# Hyperbolic motions

1. Feb 12, 2010

### Altabeh

Hi everybody

I've been lately a little bit concerned over the hyperbolic motions that have the following equations in (ct,x)-space:

$$\frac{x^2}{(c^2/a)^2}-\frac{(ct)^2}{(c^2/a)^2}=1$$.

We know that events horizons are the lines that form a 45-degree angle by both ct- and x-axis. So what does actually assure us that here, for instance, for t=0, $$x=\pm c^2/a$$ lie inside events horizens? Is this just because $$a$$ can't in magnitude gets higher than $$c$$?

AB

2. Feb 12, 2010

### George Jones

Staff Emeritus
No.

$$x^2 - \left(ct\right)^2 = \left( \frac{c^2}{a} \right)^2,$$

so $a \rightarrow \infty$ gives the horizons. For $t=0$, any value of $x$ except $x = 0$ lies inside the horizons.

3. Feb 12, 2010

### Altabeh

Yeah, I got it!

Thanks

4. Feb 12, 2010

### George Jones

Staff Emeritus
Also, differentiating

$$x^2 - \left(ct\right)^2 = \left( \frac{c^2}{a} \right)^2,$$

gives

$$\frac{dx}{dt} = c \frac{ct}{x}.[/itex] Consequently, [tex]-c < \frac{dx}{dt} < c$$

gives that $\left(ct , x \right)$ lies inside the horizons.

5. Feb 12, 2010

### Altabeh

Could you explain this a little bit more?

6. Feb 12, 2010

### DrGreg

Combine the following and what do you get?