Space traveler and time dilation

[andyp2010]

Maybe some one can help clear up a problem. According to Wikipedias article on time dilation

“In the case that the observers are in relative uniform motion, and far away from any gravitational mass, the point of view of each will be that the other's (moving) clock is ticking at a slower rate than the local clock.”

Which I can accept but this does not seem consistent with the “time slowing down for a space traveler” situation. If the space traveler is moving fast relative to the Earth the people on Earth will see that time is passing more slowly for traveler. The traveler will think that the time on the Earth is moving more slowly. So what happens when he gets back to Earth he’s been observing that Earth time has been moving more slowly so he would be older that Earth time would suggest. But also for the Earth people the travelers time would have been passing more slowly so he should be younger. Which is a contradiction.

 

[Dmitry67]

Congratulatons, you had rediscovered the http://en.wikipedia.org/wiki/Twin_paradox

 

[phyzguy]

It's called the "twin paradox", and the resolution is well understood. Here's a good place to start:

http://en.wikipedia.org/wiki/Twin_paradox

 

[George Jones]

If the space traveler is moving fast relative to the Earth the people on Earth will see that time is passing more slowly for traveler. The traveler will think that the time on the Earth is moving more slowly. So what happens when he gets back to Earth he’s been observing that Earth time has been moving more slowly so he would be older that Earth time would suggest.

Yes as the traveler moves away from Earth, he sees visually time on the Earth passing more slowly, but when the traveler is on the trip back to Earth, he sees visually time on the Earth passing more quickly. The net result is that upon arrival back on Earth, the traveler finds that the Earth has aged more than he has.

See

https://www.physicsforums.com/showthread.php?p=2669956#post2669956.

 

[andyp2010]

thanks everyone. I'll get reading then.

 

[JDługosz]

What I'd like to know is: if you went "all the way around the Universe", i.e. it was a 4-D sphere, and passed your starting point again without ever having undergone acceleration, how would the paradox be resolved? A wormhole would have the same problem in GR.

 

[ExecNight]

Ah "the twins".. I don't know about the time dilation or the paradox.

I don't know their acceleration or anything but this topic sure never gets old =)

 

[Dmitry67]

What I'd like to know is: if you went "all the way around the Universe", i.e. it was a 4-D sphere, and passed your starting point again without ever having undergone acceleration, how would the paradox be resolved? A wormhole would have the same problem in GR.

As Universe is curved, then that can be answered in SR framework, only n GR framework. But the situation is not symmetric in GR because the world line of such observer is more curved.

 

[yogi]

What I'd like to know is: if you went "all the way around the Universe", i.e. it was a 4-D sphere, and passed your starting point again without ever having undergone acceleration, how would the paradox be resolved? A wormhole would have the same problem in GR.

You don't really need to go around the universe to resolve or simulate the problem - there is one round trip voyage where both twins remain in their own inertial frame for the entire round trip - yet their clocks show different lapsed times when the traveling twin returns - no general relativity involved and no curvature - a polar orbiting satellite will do - simple construct a 100 mile high tower on the South Pole and put a satellite in polar orbit at an elevation of 100 miles - a clock on top of the tower remains fixed in the non rotating Earth centered inertial reference frame and the clock on board the satellite remains in the inertial frame of the orbiting satellite - start the clocks as the satellite passes the tower and stop them when it passes by after completing one orbit - the two clocks will not have logged the same amount of time.

 

[matheinste]

You don't really need to go around the universe to resolve or simulate the problem - there is one round trip voyage where both twins remain in their own inertial frame for the entire round trip.

In SR if the twins part and reunite then at least one of them has been moving non-inertially at some time.

What sort of motion a round trip of the universe entails I do not know. It depends on the global geometry of the universe.

Matheinste.

 

[MikeLizzi]

Yes as the traveler moves away from Earth, he sees visually time on the Earth passing more slowly, but when the traveler is on the trip back to Earth, he sees visually time on the Earth passing more quickly. The net result is that upon arrival back on Earth, the traveler finds that the Earth has aged more than he has.

Gorge Jones,
You have posted this comment several times in this forum, now. It is absolutely wrong. If you believe it, you have a serious misunderstanding of Special Relativity.

I am shocked that the other members of this forum are letting you get away with this.

 

[Ich]

George Jones' comment is not wrong. Maybe you are confusing relativistic doppler effect and time dilation? George explicitly talks about the former.

 

[phyzguy]

George Jones and Ich are correct. Attached is a picture I downloaded from among the plethora of explanations on the web. You can see that on the way out, the twin on the rocket sees two years pass on Earth, while four years pass for him. On the way back, he sees 8 years pass on Earth while four years pass for him. So when he returns, 10 years have passed on Earth while 8 years have passed for him, so the twin on the rocket is younger.

 

[Aaron_Shaw]

George Jones and Ich are correct. Attached is a picture I downloaded from among the plethora of explanations on the web. You can see that on the way out, the twin on the rocket sees two years pass on Earth, while four years pass for him. On the way back, he sees 8 years pass on Earth while four years pass for him. So when he returns, 10 years have passed on Earth while 8 years have passed for him, so the twin on the rocket is younger.

If you do calculations to compensate for doppler effect, then you'll see that time passes slower in both directions (outbound and inbound). It's just that at the turnaround point at half way and decelleration to stop back on Earth that perceptions of simultaneity will shift significantly in a small amount of time. This accounts for a huge 'jump' forwards in time on earth, as seen by the traveller, as he turns around.

Thus, even though Earth's clocks have been ticking slower on both the outbound AND inbound journey, the 'jump' forwards in time as simultaneity shifts means that the earthlings will have still aged more.

 

[George Jones]

Gorge Jones,
You have posted this comment several times in this forum, now. It is absolutely wrong.

How is it wrong?

This accounts for a huge 'jump' forwards in time on earth, as seen by the traveller, as he turns around.

A huge jump in coordinates is not actually seen visually by the traveler. There is a big difference between what is seen visually, and what happens to coordinates.

 

[phyzguy]

A huge jump in coordinates is not actually seen visually by the traveler. There is a big difference between what is seen visually, and what happens to coordinates.

This is absolutely correct. If the twin on the rocket had a telescope trained on Earth, he would see no jump. He would simply see things start to happen faster as he turned around, because he is now starting to "catch up" with the outward propagating wave fronts coming from Earth. Look again at the space-time diagram I posted a few posts ago.

 

[DrGreg]

You don't really need to go around the universe to resolve or simulate the problem - there is one round trip voyage where both twins remain in their own inertial frame for the entire round trip - yet their clocks show different lapsed times when the traveling twin returns - no general relativity involved and no curvature - a polar orbiting satellite will do - simple construct a 100 mile high tower on the South Pole and put a satellite in polar orbit at an elevation of 100 miles - a clock on top of the tower remains fixed in the non rotating Earth centered inertial reference frame and the clock on board the satellite remains in the inertial frame of the orbiting satellite - start the clocks as the satellite passes the tower and stop them when it passes by after completing one orbit - the two clocks will not have logged the same amount of time.

This analogy doesn't work in either special or general relativity.

In special relativity, ignoring gravity, the south pole pole would be inertial and the "satellite" would be accelerating.

In general relativity, the south pole pole is accelerating and the satellite is inertial.

 

[Aaron_Shaw]

This is absolutely correct. If the twin on the rocket had a telescope trained on Earth, he would see no jump. He would simply see things start to happen faster as he turned around, because he is now starting to "catch up" with the outward propagating wave fronts coming from Earth. Look again at the space-time diagram I posted a few posts ago.

I'm basing this on a model of what happened after calculating to remove the doppler effect and time it takes for light to travel, as i stated originally (probably not very well). I find that when people are trying to understand, the doppler effect and all that confuses the issue. People typically want to know what is 'really' happening, rather than what it looks like by the time the light has reached you.

So i guess what i mean is that if you plotted space and time coordinates on a graph for the whole journey then what the traveller would see is the Earth taking a large jump forwards in time at the turnaround point (assuming its instantaneous).

 

[zetafunction]

my question is, can a Wick rotation be made 'physical' ??

i mean you are in a metric t^{2}-x^{2} space and time

and you 'rotate' your reference system to get a new metric t^{2}+x^{2} which is purely Euclidean and there is no distinction between space and time

 

[JesseM]

I'm basing this on a model of what happened after calculating to remove the doppler effect and time it takes for light to travel, as i stated originally (probably not very well). I find that when people are trying to understand, the doppler effect and all that confuses the issue. People typically want to know what is 'really' happening, rather than what it looks like by the time the light has reached you.

But that isn't what's "really happening" in any meaningful sense, it's just what's happening in one particular non-inertial rest frame for the traveling twin--specifically one constructed in such a way that the definition of simultaneity in this non-inertial frame always matches up with the definition of simultaneity that would be used in the traveler's instantaneously co-moving inertial rest frame at that instant. Unlike with inertial frames, though, there isn't anyone "correct" way to construct a non-inertial rest frame for a non-inertial observer, there are an infinite number of different coordinate systems you could construct for such an observer and none of them would be considered physically "preferred".

 

[Ich]

So i guess what i mean is that if you plotted space and time coordinates on a graph for the whole journey then what the traveller would see is the Earth taking a large jump forwards in time at the turnaround point (assuming its instantaneous).

SR does not support coordinate systems with a "jump forward in time". The very notion of such a jump is IMO much more misleading that the additional mentioning of visual effects. Especially if stated in a post supposed to describe what's 'really' happening. This is exactly the kind of voodoo that laymen love to hear, but it burns out their brains. It'd destroy my brain too if I had to understand how such a 'real' jump forward in time is caused by an acceleration manoever of a probe some lightyears away.

 

[Aaron_Shaw]

SR does not support coordinate systems with a "jump forward in time". The very notion of such a jump is IMO much more misleading that the additional mentioning of visual effects. Especially if stated in a post supposed to describe what's 'really' happening. This is exactly the kind of voodoo that laymen love to hear, but it burns out their brains. It'd destroy my brain too if I had to understand how such a 'real' jump forward in time is caused by an acceleration manoever of a probe some lightyears away.

Well i could be way off then, but I'm getting at the point that on the outbound journey the Earth clock runs slower, and it also runs slower on the inbound journey. If this is the case then how does the Earth clock end up being older than the traveling clock? Because at the periods in between the two non accelerating frame journeys, where acceleration occurs, the simultaneity wil shift as the speed isn't staying constant. So at time right BEFORE turn around the Earth clock might read 25 seconds, but at the moment right AFTER turnaround (an instantaneous one) the Earth clock will read significantly more than the expected 25 seconds, for example, 180.

Of course the traveller would SEE the clock read something different. But if the traveller knows he is X light years away, and he knows the speed of light, he would conclude that the clock reading he can currently see actually occurred a certain amount of time earlier.

So baring that in mind, he might conclude that the two clocks read as follows:

Travelling clock - Earth clock
0 0
10 5
20 10
30 15
40 20
50 25

TURNAROUND

60 180
70 185
80 190
90 195
100 200


If this is complete rubbish then I'm keen to figure it out properly. But the way i currently see it is that it's 2 journeys which can be modeled by SR. The 2 journeys don't result in a paradox when the traveller is back at the start because in between there has been a part of the journey not handled by SR, where the idea of simultaneity between the two frames has changed due to the change in relative velocities.

I'm aware, btw, that my use of terminology is not proper. I'm still getting used to it.

 

[Ich]

If this is complete rubbish then I'm keen to figure it out properly.

No, it's not rubbish. It's a change of coordinate systems. You understand it clearly, different notions of simultaneity point to different events on the Earth's worldline as happening "now".
But such a thing cannot happen in a single inertial system. It's not part of SR.
And a "jump forward in time" is nothing real. Rather, different numbers are assigned to events. It's more to do with bookkeeping than time warps.

i'm getting at the point that on the outbound journey the Earth clock runs slower, and it also runs slower on the inbound journey. If this is the case then how does the Earth clock end up being older than the traveling clock?

Forget about "clocks running slower". That's Lorentz Ether language. Of course, A cannot run slower than B while B runs slower than A. This sort of language is incompatible with SR.
I know that's how they teach it. Forget it. Look at your diagrams. There's a triangle, and there's a triangle inequality (in our case the non-straight path being shorter). That's true no matter in which frame you view it.

Your math is correct. But there's no such things as slowing clocks, contracting metersticks, or planets jumping forward in time. SR is about relations of objects, not changes happening to them.
Take the geometric viewpoint, there are projections, slices, different paths (time dilation, length contraction, twin paradox), not broken clocks.

 

[Aaron_Shaw]

No, it's not rubbish. It's a change of coordinate systems. You understand it clearly, different notions of simultaneity point to different events on the Earth's worldline as happening "now".
But such a thing cannot happen in a single inertial system. It's not part of SR.
And a "jump forward in time" is nothing real. Rather, different numbers are assigned to events. It's more to do with bookkeeping than time warps.

Forget about "clocks running slower". That's Lorentz Ether language. Of course, A cannot run slower than B while B runs slower than A. This sort of language is incompatible with SR.
I know that's how they teach it. Forget it. Look at your diagrams. There's a triangle, and there's a triangle inequality (in our case the non-straight path being shorter). That's true no matter in which frame you view it.

Your math is correct. But there's no such things as slowing clocks, contracting metersticks, or planets jumping forward in time. SR is about relations of objects, not changes happening to them.
Take the geometric viewpoint, there are projections, slices, different paths (time dilation, length contraction, twin paradox), not broken clocks.

Ok... this is interesting. I've not heard anyone say this before, so it looks like I'm going to have to consider things differently; after i get some sleep : )

Thanks.

P.s. sorry if I've hijacked your thread, OP, it looked like it was winding up.

 

[yogi]

This analogy doesn't work in either special or general relativity.

In special relativity, ignoring gravity, the south pole pole would be inertial and the "satellite" would be accelerating.

In general relativity, the south pole pole is accelerating and the satellite is inertial.

A satellite in orbit is a perfectly good inertial frame - the traveling twin stays in orbit - all clocks are at the same gravitational potential (100 miles above the earth) - so there is no general relativity issues and there are no accelerations once the orbit is established and the first measurement is taken on the flyby.

 

[Al68]

SR does not support coordinate systems with a "jump forward in time".

That's right, the "jump forward in time" is in a non-inertial reference frame and is due to the equally impossible instantaneous turnaround.

The "non-real" instantaneous turnaround and the resulting "non-real" time jump are usually specified just to make the math simpler.

In a realistic turnaround, the coordinate time on Earth could be calculated at intervals during the turnaround using the lorentz transformations and a series of co-moving (to the ship) inertial frames. And we could even make those intervals infinitesimally small. This would result in Earth's clock "running fast" in the ship's accelerated frame. We could call it "gravitational time dilation". Oh, wait...Einstein beat us to it.

 

[Ich]

This would result in Earth's clock "running fast" in the ship's accelerated frame. We could call it "gravitational time dilation". Oh, wait...Einstein beat us to it.

Yes. You just need a way to express physics in arbitrary coordinate systems and transform between them. Something like general covariance.
That's why I say that accelerating frames are not part of SR, even if spacetime is flat.

 

[Al68]

Yes. You just need a way to express physics in arbitrary coordinate systems and transform between them. Something like general covariance.
That's why I say that accelerating frames are not part of SR, even if spacetime is flat.

Accelerating frames weren't part of SR originally, but predate GR. Einstein derived gravitational time dilation in 1907 (I think) by applying SR to accelerating frames.

I think this is why many consider accelerated frames and gravitational time dilation part of SR, since they were used by Einstein with SR before GR existed.

 

[Ich]

I think this is why many consider accelerated frames and gravitational time dilation part of SR, since they were used by Einstein with SR before GR existed.

I think the main reason is that there is no new physics added if you keep spacetime flat. Only the description changes. So why call it a different theory then?
But officially, the respective principles of covariance are determining the names of the theories. SR deals with transformations between standard inertial frames, GR with general coordinate transformations.
That distinction makes sense, I think. Whenever you talk about SR, you know exactly what the coordinates mean. In GR, you don't.

 

[yogi]

IMO - any analysis that results from dependence upon accelerating frames and the like is going to cloud the reality of what is properly explained by SR - this can be done by using the one way trip and doubling the result - or an orbiting satellite round trip if someone inists that the traveling clock must be returned to the start point to make comparisons (not true, but frequently asserted). Einstein confused a lot of his followers when he published his 1918 article that explained the clock paradox using a pseudo G field that gives the same answer to the aging difference - but for the wrong reason. If the problem can be simply solved without resorting to diversions that involve turn around accelerations, shifting planes of simultanety, jumping clock times and changing inertial frames etc, all such devices are bound to lead to a misunderstanding of what is really taking place.

 

[JesseM]

A satellite in orbit is a perfectly good inertial frame

The only frames in curved spacetime that qualify as "inertial" are local ones defined on a very small (technically it must be infinitesimally small) patch of spacetime, if you're talking about a coordinate system covering a large spatial or region or a long time interval (like a significant proportion of an orbit), then tidal effects would be detectable in this region so the frame can't be inertial. Do you disagree? This is a very standard idea, any textbook discussing the equivalence principle should make clear it only holds in a very small region of both space and time. For example, read the last section of the txt http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_principle/index.html , the section titled "Tidal forces, and a more precise definition", where they write:

Realizing that what matters are the size of the region, and the duration of our observations, we are led to a formulation in which the equivalence principle is not just a useful approximation, but exactly true: Within an infinitely small ("infinitesimal") spacetime region, one can always find a reference frame - an infinitely small elevator cabin, observed over an infinitely brief period of time - in which the laws of physics are the same as in special relativity. By choosing a suitably small elevator and a suitably brief period of observation, one can keep the difference between the laws of physics in that cabin and those of special relativity arbitrarily small.

 

[yogi]

Hi Jesse, long time no chat. In the context of the twin trip, the analogy is simply a way to illustrate the idea that accelerations (in the fame of the traveler) can be virtually eliminated if the trajectory is bent by a G source that will precisely return the traveler to the starting point). As you will no doubt recognize, this is simply a one orbit version of a GPS satellite - both clocks remain in separate inertial frames during the entire round trip (one orbit). Its true that on a real satellite there are minute tidal affects that would slightly alter the traveling clock frequency - but if the South Pole tower in the thought experiment is extended to 20,000, km, the orbital speed is about 14,000 km/hr, and the frequency shift is about 7 nanosec/day - verified to great accuracy every day.
In one orbit, there is going to be an accurate measurement of the time difference between the two clocks when the traveler returns.

To complete the analogy, one might place a second tower at the North Pole and check the time lapse for the halfway point - in the normal twin scenario, this corresponds to the turn around point which normally involves deceleration and acceleration in the frame of the traveler - but in the orbital version, there are no accelerations except those incidental to tidal affects and the divergence of the Earth's G field.

 

[JesseM]

Hi Jesse, long time no chat. In the context of the twin trip, the analogy is simply a way to illustrate the idea that accelerations (in the fame of the traveler) can be virtually eliminated if the trajectory is bent by a G source that will precisely return the traveler to the starting point).

Yes, but you are wrong in thinking that "no accelerations/G-forces" (i.e., accelerometers floating free at any given point in a room measuring 0 at all times) is a sufficient condition to have an approximately inertial frame in GR. An inertial frame is also one where there are no measurable tidal forces, and over the course of an orbit tidal forces should be measurable, since in general relativity tidal forces only become negligible in small regions of space over short time-intervals (small regions of spacetime, not just small regions of space). I'm not actually sure of the details of how tidal forces would manifest in the scenario of a small room in orbit for a long period of time, but I bet if you set off two balls at slightly different speeds in slightly different directions from one end of the room, such that the time for them to reach the opposite walls at constant velocity would be comparable to the time for an entire orbit, you would in fact observe that the balls' paths would depart appreciably from straight paths at constant velocity over this long time period.

 

[DrGreg]

In general relativity, the south pole pole is accelerating and the satellite is inertial.

A satellite in orbit is a perfectly good inertial frame - the traveling twin stays in orbit - all clocks are at the same gravitational potential (100 miles above the earth) - so there is no general relativity issues and there are no accelerations once the orbit is established and the first measurement is taken on the flyby.

You're missing my point. I did say

"In general relativity, ... the satellite is inertial."​

which agrees with everything you said above (subject to JesseM's correct note about local frames). But I also said

"In general relativity, the south pole pole is accelerating..."​

It is undergoing proper acceleration upwards and therefore is not inertial.

 

[JDługosz]

You don't really need to go around the universe to resolve or simulate the problem - there is one round trip voyage where both twins remain in their own inertial frame for the entire round trip - yet their clocks show different lapsed times when the traveling twin returns - no general relativity involved and no curvature - a polar orbiting satellite will do - simple construct a 100 mile high tower on the South Pole and put a satellite in polar orbit at an elevation of 100 miles - a clock on top of the tower remains fixed in the non rotating Earth centered inertial reference frame and the clock on board the satellite remains in the inertial frame of the orbiting satellite - start the clocks as the satellite passes the tower and stop them when it passes by after completing one orbit - the two clocks will not have logged the same amount of time.

Make it more symmetrical by using two satellites instead, orbiting in opposite directions. Each is traveling through the same curvature and each is accelerating the same. That creates a paradox if you apply only SR. How does it manage to work out using GR? Once I understand that, I can ponder the Universe form of the question again.

 

[yogi]

You're missing my point. I did say

"In general relativity, ... the satellite is inertial."​

which agrees with everything you said above (subject to JesseM's correct note about local frames). But I also said

"In general relativity, the south pole pole is accelerating..."​

It is undergoing proper acceleration upwards and therefore is not inertial.

Yes - things on the surface of the Earth are subject to g - but for purposes of synchronizing GPS satellites, the non-rotating Earth centered reference frame is taken as a basis - in the proposed thought experiment, the affect of satellite height is nullified by the height of the imaginary towers - so perhaps I should have said both clocks are at the same gravitational potential at all times.

The point of the analogy is, it is not necessary to explain the paradox as being the result of the traveling twin feeling "turn-around acceleration" (as is frequently asserted - perhaps even by myself in past posts). This is not what distinguishes the two clocks - they log different amounts of time because of the invariance of the spacetime interval. The traveler travels in both space and time - the tower twin travels in time only.

 

[JesseM]

The point of the analogy is, it is not necessary to explain the paradox as being the result of the traveling twin feeling "turn-around acceleration"

But you can explain it in terms of one moving inertially while the other doesn't, and the question of which one turns can be decided by the turn-around acceleration.

This is not what distinguishes the two clocks - they log different amounts of time because of the invariance of the spacetime interval. The traveler travels in both space and time - the tower twin travels in time only.

But in a GR case, you can come up with a coordinate system where the first remains at a single point in space (traveling 'in time only', not moving in space) and a different coordinate system where the second is the one that remains at a single point in space, both coordinate systems would be equally valid in GR, there'd be no physical basis for saying one "really" moved in space while the other didn't.

 

[Jocko Homo]

Forget about "clocks running slower". That's Lorentz Ether language. Of course, A cannot run slower than B while B runs slower than A. This sort of language is incompatible with SR.
I know that's how they teach it. Forget it. Look at your diagrams. There's a triangle, and there's a triangle inequality (in our case the non-straight path being shorter). That's true no matter in which frame you view it.

Your math is correct. But there's no such things as slowing clocks, contracting metersticks, or planets jumping forward in time. SR is about relations of objects, not changes happening to them.
Take the geometric viewpoint, there are projections, slices, different paths (time dilation, length contraction, twin paradox), not broken clocks.

While Minkowski diagrams provide a powerful abstraction to the physical reality of Special Relativity, I think "clocks running slower" is a fair characterization considering the accelerating twin really does come back younger...

 

[yogi]

Make it more symmetrical by using two satellites instead, orbiting in opposite directions. Each is traveling through the same curvature and each is accelerating the same. That creates a paradox if you apply only SR. How does it manage to work out using GR? Once I understand that, I can ponder the Universe form of the question again.

If you have two circular orbiting satellites at the same height traveling in opposite directions and set both clocks to zero as they pass, when they meet again, both clocks will read the same

 

[yogi]

But you can explain it in terms of one moving inertially while the other doesn't, and the question of which one turns can be decided by the turn-around acceleration.

It will give you the right answer - but you can also get the correct age difference by doing a one way trip - setting the clocks to zero on the initial flyby and stopping the traveling clock at the instant it flies by the destination - you now have a one way trip with no accelerations and no turn around acceleration - to get the result you simple double the time difference for the one way trip - my point is not that other methods give wrong results - but rather, SR age differences are fundamentally a relative velocity problem that has been misappropriated to GR

But in a GR case, you can come up with a coordinate system where the first remains at a single point in space (traveling 'in time only', not moving in space) and a different coordinate system where the second is the one that remains at a single point in space, both coordinate systems would be equally valid in GR, there'd be no physical basis for saying one "really" moved in space while the other didn't.

Agreed, travel with respect to space is meaningless - I guess it should be clarified as travel with respect to space as measured in the fame which is chosen to be at rest

 

[Al68]

IMO - any analysis that results from dependence upon accelerating frames and the like is going to cloud the reality of what is properly explained by SR - this can be done by using the one way trip and doubling the result...

I agree that if the problem was presented and explained as two one way trips, most of the confusion would be completely eliminated. But the ship is still non-inertial for an (entire) one way trip.

Einstein confused a lot of his followers when he published his 1918 article that explained the clock paradox using a pseudo G field that gives the same answer to the aging difference - but for the wrong reason.

Why would the reason be wrong? The reason for the aging difference is exactly the same in Einstein's 1918 paper as in standard resolutions. The only difference is he uses realistic acceleration (with Earth clock running fast in ship frame) instead of an instantaneous turnaround (earth clock "jumps ahead").

 

[matheinste]

In SR if the twins part and reunite then at least one of them has been moving non-inertially at some time.


Matheinste.

If you have two circular orbiting satellites at the same height traveling in opposite directions and set both clocks to zero as they pass, when they meet again, both clocks will read the same

Assuming that they move symmetrically, that is, they meet again for the first time at the opposite "pole" to where they started as referred to the orbited body,then isn't this a case of both objects traveling non-inertially but both traveling equal spacetime paths and so clocks traveling with them will record the same elapsed proper time.

Matheinste.

 

[Ich]

While Minkowski diagrams provide a powerful abstraction to the physical reality of Special Relativity, I think "clocks running slower" is a fair characterization considering the accelerating twin really does come back younger...

IMHO, time dilation is a convenient memory and calculation aid, but has no (or even less) value for understanding SR. It should be used by those only who understand SR and know how, when, and why time dilation can be applied.
If you confront beginners with that phrase ("clocks run slower"), you inevitably push them down the Lorentzian road, where SR is the oncoming traffic. However, it is common praxis in education, and you see the result here every week with a new thread about the twin paradox.

 

[JesseM]

Agreed, travel with respect to space is meaningless - I guess it should be clarified as travel with respect to space as measured in the fame which is chosen to be at rest

OK, but in the GR case where you're dealing with non-inertial frames either way, depending on the frame you choose it may not be the twin that "travels with respect to space" who ages less, it could be the other twin.

 

[Mike_Fontenot]

Here are a couple of my responses to essentially the same question that was asked on an earlier thread:
______________________________________________________

During the constant-speed legs of the trip, BOTH twins conclude that the other twin is ageing slower. But when the trip is over, they both agree that the stay-at-home twin is older. How is that possible?

It's possible because, during the turnaround, the traveler will conclude that the home twin quickly ages, with very little ageing of the traveler. The home twin concludes that neither of them ages much during the turnaround. When you add up all these segments of ageing, you get the result that the home twin is older (and both twins exactly agree on that).

Years ago, I derived a simple equation (called the "CADO" equation) that explicitly gives the ageing of the home twin during accelerations by the traveler (according to the traveler). The equation is especially easy to use for idealized traveling twin problems with instantaneous speed changes. But it also works for finite accelerations. I've got a detailed example with +-1g accelerations on my webpage:

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

And I've published a paper giving the derivation of the CADO equation:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.
____________________________________________

Here's a brief description of my "CADO" equation:
__________________________________________________ __

Years ago, I derived a simple equation that relates the current ages of the twins, ACCORDING TO EACH TWIN. Over the years, I have found it to be very useful. To save writing, I write "the current age of a distant object", where the "distant object" is the stay-at-home twin, as the "CADO". The CADO has a value for each age t of the traveling twin, written CADO(t). The traveler and the stay-at-home twin come to DIFFERENT conclusions about CADO(t), at any given age t of the traveler. Denote the traveler's conclusion as CADO_T(t), and the stay-at-home twin's conclusion as CADO_H(t). (And in both cases, remember that CADO(t) is the age of the home twin, and t is the age of the traveler).

My simple equation says that

CADO_T(t) = CADO_H(t) - L*v/(c*c),

where

L is their current distance apart, in lightyears,
according to the home twin,

and

v is their current relative speed, in lightyears/year,
according to the home twin. v is positive
when the twins are moving apart.

(Although the dependence is not shown explicitely in the above equation, the quantities L and v are themselves functions of t, the age of the traveler).

The factor (c*c) has value 1 for these units, and is needed only to make the dimensionality correct.

The equation explicitly shows how the home twin's age will change abruptly (according to the traveler, not the home twin), whenever the relative speed changes abruptly.

For example, suppose the home twin believes that she is 40 when the traveler is 20, immediately before he turns around. So CADO_H(20-) = 40. (Denote his age immediately before the turnaround as t = 20-, and immediately after the turnaround as t = 20+.)

Suppose they are 30 ly apart (according to the home twin), and that their relative speed is +0.9 ly/y (i.e., 0.9c), when the traveler's age is 20-. Then the traveler will conclude that the home twin is

CADO_T(20-) = 40 - 0.9*30 = 13

years old immediately before his turnaround. Immediately after his turnaround (assumed here to occur in zero time), their relative speed is -0.9 ly/y. The home twin concludes that their distance apart doesn't change during the turnaround: it's still 30 ly. And the home twin concludes that neither of them ages during the turnaround, so that CADO_H(20+) is still 40.

But according to the traveler,

CADO_T(20+) = 40 - (-0.9)*30 = 67,

so he concludes that his twin ages 54 years during his instantaneous turnaround.

The equation works for arbitrary accelerations, not just the idealized instantaneous speed change assumed above. I've got an example with +-1g accelerations on my web page:

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

The derivation of the equation is given in my paper

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

Mike Fontenot

 

[Al68]

Years ago, I derived a simple equation (called the "CADO" equation) that explicitly gives the ageing of the home twin during accelerations by the traveler (according to the traveler). The equation is especially easy to use for idealized traveling twin problems with instantaneous speed changes. But it also works for finite accelerations. I've got a detailed example with +-1g accelerations on my webpage:

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

And I've published a paper giving the derivation of the CADO equation:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629.

I didn't carefully analyze your web page, but will assume the results match standard lorentz transformations for a co-moving inertial frame for any given time on the accelerated clock. And I did note the phrase "the bizarre behavior of the CADO, for an accelerating observer, must be regarded as being fully meaningful and real."

I would also note that if the traveler is accelerating in a direction away from earth, the results could be even more bizarre. For example, from the accelerated traveler's point of view, the funeral of a person on Earth could be followed by that person's wedding.

I would have to say that people rising from the grave would qualify as not "fully meaningful and real" by any reasonable definition.

Note that I'm not disputing that result. We could obtain the same result by performing lorentz transformations directly during inertial motion before and after acceleration, if the ship were distant from Earth and the acceleration was away from earth. But most would disagree that the result represented reality in any sense other than assigning coordinates.

 

[stevmg]

George Jones and Ich are correct. Attached is a picture I downloaded from among the plethora of explanations on the web. You can see that on the way out, the twin on the rocket sees two years pass on Earth, while four years pass for him. On the way back, he sees 8 years pass on Earth while four years pass for him. So when he returns, 10 years have passed on Earth while 8 years have passed for him, so the twin on the rocket is younger.

JesseM gave an example using SR why the "moving" twin ages less quickly than the earthbound twin. He does it both ways - using the Earth as the inertial frame and then the spaceship as the inertial frame. In both ways the Earth twin aged more than the traveling twin. Here is the post:

https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

 

[stevmg]

George Jones and Ich are correct. Attached is a picture I downloaded from among the plethora of explanations on the web. You can see that on the way out, the twin on the rocket sees two years pass on Earth, while four years pass for him. On the way back, he sees 8 years pass on Earth while four years pass for him. So when he returns, 10 years have passed on Earth while 8 years have passed for him, so the twin on the rocket is younger.

How do I post a picture like you did on this post
https://www.physicsforums.com/showpost.php?p=2753642&postcount=13

H-E-L-P

Steve G
Melbourne FL

 

[DrGreg]

How do I post a picture like you did on this post
https://www.physicsforums.com/showpost.php?p=2753642&postcount=13

While you are editing your post (in the "advanced" editor -- which you get automatically if you use the QUOTE button) press the "Manage Attachments" button not far underneath.

 

[Mike_Fontenot]

[...]
I would also note that if the traveler is accelerating in a direction away from earth, the results could be even more bizarre. For example, from the accelerated traveler's point of view, the funeral of a person on Earth could be followed by that person's wedding.

Yes, that behavior is easy to see, directly from the form of the CADO equation itself, when there are instantanious changes in the velocity v, such that v gets more positive, or less negative (v is positive when their separation is increasing). And this effect isn't limited to instantaneous velocity changes...the voyage detailed in the web page includes segments where the traveler must conclude that the home twin gets younger, with only 1g accelerations involved.

But the web page, and my paper, also make it clear that this phenomenon in no way influences the perception, by the home twin, of the normal progression of time. It is somewhat analogous to your ability to reverse the direction of a movie projector...your action doesn't bother the actors, or make any changes to the frames of the film.

I would have to say that people rising from the grave would qualify as not "fully meaningful and real" by any reasonable definition.

The critical point to understand (which is elaborated in detail in the paper), is that the traveler CAN ADOPT NO OTHER CONCLUSION, IF HE WISHES TO AVOID CONTRADICTING HIS OWN ELEMENTARY (AND CORRECT) MEASUREMENTS. It is in that sense that I use the phrase "real and meaningful". If elementary, correct measurements are not "real and meaningful", how can anyone ever hope to do any physics?

Mike Fontenot