How Does Acceleration Affect the Twin Paradox?

[BlackyTheCat]

A spaceship leaves earth. One twin stays back, the other is on the ship. The ship accelerates for 5 years with a constant acceleration (the 5 years are in the reference frame of the ship), then it decelerates for 5 years. Then, it turns around and does the same thing again. All the accelerations and decelerations are at the same rate. When he returns, he is 40 years old. How old is the twin who stayed on Earth?I know the following equations that could be relevant:
a = (1-v²/c²)^3/2/(1+v*u/c²)*a'
Where a is the acceleration in the resting frame, a' in the moving frame, v is the velocity of the frame that is moving, and u is the velocity of the object moving within the moving frame.
Also, time dilatation formula.I know how to solve the normal twin paradox with time dilatation. Here, I assume that I should use the acceleration formula above somehow. However, I am at a loss at how. I do not know any velocities. Also, even if I could for example calculate a, I don't see how it would help. I could get gamma from it I guess, but for that I would already need gamma before, so it would be a bit redundant.

I see that if I solve the problem for one step (one acceleration or deceleration), I am done, since it is symmetric though.

 

[PeroK]

A spaceship leaves earth. One twin stays back, the other is on the ship. The ship accelerates for 5 years with a constant acceleration (the 5 years are in the reference frame of the ship), then it decelerates for 5 years. Then, it turns around and does the same thing again. All the accelerations and decelerations are at the same rate. When he returns, he is 40 years old. How old is the twin who stayed on Earth?I know the following equations that could be relevant:
a = (1-v²/c²)^3/2/(1+v*u/c²)*a'
Where a is the acceleration in the resting frame, a' in the moving frame, v is the velocity of the frame that is moving, and u is the velocity of the object moving within the moving frame.
Also, time dilatation formula.I know how to solve the normal twin paradox with time dilatation. Here, I assume that I should use the acceleration formula above somehow. However, I am at a loss at how. I do not know any velocities. Also, even if I could for example calculate a, I don't see how it would help. I could get gamma from it I guess, but for that I would already need gamma before, so it would be a bit redundant.

I see that if I solve the problem for one step (one acceleration or deceleration), I am done, since it is symmetric though.

Are you sure you have all the information? If the acceleration is small, so that speeds are non-relativistic, then the twins would be approximately the same age.

 

[BlackyTheCat]

Are you sure you have all the information? If the acceleration is small, so that speeds are non-relativistic, then the twins would be approximately the same age.

Yes, that is all the information given. I am assuming that they travel at relativistic speeds, since our problem sets for the last 2 weeks were on this topic, and the course is a 2nd year theoretical physics class, so I think the non-relativistic case would be silly to ask for. (Yes, the problem statements are often set terribly and there is implicit information like that you should consider relativistic speeds and alike...)

 

[PeroK]

Yes, that is all the information given. I am assuming that they travel at relativistic speeds, since our problem sets for the last 2 weeks were on this topic, and the course is a 2nd year theoretical physics class, so I think the non-relativistic case would be silly to ask for. (Yes, the problem statements are often set terribly and there is implicit information like that you should consider relativistic speeds and alike...)

Okay, but that's not the point. The point is that the time dilation will depend on the velocity attained, which depends on the acceleration. If the acceleration is low, time dilation will be small, and if it's high enough, time dilation could be very great indeed.

Alternatively, your answer will be a function of $a$, the acceleration in the reference frame of the ship.

 

[BlackyTheCat]

Okay, but that's not the point. The point is that the time dilation will depend on the velocity attained, which depends on the acceleration. If the acceleration is low, time dilation will be small, and if it's high enough, time dilation could be very great indeed.

Alternatively, your answer will be a function of $a$, the acceleration in the reference frame of the ship.

Okay, so I was now confused too and checked the problem statement again. Turns out, they changed the problem statement about an hour ago, now stating the acceleration a' in the ships frame of reference is: a'=30 m/s^2. This seems to me like a large enough acceleration to cause some time dilatation over time.

My idea right now is to write v'(t) as a function of a' (v'=a'*t), and then plug this into the formula for the acceleration that I gave before. Then I'd have a time dependent acceleration of the ship in the Earths frame of reference. Does this sound about right?

I don't see, however, how I would then get to the velocity of the ship in the resting frame in order to get some gamma out of that...

 

[PeroK]

Okay, so I was now confused too and checked the problem statement again. Turns out, they changed the problem statement about an hour ago, now stating the acceleration a' in the ships frame of reference is: a'=30 m/s^2. This seems to me like a large enough acceleration to cause some time dilatation over time.

My idea right now is to write v'(t) as a function of a' (v'=a'*t), and then plug this into the formula for the acceleration that I gave before. Then I'd have a time dependent acceleration of the ship in the Earths frame of reference. Does this sound about right?

I don't see, however, how I would then get to the velocity of the ship in the resting frame in order to get some gamma out of that...

Try to get a formula for v in terms of a' and t.

You can get a formula relating t and $\tau$ the ship's proper time from that.

 

[BlackyTheCat]

Try to get a formula for v in terms of a' and t.

You can get a formula relating t and $\tau$ the ship's proper time from that.

Thanks a lot, I was able to solve it now!