Time dilation puzzle

[md65536]

I was thinking about misleading intuition of the role of acceleration in the twin paradox, so I came up with this puzzle to test intuition about it:

 

There are 2 observers, Cat and Mouse, starting at relative rest. Mouse instantly accelerates and takes off in a random direction, traveling with a speed of v. After some time, Cat instantly accelerates and chases, approaching at a speed of -v relative to Mouse. When Cat reaches Mouse, Cat remains inertial, while Mouse again takes off in some direction at speed v relative to Cat. Again after some time, Cat chases at relative speed -v. This repeats.

 

If [math]v=\frac{\sqrt{3}}{2}c[/math], so that [math]\gamma=2[/math], and this goes on for 1 year according to Cat's clock, how much time has passed on Mouse's clock (assume they're reunited at that point)?

 

Is additional information needed to answer the question?

 

 

Additional questions:

 

1) If they started on Earth, and Mouse reverses direction each time Cat catches up, and if Cat waits one day each iteration before chasing, roughly where will they end up at the end of the year?

 

2) In the case of question (1), who experiences more proper acceleration? If instead of heading backwards, Mouse heads off always in the same direction, at v, then who experiences more proper acceleration? How does this modification affect the relative clock timing?

Edited August 29, 2013 by md65536

[victorqedu]

What is ?

[md65536]

What is ?

The Lorentz factor when v = 0.866c.

[Janus]

I was thinking about misleading intuition of the role of acceleration in the twin paradox, so I came up with this puzzle to test intuition about it:

 

There are 2 observers, Cat and Mouse, starting at relative rest. Mouse instantly accelerates and takes off in a random direction, traveling with a speed of v. After some time, Cat instantly accelerates and chases, approaching at a speed of -v relative to Mouse. When Cat reaches Mouse, Cat remains inertial, while Mouse again takes off in some direction at speed v relative to Cat. Again after some time, Cat chases at relative speed -v. This repeats.

 

If [math]v=\frac{\sqrt{3}}{2}c[/math], so that [math]\gamma=2[/math], and this goes on for 1 year according to Cat's clock, how much time has passed on Mouse's clock (assume they're reunited at that point)?

 

Is additional information needed to answer the question?

 

 

Additional questions:

 

1) If they started on Earth, and Mouse reverses direction each time Cat catches up, and if Cat waits one day each iteration before chasing, roughly where will they end up at the end of the year?

 

 

A couple of clarifying points.

 

When you say that the cat chases after the mouse at -v, I assume that you mean that the relative velocity between mouse and cat is -v as measured by either the cat and mouse. In which case, the closing velocity between the two in the earth frame is 0.123714c and the Cat's velocity relative to the Earth frame is 0.989739c. Also when you say that the mouse reverses direction, it is traveling at 0.866c towards the Earth relative to the Earth.

 

When you say that the cat waits for one day before persuing the mouse, is that according to the Earth clock or his own clock? At the start this doesn't matter because the Earth and cat share the same rest frame, but after that it does matter.

 

Case in point:

 

At the start mose races away at 0.866c for 1 day. After which, it is 0.866c light days from the cat and the mouse clock reads 1/2 day. ( i'm going to work this out from the Earth frame because it is the easiest one to do so.)

 

The cat starts after him at 0.989739c and takes 8 days to catch up. At this time, the mouse clock will read 4.5 days and the cat clock will read 2.14309 days (1 day of waiting and 1.14309 days to catch up to the mouse.) the cat and mouse will be 7.91791 light days from Earth.

 

Mouse turns around and heads back to Earth (I assume relative to the Earth and not relative to the Cat.)

 

If the cat waits one Earth day before turning around, the distance between mouse and cat will be 1.85574 light days apart. at which time the Mouse clock reads 5 days and the cat clock reads 2.28598 days. the cat turns around to chase the mouse. It will take it will take 15 days to catch up, at which time the mouse clock will read 12 days. and the cat clock reads 4.42927 days. The cat and mouse will now be 5.93844 light days from the Earth ( in the opposite direction from which they started). The earth frame will have recorded 8.85854 days from when the mouse first left Earth.

 

If on the other hand, the cat waits for 1 day to pass on his clock before chasing the mouse, the distance between them before he takes up the chase again will be ~13 light days. At which time his clock will read 3.14309 days and the mouse clock reads 8 days. It will take 105 days for the cat to catch the mouse, at which time, his clock will read 18.1546 days and the mouse clock will read 60.5 days. Cat and mouse will now be 91 light days from Earth. (in the same direction as the first leg.) The time in the Earth frame will be 121 days from when the mouse first left Earth.

[md65536]

When you say that the cat chases after the mouse at -v, I assume that you mean that the relative velocity between mouse and cat is -v as measured by either the cat and mouse. In which case, the closing velocity between the two in the earth frame is 0.123714c and the Cat's velocity relative to the Earth frame is 0.989739c. Also when you say that the mouse reverses direction, it is traveling at 0.866c towards the Earth relative to the Earth.

 

When you say that the cat waits for one day before persuing the mouse, is that according to the Earth clock or his own clock? At the start this doesn't matter because the Earth and cat share the same rest frame, but after that it does matter.

 

Case in point:

 

At the start mose races away at 0.866c for 1 day. After which, it is 0.866c light days from the cat and the mouse clock reads 1/2 day. ( i'm going to work this out from the Earth frame because it is the easiest one to do so.)

 

The cat starts after him at 0.989739c and takes 8 days to catch up. At this time, the mouse clock will read 4.5 days and the cat clock will read 2.14309 days (1 day of waiting and 1.14309 days to catch up to the mouse.) the cat and mouse will be 7.91791 light days from Earth.

 

Mouse turns around and heads back to Earth (I assume relative to the Earth and not relative to the Cat.)

+/- v here is always the velocity of Cat and Mouse relative to each other.

When I say Mouse reverses direction I mean that it travels at v relative to Cat, in the direction that Cat has just come from. I didn't realize it at the time, but this is not worded intuitively, once you calculate its change in velocity consistent with everything else that's written. "Mouse reverses" might better be replaced by "Mouse moves in the opposite direction".

 

> Also when you say that the mouse reverses direction, it is traveling at 0.866c towards the Earth relative to the Earth.

 

No, I didn't make it clear enough but none of the velocities are relative to Earth. I mean that if Mouse was approaching at speed |v| along Cat's positive x-axis, then after they meet it recedes at speed |v| along Cat's negative x-axis.

 

 

All times are according to local clocks... Cat waits 24 of its own hours.

 

The "additional questions" don't change any of the details of the main question, except for specifying particular directions instead of "random".

 

 

Does this clarify enough? Sorry, as usual I was sloppy with the details.

 

 

With this would you like to change your answer? The first part of your answer I think was interpreted correctly but I think the answer is wrong (8 days to catch up, I don't think is right according to any of the 3 observers).

 

 

 

By the way: The puzzle should be solvable using relativistic reasoning alone, essentially without math. I know there are some people on these forums who want to understand SR and/or the twin paradox intuitively without the math... well if such an understanding is possible then this puzzle should showcase it. That is, if the solution I have in mind is correct.

Edited September 2, 2013 by md65536

[pears]

By the way: The puzzle should be solvable using relativistic reasoning alone, essentially without math. I know there are some people on these forums who want to understand SR and/or the twin paradox intuitively without the math... well if such an understanding is possible then this puzzle should showcase it. That is, if the solution I have in mind is correct.

 

I'm glad about this. I never really grasped the reasoning behind the twins paradox except that it is something to do with the system being assymmetric in some way because of acceleration, and only really ever found conflicting information when reading up about it online. Looking forward to seeing the solution.

[Toffo]

I was thinking about misleading intuition of the role of acceleration in the twin paradox, so I came up with this puzzle to test intuition about it:

 

There are 2 observers, Cat and Mouse, starting at relative rest. Mouse instantly accelerates and takes off in a random direction, traveling with a speed of v. After some time, Cat instantly accelerates and chases, approaching at a speed of -v relative to Mouse. When Cat reaches Mouse, Cat remains inertial, while Mouse again takes off in some direction at speed v relative to Cat. Again after some time, Cat chases at relative speed -v. This repeats.

 

If [math]v=\frac{\sqrt{3}}{2}c[/math], so that [math]\gamma=2[/math], and this goes on for 1 year according to Cat's clock, how much time has passed on Mouse's clock (assume they're reunited at that point)?

 

Is additional information needed to answer the question?

 

Mouse says: "Cat travels away at speed v, then cat travels back at speed v, so cat ages at half the normal rate (because v is such that time dilation factor is 1/2), my own aging rate is the normal rate."

Edited September 2, 2013 by Toffo

[victorqedu]

The answer to the original problem, when mouse changes directions at random is 2 years(in the mouse frame of reference).

I got to 2 years ignoring earth and considering the mouse as an inertial frame with Lorentz factor = 2 with regard to the cat inertial frame.
When the angle is changed acceleration is produced and the inertial frame is changed but because the speed remains constant the frames are equivalent.

This is a proof of the fact that acceleration is not directly involved in space-time deformation.

I have ignored the time between the start of the mouse and the start of the cat since in the original setup is unknown.

Is this correct?

Edited September 2, 2013 by victorqedu

[Janus]

+/- v here is always the velocity of Cat and Mouse relative to each other.

When I say Mouse reverses direction I mean that it travels at v relative to Cat, in the direction that Cat has just come from. I didn't realize it at the time, but this is not worded intuitively, once you calculate its change in velocity consistent with everything else that's written. "Mouse reverses" might better be replaced by "Mouse moves in the opposite direction".

 

 

Then it doesn't make sense to say that when the cat catches up to the Mouse, the Mouse "moves in the opposite direction". All that happens is that the cat passes the Mouse. Sure, the from the cat's perspective,the mouse goes from approaching at 0.866 c from one direction, and then receding in the other, but the Mouse does not under go any acceleration at this time.

 

In this scenerio, What the Earth frame sees is the cat passing the mouse, then after one day the cat comes to a rest, and waits for the mouse to catch up.

 

Then the mouse will accelerate back towards the Earth and we repeat in the other direction.

Or

If you want things as perceived by the cat to be consistent, (mouse approaching from one direction and receding in the other), the cat just lets the mouse pass him at 0.866c, then one day later accelerates back up to speed in order to catch it again. in this last case, the mouse never needs to accelerate again after the intial acceleration up to 0.866c

 

 

Is either of these the scenerio you want to examine?

Edited September 2, 2013 by Janus

[md65536]

I'm glad about this. I never really grasped the reasoning behind the twins paradox except that it is something to do with the system being assymmetric in some way because of acceleration, and only really ever found conflicting information when reading up about it online. Looking forward to seeing the solution.

A warning though, I think that the "intuitive" solutions might do nothing to help make the paradox more understandable. Also, to verify or prove the intuitive solutions should take some maths. Personally, I think that an intuitive understanding comes from figuring out the details (using math) for all the different observers and seeing how they're all mutually consistent, but then hopefully that understanding can be applied in cases like this.

Mouse says: "Cat travels away at speed v, then cat travels back at speed v, so cat ages at half the normal rate (because v is such that time dilation factor is 1/2), my own aging rate is the normal rate."

I believe this is the correct answer. Is it possible to show that the other details don't matter (Mouse's own accelerations for example), or that "what Cat measures" is consistent?

[pears]

OK fair enough. I'll have my pen and paper at the ready.

[md65536]

The answer to the original problem, when mouse changes directions at random is 2 years(in the mouse frame of reference).

I got to 2 years ignoring earth and considering the mouse as an inertial frame with Lorentz factor = 2 with regard to the cat inertial frame.

When the angle is changed acceleration is produced and the inertial frame is changed but because the speed remains constant the frames are equivalent.

This is a proof of the fact that acceleration is not directly involved in space-time deformation.

I have ignored the time between the start of the mouse and the start of the cat since in the original setup is unknown.

Is this correct?

The answer's right and the reasoning is interesting and I hadn't thought of it that way. You say that when Mouse accelerates the speed remains constant... can the same thing be said for Cat when it accelerates?

Then it doesn't make sense to say that when the cat catches up to the Mouse, the Mouse "moves in the opposite direction". All that happens is that the cat passes the Mouse. Sure, the from the cat's perspective,the mouse goes from approaching at 0.866 c from one direction, and then receding in the other, but the Mouse does not under go any acceleration at this time.

 

In this scenerio, What the Earth frame sees is the cat passing the mouse, then after one day the cat comes to a rest, and waits for the mouse to catch up.

 

Then the mouse will accelerate back towards the Earth and we repeat in the other direction.

Or

If you want things as perceived by the cat to be consistent, (mouse approaching from one direction and receding in the other), the cat just lets the mouse pass him at 0.866c, then one day later accelerates back up to speed in order to catch it again. in this last case, the mouse never needs to accelerate again after the intial acceleration up to 0.866c

 

 

Is either of these the scenerio you want to examine?

Yes, the second scenario is what I was describing for additional question (1). When I first wrote it out I hadn't realized that Mouse remains inertial after its first acceleration! With this version, it is easy to make an analogy to the twin paradox, with Mouse being the inertial twin. Then the puzzle becomes figuring out whether it matters if Mouse keeps accelerating between each iteration of the twin paradox procedure.

 

One way to squeeze some sense out of "Mouse reverses direction" is to consider it like this: When Cat joins Mouse, Mouse could instantly accelerate to be at rest with Cat, and then instantly accelerate in the opposite direction. (And then seeing that the result of Mouse instantly leaving and returning to its old inertial frame is the same as if Mouse simply remained in that inertial frame.)

Edited September 2, 2013 by md65536

[Toffo]

I believe this is the correct answer. Is it possible to show that the other details don't matter (Mouse's own accelerations for example), or that "what Cat measures" is consistent?

Mouse thinks about the equivalence principle: "If I am sitting on a chair in a gravity field, or in an accelerating rocket, and the cat has just fallen off his chair next to me, then me and the cat are experiencing the same gravitational time dilation. Or, to put it more briefly: me and the cat are near each other, therefore I can ignore my own acceleration."

[Iggy]

I believe this is the correct answer. Is it possible to show that the other details don't matter (Mouse's own accelerations for example), or that "what Cat measures" is consistent?

For "what cat measures is consistent", what mouse measures is:

 

[math]\tau_M = \tau_C \left( \frac{1}{\sqrt{1-{v_C}^2}} \right)[/math]

 

so we want to get the same equation from cat's perspective.

 

Cat also measures mouse's clock slow:

 

[math]\tau_M = \tau_C \sqrt{1-{v_M}^2}[/math]

 

But also measures a jump in mouse's clock from the relativity of simultaneity by:

 

[math]\frac{(x)(v_M)}{\sqrt{1-{v_M}^2}}[/math]

 

if you use velocity and time rather than distance, x, it would be:

 

[math]\tau_C \frac{{v_M}^2}{\sqrt{1-{v_M}^2}}[/math]

 

Add the two terms (the time dilated clock and the jump in simultaneity):

 

[math]\tau_M = \tau_C \sqrt{1-{v_M}^2} + \tau_C \frac{{v_M}^2}{\sqrt{1-{v_M}^2}}[/math]

 

by the principle of reciprocity, v_M = v_C

 

[math]\tau_M = \tau_C \sqrt{1-{v_C}^2} + \tau_C \frac{{v_C}^2}{\sqrt{1-{v_C}^2}}[/math]

 

A few steps in algebra will take you from the above equation to this one:

 

[math]\tau_M = \tau_C \left( \frac{1}{\sqrt{1-{v_C}^2}} \right)[/math]

 

which is what we were looking to show from Cat's perspective. It's just the jump in simultaneity that makes up the difference because, like Toffo said, if it happens at a distance then it matters. Mouse accelerates when they are near, and Cat when they are not.

[md65536]

Excellent answers, better than what I had in mind. As usual the maths can verify any intuitive understanding and drive further understanding.

 

For example, while "nearby acceleration" doesn't make a big difference, it doesn't mean that nearby motion can be ignored. For example if Cat waited only 1 nanosecond before each chase, it would stay within a meter of Mouse, but Mouse would still age at twice the rate. Each chase would contribute only a tiny amount to the difference in age, but there would be quadrillions of them over a year. Or in other words, with high relative speed and a long duration there will be time dilation.

 

 

 

Each time that Cat and Mouse separate (inertially), and then Cat accelerates toward Mouse, is essentially an iteration of the Twin Paradox, with Mouse being the inertial twin. What happens in between happens when they're together, for a negligible time, and contributes nothing to the time dilation. I can't explain it better than that but Toffo has and Iggy shows why it is.

 

For additional question (1):

If Cat spends one day of proper time receding from Mouse at v, then approaches at -v, it will take one day to catch up. As Janus pointed out, in this variation Mouse remains inertial when Cat catches up, and simply lets Cat pass by it.

 

Where are they? 1 year proper time has passed for Cat, and 2 years for Mouse. According to Mouse, Earth has receded for 2 years at 0.866c and they are 1.732 LY away from Earth. According to Mouse, Earth has also aged at half the rate (1 year).

 

For an Earthbound observer, it is Mouse that has aged at half the rate. After 1 year of proper time for Cat, which is 2 for Mouse, Earth ages 4 years. Mouse has spent 4 years at 0.866c and is 3.464 LY away.

 

From Cat's perspective, it is at the same location as Mouse but currently traveling at 0.990c relative to Earth (composition of velocities). If we take the distance according to Mouse of 1.732 LY, consider that a rest distance in Mouse's frame (okay to do I think?), then for Cat this is length contracted by a factor of gamma (=2, since Cat is traveling at v=0.866c relative to Mouse), so it is 0.866 LY from Earth in its frame.

 

 

For question (2):

In (1), Mouse has only experienced the initial acceleration away from Earth, while Cat keeps accelerating on each chase. In (2), Mouse and Cat each accelerate to a velocity of 0.990c (composition of velocities) relative to their previous rest frame, on each chase. The proper acceleration is comparable, though technically Mouse accelerated less on its first leg; whoever accelerated last experienced the most overall proper acceleration. Technically when Cat catches up to Mouse (at the end of 1 year), that's still Cat.

 

 

The choice of direction that Mouse takes each time doesn't make a difference in the relative timing and distance between Cat and Mouse, but makes a huge difference relative to Earth. In (2), they've essentially been accelerating away from Earth to near-c, 183 times I think, over the Cat's year. Before a Cat-month is up, the Lorentz factor relative to Earth is in the millions. According to Earth, Cat and Mouse's clocks are so dilated and they're traveling at very close to c, the universe will probably end before Cat ages a full year. Meanwhile according to Cat, Earth time has ground to a halt, but after 1 year Cat is still traveling at its fastest speed and the distance to Earth is extremely length contracted... Earth has been receding at very near c for a year but it remains (slightly) less than one light year away. For Mouse: each time it accelerates it is traveling faster away from Earth than Cat is, yet it still ages 2 years while Cat ages 1, and it is slightly less than 2 LY away from Earth.

 

 

I may have made some mistakes.

Theory and its maths comes from observations, and intuitive understanding comes from all 3, but trying to answer using minimal math can be unreliable.

Edited September 3, 2013 by md65536

[Iggy]

Where are they? 1 year proper time has passed for Cat, and 2 years for Mouse. According to Mouse, Earth has receded for 2 years at 0.866c and they are 1.732 LY away from Earth. According to Mouse, Earth has also aged at half the rate (1 year).

 

For an Earthbound observer, it is Mouse that has aged at half the rate. After 1 year of proper time for Cat, which is 2 for Mouse, Earth ages 4 years. Mouse has spent 4 years at 0.866c and is 3.464 LY away.

 

From Cat's perspective, it is at the same location as Mouse but currently traveling at 0.990c relative to Earth (composition of velocities). If we take the distance according to Mouse of 1.732 LY, consider that a rest distance in Mouse's frame (okay to do I think?), then for Cat this is length contracted by a factor of gamma (=2, since Cat is traveling at v=0.866c relative to Mouse), so it is 0.866 LY from Earth in its frame.

That all looks ok except the last paragraph. Where you say "ok to do I think?", I believe not. In Mouse's frame both the earth and cat are moving so you shouldn't be able to use the length contraction formula between them. One of them would have to be stationary in the frame. You could use earth's frame, and of course we could just use the velocity you give and the distance you give in the length contraction formula:

 

[math]L = L_0 \sqrt{1-v^2} = 3.464 \sqrt{1 - 0.990^2} = 0.489 \ ly[/math]

 

But I bet we can get a more general formula and have more fun with it...

 

 

Simplifying the velocity addition formula ends up saying that the cat's velocity in earth's frame is [math]v_c = (2 v_m)/(1 + {v_m}^2 )[/math] where v_m is the velocity of the mouse relative to earth and the velocity of the cat relative to the mouse (which is to say [math]v_m = \sqrt{3}/2[/math]). So if we plug that formula into the length contraction formula:

 

Length contraction formula:

 

[math]L_C = L_E \sqrt{1-{v_c}^2}[/math]

 

Substituting our [math]v_c[/math] formula for [math]v_c[/math]:

 

[math]L_C = L_E \sqrt{1- \left( \frac{2v_m}{1+{v_m}^2} \right)^2}[/math]

 

the above simplifies to:

 

[math]L_C = L_E \left( \frac{1-{v_m}^2}{{v_m}^2+1} \right) [/math]

 

and since v_m is [math]\sqrt{3}/2[/math] and [math]L_E[/math] (the distance in earth's frame) is [math]2 \sqrt(3)[/math] the distance in Cat's frame would be:

 

[math]L_C = (2)(\sqrt{3}) \left( \frac{1-0.75}{0.75+1} \right) [/math]

 

or

 

[math]L_C = \frac{2 \sqrt{3}}{7}[/math]

 

Sweet!

 

[math]L_C = 0.49487 \ ly[/math]

 

 

That should make sense too because if Earth travels 0.49487 light years in the cat's frame in 0.5 years then the velocity is 0.49487 / 0.5 = 0.9897, and that is indeed the Earth / Cat relative velocity.

Edited September 4, 2013 by Iggy

[md65536]

That all looks ok except the last paragraph. Where you say "ok to do I think?", I believe not. In Mouse's frame both the earth and cat are moving so you shouldn't be able to use the length contraction formula between them. One of them would have to be stationary in the frame.

Oops, thanks for the correction! The lesson might be that proper use of math trumps intuition every time.

 

Another way to see that I was wrong, is if we placed a flag in space at that time at Earth's location but at rest in Mouse's frame, the distance between Mouse and flag would be the rest distance I was speaking of. However, though Earth passes the flag at that moment in Mouse's frame, it wouldn't pass at the same moment in Cat's frame.

 

That should make sense too because if Earth travels 0.49487 light years in the cat's frame in 0.5 years then the velocity is 0.49487 / 0.5 = 0.9897, and that is indeed the Earth / Cat relative velocity.

Is the 0.5 years because Cat spends half its year traveling and half at rest relative to Earth?

[Iggy]

Oops, thanks for the correction! The lesson might be that proper use of math trumps intuition every time.

 

Another way to see that I was wrong, is if we placed a flag in space at that time at Earth's location but at rest in Mouse's frame, the distance between Mouse and flag would be the rest distance I was speaking of. However, though Earth passes the flag at that moment in Mouse's frame, it wouldn't pass at the same moment in Cat's frame.

Exactly how I thought about it. Earth would be at different positions for the two present instants of the measurement event. Yup.

 

Is the 0.5 years because Cat spends half its year traveling and half at rest relative to Earth?

Right. 0.5 resting w/ earth and .5 catching the mouse.

 

And, earth should spend seven times as long (.5*7) for the catching phase since v=0.9897 has a gamma factor of 7. So, 3.5y + 0.5y makes four years total earth time.

 

Yep, yep.