# Accelerating away from earth, time such that a light beam won't catch up with you

1. Nov 18, 2012

### zardiac

1. The problem statement, all variables and given/known data
You start at t=0 at rest on earth and accelerate with uniform acceleration a away form earth.
Find a point in time $t_0$ such that when a beam emitted from earth at $t>t_0$won't catch up.

2. Relevant equations
$x(t)=c^2/a(\sqrt{1+\frac{a^2}{c^2}t^2}-1)$

3. The attempt at a solution
I think that light travel with velocity c. So if the beam is emitted at $t=t_1$ then at time t, the beam have traveled $c(t-t_1)$. So I try to find the solution for $x(t)=c(t-t_1)$, and I end up with the following expression for $t$:
$t=\frac{a}{2c}\frac{t_1(2-a/c t_1)}{(a/c - a^2/c^2 t_1)}$

According to this the time would be negatic in the intervall $t_1=c/a$ and $t_1=2c/a$ So I think in this intevall the beam won't be able to catch up, but after $t_1=2c/a$ the time becomes positive again, which I don't know how to interpret.
Am I approaching this problem the wrong way?

2. Nov 18, 2012

### voko

Think about it this way: will you ever accelerate to a speed greater than that of light? If not, how can you possibly outrun light?

3. Nov 18, 2012

### zardiac

Well it is problem 3.9 in D'inverno Introducing Einsteins relativity. I agree that it seem impossible but the problem statement is that if you get a large enough headstart the light won't catch up.

4. Nov 18, 2012

### voko

What assumptions are you supposed to make?

5. Nov 18, 2012

### SteamKing

Staff Emeritus
negatic?

6. Nov 19, 2012

### vela

Staff Emeritus
You should be able to show that your world line is a hyperbola. Find its asymptotes.

7. Nov 19, 2012

### TSny

Note t approaches ∞ as the denominator on the right approaches 0.

8. Nov 26, 2012

### Myslius

Let's say you keep uniform acceleration a, relative to the stationary observer. After c/a time you will be moving at the speed of light. To keep uniform acceleration you need infinite amount of energy. I think the answer is c/a, just the problem is that you can't keep uniform acceleration.

9. Nov 26, 2012

### zardiac

10. Nov 28, 2012

### vela

Staff Emeritus
You've misinterpreted the problem. The acceleration is uniform relative to the moving observer. As you noted, you can't have a uniform acceleration relative to the stationary observer indefinitely.