- #1
mersecske
- 186
- 0
Hi everybody,
The framework of infitesimally thin shells
is the well known Israel junction formalism.
Let us suppose motion of a thin in Schwarzschild spacetime.
I mean both side of the shell is desribed by Schwarzschild metric.
Let us suppose that the shell consists non-massless particles,
in this case the hypersurface of the shell has to be timelike.
The four velocity of the shell u^a = (tdot,rdot,0,0),
where t and r is Schwarzschild coordinates, dot means d/dtau,
where tau is the proper time on the shell measured by co-moving observer.
This formula is used every papers below and above the horizon.
I am a little bit confused, because t is timelike coordinate just outside the horizon!
Below the horizon, with the same Schwarzild metric,
the metric has signature (+-++) instead of (-+++)
Can we use the Schwarzschild metric under the horizon?
Lots of papers have used.
Let us suppose the metric in the form:
ds^2 = -F(r)*dt^2 + dr^2/F(r) + r^2*domega^2
In this case u_a = (-F*tdot,rdot/F,0,0) therefore
-1 = u^a*u_a = -F*tdot^2 + rdot^2/F
tdot can be eliminated from this equation,
but this equation determine only the square of tdot,
therefore we don't know its signum!
(rdot^2 can be derived independent of the signum other way)
Normally you think that tdot is positive, but not always!
For example Eid and Langer 2000 said
that tdot is positive above the horizon,
but can change sign under the horizon (but can be positive and negative also).
Its ok, but how can I determine the sign in a situation?
Why is positive for example above the horizon?
The framework of infitesimally thin shells
is the well known Israel junction formalism.
Let us suppose motion of a thin in Schwarzschild spacetime.
I mean both side of the shell is desribed by Schwarzschild metric.
Let us suppose that the shell consists non-massless particles,
in this case the hypersurface of the shell has to be timelike.
The four velocity of the shell u^a = (tdot,rdot,0,0),
where t and r is Schwarzschild coordinates, dot means d/dtau,
where tau is the proper time on the shell measured by co-moving observer.
This formula is used every papers below and above the horizon.
I am a little bit confused, because t is timelike coordinate just outside the horizon!
Below the horizon, with the same Schwarzild metric,
the metric has signature (+-++) instead of (-+++)
Can we use the Schwarzschild metric under the horizon?
Lots of papers have used.
Let us suppose the metric in the form:
ds^2 = -F(r)*dt^2 + dr^2/F(r) + r^2*domega^2
In this case u_a = (-F*tdot,rdot/F,0,0) therefore
-1 = u^a*u_a = -F*tdot^2 + rdot^2/F
tdot can be eliminated from this equation,
but this equation determine only the square of tdot,
therefore we don't know its signum!
(rdot^2 can be derived independent of the signum other way)
Normally you think that tdot is positive, but not always!
For example Eid and Langer 2000 said
that tdot is positive above the horizon,
but can change sign under the horizon (but can be positive and negative also).
Its ok, but how can I determine the sign in a situation?
Why is positive for example above the horizon?