective clocks will read the same as they pass each other. it doesn't matter if they were under high acceleration for a short time or low acceleration for a long time (as long as both ship's do the same thing).
    I see this as a model where each ship starts from rest, accelerates to a cruise, stays there for a while, and then perhaps (de)accelerates back to rest with respect to F. During the acceleration phases, each ship sees the other's clock running fast, and during the cruise phase, each ship sees the other one running slow. The idea is that the two effects cancel so that both clocks agree when they meet at F. Have I understood your point correctly?

    This makes complete sense to me, but I don't quite see how it helps. If we assume that each acceleration period is over a fixed distance (say it takes one light year of travel to get from rest at the start points (which are also in frame F) to cruise, and another light year to get back to rest again), then the effect on the clocks of the acceleration should be fixed, regardless of the length of the cruise. There will be a length of cruise at which the effects will cancel, and the ships' clocks would agree. But if you have a longer cruise time, the ships will see each other's clocks run slow for longer, with a greater cumulative effect on the whole journey, whilst the acceleration phases are unchanged. The two should no longer cancel. If the journey is made sufficiently long, the contribution of the acceleration phases must surely become insignificant in comparison to that of the cruise phase?
    Reply With Quote  
     

  8. #8  
    Member
    Join Date
    Jul 2013
    Location
    Scotland
    Posts
    40
    Thanks for your reply!

    Quote Originally Posted by AndrewC View Post
    But the acceleration DOESN'T "tend to zero". In a realistic scenario is there, and you cannot use hacks to do away with it. (More hacks are at the end of your post).
    I should explain, before you lose patience with me, that I am not trying to argue that relativity is wrong, or in any way go against the generally accepted view of it.

    I am studying this at an undergraduate level, and find that I can plug numbers into equations and provide answers to nicely crafted exercise questions. But if I cannot confidently counter my own challenges to relativity's predictions, or apply my understanding to situations beyond those neat little exercises, then I don't understand it well enough! Even though I know I'm only scratching the surface of the subject at my level!

    Please see my reply to Janus.
    Reply With Quote  
     

  9. #9  
    Member Janus's Avatar
    Join Date
    Jan 2013
    Location
    Portland, OR
    Posts
    96
    Quote Originally Posted by lesaid View Post
    Thank you for your reply. Your points, and those from AndrewC have raised a number of things for me to think about, and to relate to what I think I already understand. I can only spend limited time on this so my replies will be a day or two in coming - please be patient!

    Rather than try to respond everything at once, I'd like just now to take forward one particular point that you and I think, AndrewC both raised, in different ways. And then will follow up on one or two other points over the next few days.



    I see this as a model where each ship starts from rest, accelerates to a cruise, stays there for a while, and then perhaps (de)accelerates back to rest with respect to F. During the acceleration phases, each ship sees the other's clock running fast, and during the cruise phase, each ship sees the other one running slow. The idea is that the two effects cancel so that both clocks agree when they meet at F. Have I understood your point correctly?

    This makes complete sense to me, but I don't quite see how it helps. If we assume that each acceleration period is over a fixed distance (say it takes one light year of travel to get from rest at the start points (which are also in frame F) to cruise, and another light year to get back to rest again), then the effect on the clocks of the acceleration should be fixed, regardless of the length of the cruise. There will be a length of cruise at which the effects will cancel, and the ships' clocks would agree. But if you have a longer cruise time, the ships will see each other's clocks run slow for longer, with a greater cumulative effect on the whole journey, whilst the acceleration phases are unchanged. The two should no longer cancel. If the journey is made sufficiently long, the contribution of the acceleration phases must surely become insignificant in comparison to that of the cruise phase?
    You missed the part where the rate at which the other ship's clock runs fast according to the the accelerating ship in depends on the distance between the ships. So if the ships are 2 light years apart, and accelerate for 1 light year and coast for light 1 light year, they will note that that the amount the other ship's clock ran fast exactly compensates for the time it ran slow. If they start 10 light years apart, and spend 9 light years of it coasting, even though they accelerate for the same distance at the the same rate as before, because the other ship is further away from them during the acceleration phase, the other ship's clock will run that much faster according to them with the 2 ly separation. Again, they will find that the two effect cancel out. (when applied to this symmetrical situation. In non-symmetrical cases, such as the original twin paradox version, one ship's clock can tick off less total time than the others)

    This additional time dilation in an accelerated frame is not restricted to just between the ships either. Observers will note it in their own ships also. A clock in the tail of the ship will run slow compared to one in the nose. At the same acceleration, the longer the ship, the greater the difference in tick rate between the nose and tail. If you strung a line of clocks from nose to tail, the further apart any pair of the clocks are, the greater the difference in their tick rate.
    Reply With Quote  
     

  10. #10  
    Member
    Join Date
    Jul 2013
    Location
    Scotland
    Posts
    40
    Quote Originally Posted by Janus View Post
    You missed the part where the rate at which the other ship's clock runs fast according to the the accelerating ship in depends on the distance between the ships.
    That notion is completely new to me - though there was a reference to it that I didn't pick up on in an earlier post here ! That could explain a lot of my confusion! But it raises an immediate question in my mind. When the observing ship measures the rate of the other ship's clock - the measurement, as I understand it, isn't really being made from the 'ship' - it is being made from the rest frame of the ship. Should we not get the same observation of the other ship's clock from any observer in that frame? The measurement depending on distance implies that different observers in the same frame will observe different relative clock rates? Is that really correct?

    I haven't studied accelerating frames in SR in any depth (yet), so this is new to me. I'm off to do some reading!

    Thank you!
    Reply With Quote  
     

  11. #11  
    Member
    Join Date
    Jul 2013
    Location
    Scotland
    Posts
    40
    Quote Originally Posted by Janus View Post
    This additional time dilation in an accelerated frame is not restricted to just between the ships either. Observers will note it in their own ships also. A clock in the tail of the ship will run slow compared to one in the nose. At the same acceleration, the longer the ship, the greater the difference in tick rate between the nose and tail. If you strung a line of clocks from nose to tail, the further apart any pair of the clocks are, the greater the difference in their tick rate.
    Also occurs to me that the nose and the tail of the ship are both subject to the same acceleration. A remote observer at the end of a ship's journey in this thought experiment is in free fall with the ship accelerating as it approaches. I'm not quite sure how the logic you gave for the 'nose/tail' scenario applies in that circumstance?


    EDIT : this is what happens when I reply too quickly! I'm still thinking about how this effect would vary between two ships accelerating in the same way (i.e. just a very long and constant distance from 'nose' to 'tail' of a conceptual huge 'ship'), against one where the ships are accelerating towards each other so the distance is changing while both are accelerating, and where one observer is in free fall!
    Last edited by lesaid; 05-26-2018 at 06:28 PM.
    Reply With Quote  
     

  12. #12  
    Member Janus's Avatar
    Join Date
    Jan 2013
    Location
    Portland, OR
    Posts
    96
    Quote Originally Posted by lesaid View Post
    Also occurs to me that the nose and the tail of the ship are both subject to the same acceleration. A remote observer at the end of a ship's journey in this thought experiment is in free fall with the ship accelerating as it approaches. I'm not quite sure how the logic you gave for the 'nose/tail' scenario applies in that circumstance?
    But according the the far off free fall observer, the ship is constantly changing its relative velocity, and thus is constantly shrinking in length due to length contraction. ( if it started at 0 relative velocity and ended at 0.866 c relative velocity, our free fall observer would see it contract to 1/2 its original length during the acceleration. But this means that over the course of the acceleration, the rear of the ship "catches up" to the front of the ship, and must have a higher mean velocity relative to our observer. A higher mean velocity means a greater time dilation factor, and a clock at the rear of the ship would have to run slower than one at the front according to our free-fall observer.

    Another way to look at it is the Doppler shift argument. Doppler shift is due to the difference in velocity between source and receiver at the moments of transmission and reception. So for example, if the source changes velocity after transmission, this has no effect on the Doppler shift measured by the receiver in the signal sent before the change in velocity. If however, the receiver changes velocity between transmission and reception this will effect the Doppler shift it measures.

    Imagine light leaving the rear clock and heading towards the front. At the moment of transmission, the front and back have the same velocity* But in the time it takes the light to travel from rear to front, the ship has changed velocity, so now the front is moving at a different velocity than the rear was at the moment of transmission with the velocity increase being away form the rear. This will result in the front seeing a red-shift in the light coming from the rear. Conversely for for light coming from the front and seen by the rear, a blue shift will be measured.

    Now typically with Doppler shift, we are dealing with a source moving towards or away from the receiver, and a component of the shift is due to the changing distance. To work out the time dilation from the Doppler shift you factor out the component due to the changing distance. With the accelerating rocket, the distance between tail and nose is not changing ( at least in the rocket frame) and all the Doppler shift is attributed to time dilation.

    * As noted above, If being viewed by an inertial observer, there is a change in nose to tail distance. So at the moment of transmission, the Nose and tail won't have exactly the same velocity, but this is a minor correction compared to the difference in velocity change between the moments of transmission and reception.

    The change in time in clocks due to acceleration is akin to the change of relative position of objects when you face in a different direction. If you are standing facing two objects with one object 1m in front of you and the other 10m in front of you, and then do a 180 turn, they go from 1 and 10m in front of to 1 and 10m behind you. An a apparent shift of 2m for the nearer object and 20m for the further object. The time-shift in clocks can likewise be attributed to a rotation in space-time ( rather than just a rotation in space like in the example)
    lesaid likes this.
    Reply With Quote  
     

  13. #13  
    Senior Member
    Join Date
    Sep 2017
    Posts
    278
    I'll give you the most elementary explanation that clarifies why A and F are not symmetric:

    F measures the distance between him and A , in his frame as being {tex]L[/tex]. The time , on F's clock for A to get to F is where is A's speed measured by F.
    But A measures the distance to F as being length contracted . Therefore, A's clock, when it meets F shows the time
    .
    The asymmetry was caused by the fact that you gave A a "running start", i.e. you accelerated A up to speed
    This is what Janus has been telling you, acceleration produces clock de-synchronization.
    Reply With Quote  
     

  14. #14  
    Member
    Join Date
    Jul 2013
    Location
    Scotland
    Posts
    40
    OK, sorry it's taken me a while, but I at last have this straight (I think!). What turned it around for me was to take the end event as the origin of the various frames, rather than trying to find a common start point (obvious with hindsight!) and I now have a picture where the maths is consistent and clear, and all the spacetime separations match between the pairs of events across the frames.

    I cannot claim to have my head around this intuitively, but I can see what the maths says, and think I understand it.


    I have also done the same for the 'regular' twin scenario, and also have everything matching, as I expected.

    This leaves me with one conundrum however. Going to the regular twin 'paradox', the calculations seem clear and consistent, while completely ignoring the effect of the acceleration (other than fact of a change of frame). The nature, duration and scale of acceleration, and the distance from 'home' at which it occurs has to be irrelevant, or I would have needed to include it somehow in the calculations to get consistent results.

    This leads me to believe that the ONLY thing that is needed to account for the 'standard' twin paradox has to be the change of frame. Accelerations would change the calculations, but must change them in a way that always cancels out, so that they have no net effect on the relative times experienced by the two twins.

    But this position seems, at least superficially, to be at odds with earlier responses in this thread that stress the role of acceleration. Am I still missing something?

    My next challenge is to get a real understanding of accelerating frames in SR!
    Reply With Quote  
     

  15. #15  
    Member
    Join Date
    Jul 2013
    Location
    Scotland
    Posts
    40
    Thank you for this comprehensive post (Edit : replying to post #12 from Janus)

    I am now having a go at setting up some example scenarios for accelerating frames/observers. Some of what you have said is new to me though it all makes sense to me as I read it. I'm not going to ask follow-up questions just now, until I have thought about it more myself and tried to represent some of these situations mathematically.

    The idea that the three accelerations (start, turnaround and finish) in the 'twin paradox' might have to combine in a way that does not influence the relative aging of the twins over the whole journey intrigues me (see my other post today). If this is in fact correct, I want to understand how it comes to be so and if I can, prove it.
    Last edited by lesaid; 06-06-2018 at 02:01 PM.
    Reply With Quote  
     

  16. #16  
    Senior Member
    Join Date
    Sep 2017
    Posts
    278
    Quote Originally Posted by lesaid View Post
    OK, sorry it's taken me a while, but I at last have this straight (I think!). What turned it around for me was to take the end event as the origin of the various frames, rather than trying to find a common start point (obvious with hindsight!) and I now have a picture where the maths is consistent and clear, and all the spacetime separations match between the pairs of events across the frames.

    I cannot claim to have my head around this intuitively, but I can see what the maths says, and think I understand it.


    I have also done the same for the 'regular' twin scenario, and also have everything matching, as I expected.

    This leaves me with one conundrum however. Going to the regular twin 'paradox', the calculations seem clear and consistent, while completely ignoring the effect of the acceleration (other than fact of a change of frame). The nature, duration and scale of acceleration, and the distance from 'home' at which it occurs has to be irrelevant, or I would have needed to include it somehow in the calculations to get consistent results.

    This leads me to believe that the ONLY thing that is needed to account for the 'standard' twin paradox has to be the change of frame. Accelerations would change the calculations, but must change them in a way that always cancels out, so that they have no net effect on the relative times experienced by the two twins.

    But this position seems, at least superficially, to be at odds with earlier responses in this thread that stress the role of acceleration. Am I still missing something?

    My next challenge is to get a real understanding of accelerating frames in SR!
    You will need to read the references on the two sides of the argument, see here.
    The side that argues that acceleration has nothing to do with it totally misses the point that I just made to you earlier in that one twin had to be accelerated in order to get up to speed. Acceleration is intrinsic in creating the disparity between the twins. Until you understand that, you will not understand the issue.
    Last edited by AndrewC; 06-06-2018 at 04:33 PM.
    Reply With Quote  
     

  17. #17  
    Senior Member
    Join Date
    Sep 2017
    Posts
    278
    Quote Originally Posted by lesaid View Post

    My next challenge is to get a real understanding of accelerating frames in SR!
    This is an excellent starting point.
    lesaid likes this.
    Reply With Quote  
     

  18. #18  
    Member
    Join Date
    Jul 2013
    Location
    Scotland
    Posts
    40
    Quote Originally Posted by AndrewC View Post
    You will need to read the references on the two sides of the argument, see here.
    The side that argues that acceleration has nothing to do with it totally misses the point that I just made to you earlier in that one twin had to be accelerated in order to get up to speed. Acceleration is intrinsic in creating the disparity between the twins. Until you understand that, you will not understand the issue.
    I think I do understand that, though I don't see how that applies in all formulations of the paradox.

    I also have a long string of questions now in mind that I want to get to the bottom of - such as
    • understanding the effects of modelling the twin scenario with differing proportions of acceleration and constant speed motion - from the one I've been looking at where the acceleration is presumed instantaneous, to one where there is no free fall portion and the entire journey consists of long, sustained accelerations, and where there is a mixture of the two.
    • the difference between an object reversing direction using a rocket to accelerate, and one doing a half-orbit around a mass - i.e. following a free-fall geodesic without experiencing acceleration to reverse direction, and how those two scenarios fit with the equivalence principle (the outcome - reversal of direction - is the same, but one experiences acceleration while the other does not). And the role of GR time dilation as the gravity well is traversed. And how those two scenarios compare in situations like the twin paradox. I've wondered about this kind of thing for a while, and saw it was briefly referred to at the end of the Wiki link you gave me, but only in passing.
    and more - I could go on!

    But, rather than talking round in circles at an intuitive level (my intuition is already struggling!), I feel I first need to understand acceleration in SR properly, so I'll get reading, and come back if things don't come clear after that! Will take a little while !!

    Thank you
    Reply With Quote  
     

  19. #19  
    Member
    Join Date
    Jul 2013
    Location
    Scotland
    Posts
    40
    Quote Originally Posted by AndrewC View Post
    This is an excellent starting point.
    Thank you, that looks exactly what I need - I'm reading !!!
    Reply With Quote  
     

Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •