Aleberto69 said:
TY very much,
could you suggest a good textbook which address those calculation and deduce the formula you have used?
TY very much in advance
Here's a long-winded derivation.
If the rear of a rocket has acceleration g in a frame in which it is (momentarily) at rest, then it has acceleration g_v = \frac{g}{\gamma^3} in a frame in which it is traveling at v. You can figure this out by using the inverse Lorentz transforms:
- x' = \gamma (x - v t)
- t' = \gamma(t - \frac{vx}{c^2})
Now, in the primed frame, assume that position of the rocket is given by: x' \approx \frac{1}{2} g (t')^2. (\approx means I'm ignoring terms of order (t')^3 or higher.) So in this frame, the rocket is accelerating from rest at rate g. In the unprimed frame, assume that x \approx vt + \frac{1}{2} a t^2. In this frame it's accelerating from speed v at rate a. We want to solve for a. Plugging things into our equations, we have:
- \frac{1}{2} g (t')^2 \approx \gamma \frac{1}{2} a t^2. This implies that a = g (\frac{t'}{t})^2 \frac{1}{\gamma}.
- t' \approx \gamma (t - \frac{v^2}{c^2} t) = \gamma(1-\frac{v^2}{c^2}) t = \frac{t}{\gamma}. this implies that \frac{t'}{t} \approx \frac{1}{\gamma}
Putting 1 and 2 together gives us: a = g \frac{1}{\gamma^3}
Since a = \frac{dv}{dt}, this means \frac{dv}{dt} = g \frac{1}{\gamma^3}. We can invert this to get: \frac{dt}{dv} = \frac{1}{g} \gamma^3
So we integrate to get: t = \int \frac{1}{g} \gamma^3 dv = \frac{1}{g} \int (1-\frac{v^2}{c^2})^{-\frac{3}{2}} dv
To do this integral, we do a variable substitution: v = c tanh(\frac{gT}{c}), so dv = g sech^2(\frac{gT}{c}) dT (tanh is hyperbolic tangent, sech is hyperbolic secant.) Then 1-\frac{v^2}{c^2} = 1 - tanh^2(\frac{gT}{c}) = sech^2(\frac{gT}{c}). So the integral becomes:
t = \int \frac{1}{sech(\frac{gT}{c})} dT = \int cosh(\frac{gT}{c}) dT = \frac{c}{g} sinh(\frac{gT}{c})
We can integrate v to get x, and we find:
x = \int v dt = \int c tanh(\frac{gT}{c}) cosh(\frac{gT}{c}) dT = c \int sinh(\frac{gT}{c}) dT = \frac{c^2}{g} cosh(\frac{gT}{c})
(To simplify the expression, I've chosen the origin of the x-axis so that the rocket is initially at x = \frac{c^2}{g} rather than x=0)
We can also compute:
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = cosh(\frac{gT}{c}) = \frac{gx}{c^2}
\gamma v = cosh(\frac{gT}{c}) c tanh(\frac{gT}{c}) = c sinh(\frac{gT}{c}) = \frac{gt}{c}
And we show that x and t obey the equation: x^2 - c^2 t^2 = \frac{c^4}{g^2}
All right. Now, let's try to figure out a coordinate system that can be used on board the rocket. We already have a time coordinate, T, but what's its significance? Well, since the proper time for a clock at the rear of the rocket is given by: \tau = \int \sqrt{1-\frac{v^2}{c^2}} dt, we can change variables from t to T using t = \frac{c}{g} sinh(\frac{gT}{c}). So dt = cosh(\frac{gT}{c}) dT. Using our formula for v: v = c tanh(\frac{gT}{c}) so \sqrt{1-\frac{v^2}{c^2}} = \sqrt{1-tanh(\frac{gT}{c})^2} = sech(\frac{gT}{c}) = \frac{1}{cosh(\frac{gT}{c})}. So our integral for proper time becomes:
\tau = \int dT = T
So T is just the time on the rear clock.
For our spatial coordinate, we want it to be the case that the length of the rocket stays the same, as viewed by someone on board the rocket. Otherwise, the rocket would either be ripped apart or crushed. Let L be the length of the rocket as measured in the rocket's accelerated frame. We're going to assume that this is also the length of the rocket as measured in the inertial coordinate system in which the rocket is (momentarily) at rest. So let's suppose that the rocket has traveled for a while, and the rear of the rocket is at location x at time t. The rocket is traveling at speed v. Let e_1 be the event (moment in space and time) when the rear of the rocket is traveling at speed v. Let e_2 be the event at the front of the rocket that is simultaneous with e_1 in the instantaneous rest frame of the rocket.
- Let (x,t) be the coordinates of e_1 in the Earth frame.
- Let (x', t') be the coordinates in the instantaneous rest frame of the rocket.
- Let (x_2, t_2) be the coordinates of e_2 in the Earth frame.
- Let (x_2', t_2') be the coordinates in the rocket frame.
Since e_1 and e_2 are simultaneous in the rocket frame, that means: t_2' = t'.
Since the rocket has length L in its own rest frame, this means: x_2' = x' + L
Now, we can use the inverse Lorentz transformations to find x_2 and t_2:
x = \gamma (x' + v t')
t = \gamma(t' + \frac{vx'}{c^2})
x_2 = \gamma (x_2' + v t_2') = \gamma(x' + L + v t') = x + \gamma L
t_2 = \gamma(t_2' + \frac{v x_2'}{c^2}) = \gamma(t' + \frac{v}{c^2} (x' + L)) = t + \frac{\gamma vL}{c^2}
But we know from above for our constant acceleration rocket, \gamma = \frac{gx}{c^2}. So
x_2 = x (1 + \frac{gL}{c^2})
We also know that \gamma v = \frac{gt}{c}. So
t_2 = t + \frac{gt}{c} \frac{L}{c^2} = t (1 + \frac{gL}{c^2})
Now, we see an amazing, or possibly not amazing relationship between x_2 and t_2:
(x_2)^2 - c^2 (t_2)^2 = x^2 (1+ \frac{gL}{c^2}) - c^2 t^2 (1+\frac{gL}{c^2})^2 = (x^2 - c^2 t^2) (1+\frac{gL}{c^2})^2
But x^2 - c^2 t^2 = \frac{c^4}{g^2}, so
(x_2)^2 - c^2 (t_2)^2 = \frac{c^4}{g^2} (1+\frac{gL}{c^2})^2 = (\frac{c^2}{g} + L)^2
Now, remember that I told you that the initial location of the rear of the rocket was at \frac{c^2}{g}? The initial location of the front of the rocket was \frac{c^2}{g} + L. Letting X_{rear} be the initial location of the rear of the rocket, and X_{front} be the initial location of the front of the rocket, then we find that
x^2 - c^2 t^2 = \frac{c^4}{g^2} = (X_{rear})^2
(x_2)^2 - c^2 (t_2)^2 = (X_{front})^2
What this means is that the front and rear are following the same sort of accelerated trajectory, but with
different accelerations:
x = X_{rear} cosh(\frac{gT}{c})
x_2 = X_{front} cosh(\frac{gT}{c})
In general, if a section of the rocket starts off at location X, then its position later will be given by:
x = X cosh(\frac{gT}{c})
with the time given by:
t = \frac{1}{c} X sinh(\frac{gT}{c})
The interesting thing about this is that even though the acceleration experienced by the rear of the rocket is g_{rear} = \frac{c^2}{X_{rear}}, the acceleration experienced by the front of the rocket is smaller: g_{front} = \frac{c^2}{X_{front}}. The proper time computed earlier for a clock in the rear of the rocket is: \tau = T. But the proper time for the front of the rocket is \tau = \frac{g T X_{front}}{c^2}. So clocks in the front run faster than clocks in the rear. (proper time is only equal to T at the rear of the rocket).