Special relativity - constant acceleration

[Aleolomorfo]

Homework Statement

A rocket is flying away from the Earth with a costant acceleration $a$ in a frame in which the rocket is at rest. Finding the equation of motion (the relation between the distance from the Earth and the terrestrial time). Showing that there is a maximum time $T$ after the departure, after which it's impossible to send messages from the Earth which can reach the rocket; calculating the relation between $T$ and $a$.

Homework Equations

$$a^\mu=\biggr(\gamma\frac{d\gamma}{dt},\gamma\frac{d(\gamma\vec{v})}{dt}\biggr)$$

The Attempt at a Solution

I've done the first part without problems and the result is:
$$x=\frac{\sqrt{1+a^2t^2}-1}{a}$$
But for the second part I don't know what to do, I need a hint.

 

[Orodruin]

What is the world line of a signal that is sent from the Earth at the speed of light? What is the condition for the event at which this signal arrives to the rocket?

 

[Aleolomorfo]

What is the world line of a signal that is sent from the Earth at the speed of light? What is the condition for the event at which this signal arrives to the rocket?

The world line is a straight line with an inclination of 45° in t-x plane, and the condition is that $ds^2=0$. So $t^2-x^2=0$ and I can substitute the relation for $x$ found in the first part, but after calcus I've found $\sqrt{1+a^2t^2}=1$. If $a^2t^2 << 1$ the relation is approximately true, so $t<<\frac{1}{a}$. I'm not so sure this is correct, where is the mistake?

 

[Orodruin]

So $t^2-x^2=0$

This is not correct. The world line does not pass the event $(t,x) = (0,0)$ as it is released at time $t = T$ from $x = 0$. Note that $ds^2 = 0$ will give you a differential equation for $x$ as a function of $t$. To get the right world line, you also should take into account that the direction of the signal needs to be towards the rocket (this will fix the direction in which the signal travels).

I've found $\sqrt{1+a^2t^2}=1$.

This can never be satisfied as $\sqrt{1+a^2t^2} > 1$ if $t \neq 0$. What you have computed is when a light signal sent at the same time as the rocket departs meets the rocket. Since the rocket can never reach light speed or faster, the only solution is $t = 0$. You need to use a world line of a signal that is sent at a later time $t = T$.

 

[Aleolomorfo]

This is not correct. The world line does not pass the event $(t,x) = (0,0)$ as it is released at time $t = T$ from $x = 0$. Note that $ds^2 = 0$ will give you a differential equation for $x$ as a function of $t$. To get the right world line, you also should take into account that the direction of the signal needs to be towards the rocket (this will fix the direction in which the signal travels).This can never be satisfied as $\sqrt{1+a^2t^2} > 1$ if $t \neq 0$. What you have computed is when a light signal sent at the same time as the rocket departs meets the rocket. Since the rocket can never reach light speed or faster, the only solution is $t = 0$. You need to use a world line of a signal that is sent at a later time $t = T$.

Thank you, I have seen my mistake but I still have difficulties in understanding what I have to do. Do I have to impose $(t-T)^2-x^2=0$? And then what do I have to do? It seems that I just obtain a formula for $T$ and nothing else.

 

[Orodruin]

Your equation for the world-line of the signal implies that $x = \pm (t - T)$. Which of the signs is compatible with the signal being sent in the same direction as the rocket? (If you send it in the other direction it will obviously never catch it.)

What is the resulting requirement for the signal catching the rocket? If that requirement can be satisfied for a fixed $T$, then the signal will reach the rocket. If it cannot be satisfied, then the signal can never reach the rocket.

 

[Aleolomorfo]

Your equation for the world-line of the signal implies that $x = \pm (t - T)$. Which of the signs is compatible with the signal being sent in the same direction as the rocket? (If you send it in the other direction it will obviously never catch it.)

What is the resulting requirement for the signal catching the rocket? If that requirement can be satisfied for a fixed $T$, then the signal will reach the rocket. If it cannot be satisfied, then the signal can never reach the rocket.

The correct sign is plus: $x=t-T$. I think the requirment is the intersection:
$$t-T=\frac{\sqrt{1+a^2t^2}-1}{a}$$
$$T=t+\frac{1-\sqrt{1+a^2t^2}}{a}$$

I got that over a certain value of $T$ there is no intersection and so the message can't reach the rocket, but how can I extrapolate the result from the equation above. I know deeply that the result is in front of my eyes but I can't really see it.

 

[Orodruin]

What value of $T$ corresponds to $t \to \infty$?

Alternatively, solve the equation for $t$, not $T$. This $t$ is the time that the signal reaches the rocket. If there is no solution, then the signal never reaches the rocket. (Beware of false solutions introduced by squaring stuff.)

 

[Aleolomorfo]

What value of $T$ corresponds to $t \to \infty$?

Alternatively, solve the equation for $t$, not $T$. This $t$ is the time that the signal reaches the rocket. If there is no solution, then the signal never reaches the rocket. (Beware of false solutions introduced by squaring stuff.)

The limit is $1/a$ (I hope the calcus is not wrong). So I can say: if the time at which I send the signal is grater than $1/a$ it will never reach the rocket, otherwise it will, right? I think (and hope) this is the correct answer.