Count Iblis
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C'mon, this is supposed to be the best theoretical physics forum on the entire internet. "Google relativistic calculator"
Let's work out the case of the photon rocket. Suppose in some frame the rocket is initially at rest in space. The four momentum is P1 = (M1,0). Then it burns anti-matter fuel and it's four momentum becomes P2= (M2 gamma, M2 gamma v). Conservation of four momentum:
P1 = P2 + Pf
where Pf is the total four momentum of the emitted photons and is thus of the form (E, -E). This means that:
Pf = P1 - P2
Square both sides and use that
Pf^2 = 0,
P1^2 = M1^2
P2^2 = M2^2
P1 dot P2 = gamma M1*M2
So we have:
0 = M1^2 + M2^2 - 2 gamma M1 M2 ----->
gamma = 1/2 [X + X^(-1)]
where X is the ratio of the final and initial mass
Note that if we use massive particles instead of photons Pf^2 would be strictly larger than zero and you would get a smaller gamma factor for the same initial/final mass ratio. So, the photon rocket is the best we can get.
Now, if we want to return to Earth we must put X= (M1/M2)^(1/4), where M1 is the initial mass of the rocket (which includes the fuel) and M2 the final mass. This is because we must accelerate to the gamma factor, and then change the direction of the velocity, which is equivalent to changing the velocity to zero and then back to the sama gamma factor but with the velocity in the opposite direction.
Then, when we reach Earth we must reduce the velocity to zero. If we want to travel at the same gamma factor during the trip, then the mass ratio's before and after the boosts must be the same each time, so X^4 = M1/M2
Now, we can play the following game. Suppose we have an anti-matter factory that produces anti-matter at a constant rate. We want to travel to some far away place, so we need a lot of antimatter. But, unfortunately, that takes a long time. An obvious strategy is to use some of the produced anti-matter to make small excursions. When we return form an excursion more time has passed in the frame of the factory, so we have a lot of anti-matter. If we do this right, we have more anti-matter than we would have had, had we stayed home despite using some for the excursion.
Now, if I remember correctly, it turns out that you can reduce the proper time you need to wait before you have the desired amnount of anti-matter be a factor of order Log(T/t), where T is the time needed to produce the anti-matter in the rest frame of the factory and t is the time needed to produce an amount of anti-matter equal to the mass of the rocket.
Let's work out the case of the photon rocket. Suppose in some frame the rocket is initially at rest in space. The four momentum is P1 = (M1,0). Then it burns anti-matter fuel and it's four momentum becomes P2= (M2 gamma, M2 gamma v). Conservation of four momentum:
P1 = P2 + Pf
where Pf is the total four momentum of the emitted photons and is thus of the form (E, -E). This means that:
Pf = P1 - P2
Square both sides and use that
Pf^2 = 0,
P1^2 = M1^2
P2^2 = M2^2
P1 dot P2 = gamma M1*M2
So we have:
0 = M1^2 + M2^2 - 2 gamma M1 M2 ----->
gamma = 1/2 [X + X^(-1)]
where X is the ratio of the final and initial mass
Note that if we use massive particles instead of photons Pf^2 would be strictly larger than zero and you would get a smaller gamma factor for the same initial/final mass ratio. So, the photon rocket is the best we can get.
Now, if we want to return to Earth we must put X= (M1/M2)^(1/4), where M1 is the initial mass of the rocket (which includes the fuel) and M2 the final mass. This is because we must accelerate to the gamma factor, and then change the direction of the velocity, which is equivalent to changing the velocity to zero and then back to the sama gamma factor but with the velocity in the opposite direction.
Then, when we reach Earth we must reduce the velocity to zero. If we want to travel at the same gamma factor during the trip, then the mass ratio's before and after the boosts must be the same each time, so X^4 = M1/M2
Now, we can play the following game. Suppose we have an anti-matter factory that produces anti-matter at a constant rate. We want to travel to some far away place, so we need a lot of antimatter. But, unfortunately, that takes a long time. An obvious strategy is to use some of the produced anti-matter to make small excursions. When we return form an excursion more time has passed in the frame of the factory, so we have a lot of anti-matter. If we do this right, we have more anti-matter than we would have had, had we stayed home despite using some for the excursion.
Now, if I remember correctly, it turns out that you can reduce the proper time you need to wait before you have the desired amnount of anti-matter be a factor of order Log(T/t), where T is the time needed to produce the anti-matter in the rest frame of the factory and t is the time needed to produce an amount of anti-matter equal to the mass of the rocket.
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