Tom Roberts wrote:
To an accelerated observer, the horizon has no "color", as no
radiation
from it can ever reach the observer. An horizon is just the boundary
between a region that is causally-connected to the observer and a
region
that is not. It has no more "existence" or "reality" than the
international date line -- but like that date line it has observable
consequences.

For instance, no local experiment can determine when a
horizon is crossed; global observations are required.
No shipboard clock can determine when the date line is
crossed; observations of shorebound clocks are required.

Maybe I am wrong - but is there not some Unruh radiation coming from
(near) the horizon in the same was that some BH radiation
comes from (near) a black hole horizon? I know the two are different,
but asking is never dangerous...

The question would then be how this "color" (due to Unruh radiation)
varies over the surface of the Rindler horizon. Charles and I would
guess
that the radiation must get darker and redder at higher angles.
True or not?

(David McAnally) writes:

The trajectory of the accelerating observer is given by

t = 1/g sinh(gs),

x = 1/g cosh(gs) - 1/g,

y = 0,

z = 0,

where g is the constant acceleration, and is in the positive x direction,
the observer is stationary at the origin at time t = 0, s is the proper
time that has passed for the observer since t = 0, and c has been set
equal to 1.

The light that reaches the observer at proper time s starts at t = 0 from
a sphere with centre (1/g cosh(gs) - 1/g, 0, 0) and radius 1/g sinh(gs).
As s approaches infinity, the radius approaches infinity, and the sphere
approaches the plane x = - 1/g. This plane is the Rindler horizon.

In a more general case, the trajectory of the accelerating observer is
given by

t = 1/g cosh(K) sinh(gs),

x = 1/g cosh(gs) - 1/g,

y = 1/g sinh(K) sinh(gs),

z = 0,

where

g is the constant acceleration relative to the observer's
instantaneous rest frame, and is in a constant direction relative
to the observer,

the observer is at the origin at time t = 0, travelling in the y
direction with speed tanh(K), and accelerating in the positive x
direction,

s is the proper time that has passed for the observer since t = 0,

and c has been set equal to 1.

This is the most general form of the trajectory, up to rotations in space,
and translations in space and time.

When the proper time is s, then the velocity is

(sech(K) tanh(gs), tanh(K), 0).

The light that reaches the observer at proper time s starts at t = T from
a sphere with centre (1/g cosh(gs) - 1/g, 1/g sinh(K) sinh(gs), 0) and
radius 1/g cosh(K) sinh(gs) - T. This sphere passes through the point
(1/g cosh(gs) - 1/g - 1/g sinh(gs) + T sech(K), T tanh(K), 0), which
approaches (T sech(K) - 1/g, T tanh(K), 0) as s approaches infinity. As
s approaches infinity, the radius approaches infinity, and the sphere
approaches the plane orthogonal to (sech(K), tanh(K), 0), and passing
through (T sech(K) - 1/g, T tanh(K), 0). The equation of the plane is

x sech(K) + y tanh(K) = T - 1/g sech(K).

This plane is the Rindler horizon at t = T.

David

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