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Twin paradox math problem


md65536
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The twin paradox doesn't require an inertial twin. Suppose two twins left Earth at the same time and returned at the same time, each traveling a different but constant speed relative to Earth. Whom does your intuition say traveled a longer distance, the twin who ages more, or less?

Don't read the following puzzle if you want to think about it first.

Spoiler

Two twins leave Earth at the same time and return at the same time, each traveling at a respective constant speed relative to Earth. Each measures the distance she traveled, and the time it takes. They find that one aged twice as much as the other, and they traveled the same distance. What were their speeds?

 

 

Bonus question: What speed should a twin go to maximize the distance it travels, for such a round trip?

 

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4 hours ago, md65536 said:

The twin paradox doesn't require an inertial twin.

Yes, it does. The twin paradox has a particular form of how it is framed. If you present a different scenario, it’s not the twin paradox, it’s something else.

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all reference frames in your scenario are inertial frames the Earth at constant velocity and the two twins. Your scenario has three reference frames involved so I guess you could call it the triplet problem lol. However as Swansont noted you require valid reference frames.

edit I should note that during the turnaround acceleration the two twins would no longer be at constant velocity

Edited by Mordred
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55 minutes ago, Mordred said:

all reference frames in your scenario are inertial frames the Earth at constant velocity and the two twins. Your scenario has three reference frames involved so I guess you could call it the triplet problem lol. However as Swansont noted you require valid reference frames.

edit I should note that during the turnaround acceleration the two twins would no longer be at constant velocity

He said constant speed. Orbiting the earth is a possible approach to the problem; the general framing does not eliminate gravitational effects, and earth rotation can be used (since westbound clocks speed up). Whether or not that was the intent is unknown.

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6 hours ago, swansont said:

Yes, it does. The twin paradox has a particular form of how it is framed. If you present a different scenario, it’s not the twin paradox, it’s something else.

I disagree. If the "stay at home twin" moves around, even at high speed (but slower enough than the other twin), the different aging still happens. Ceasing to be inertial doesn't make the effect go away.

Further, if any two worldlines intersect at two separate events, the proper time along the two paths will generally be different, and there is differential aging. There are variations to the twin paradox, but if "twin paradox" means only the case where one twin is inertial, then instead of "twin paradox" I mean the seemingly paradoxical different aging of twins that depart and meet again, having traveled at different speeds, where each measures the other's clock as ticking slower than their own when erroneously neglecting relativity of simultaneity.

1 hour ago, swansont said:

He said constant speed. Orbiting the earth is a possible approach to the problem; the general framing does not eliminate gravitational effects, and earth rotation can be used (since westbound clocks speed up). Whether or not that was the intent is unknown.

No, that's not the intent. The two twins could each travel in a giant circle in empty space, or they could leave in a straight line and return in a straight line with a sharp turn around or even come to a stop if they spend negligible time accelerating, their direction doesn't matter, only speed, and a path that will let them meet again.

2 hours ago, Mordred said:

all reference frames in your scenario are inertial frames the Earth at constant velocity and the two twins. Your scenario has three reference frames involved so I guess you could call it the triplet problem lol. However as Swansont noted you require valid reference frames.

edit I should note that during the turnaround acceleration the two twins would no longer be at constant velocity

Well of course they all have valid reference frames. If you want to do it with inertial frames, each traveling twin would require a minimum of two.

Earth is used as a reference for the requirement that their speeds are constant, because their speeds are not constant relative to each other all the time, in other reference frames. I choose that to make the description and calculations as simple as possible. You could have the twins leave from one event and meet up at some other event anywhere, it wouldn't have to be at the same place... except of course that you could join the two events with an inertial observer, in whose frame the events are at the same place. However, no physical observer actually needs to inertially intersect the two events in order for the twins' different aging to occur.

23 minutes ago, Mordred said:

Considering orbits are either circular or elliptical which involves continuous direction changes that could get quite complex. 

Direction doesn't matter, the different aging can be calculated from time and relative speed alone.

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8 minutes ago, md65536 said:

I disagree. If the "stay at home twin" moves around, even at high speed (but slower enough than the other twin), the different aging still happens. Ceasing to be inertial doesn't make the effect go away.

Further, if any two worldlines intersect at two separate events, the proper time along the two paths will generally be different, and there is differential aging. There are variations to the twin paradox, but if "twin paradox" means only the case where one twin is inertial, then instead of "twin paradox" I mean the seemingly paradoxical different aging of twins that depart and meet again, having traveled at different speeds, where each measures the other's clock as ticking slower than their own when erroneously neglecting relativity of simultaneity.

No, that's not the intent. The two twins could each travel in a giant circle in empty space, or they could leave in a straight line and return in a straight line with a sharp turn around or even come to a stop if they spend negligible time accelerating, their direction doesn't matter, only speed, and a path that will let them meet again..

Ok let's get rid of this ceasing to be inertial all frames under GR are inertial. Secondly the sharp turnaround is a hyperbolic rotation which isn't constant velocity.

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12 minutes ago, Mordred said:

Ok let's get rid of this ceasing to be inertial all frames under GR are inertial. Secondly the sharp turnaround is a hyperbolic rotation which isn't constant velocity.

I never said constant velocity, I said constant speed *; swansont already pointed that out. GR is not needed for this, the basic twin paradox is an SR problem and so is this variation. You can assume flat spacetime and ignore any complications not mentioned in the problem. I didn't think it was such a difficult problem. Assume the simplest paths that won't change the answer. I think the simplest is probably "straight line away, straight line back, negligible turn-around time".

 

* That wasn't meant as a trick to the problem, just a simplification of the description. I usually make that simplification automatically because it's burned into my brain that the direction doesn't matter. I was about to say, "I should have said constant velocity outbound and constant with the same magnitude back", but on second thought, I think it's more important to not get stuck thinking that a straight-line path somehow matters.

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The turnaround is useful in the twin paradox as part of its solution. It shows the symmetry breaking aspects of the Lorentz transforms.

Recall in SR you can arbitrarily set any observer in any inertial frame to zero.  

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3 minutes ago, Mordred said:

The turnaround is useful in the twin paradox as part of its solution. It shows the symmetry breaking aspects of the Lorentz transforms.

Recall in SR you can arbitrarily set any observer in any inertial frame to zero.  

Then assume a negligible turnaround time. Or a giant circle, all turn around time (but constant speed), the answer's the same.

This problem doesn't involve calculations involving acceleration. The maths don't get any more complicated than the Lorentz factor.

2 hours ago, Mordred said:

Your scenario has three reference frames involved so I guess you could call it the triplet problem lol.

Treating it like a triplet problem is probably the easiest way to solve it.

11 hours ago, md65536 said:

Each measures the distance she traveled

Is this too ambiguous?

If a twin makes a round-trip journey that is 2 light years as measured on Earth, with constant gamma=2, is it unambiguous to say that the twin measured traveling a distance of one light year? Or do you have to technically include "one light year relative to Earth"? Or is this really mixing frames, and if using the twin's frame to measure distance, it doesn't travel at all, you'd have to say that the Earth traveled one light year?

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You can simplify it by treating the Earth frame as the rest frame under Lorentz then run a comparison to each twin relative to the Earth frame. Have both twins run the same path along the x axis, they both have the same turnaround but at different instantaneous velocities during turnaround due to both twins having different velocities for the main flight paths.

Yes each reference frame will have its own time dilation factor but we only need the ratios of differences to the Earth reference frame of the two twins.

The Earth triplet being the only reference frame that doesn't undergo acceleration.

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2 hours ago, md65536 said:

I disagree. If the "stay at home twin" moves around, even at high speed (but slower enough than the other twin), the different aging still happens. Ceasing to be inertial doesn't make the effect go away.

You misunderstand. The twin paradox is a name given to a particular problem. This is an issue of history.

It has nothing to do with whether or not you can come up with other physics problems, though I might disagree they are paradoxes

2 hours ago, md65536 said:

No, that's not the intent. The two twins could each travel in a giant circle in empty space, or they could leave in a straight line and return in a straight line with a sharp turn around or even come to a stop if they spend negligible time accelerating, their direction doesn't matter, only speed, and a path that will let them meet again.

Well, you specified earth, which isn’t an inertial frame, and you went out of your way to point out that non-inertial frames would be involved.

 

 

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1 hour ago, Mordred said:

You can simplify it by treating the Earth frame as the rest frame under Lorentz then run a comparison to each twin relative to the Earth frame. Have both twins run the same path along the x axis, they both have the same turnaround but at different instantaneous velocities during turnaround due to both twins having different velocities for the main flight paths.

Yes each reference frame will have its own time dilation factor but we only need the ratios of differences to the Earth reference frame of the two twins.

The Earth triplet being the only reference frame that doesn't undergo acceleration.

Sounds good. Yes, I was assuming the Earth approximates an inertial frame, as with the basic twin paradox setup.

The twins don't have the same turnaround point. They travel at different speeds, but start together and end together. They travel different distances as measured from Earth. Each twin measures the distance that they travel as length contracted by a respective Lorentz factor, and I've contrived these length-contracted distances to be the same, ie. the distances they each separately measure.

They would turn around simultaneously according to the Earth frame.

Definitely running them on the same x-axis is a good idea. Over the whole trip, the faster twin ages half the slower one, but I don't think the Lorentz factor of one twin moving relative to the other is equal to 2 during the outbound or inbound leg. Each twin measures the other's clock as slower during the inertial sections. I think that each twin, just before its turnaround, says the other twin has not yet turned around, and just after turnaround it says the other twin turned around some time ago. It seems that would be the case in any situation where the twins turn around simultaneously in the Earth frame, whatever their speeds.

 

1 hour ago, swansont said:

Well, you specified earth, which isn’t an inertial frame, and you went out of your way to point out that non-inertial frames would be involved.

"In physics, the twin paradox is a thought experiment in special relativity involving identical twins, one of whom makes a journey into space in a high-speed rocket and returns home to find that the twin who remained on Earth has aged more." Do you agree? Are you going to argue both that the twin paradox requires an inertial twin, and that the Earth twin is not an inertial twin?

I don't need to go out of my way to point out that non-inertial frames are involved. If any two twins separate and then re-unite, in SR, at least one of the twins is non-inertial.

1 hour ago, swansont said:

 though I might disagree they are paradoxes
 

We can all agree they're not paradoxes, I should have put the word in quotes. I assumed understanding the resolution of the "paradox", my puzzle isn't intended be a new paradox that seems to not make sense.

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md65536;

In the 1905 paper, clock B moved away at constant speed from clock A and returned, with clock B recording less time than clock A, even though they were synchronized initially. The thought experiment was done in an SR environment.

Two clocks A and B are sent away from location U, in the direction x, at constant speeds a and b respectively. They reverse direction in zero time simultaneously relative to U and return at constant speeds a and b respectively. Each clock does not experience acceleration for the entire trip.
This is irrelevant in SR, since gamma (Lorentz factor) only contains 1st order terms v/c, and nothing for acceleration.


For political correctness, two clocks can be used, one outbound and one inbound, exchanging data at reversal.

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Yes, that all sounds good. I definitely like the idea of switching clocks at the turnaround, because then you're dealing with their different frames of reference instead of thinking of something "happening" to the twin or single clock. Political correctness I suppose is used to avoid arguments about the role of acceleration in the setup. But, politics can be avoided too if we just say "assume that the clock postulate holds", which is a given. http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

However, no one has yet taken a stab at the puzzle, by calculating the speeds of the two twins or clocks, given that they both measured (in their respective frames) having traveled the same distance relative to Earth, and that one aged twice as much as the other over the trip.

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md65536;

However, no one has yet taken a stab at the puzzle, by calculating the speeds of the two twins or clocks, given that they both measured (in their respective frames) having traveled the same distance relative to Earth, and that one aged twice as much as the other over the trip.

In the graphic, the arc of radius 1 represents one unit of ct (sec, yr., etc).

The red curve is the perception of distance traveled by the observer based on his speed and local time In the interesting scenario of an anaut moving outward from the center of a spherical system of objects, over a range of speeds, the sphere has a maximum radius at v/c=.707c, and two different speeds for each calculated distance.

Relative to your puzzle, the green curve is 1/2 the local time for any given speed.

Any pair satisfies 'equal distance'. Only one satisfies the '2:1 aging'.

A horizontal line between the intersection of the green line and blue diagonal (light speed) projected downward to the red should provide the answer.

A speed is .4472 and B speed is .8944.

What do you think?

295644793_universeradius.gif.28b9c0916dc6a572800c8d5041825d60.gif

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Perfect! You've also answered the bonus question.

d=v*t' (or v*tau) measured in the traveler's frame. Half the proper time during the trip means double the speed, to keep the distances the same. Half the proper time also means double gamma. So 2v_a = v_b, and 2/sqrt(1-v_a^2) = 1/sqrt(1-v_b^2), solve and you get 1/sqrt(5), 2/sqrt(5), which is what you got.

The red curve shows something I'd hoped might be intuitive.

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  • 1 month later...
On 9/1/2019 at 8:48 AM, md65536 said:

The twin paradox doesn't require an inertial twin. Suppose two twins left Earth at the same time and returned at the same time, each traveling a different but constant speed relative to Earth. Whom does your intuition say traveled a longer distance, the twin who ages more, or less?

Don't read the following puzzle if you want to think about it first.

  Reveal hidden contents

Two twins leave Earth at the same time and return at the same time, each traveling at a respective constant speed relative to Earth. Each measures the distance she traveled, and the time it takes. They find that one aged twice as much as the other, and they traveled the same distance. What were their speeds?

 

 

Bonus question: What speed should a twin go to maximize the distance it travels, for such a round trip?

 

 

On 9/4/2019 at 7:42 PM, md65536 said:

However, no one has yet taken a stab at the puzzle, by calculating the speeds of the two twins or clocks, given that they both measured (in their respective frames) having traveled the same distance relative to Earth, and that one aged twice as much as the other over the trip.

 

So they both left at the same time, traveled the same distance at different speeds but both returned at the same time?

 

What am I missing?

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1 hour ago, between3and26characterslon said:

So they both left at the same time, traveled the same distance at different speeds but both returned at the same time?

 

What am I missing?

You left out that they measure the distance in their respective frames, which are different.

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  • 1 month later...
On 9/2/2019 at 12:34 AM, Mordred said:

Considering orbits are either circular or elliptical which involves continuous direction changes that could get quite complex. 

Not in this case.  The time will be computed by

[math]\tau = \int_{t_0}^{t_1} \sqrt{1-\frac{|v(t)|^2}{c^2}} dt,[/math]

and if |v| is constant, then it does not matter how the trajectory itself is curved, you get 

[math]\tau = \sqrt{1-\frac{|v|^2}{c^2}} (t_1-t_0)[/math]

without any necessity to care about frames and so on. 

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Try that using four velocity there is a methodology to calculate the propertime with four velocity in this case and in the case of the turnaround twin. 

 Though you can get a first order approximation from your proposal.

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16 hours ago, Mordred said:

Try that using four velocity there is a methodology to calculate the propertime with four velocity in this case and in the case of the turnaround twin. 

 Though you can get a first order approximation from your proposal.

It gives an exact solution in SR.

The problem's been solved. The proper time of the traveling twins can be found with time dilation alone. You're still suggesting complications, but no solutions, and a simple solution is already given. What don't you understand about it?

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Google hyperbolic rotation of the the Minkowskii group. You should get a hyperbolic curve of [math]\frac {c^4}{g^2}[/math] if you use the proper four momentum

As mentioned acceleration causes rapidity of the the Minkowskii  group. This can be shown through a rotation or boost. A boost is a type of rotation.

 That is part of the solution for resolving the twin paradox. The twin that undergoes rapidity is the one that ages differently and breaks the symmetry of the Lorentz transforms.

You do not seem to understand that the Minkowskii group relies upon the symmetry relation 

[math]\mu \cdot \nu=\nu \cdot \nu [/math] this is under constant velocity. When you undergo acceleration you perform a rotation of that relation 

Ok lets do the standard twin paradox

first start with the four velocity

[math]\mu^\mu=\frac{dx^\mu}{d\tau}=c\frac{dt}{dx\tau},\frac{dx}{d\tau},\frac{dy}{d\tau})[/math]

then take

[math]ds^2=\eta_{\mu\nu}dx^\mu dx^\nu[/math] you have 16 terms [math] ds^2=\eta_{00}(dx^0)^2+\eta_{01}dx^0dx^1+\eta_{02}dx^0dx^2+....[/math] (16 terms)

and [math]ds^2=-c^2d\tau+dx^2+dy^2+dz^2[/math]

with that you get

[math]\eta_{\mu\nu}\mu^\mu\mu^nu=\mu^mu\mu_mu=-c^2[/math]

the four velocity has constant length******

differentiating this gives (with [math]\dot{\mu}^\mu=\frac{d\mu^\mu}{d\tau})[/math]

[math]\frac{d}{d\tau}(\mu^\mu \mu_mu)=0=2\dot{\mu}^\mu\mu_\mu[/math]

defining acceleration four vector

[math]a^\mu=\dot{\mu}^\mu[/math]

[math]\eta_{\mu\nu}a^\mu\mu^\nu=0[/math]

now consider a particle moving in constant acceleration g in the x^1 direction. The velocity and acceleration four vectors becomes

[math]c\frac{dt}{d\tau}=\mu^0, \frac{dx^1}{d\tau}=\mu^1,\frac{d\mu_0}{d\tau}=a^0,\frac{d\mu^1}{d\tau}=a^1[/math]

in addition

[math] a^\mu a_\mu=-(a^0)^2+(a^1)^2=g^2[/math] which defines constant acceleration g.

[math]a^0=\frac{g}{c}\mu^1,a^0=\frac{g}{c}\mu^0[/math] from which [math]\frac{da^0}{d\tau}=\frac{g}{c}\frac{d\mu^1}{d\tau}=\frac{g}{c}a^1=\frac{g^2}{c^2}\mu^0[/math]

hence

[math]\frac{d^2\mu^0}{d\tau^2}=\frac{g^2}{c^2}\mu^0[/math]

and

[math]\frac{d^2\mu^1}{d\tau^2}=\frac{g^2}{c^2}\mu^1[/math]

becomes for last [math]\mu^1=A^1e^g{\tau/c}+Be^{-g\tau/c}[/math]

[math]u^1=dx/d\tau=csinh(g\tau/c)[/math]

[math]a^0=d\mu^0/d\tau=gsinh(g\tau/c)[/math]

[math]\mu^0=c\frac{dt}{d\tau}=cosh(g\tau/c)[/math]

[math] x=\frac{c^2}{g}cosh(g\tau/c),,,,ct=\frac{c^2}{g}(sinhg\tau/c)[/math]

the space and time coordinates then fall on [math] x^2-c^2t^2=\frac{c^4}{g^2}[/math]

There is your rapidity curve for the turnaround acceleration...you can also see that during turnaround the transformations differ from the typical Lorentz transforms under constant velocity

Excerpt from Lewis Ryder "Introduction to General Relativity" for further details 

 

 

 

 

Edited by Mordred
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Hahaha, well I look forward to your solution to the problem. It looks like a lot of equations to solve!

3 hours ago, Mordred said:

 That is part of the solution for resolving the twin paradox. The twin that undergoes rapidity is the one that ages differently and breaks the symmetry of the Lorentz transforms.

The solution to the problem of this thread doesn't need the twin paradox to be resolved. You can calculate what's needed from one frame (the inertial one is the easiest). The calculations in any other frame are consistent but there's no need to verify that.

"the one that ages differently" makes no sense. They all age differently from each other. That's like taking two frames with relative velocity and saying that only one of them is moving relative to the other. Besides, they all age "similarly", at 1s/s; the differences are only relative to each other.

Also you're using 'rapidity' wrong. It's a measure of rate of motion, not an acceleration. If the twins are moving relative to each other, they both have rapidity relative to the other, just like they both have relative velocity.

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