Horizon for Constant Acceleration - Special Relativity

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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.
 
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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
 
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Michael_1812 said:
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.