Warning: I am not a GR expert, I know more about SR than GR.
DaleSpam said:
In the usual "windowless room" thought experiment it is stated that there is no experiment which can be performed which will determine if the room is undergoing "uniform accleration" far from any massive body or is in a "uniform gravitational field". The two situations are therefore said to be "equivalent". Also, since the spacetime in the uniform acceleration case is obviously flat the equivalent spacetime in the uniform gravitation must also be flat.
The equivalence principle is usually considered in a
small room so that tidal effects can be ignored. In a
large room, what does “uniform acceleration” mean? Well, an observer who is static (relative to the room) should experience proper acceleration that is constant over time. Beyond that:
(a) Do you want
all static objects to experience the
same proper acceleration as each other?
(b) Do you want the room to remain a constant size (according to its occupants)?
Conditions (a) and (b) are incompatible. (a) is the acceleration in paradox[/url]. (b) is
Born rigid acceleration.
I believe I am correct is saying that in relativity “uniform acceleration” is defined to be (b), even though that may seem a counterintuitive name. This also raises the question of how does an observer in the room measure the room’s size. Radar won’t give a valid answer because the observer is accelerating. What the observer must do is jump in the air and make radar measurements while airborne. Whilst airborne, the observer is an inertial observer (ignoring air resistance). The measurement deemed to be simultaneous with the top of the jump, when the observer is stationary relative to the room, is deemed to be the valid measurement for the accelerating frame at that instant. The jumping observer defines a “co-moving inertial frame” at the top of the jump. Simultaneity, distance and time for the accelerating frame are all defined via the appropriate co-moving inertial frame, i.e. a jumping observer. The proper acceleration is also the acceleration of the room relative to the co-moving inertial frame.
If you have a number of jumping observers (reaching their peaks at different times) they can use the Lorentz transform to convert between each other’s points of view.
gel said:
For a gravitation acceleration of a in the -ve x direction at the origin, I get the following metric
<br />
d\tau^2 = \left(1+c^{-2}ax\right)^2dt^2-c^{-2}(dx^2+dy^2+dz^2)<br />
where \tau is the proper time.
It is non-uniform, the gravitational acceleration at coordinate x is \frac{a}{1+c^{-2}ax}. Note, it blows up at x=-c2/a.
And, you can transform to inertial coords by
<br />
x'=(x+a^{-1}c^2)\cosh(c^{-1}at)-a^{-1}c^{2}, t'=c^{-1}(x+a^{-1}c^2)\sinh(c^{-1}at), y'=y,z'=z.<br />
This is correct. (t, x, y, z) are
Rindler coordinates, relative to the room. (t', x', y', z') are inertial coordinates. In fact, they are the co-moving inertial frame at all the events where t = 0.
More generally at any event
E where t = t_0, it can be proved that the co-moving inertial frame is given by
x'_E = \left( x + \frac{c^2}{a} \right) \cosh \frac {a (t - t_0)} {c} - \frac{c^2}{a}
t'_E = \frac {1}{c} \left( x + \frac{c^2}{a} \right) \sinh \frac {a (t - t_0)} {c}
t is the proper time of the observer at x = 0. For any other object stationary in the room, the proper time is
\tau = t \left(1 + \frac {ax}{c^2} \right)
This illustrates gravitational time dilation. Two events with the same t coordinate are simultaneous in the room’s frame (i.e. in the co-moving inertial frame).
gel said:
btw, I think this is what's called Born rigid motion. It's flat because the curvature tensor vanishes, and can be transformed to Minkowski space.
This is also true, Any equation for the metric that can be transformed by change of variable into the standard Minkowski equation
must be flat, by definition.
How to calculate the proper acceleration.
Use the equations I gave above for the co-moving inertial frame. You can simplify the maths a bit by moving the spatial origin (in both frames) to get
x'_E = x \cosh \frac {a (t - t_0)} {c}
t'_E = \frac {x}{c} \sinh \frac {a (t - t_0)} {c}
For a constant value of x, calculate dx' / dt and dt' / dt. Divide these to get the velocity dx' / dt' measured in the co-moving frame. Differentiate this with respect to t and divide by dt' / dt to get the acceleration d^2x' / dt'^2 measured in the co-moving frame. Finally put t = t_0 to get the event when the inertial frame is co-moving; this, by definition, is the proper acceleration, which should come out as c^2 / x, which, after the change of origin, is equivalent to the value given by
gel in post #4.
On a final note, uniform acceleration gives rise to an event horizon. Any photons in the region
x' + \frac{c^2}{a} \leq c t'
(in
gel’s original coordinates) will never reach the accelerating room.
