## Optimizing the Twin Paradox

**Moderators:** gmalivuk, Moderators General, Prelates

### Optimizing the Twin Paradox

This is a problem of my own invention, and I think I've figured out how to solve it, but I thought y'all might be interested in looking at it too, as there is a (very) high chance I did something completely wrong.

A spaceship has a constant acceleration of 9.8 m/s^2 (ignore relativistic mass and stuff, for now). This space ship accelerates, reaches a target speed of k m/s, maintains that speed for a certain period of time, and then decelerates at 9.8 m/s^2 to rest again. From a stationary POV, you've traveled for y seconds (or some other unit of time). Given y, find the value of k which minimizes the passing of time from the ship's POV.

The next level of this problem says that the ship weighs 1 kg in total, has a 9.8-N engine, and magically does not lose mass from fuel while accelerating. Relativistic mass now comes into play, but frankly, I think I'm just getting confuddled in the time dilation formula and in my reference frames.

A spaceship has a constant acceleration of 9.8 m/s^2 (ignore relativistic mass and stuff, for now). This space ship accelerates, reaches a target speed of k m/s, maintains that speed for a certain period of time, and then decelerates at 9.8 m/s^2 to rest again. From a stationary POV, you've traveled for y seconds (or some other unit of time). Given y, find the value of k which minimizes the passing of time from the ship's POV.

The next level of this problem says that the ship weighs 1 kg in total, has a 9.8-N engine, and magically does not lose mass from fuel while accelerating. Relativistic mass now comes into play, but frankly, I think I'm just getting confuddled in the time dilation formula and in my reference frames.

Happy hollandaise!

"The universe is a figment of its own imagination" -Douglas Adams

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### Re: Optimizing the Twin Paradox

scikidus wrote:This is a problem of my own invention, and I think I've figured out how to solve it, but I thought y'all might be interested in looking at it too, as there is a (very) high chance I did something completely wrong.

A spaceship has a constant acceleration of 9.8 m/s^2 (ignore relativistic mass and stuff, for now). This space ship accelerates, reaches a target speed of k m/s, maintains that speed for a certain period of time, and then decelerates at 9.8 m/s^2 to rest again. From a stationary POV, you've traveled for y seconds (or some other unit of time). Given y, find the value of k which minimizes the passing of time from the ship's POV.

The next level of this problem says that the ship weighs 1 kg in total, has a 9.8-N engine, and magically does not lose mass from fuel while accelerating. Relativistic mass now comes into play, but frankly, I think I'm just getting confuddled in the time dilation formula and in my reference frames.

K=0. Unless you allow moving backwards in time at speeds faster than c

I kind of know what you're thinking, but it seems like you're going to heck of a time finding a problem whose answer isn't 0 or c.

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### Re: Optimizing the Twin Paradox

How does this affect time dilation?scikidus wrote:Relativistic mass now comes into play

No, because by going any speed at all it's possible to experience less than the full y seconds you'd experience with k=0.Charlie! wrote:K=0.

In reality, k is whatever speed you could get up to in half of y, and then you turn around and start decelerating. Any coast time is time spent at a slower speed than you could be going if you'd continued accelerating.

### Re: Optimizing the Twin Paradox

Because at the beginning, when you're not dealing with relativistic speeds, 9.8-N on a 1kg object gives an acceleration of 9.8 m/s^2, but once the mass of the ship increases, those 9.8-N of force won't go as long a way towards accelerating the ship, right?gmalivuk wrote:How does this affect time dilation?scikidus wrote:Relativistic mass now comes into play

That makes sense. A lot of sense. I'm currently *headdesk*ing because I should've seen that.gmalivuk wrote:In reality, k is whatever speed you could get up to in half of y, and then you turn around and start decelerating. Any coast time is time spent at a slower speed than you could be going if you'd continued accelerating.

With constant acceleration, the equation for k then is just (9.8 m/s^2)*y/2, right? You'd then find the time dilation over the trip as , or: [url]http://www.wolframalpha.com/input/?i=%28x*sqrt%281-x^2%2F299792458^2%29%2B299792458*sin^%28-1%29%28x%2F299792458%29%29[/url]

Happy hollandaise!

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### Re: Optimizing the Twin Paradox

scikidus wrote:ignore relativistic mass and stuff, for now).

If you have relativistic velocities, you have relativistic energies. There is no regime of approximation in which time dilation happens, but simple F=ma still holds.

I further advise against ever using relativistic mass, but I think that's been discussed at length already.

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### Re: Optimizing the Twin Paradox

scikidus wrote:A spaceship has a constant acceleration of 9.8 m/s^2 (ignore relativistic mass and stuff, for now). This space ship accelerates, reaches a target speed of k m/s, maintains that speed for a certain period of time, and then decelerates at 9.8 m/s^2 to rest again. From a stationary POV, you've traveled for y seconds (or some other unit of time). Given y, find the value of k which minimizes the passing of time from the ship's POV.

The next level of this problem says that the ship weighs 1 kg in total, has a 9.8-N engine, and magically does not lose mass from fuel while accelerating. Relativistic mass now comes into play, but frankly, I think I'm just getting confuddled in the time dilation formula and in my reference frames.

These two setups are identical, assuming by 9.8 m/s

^{2}constant acceleration you mean that the acceleration is always 9.8 m/s

^{2}as felt by the passengers of the spaceship (i.e., as measured in a reference frame in which the spaceship is instantaneously at rest).

If you're using the drive to generate artificial gravity, that's what you would do.

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### Re: Optimizing the Twin Paradox

scikidus wrote:Because at the beginning, when you're not dealing with relativistic speeds, 9.8-N on a 1kg object gives an acceleration of 9.8 m/s^2, but once the mass of the ship increases, those 9.8-N of force won't go as long a way towards accelerating the ship, right?

Yes it will. You are using relativistic mass wrong. To the inhabitants on the ship their mass have not increased. A force of mg will give you an acceleration of g and you will feel one g, always (m being center-of-mass mass (rest mass) not relativistic mass, which we should not use). To an outside observer you will of course accelerate slower because of your increased mass, that's why it takes you forever to reach lightspeed.

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### Re: Optimizing the Twin Paradox

Personally, I feel that bringing mass dilation into the picture here only serves to distract... think of it like this: in order to be simulating 1g for the people on the ship, the ship's going to be accelerating at 9.8 ship-meters per ship-second squared. As the ship gets faster, our outside stationary observer will be seeing the ship-metres as getting shorter, and the ship-seconds as getting longer... so the acceleration of the ship as seen by our outside observer will be decreasing (limiting to 0 as the spaceship approaches c).

For the ship to accelerate at a constant speed as seen from an outside observer, the acceleration as viewed by the ship would have to increase over time, limiting to infinity as the ship approaches c.

For the ship to accelerate at a constant speed as seen from an outside observer, the acceleration as viewed by the ship would have to increase over time, limiting to infinity as the ship approaches c.

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`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### Re: Optimizing the Twin Paradox

Alright, I see why from a stationary observer's POV you'd slow, but what about from the ship's perspective?

Also, since you're obviously never going to hit light speed, what does the graph look like for velocity of the ship vs time?

Also, since you're obviously never going to hit light speed, what does the graph look like for velocity of the ship vs time?

Happy hollandaise!

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### Re: Optimizing the Twin Paradox

Well, that's a bit more complicated, since the ship's PoV isn't an inertial frame, but one with acceleration. But one thing you can do is consider a series of inertial frames, in each of which, the ship is temporarily stationary at a particular time. To picture it, see an observer moving from behind the ship, faster than the ship, but at a constant velocity... it would be catching up to the ship (from the observer's point of view, the ship would be moving backwards towards it), but then the ship will speed up to eventually match the observer's speed (come to a rest in the observer's frame - for simplicity, say that it comes to rest right next to the observer... it doesn't actually affect anything when it comes to the observer's PoV, since frames are translation-invariant in SR, but it means we can talk about the place where the rocket comes to rest in this frame a bit easier) and then accelerates away.

So, we'll consider such observers that meet the rocket 1s ship-time after launch, 2s ship-time after launch, etc. So from the rocket's PoV, it's meeting one every second. From the PoV of the place it was launched, they're getting slower and slower, due to time dilation increasing.

Now, the point is, in the second after launch time, as measured by the ship, it will have accelerated to 9.8 m/s. So the first observer will be moving at 9.8 m/s relative to the launch point. Then, similarly, the second observer is moving at 9.8 m/s relative to the first. And so on. To find out how fast more separated observers see each other, we need to use the velocity-addition formula, which says that the combined velocity is slightly less than the sum of the velocities... in particular, such that the sum of two slower-than-light velocities is always slower than light. We can continually add more velocities to our total, but it will only approach c, never reach it.

For instance, after a million seconds, the millionth observer is moving about 9.797x10

And the good part is that it's symmetric - all our observers are inertial, they just happen to coincide with this rocket's path. So the millionth observer will also see the launch point as moving back at 9.797x10

I really should start just linking to my old explanations every time things like this come up, rather than typing it all up all over again...

So, we'll consider such observers that meet the rocket 1s ship-time after launch, 2s ship-time after launch, etc. So from the rocket's PoV, it's meeting one every second. From the PoV of the place it was launched, they're getting slower and slower, due to time dilation increasing.

Now, the point is, in the second after launch time, as measured by the ship, it will have accelerated to 9.8 m/s. So the first observer will be moving at 9.8 m/s relative to the launch point. Then, similarly, the second observer is moving at 9.8 m/s relative to the first. And so on. To find out how fast more separated observers see each other, we need to use the velocity-addition formula, which says that the combined velocity is slightly less than the sum of the velocities... in particular, such that the sum of two slower-than-light velocities is always slower than light. We can continually add more velocities to our total, but it will only approach c, never reach it.

For instance, after a million seconds, the millionth observer is moving about 9.797x10

^{6}m/s relative to the launch point, not 9.800x10^{6}. Not sure how accurate that value is with all the accumulated rounding errors, but all the same, it's smaller. Not much smaller, since we're only just starting to get into relativistic speeds (that's only about 3% of c) but still, smaller.And the good part is that it's symmetric - all our observers are inertial, they just happen to coincide with this rocket's path. So the millionth observer will also see the launch point as moving back at 9.797x10

^{6}m/s. And when the rocket catches up to that millionth observer, one million seconds after launching, it will agree with that observer, for the split second where their speeds match. So the rocket will see the launch point in the distance get faster, but never cross c.I really should start just linking to my old explanations every time things like this come up, rather than typing it all up all over again...

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`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

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### Re: Optimizing the Twin Paradox

Velocity versus whose time?scikidus wrote:what does the graph look like for velocity of the ship vs time?

In any case, the equations you want are at http://www.xs4all.nl/~johanw/PhysFAQ/Re ... ocket.html

### Re: Optimizing the Twin Paradox

For instance, after a million seconds, the millionth observer is moving about 9.797x10^{6}m/s relative to the launch point, not 9.800x10^{6}. Not sure how accurate that value is with all the accumulated rounding errors, but all the same, it's smaller.

It's pretty good. Using the equation v = c * tanh(at/c), I get 9796510.76(6976) m/s, at proper time T = 10^6, using a = 9.80 m/s^2 and c = 299792458 m/s.

### Re: Optimizing the Twin Paradox

gmalivuk wrote:How does this affect time dilation?scikidus wrote:Relativistic mass now comes into playNo, because by going any speed at all it's possible to experience less than the full y seconds you'd experience with k=0.Charlie! wrote:K=0.

In reality, k is whatever speed you could get up to in half of y, and then you turn around and start decelerating. Any coast time is time spent at a slower speed than you could be going if you'd continued accelerating.

I was responding to the first problem - "what is the speed you should accelerate to so that you experience the shortest time."

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### Re: Optimizing the Twin Paradox

Yes, I know. And k=0 is the speed you should accelerate to in order to experience the *longest* time, since any nonzero speed will result in a shorter time from the shipboard perspective.

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