The twin paradox

[michel123456]

No. Moving clocks run slow. Even though the frequency of the arriving signal increases when the sources are moving toward each other, the moving clock still runs slow

 

http://www.einsteins-theory-of-relativity-4engineers.com/twin-paradox-2.html

The twin paradox again.

The paradox is not that the twins have a different age at the end, the paradox is that both twins should observe the other aging less because motion is relative.

 

In all examples the one who moves observes the one at rest aging more than him (instead of less) and that does not solve the paradox, that IS the paradox.

WE decided right from the beginning which one moves and which one is at rest and that is contrary to the relativity of motion.

 

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As a side-remark, in this last diagram from your link, Pam traveled 7,5LY away in 4 years! (that is almost twice c)

Of course in 4 of her years and not in 4 "earth's years" because she was supposed to travel at 0,6C..

.

[swansont]

The twin paradox again.

I am happy to move this to a new thread if you want to discuss the details of the twin paradox.

[michel123456]

I hope the minus rep does not come from an error in my post.

[michel123456]

--------------

Error(s)? please correct me:

 

here the diagram again

 

 

Doesn't that mean that Pam "made it" to a point A lying 7,5 LY away in only 4 of her years?

Or do I read wrong?

Edited April 30, 2013 by michel123456

[swansont]

michel123456, on 30 Apr 2013 - 02:50, said:

 

Doesn't that mean that Pam "made it" to a point A lying 7,5 LY away in only 4 of her years?

Or do I read wrong?

It's not 7.5 LY in her frame of reference.

 

The twin paradox again.

The paradox is not that the twins have a different age at the end, the paradox is that both twins should observe the other aging less because motion is relative.

They do see each other aging less, up until the point that someone accelerates. And someone has to accelerate in order to for them to reunite.

 

 

[tar]

swansont,

 

Perhaps it should move to twins paradox, but from my point of view, it has more to do with philosophy in terms of the perspective that we truely hold. If we have a here and now, that we consider the moment we are in, we are bound by that moment, in that we have no "other" perspective, from which we can witness, or calculate, both twins at once.

 

We can either put ourselves in the shoes of the other, in which case the relative velocities, and the ticking of time are exactly the same in both cases, or we can stay with one or the other for the whole duration and witness the red and blue shift of the other, the slow arrival of messages on the way out trip, the rapid arrival on the way back, and the reunion years later, when both would report the exact same experience of the other.

 

There is no vantage point, outside of time and space, from which we can view both at once.

There is no vantage point, outside of time and space, that can track both, simultaneously, receiving instant information about position and the ticking of each clock.

Both clocks tick, and never leave the others reality. It all has to add back, with no incongruities, in both cases.

 

Regards, TAR2

Consider a third triplet with a ship and a velocity and a course that would keep him exactly the same distance from both of the others. He would always have both the others in view at all times, and be the middle man.

 

Then consider a fourth, inbetween the first and third, and a fifth inbetween the first and forth...and so on. Hypothetically you could make an argument, that nobody's clock ever ticked slower, and they all just keep ticking along.

Or consider both twins counting the pulses of a distant pulsar. When the twin returned and they gave each other the count, it would be exactly the same.

[michel123456]

It's not 7.5 LY in her frame of reference.

Yes, I know that. But anyway, traveling at 0,6c she reached a planet that we observe at 7,5 LY from us. It means that an object that travel at velocity less than c can reach an object very (very) far away. I wonder where she could go in 4 of her years traveling at 0,99c. (edit) I even wonder whether at c it isn't a trivial solution to the horizon problem.

They do see each other aging less, up until the point that someone accelerates. And someone has to accelerate in order to for them to reunite.

Oh, acceleration matters then. I am happy to hear that.

I am sure you realize that from the exact beginning (at point o) , when paths diverge, acceleration is in action already. You cannot change FOR without acceleration. Each change of angle in this diagram represent acceleration.

 

 

Edited April 30, 2013 by michel123456

[swansont]

Yes, I know that. But anyway, traveling at 0,6c she reached a planet that we observe at 7,5 LY from us. It means that an object that travel at velocity less than c can reach an object very (very) far away. I wonder where she could go in 4 of her years traveling at 0,99c. (edit) I even wonder whether at c it isn't a trivial solution to the horizon problem.

 

There was a recent thread on mixing frames where this was discussed.

 

Oh, acceleration matters then. I am happy to hear that.

I am sure you realize that from the exact beginning (at point o) , when paths diverge, acceleration is in action already. You cannot change FOR without acceleration. Each change of angle in this diagram represent acceleration.

 

The initial acceleration happened before the initial synchronization of the clock, i.e. at the beginning of the measurement they are already in different frames.

[michel123456]

So are we.

(...) And someone has to accelerate in order to for them to reunite.

O.K.

 

Does that mean that all the "paradox" lies in the turning point?

[swansont]

The paradox appears because one is trying to apply a model that applies to inertial systems to a non-inertial event. When one properly analyzes the experiment, there is no paradox.

[michel123456]

The paradox appears because one is trying to apply a model that applies to inertial systems to a non-inertial event. When one properly analyzes the experiment, there is no paradox.

Yes.

 

Do you have a link to a proper analysis?

[imatfaal]

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

[tar]

But what about the pulsar. Would both twins not count the same amount of pulses?

[md65536]

We can either put ourselves in the shoes of the other, in which case the relative velocities, and the ticking of time are exactly the same in both cases, or we can stay with one or the other for the whole duration and witness the red and blue shift of the other, the slow arrival of messages on the way out trip, the rapid arrival on the way back, and the reunion years later, when both would report the exact same experience of the other.

No, they don't experience the same thing. In the typical twin setup, the twin that turns around experiences its change of direction halfway through the round trip. The inertial twin only experiences or observes this much closer to the end of the round trip, due to delay of light.

Consider a third triplet with a ship and a velocity and a course that would keep him exactly the same distance from both of the others. He would always have both the others in view at all times, and be the middle man.

 

Then consider a fourth, inbetween the first and third, and a fifth inbetween the first and forth...and so on. Hypothetically you could make an argument, that nobody's clock ever ticked slower, and they all just keep ticking along.

You could do this by having N observers all travel for a fixed time, then return for an equal amount of time, where the N+1'th observer travels at half the velocity of the N'th. Then you could easily calculate their relative aging, all in agreement with SR (where your hypothetical argument is wrong).

 

The 21st observer would travel at about a millionth the speed of the 1st observer. If the first is very close to c, the 21st would be close to 80km/hr, and would experience negligible relativistic effects.

 

Edit: Even the 7th traveling observer at about 1.5% c would measure less than half a second difference per hour relative to the inertial observer. Relativistic effects diminish quickly as velocity decreases.

Or consider both twins counting the pulses of a distant pulsar. When the twin returned and they gave each other the count, it would be exactly the same.

Yes. They'd also count an identical number of observed ticks of any clock. They'd both agree that Twin A's clock ticked a certain number t, and that Twin B's ticked a certain number t', but t and t' are not equal in the typical twin experiment.

WE decided right from the beginning which one moves and which one is at rest and that is contrary to the relativity of motion.

Check out http://en.wikipedia.org/wiki/Twin_paradox#Viewpoint_of_the_traveling_twin

for a description of using the equivalence principle to consider the traveling twin to be stationary. This is done in terms of an equivalent gravitational field (the same twin experiences proper acceleration either way).

 

Then continue reading the next section http://en.wikipedia.org/wiki/Twin_paradox#Difference_in_elapsed_time_as_a_result_of_differences_in_twins.27_spacetime_paths to see the calculations involved in considering acceleration over a non-negligible time. In my opinion this is a more complex situation that does nothing to help understand the twin paradox, but does show that the simplest cases are consistent with more complex ones in SR.

 

Notice that there are 6 phases in this version, 4 of which involve acceleration, but that the 2 other phases are exactly the same as the simple case where acceleration is neglected. In other words, the acceleration phases do not affect what happens during the constant v phases, and the twin paradox is readily apparent with only those constant v phases.

Edited May 1, 2013 by md65536

[michel123456]

It's not 7.5 LY in her frame of reference.

Then the labeling 7.5LY counts only for Earth's FOR. For the traveler, the labeling should be much less (2.4 LY). That corresponds to a clockwise rotation of the X axis. Then i suppose the Y axis for the traveler should also rotate clockwise. And then signals as observed in the traveler's FOR should not be represented at 45 degrees in the diagram.

 

 

Edited May 1, 2013 by michel123456

[Iggy]

Then the labeling 7.5LY counts only for Earth's FOR. For the traveler, the labeling should be much less (2.4 LY). That corresponds to a clockwise rotation of the X axis. Then i suppose the Y axis for the traveler should also rotate clockwise. And then signals as observed in the traveler's FOR should not be represented at 45 degrees in the diagram.

 

 

 

You have to use the Lorentz transforms. It isn't a straightforward rotation. Here are all three frames.

 

The arrow you have pointing down corresponds to the arrows I've marked "present instant". It doesn't rotate like you would intuitively think.

 

 

 

From red's perspective on the first leg of the trip:

 

 

 

 

 

From red's perspective on the second leg of the trip (after red accelerates):

 

 

 

 

 

From blue's perspective (blue doesn't accelerate -- only has one frame):

 

 

 

 

 

You can use the Lorentz transforms to calculate where E1 and E2 are in any frame.

 

this page and this image will give you an idea of why a simple Euclidean rotation doesn't work.

Edited May 1, 2013 by Iggy

[tar]

michel123456,

 

Not sure I am wrong about the two experiencing the same thing. If red is considered stationary, blue makes the turnaround, and red does not notice this til later.

 

If t and t' are not equal in the twin experiment, then similarly t' and t are not equal. Being that they are unequal in the symetrically opposing ways during the whole adventure, and they started out equal and ended up equal, there would be no reason to call one twin older than the other. And being that they both experienced the same number of ticks of the pulsar, I would argue they are both the same age at the end of the adventure.

 

Regards, TAR2

Can not red be exchanged with blue and blue with red in any of the above diagrams?

[michel123456]

You have to use the Lorentz transforms. It isn't a straightforward rotation. Here are all three frames.

 

The arrow you have pointing down corresponds to the arrows I've marked "present instant". It doesn't rotate like you would intuitively think.

 

 

 

From red's perspective on the first leg of the trip:

 

 

 

From red's perspective on the second leg of the trip (after red accelerates):

 

 

 

 

 

From blue's perspective (blue doesn't accelerate -- only has one frame):

 

 

 

You can use the Lorentz transforms to calculate where E1 and E2 are in any frame.

 

this page and this image will give you an idea of why a simple Euclidean rotation doesn't work.

What I say is that none of your diagrams show the distance that the red has observed having traveled. From the red FOR, he didn't travel 7,5 LY (as pointed by Swansont).

[Iggy]

What I say is that none of your diagrams show the distance that the red has observed having traveled. From the red FOR, he didn't travel 7,5 LY (as pointed by Swansont).

 

red traveled zero lightyears on the first leg of the trip in red's frame of reference for the first leg of the trip. This is red's perspective on the first leg of the trip:

 

 

 

 

v = 0c, tau = 4

 

for 4 of red's years, red traveled zero distance from himself. You may be wondering how far red travels according to blue? That was 3 lightyears in 5 years. The last diagram.

 

Otherwise you are asking how far red traveled in the frame of one leg of the trip according to the time he measures during a different leg of the trip, and there you're mixing frames. The answer is 7.5 lightyears in 4 years. For example, in the second leg frame he traveled 7.5 lightyears during the first leg, and in the first leg frame he measured 4 years during the first leg. The distance comes from one frame and the time from another. So, it's mixing frames.

 

It is possible to combine two spacetime diagrams into one image. The link I gave explains how to do it. It ends up looking like this:

 

I could combine one leg of red and the whole of blue into a single image if you like, but times from one frame can't be mixed with distances from another into a single coordinate system.

 

 

EDIT:

 

I had to edit the diagrams. The angles were slightly off on the present instants. Appalling trig skills

Edited May 1, 2013 by Iggy

[michel123456]

(...)

Otherwise you are asking how far red traveled in the frame of one leg of the trip according to the time he measures during a different leg of the trip, and there you're mixing frames. The answer is 7.5 lightyears in 4 years. For example, in the second leg frame he traveled 7.5 lightyears during the first leg, and in the first leg frame he measured 4 years during the first leg. The distance comes from one frame and the time from another. So, it's mixing frames. (...)

Exactly, it is mixing frames, like in the following graphs.

 

 

 

[Iggy]

Exactly, it is mixing frames, like in the following graphs.

 

<...>

 

I don't know what that means. I don't understand the markings on the second image, but each image has the correct world lines and the correctly labeled axes to show a single frame of reference.

 

Perhaps this would help...

 

 

 

 

At t = 4 red turns toward blue (red accelerates). Pretend that red was with someone who did not turn (call him black). Black kept going on red's initial course -- staying at rest in this frame of reference. According to black's proper time (according to black's clock) red would meet up with blue at t = 12.5. He would figure that red traveled 7.5 lightyears (according to black's ruler) in 8.5 minutes (according to black's clock). He would notice (or he could calculate using the time dilation formula) that red's clock advanced 4 years between the turn and the meet up with blue.

 

If that doesn't help then perhaps you can explain in detail what the issue is.

Edited May 1, 2013 by Iggy

[md65536]

There seems to be two separate conversations here, based on two separate ideas brought up by OP...

Not sure I am wrong about the two experiencing the same thing. If red is considered stationary, blue makes the turnaround, and red does not notice this til later.

If t and t' are not equal in the twin experiment, then similarly t' and t are not equal. Being that they are unequal in the symetrically opposing ways during the whole adventure, and they started out equal and ended up equal, there would be no reason to call one twin older than the other. And being that they both experienced the same number of ticks of the pulsar, I would argue they are both the same age at the end of the adventure.

Edit: I should have used [latex]\tau[/latex] instead of t' to make it clear I'm talking about proper time, on which all observers agree. Consistency would make this less confusing. :S

Suppose that t=2 and t'=1. Then t and t' are not equal the same way that t' and t are not equal, but switching their order does not change their values. They're not symmetrical. One is definitively older.

Say that t=2 and t'=1 and the pulsar ticked 10 times.
Then the inertial twin measured 2 ticks of its own clock, 1 tick of its twin's, and 10 ticks of the pulsars. In this frame, the pulsar appeared to tick at a rate of 5 ticks per tick of proper time (measured on a local clock).

The traveling twin measured 1 ticks of its own clock, 2 ticks of the other twin's, and 10 ticks of the pulsars. In this frame, the pulsar appeared to tick on average at a rate of 10 ticks per tick of local time.

The twins age according to local clocks, which tick at different rates relative to the other twin's, so we say the pulsar ticks at different rates in different frames, not that the twins age the same according to some universal clock.

You can also replace "tick" with "second" (or "year") if it's more intuitive.

You could do this by having N observers all travel for a fixed time, then return for an equal amount of time, where the N+1'th observer travels at half the velocity of the N'th. Then you could easily calculate their relative aging, all in agreement with SR (where your hypothetical argument is wrong).

No one seemed to care but I was sloppy here and feel compelled to correct myself.

Each of the observers will age a different amount in this experiment, and so there's no sense in having them travel at different speeds for the same fixed time, unless we measure that time in the inertial (stay at home) twin's frame. By that observer's clock, all others would turn around at a fixed time, ie. simultaneously. For the other observers, the turn-around wouldn't be simultaneous; each traveler would turn at a different time. So in the other observers' frames, they wouldn't measure being halfway between Earth and the next faster observer. As usual everything's consistent in SR; the observers wouldn't measure being half the speed as the next, because if an observer's speed relative to Earth is v, then the next faster observer will have a speed of a composition of v and v (Edit: wrong once again!) say vb relative to this observer, where the composition of v and vb is 2*v, so vb doesn't equal v, (vb should be greater)... we must use relativistic composition of velocities.

tl;dr What I described is measured only in the inertial observer's frame. It seems like to do what you described you would always have to choose a single observer for which the symmetry holds.

Edited May 1, 2013 by md65536

[phyti]

--------------

Error(s)? please correct me:

 

here the diagram again

 

 

Doesn't that mean that Pam "made it" to a point A lying 7,5 LY away in only 4 of her years?

Or do I read wrong?

The earth distance= 7.5 ly. Speed =.6c. Expected travel time= 7.5/.6= 12.5 y.

gamma =1.25. Time dilation for Pam = 12.5/1.25=10 y. Since her clock and her sense of time indicate 10y, she concludes the interval of 7.5 ly has contracted to 7.5/1.25= 6 ly.

 

The paradox is based on a false assumption that each twin sees the others time dilation. They do not. What they see is relativistic doppler effects, because clocks are frequencies.That's why the effects are reciprocal, and depend on converging or diverging speeds.

[Iggy]

 

You could do this by having N observers all travel for a fixed time, then return for an equal amount of time, where the N+1'th observer travels at half the velocity of the N'th. Then you could easily calculate their relative aging, all in agreement with SR (where your hypothetical argument is wrong).

 

The 21st observer would travel at about a millionth the speed of the 1st observer. If the first is very close to c, the 21st would be close to 80km/hr, and would experience negligible relativistic effects.

No one seemed to care but I was sloppy here and feel compelled to correct myself.

 

Each of the observers will age a different amount in this experiment, and so there's no sense in having them travel at different speeds for the same fixed time, unless we measure that time in the inertial (stay at home) twin's frame.

 

You could have pulled that off

 

"All travel for a fixed time" sounded for sure like you meant by earth's clock. Made perfect sense to me that way. You should have run with it

 

 

 

 

---

 

 

 

 

The earth distance= 7.5 ly. Speed =.6c. Expected travel time= 7.5/.6= 12.5 y.

gamma =1.25. Time dilation for Pam = 12.5/1.25=10 y. Since her clock and her sense of time indicate 10y, she concludes the interval of 7.5 ly has contracted to 7.5/1.25= 6 ly.

 

I think Pam is the red line in that pic. [math]v = \frac{7.5}{8.5}c[/math]... [math]\gamma = \frac{17}{8}c[/math]

Edited May 1, 2013 by Iggy

[tar]

Well,

 

 

OK, the equations work out, and agree with experiment, Still not sure though what proper time is.

 

Couple reality questions on the twin thought experiment though. If red makes a turnaround, would that not mean she would have to expend a great deal of energy to "brake" back into our FOR for that moment where she is no longer traveling at a v away, and is then traveling at a v toward us. For that moment (which could hardly be in reality a short moment) she would have to return to our frame of reference (blue's frame) and then through whatever wonderful technology, regain her .6C velocity back in blue's direction.

 

Being that the given in the experiment is that she is already traveling at .6C as she passes the Earth at the start of the experiment, there is an assumption that it took her some time (and a great deal of energy) to accelerate to that speed. The same conditions of reality would hold, during the turnaround, and she would first have to expend the energy to slow to a stop (back in blue's FOR) and then build up the tremendous speed again, for the return trip. Seems the only way to avoid a return to blue's frame of reference would be to travel in a circle, in which case, the diagrams would have to be refigured, and she would be in a constant state of acceleration, plastered against the wall of her vessel, that was opposite the center of the circle.

 

Also, not included in the thought experiment, is any effect that her bow may be experiencing, different from her stern. The light from the rest of the galaxy is extremely red shifted hitting her stern and extremely blue shifted hitting her bow. Off at right angles to her motion things would look about right, but although light is hitting both her bow and stern at C, the frequency is likely doubled in front and halved in back. This unbalance of energy hitting the craft, should have a slowing effect. As well as imbalance of heat. She would KNOW which way she was headed, and could not fool herself into thinking she was stationary.

 

Regards, TAR2