Can someone help me understand this paradox please

[rede96]

I am really trying to get my head around SR, in particular the twin paradox I have read so much about.

From my very elementary understanding of the twin paradox, the basic reason for the twin that is traveling aging less is because he accelerates and thus breaks the symmetry between the two twins.

So my logical question would be what would happen if there was no acceleration? Here is a simple thought experiment to demonstrate.

Imagine I am stood on Earth holding a clock. My friend Jack passes me in his super fast spaceship traveling at say 0.7c He also has a clock and as he passes very close to me his clock is synchronised to mine. He then continues his journey.

A second friend Jill is on a return journey from some faraway place in her super fast space ship, traveling also at 0.7c but she is coming from the opposite direction to Jack.

So naturally, Jack and Jill pass each other at some point. Jill also has a clock and as Jack and Jill pass each other, Jill’s clock is synchronised to Jack’s.

Jill then continues her journey and as she passes me on earth, I take a reading of her clock. As there has been no acceleration between the synchronisation of the clocks, does Jill’s clock read the same as mine as she passes me?

If they don’t read the same, which system (I.e. my clock time or the traveling clock’s time) would have recorded less time?

 

[JesseM]

It's really about the path though spacetime not being "straight" relative to any inertial frame, rather than acceleration per se. So here, if you draw a spacetime diagram from the perspective of an inertial frame, and look at the path through spacetime made up of Jack's outbound journey until he passes Jill and Jill's inbound journey from the moment she passes jack, this path is "bent" from the perspective of any inertial frame, and thus less proper time elapses along it than your own straight path through spacetime (a straight line through spacetime maximizes the proper time, in much the same way that in 2D space a straight line is always the path with the shortest distance...see [post=2972720]this post[/post] for more details on the analogy between paths through spacetime and paths through space). So it will be just as in the twin paradox here, Jill's clock will show a lesser time when she meets you than your own clock does.

 

[Mike_Fontenot]

[...]
As there has been no acceleration between the synchronisation of the clocks, does Jill’s clock read the same as mine as she passes me?

No. Jill's clock (or her age, if you use ages as clocks) will be less than yours, exactly like Jack's would, if he had turned around and rejoined you.

The key is understanding that when Jack and Jill pass by each other (and synchronize their watches), their two conclusions about your current age are VERY different, even though they are co-located at that instant. They could even be receiving exactly the same TV image of you at that instant, with you holding a sign giving the current reading on your watch and calendar (or your current age). But they each know that the date and time shown in that image is not really the CURRENT time there, because that image didn't travel instantaneously between you and them. They each can correct for that transit time, but when they do that calculation CORRECTLY, they will get VERY different corrections.

Mike Fontenot

 

[rede96]

It's really about the path though spacetime not being "straight" relative to any inertial frame, rather than acceleration per se. So here, if you draw a spacetime diagram from the perspective of an inertial frame, and look at the path through spacetime made up of Jack's outbound journey until he passes Jill and Jill's inbound journey from the moment she passes jack, this path is "bent" from the perspective of any inertial frame, and thus less proper time elapses along it than your own straight path through spacetime (a straight line through spacetime maximizes the proper time, in much the same way that in 2D space a straight line is always the path with the shortest distance...see [post=2972720]this post[/post] for more details on the analogy between paths through spacetime and paths through space). So it will be just as in the twin paradox here, Jill's clock will show a lesser time when she meets you than your own clock does.

Ah, I see. (I think!)
So would it be fair to say that if it were possible to travel on a 'straight' path through space-time relative to any inertial frame then there would be no time dilation?

Just an afterthought, does this mean that because we know the effects of time dilation that we know how curved space-time is?

 

[JesseM]

Ah, I see. (I think!)
So would it be fair to say that if it were possible to travel on a 'straight' path through space-time relative to any inertial frame then there would be no time dilation?

Time dilation is different from the issue of differences in total aging. Time dilation is just about the rate that a given clock is ticking at a given moment, and it depends which frame you use. If you and I are moving apart inertially (on straight paths through spacetime at different angles), then at any given moment in my frame you are aging slower than me (so at any given moment in my frame, my clock shows a greater time than yours), while at any given moment in your frame I am aging slower than you (so at any given moment in your frame, your clock shows a greater time than mine--see the relativity of simultaneity). Only in the case where we compare clocks at the same location is there an objective frame-independent answer to what each one reads "at the same moment".

Just an afterthought, does this mean that because we know the effects of time dilation that we know how curved space-time is?

Time dilation due to velocity in special relativity isn't related to curved spacetime, in fact special relativity assumes spacetime is "flat" (zero curvature) everywhere, only in general relativity where you have to take into account gravitational time dilation does curved spacetime come into play.

 

[harrylin]

I am really trying to get my head around SR, in particular the twin paradox I have read so much about.

From my very elementary understanding of the twin paradox, the basic reason for the twin that is traveling aging less is because he accelerates and thus breaks the symmetry between the two twins.

Yes. One changes his velocity while the other does not.

So my logical question would be what would happen if there was no acceleration?

What matters is if one changes velocity - which is typically done by acceleration. BTW, this was understood right from the start: "any change of velocity, or any acceleration has an absolute sense". See p.47 and on from p.50, of http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time .

Here is a simple thought experiment to demonstrate.

Imagine I am stood on Earth holding a clock. My friend Jack passes me in his super fast spaceship traveling at say 0.7c He also has a clock and as he passes very close to me his clock is synchronised to mine. He then continues his journey.

A second friend Jill is on a return journey from some faraway place in her super fast space ship, traveling also at 0.7c but she is coming from the opposite direction to Jack.

So naturally, Jack and Jill pass each other at some point. Jill also has a clock and as Jack and Jill pass each other, Jill’s clock is synchronised to Jack’s.

Jill then continues her journey and as she passes me on earth, I take a reading of her clock. As there has been no acceleration between the synchronisation of the clocks, does Jill’s clock read the same as mine as she passes me?

If they don’t read the same, which system (I.e. my clock time or the traveling clock’s time) would have recorded less time?

There is no significant difference with the standard twin paradox: as you now certainly understand, the only thing that matters is the change in velocity of the time keeping instrument(s). You can do that either by accelerating a single clock as Langevin suggested, or by using a second clock as you outlined here above.

Cheers,
Harald

 

[rede96]

Thanks everyone for your replies.

Let's see if I understand...

Jack and Jill set off on two separate journeys in the shape of a big 'baseball' diamond.

They start together (on 'home plate') and will meet at the top of the diamond ('second base').

Jack takes the left route, so goes via third base while Jill takes the right route and goes via first base. They are both traveling at the same speed.

During the journey, Jack will see Jill's clock run slower and Jill will see Jack's clock run slower.

However when they meet up again, both clocks will have 'ticked' the same amount of time and thus Jack and Jill have aged the same.

Jack and Jill now repeat this journey going in the oppisite direction back to 'home plate'. However as Jill doesn't really like baseball anyway, she takes a short cut and goes straight over the pitcher's mound to get to the 'home plate' in a straight line. Also, Jill goes slower than Jack this time so that they both meet at the 'home plate' at the same time.

This time, because Jack's path was not straight, (i.e. he went back via 'third base' again), he accelerated wrt to Jill’s path. So when they meet Jack's clock will have 'ticked' less than Jill's and thus Jack will aged less.

 

[JesseM]

During the journey, Jack will see Jill's clock run slower and Jill will see Jack's clock run slower.

Are you talking about what they see visually (which is based on the Doppler effect, not time dilation alone) or are you talking about what's true in "their own" frame of reference? If the latter, there isn't any inertial frame where Jack (or Jill) is at rest both before and after his acceleration, and the time dilation equation is only meant to apply in inertial frames.

However when they meet up again, both clocks will have 'ticked' the same amount of time and thus Jack and Jill have aged the same.

Jack and Jill now repeat this journey going in the oppisite direction back to 'home plate'. However as Jill doesn't really like baseball anyway, she takes a short cut and goes straight over the pitcher's mound to get to the 'home plate' in a straight line. Also, Jill goes slower than Jack this time so that they both meet at the 'home plate' at the same time.

This time, because Jack's path was not straight, (i.e. he went back via 'third base' again), he accelerated wrt to Jill’s path. So when they meet Jack's clock will have 'ticked' less than Jill's and thus Jack will aged less.

Yes, all this is correct. But note that when I talked about the path being straight, I meant straight in spacetime, not just space; in your example Jill's path is straight in both senses, but if she had traveled in a straight line through space but changed velocity as she did so, then her path through spacetime (as represented on a spacetime diagram where one axis is time) would not be straight.

 

[rede96]

Are you talking about what they see visually (which is based on the Doppler effect, not time dilation alone) or are you talking about what's true in "their own" frame of reference? If the latter, there isn't any inertial frame where Jack (or Jill) is at rest both before and after his acceleration, and the time dilation equation is only meant to apply in inertial frames.

Yes, all this is correct. But note that when I talked about the path being straight, I meant straight in spacetime, not just space; in your example Jill's path is straight in both senses, but if she had traveled in a straight line through space but changed velocity as she did so, then her path through spacetime (as represented on a spacetime diagram where one axis is time) would not be straight.

I was thinking that once they set off from 'home plate' they are moving away from each other. But from Jack's own frame of reference, he is moving away from Jill and visa-versa.

And as they are moving in different directions, I thought they must see each other's clocks tick at a different rate.

If you and I are moving apart inertially (on straight paths through spacetime at different angles), then at any given moment in my frame you are aging slower than me (so at any given moment in my frame, my clock shows a greater time than yours), while at any given moment in your frame I am aging slower than you (so at any given moment in your frame, your clock shows a greater time than mine--see the relativity of simultaneity).

 

[JesseM]

I was thinking that once they set off from 'home plate' they are moving away from each other. But from Jack's own frame of reference, he is moving away from Jill and visa-versa.

But then after they pass first/third base they are moving towards each other, so they don't have a single inertial frame where they're at rest throughout the journey. If you pick an inertial frame where Jack is at rest for just the first half, yes it's true that in this frame Jill's clock is ticking slower than Jack's before either changes direction, but afterwards it will be Jack's clock that ticks slower this frame, so this frame still predicts that their clocks have elapsed the same time in total when they reunite.

 

[harrylin]

Thanks everyone for your replies.

Let's see if I understand...

Jack and Jill set off on two separate journeys in the shape of a big 'baseball' diamond.

They start together (on 'home plate') and will meet at the top of the diamond ('second base').

Jack takes the left route, so goes via third base while Jill takes the right route and goes via first base. They are both traveling at the same speed.

During the journey, Jack will see Jill's clock run slower and Jill will see Jack's clock run slower.

However when they meet up again, both clocks will have 'ticked' the same amount of time and thus Jack and Jill have aged the same.

Jack and Jill now repeat this journey going in the oppisite direction back to 'home plate'. However as Jill doesn't really like baseball anyway, she takes a short cut and goes straight over the pitcher's mound to get to the 'home plate' in a straight line. Also, Jill goes slower than Jack this time so that they both meet at the 'home plate' at the same time.

This time, because Jack's path was not straight, (i.e. he went back via 'third base' again), he accelerated wrt to Jill’s path. So when they meet Jack's clock will have 'ticked' less than Jill's and thus Jack will aged less.

Quite so - they could also have followed the same trajectory through space. Again: what matters is the change in velocity. This was summarised by Langevin on p.49/50:
"those have aged the least, for whom the motion during separation was the least uniform".

Cheers,
Harald

 

[MikeLizzi]

Thanks everyone for your replies.

...

During the journey, Jack will see Jill's clock run slower and Jill will see Jack's clock run slower.

However when they meet up again, both clocks will have 'ticked' the same amount of time and thus Jack and Jill have aged the same.

.....

Ouch. If that were true, it would constitute a paradox. If they arrive at the same location having the same age, and during some portion of the trip either "saw" the other's clock as running slow then during another portion of the trip, he/she must "see" the other's clock as running fast.

 

[DarkMatterHol]

Ouch. If that were true, it would constitute a paradox. If they arrive at the same location having the same age, and during some portion of the trip either "saw" the other's clock as running slow then during another portion of the trip, he/she must "see" the other's clock as running fast.

Yes. I'd be interested in knowing just when does Jill's clock appear to be running fast. Is it during the time they are converging - when she is going from first to second?

 

[harrylin]

Ouch. If that were true, it would constitute a paradox. If they arrive at the same location having the same age, and during some portion of the trip either "saw" the other's clock as running slow then during another portion of the trip, he/she must "see" the other's clock as running fast.

It's indeed called a "paradox".
They "see" the other clock as running slow by a factor gamma in a manner of speaking, if they adopt a reference system in which they are in rest while traveling - and then at the turn-around they switch "rest system". In the new rest system the distant time is defined differently so that distant time "jumps" at the switch but this is not truly "seeing". What they truly can see is the Doppler affected signals that they receive from each other, and in the end these Doppler counts must match of course. This was already elaborated in the first paper on such an example, here is again a link:
http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time
(starting from p.50)

 

[MikeLizzi]

It's indeed called a "paradox".
They "see" the other clock as running slow by a factor gamma in a manner of speaking, if they adopt a reference system in which they are in rest while traveling - and then at the turn-around they switch "rest system". In the new rest system the distant time is defined differently so that distant time "jumps" at the switch but this is not truly "seeing". What they truly can see is the Doppler affected signals that they receive from each other, and in the end these Doppler counts must match of course. This was already elaborated in the first paper on such an example, here is again a link:
http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time
(starting from p.50)

So are you withdrawing your previous approvalof rede96’s last statement? (statement #11)

 

[MikeLizzi]

Yes. I'd be interested in knowing just when does Jill's clock appear to be running fast. Is it during the time they are converging - when she is going from first to second?

If you are a non-inertially moving observer (you are accelerating) the rate of any inertially moving clock must be faster than yours. Those who are familiar with General Relativity will recognize an analogy to an observer in a gravitational well.

If you are a non-inertially moving observer the rate of a non-inertially moving clock could be faster or slower than yours. It depends upon the magnitude and direction of the acceleration with respect to you. Again the gravitational well analogy applies.

With regards to the baseball diamond problem, it is not so easy to determine when the "other guy's" clock is moving faster. Because of Relativity of Simultaneity.

 

[rede96]

If you are a non-inertially moving observer (you are accelerating) the rate of any inertially moving clock must be faster than yours. Those who are familiar with General Relativity will recognize an analogy to an observer in a gravitational well.

If you are a non-inertially moving observer the rate of a non-inertially moving clock could be faster or slower than yours. It depends upon the magnitude and direction of the acceleration with respect to you. Again the gravitational well analogy applies.

With regards to the baseball diamond problem, it is not so easy to determine when the "other guy's" clock is moving faster. Because of Relativity of Simultaneity.

Does that mean that if the first non-inertially moving observer sees the second non-inertially moving observer's clock moving slower (because of the magnitude and direction of the acceleration) then the opposite applies. i.e. the second observer must see the first observer's clock moving faster?

 

[MikeLizzi]

Does that mean that if the first non-inertially moving observer sees the second non-inertially moving observer's clock moving slower (because of the magnitude and direction of the acceleration) then the opposite applies. i.e. the second observer must see the first observer's clock moving faster?

Well, my last post was the limit of what I currently know. I’ve studied relativistic transformations when both bodies are inertial, when one body is inertial and the other non-inertial. But I am just starting on the math when both bodies are non-inertial. As you know the standard Twins Paradox involves one inertial and one sometimes non-inertial body.

But the baseball diamond problem involves two sometimes inertial sometimes non-inertial runners. And from either runner’s point of view the other is not running symmetrically.

I could guess at an answer to your question but that is a good way to end up looking foolish. And I’ve already past my quota of looking foolish this year.

 

[harrylin]

So are you withdrawing your previous approvalof rede96’s last statement? (statement #11)

No, I elaborated on it. There are many ways of saying the same as Langevin. No explanation with mere words is perfect (also not his), but I think he did a rather good job back then, in the first publication on this topic.

 

[harrylin]

Does that mean that if the first non-inertially moving observer sees the second non-inertially moving observer's clock moving slower (because of the magnitude and direction of the acceleration) then the opposite applies. i.e. the second observer must see the first observer's clock moving faster?

Not if you described a symmetrical situation, as I thought you did - see my posts #11 and #14.

Cheers,
Harald

 

[rede96]

Not if you described a symmetrical situation, as I thought you did - see my posts #11 and #14.

Cheers,
Harald

Hi harald,

Yes I did describe a symmetrical situation. So my guess is that the clocks would be ticking at the same rates in that situation.

I was thinking of a different situation where one of them may have been traveling at a different speed and hence the clocks were not ticking at the same rate.

 

[Mike_Fontenot]

If you are a non-inertially moving observer (you are accelerating) the rate of any inertially moving clock must be faster than yours.

That's not true, if you are using the CADO reference frame for the accelerating observer's frame (which I consider to be the only acceptable choice). Perhaps you are referring to one of the other alternative reference frames that some other members of this forum believe to be acceptable, and perhaps even prefer.

I can't respond in more detail right now, because I'm currently "over-subscribed" with another pressing exercise. But, you might be able to figure out what I'm talking about, if you take a look at my webpage:

http://home.comcast.net/~mlfasf


Mike Fontenot

 

[pervect]

re:

If you are a non-inertially moving observer (you are accelerating) the rate of any inertially moving clock must be faster than yours.

It's not quite true as written. What is true that in special relativity (i.e. no gravity, no space-time-curvature), if you compare the trip time of two observers who start out and end up at the same spot by a direct clock comparison at zero distance (which is unambiguous), the inertial observer will accumulate the longest time on their clock.

 

[harrylin]

Hi harald,
Yes I did describe a symmetrical situation. So my guess is that the clocks would be ticking at the same rates in that situation.

Yes that is correct from the perspective of the frame in which the situation is symmetrical - but only from that perspective (see my reply in post #14).

I was thinking of a different situation where one of them may have been traveling at a different speed and hence the clocks were not ticking at the same rate.

Yes, and I also replied to that (in post #11).

Harald

 

[Mike_Fontenot]

re:It's not quite true as written. What is true that in special relativity (i.e. no gravity, no space-time-curvature), if you compare the trip time of two observers who start out and end up at the same spot by a direct clock comparison at zero distance (which is unambiguous), the inertial observer will accumulate the longest time on their clock.

OK. I interpreted this startement of MikeLizzi's:

"If you are a non-inertially moving observer (you are accelerating) the rate of any inertially moving clock must be faster than yours"

as a statement about the INSTANTANEOUS rate (i.e., the relative tic rate at any given INSTANT of the accelerating traveler's life). Perhaps he MEANT the relative tic rate, averaged over an ENTIRE round-trip. Two VERY different things, for an accelerating observer.

Mike Fontenot