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

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

 

[Altabeh]

Consequently,

-c &lt; \frac{dx}{dt} &lt; 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 &lt; \frac{dx}{dt} &lt; c