Horizon for Contant Acceleration

**heinz** · Jun19-09, 08:30 AM

Special relativity shows that any accelerated observer
sees an event horizon. In fact, if an observer is accelerated
by a, the horizon is at distance l=c^2/a in the direction
opposite to a.

**Michael_1812** · Jun28-09, 05:36 PM

George,
Heintz says that "if an observer is accelerated by a, the horizon is at distance l=c^2/a in the direction opposite to a."

I am trying to understand this by looking at the picture in the Fermi-Walker section of MTW, and it is not immediately apparent. Where could I look this up?

Many thanks,
Michael

**Naty1** · Jun28-09, 06:23 PM

You can try Wikipedia at
http://en.wikipedia.org/wiki/Event_h...rated_particle

**George Jones** · Jun28-09, 09:22 PM

Originally Posted by Michael_1812

George,
Heintz says that "if an observer is accelerated by a, the horizon is at distance l=c^2/a in the direction opposite to a."

I am trying to understand this by looking at the picture in the Fermi-Walker section of MTW, and it is not immediately apparent. Where could I look this up?

Many thanks,
Michael

Do you mean both horizon and horizon distance, or just horizon distance? MTW and I will not have the same spacetime coordinates until tomorrow.