- #1
Frostman
- 114
- 17
- Homework Statement
- A rocket, starting from the earth, moves away from it subjected to ##a## constant acceleration in the reference system of instantaneous rest of the rocket itself.
a) Obtain the law of motion, that is the dependence of the distance of the rocket from the earth, as a function of the earth's time ##t##.
b) Show that there is a maximum time ##T## after departure, after which it is impossible to send messages capable of reaching the rocket from the ground. Calculate the dependence of ##T## on ##a##
- Relevant Equations
- Tensor formalism
I thought I'd start by writing the problem in a tensor formalism. I have identified with ##S## the Earth and ##S'## the rocket. Since the acceleration provided is in the rocket's frame of reference, I can write the following four-vector.
$$
a'^\mu=(0, a, 0, 0)
$$
Since we are interested in the equation of motion in the frame of the earth, I go directly to the frame of reference ##S##. Applying the Lorentz transformations.
$$a^0 = \gamma(a'^0+va'^1)=\gamma v a$$$$a^1 = \gamma(a'^1+va'^0)=\gamma a$$
But I get the four-vector acceleration starting from the derivative of the four-vector velocity with respect to proper time.
$$a^\mu = \frac{d}{d\tau}(\gamma, \gamma v_x, \gamma v_y, \gamma v_z)=(\gamma \frac{d(\gamma)}{dt},\gamma \frac{d(\gamma v_x)}{dt},\gamma \frac{d(\gamma v_y)}{dt},\gamma \frac{d(\gamma v_z)}{dt})$$
In particular
$$a^0 =\gamma v a=\gamma \frac{d(\gamma)}{dt}$$$$a^1=\gamma a = \gamma \frac{d(\gamma v_x)}{dt}$$
Considering the second result:
$$\int_0^v d(\gamma v) = a\int_0^t dt$$
So we have:
$$ \frac{v}{\sqrt{1-v^2}}=at$$$$v(t)=\frac{at}{\sqrt{1+a^2t^2}}$$
By integrating speed over time we have:
$$x(t)=\frac 1a \bigg(\sqrt{1+a^2t^2}-1\bigg)$$
Can it be exhaustive?
The interesting thing is that for big times (##t\rightarrow +\infty##), the speed tends to ##1##. So it comes back to me as a result.
For the second point how could I operate? From this time ##T##, since no luminous message sent from Earth will never be able to reach the rocket, it means that the rocket is also traveling at speed ##1##. I would conclude that this time ##T## is for ##T\rightarrow +\infty##. But it seems trivial and stupid (because I have no dependence from ##a##), in what terms should I reason?
$$
a'^\mu=(0, a, 0, 0)
$$
Since we are interested in the equation of motion in the frame of the earth, I go directly to the frame of reference ##S##. Applying the Lorentz transformations.
$$a^0 = \gamma(a'^0+va'^1)=\gamma v a$$$$a^1 = \gamma(a'^1+va'^0)=\gamma a$$
But I get the four-vector acceleration starting from the derivative of the four-vector velocity with respect to proper time.
$$a^\mu = \frac{d}{d\tau}(\gamma, \gamma v_x, \gamma v_y, \gamma v_z)=(\gamma \frac{d(\gamma)}{dt},\gamma \frac{d(\gamma v_x)}{dt},\gamma \frac{d(\gamma v_y)}{dt},\gamma \frac{d(\gamma v_z)}{dt})$$
In particular
$$a^0 =\gamma v a=\gamma \frac{d(\gamma)}{dt}$$$$a^1=\gamma a = \gamma \frac{d(\gamma v_x)}{dt}$$
Considering the second result:
$$\int_0^v d(\gamma v) = a\int_0^t dt$$
So we have:
$$ \frac{v}{\sqrt{1-v^2}}=at$$$$v(t)=\frac{at}{\sqrt{1+a^2t^2}}$$
By integrating speed over time we have:
$$x(t)=\frac 1a \bigg(\sqrt{1+a^2t^2}-1\bigg)$$
Can it be exhaustive?
The interesting thing is that for big times (##t\rightarrow +\infty##), the speed tends to ##1##. So it comes back to me as a result.
For the second point how could I operate? From this time ##T##, since no luminous message sent from Earth will never be able to reach the rocket, it means that the rocket is also traveling at speed ##1##. I would conclude that this time ##T## is for ##T\rightarrow +\infty##. But it seems trivial and stupid (because I have no dependence from ##a##), in what terms should I reason?
