# Thin shell under the horizon

1. Jul 29, 2010

### mersecske

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 dont 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?

2. Jul 29, 2010

### George Jones

Staff Emeritus
Could you give references to soem specific papers that do this?

3. Jul 29, 2010

### mersecske

Last edited by a moderator: Apr 25, 2017
4. Jul 29, 2010

### mersecske

in an other topic: Thin shell velocity is greater than speed of light?

The signum is also missed in this book at equation (16.62)
From equation (16.61) only the square of tdot can be extracted,
therefore in the right side of (16.62)
instead of tdot -> +-tdot is the right expression
How can we found the right signum?
In the refered papers you can see that the signum is not always +1

5. Jul 31, 2010

### George Jones

Staff Emeritus
Above the horizon, only the plus sign is valid because 4-velocity is a future-directed timelike vector. If $dt/d\tau$ were negative above the horizon, then the material in the shell would be traveling into the past instead of into the future.
I think that this example only means to treat the case above the horizon.
Below the horizon, $t$ is a spatial coordinate, so $dt/d\tau$ can be positive or negative, just as $dr/d\tau$ can be positive or negative above the horizon.

6. Jul 31, 2010

### mersecske

Below the horizon r is time coordinate, isnt it?
In this case dr/dtau has to be positive, but this cannot be true.
Or -r is the right time coordinate?

It is possible to use metric in the Einstein theory,
which has different signature in different regions?
Because the Schwarzschild metric
has -+++ signature above the horizon,
and +-++ under the horizon

7. Aug 3, 2010

### mersecske

So -r is (one of) the time coordinate below the horizon?

8. Aug 3, 2010

### George Jones

Staff Emeritus
Yes.
r is a past-directed timelike coordinate, and -r is a future-directed timelike coordinate.
Signature doesn't have an ordering. For two common conventions for signature, see
http://en.wikipedia.org/wiki/Metric_signature.
Yes.

9. Aug 8, 2010

### mersecske

dt/dtau = +-sqrt(f(r) + rdot^2)/f(r)

Ok, the signum is + above the horizon,
because f(r) > 0 and dt/dtau has to be positive!

Below the horizon t is nomore time coordinate,
therefore the sign can be +- !

However if we assume that the shell just comes thru the horizon
and we use eddington ingoing null coordinate (v) avoiding the coordinate singularirty
-> the the velocity dv/dtau has to be continuous:

dv/dtau = (+-sqrt(f(r) + rdot^2)+rdot)/f(r)

since the sqrt term is not zero at the horizon,
the sign + has to be valid to be continuous.

But there is a radius r=R,
where sqrt(f(r) + rdot^2) = 0 in the outer Schwarzschild spacetime
(in the inner one, this is always positive)
For a dust sell, with constant mass parameters,

Below this radius we cannot use the continuity argument,
because the sqrt term is zero!
but we can use the continuity argument for the derivative!

This is right?

10. Aug 28, 2010

### mersecske

There is another issue about shells.
Authors usually define inside- and outside- region
in spherically symmetric spacetimes such a way that
in the direction of the space-like normal vector
to the hypersurface (world sheet) of the shell
(pointing into the region) the area radius increase or decrease.
However this definition is not always match to our intuition,
because under the horizon it is possible to chage inside to outside
along a time-like geodetic. This is the case when the world-line
cross one of the r=constant coordinate curves orthogonally.
Are there any sense to define inside- outside locally in general?