OUT Hyperbolic Motions | 65 Characters

[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

 

[George Jones]

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

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.

 

[Altabeh]

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.

Yeah, I got it!

Thanks

 

[George Jones]

Also, differentiating

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

gives

[tex]\frac{dx}{dt} = c \frac{ct}{x}.[/itex]<br /> <br /> Consequently,<br /> <br /> $$-c < \frac{dx}{dt} < c$$<br /> <br /> gives that $\left(ct , x \right)$ lies inside the horizons.[/tex]

 

[Altabeh]

Consequently,

$$-c < \frac{dx}{dt} < c$$

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

Could you explain this a little bit more?

 

[DrGreg]

Could you explain this a little bit more?

Combine the following and what do you get?

$$\frac{dx}{dt} = c \frac{ct}{x}.$$
$$-c < \frac{dx}{dt} < c$$