Can we use the Schwarzschild metric under the horizon?

In summary, the continuity of the velocity dv/dtau above the horizon is valid, but below the horizon the velocity is discontinuous. The radius at which the continuity breaks down is called the horizon.
  • #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?
 
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  • #2
mersecske said:
This formula is used every papers below and above the horizon.

Could you give references to soem specific papers that do this?
 
  • #4
You have linked this book:
https://www.physicsforums.com/showthread.php?t=418941
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 referred papers you can see that the signum is not always +1
 
  • #5
mersecske said:
Why is positive for example above the horizon?

Above the horizon, only the plus sign is valid because 4-velocity is a future-directed timelike vector. If [itex]dt/d\tau[/itex] were negative above the horizon, then the material in the shell would be traveling into the past instead of into the future.
mersecske said:
You have linked this book:
https://www.physicsforums.com/showthread.php?t=418941
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?

I think that this example only means to treat the case above the horizon.
mersecske said:
In the referred papers you can see that the signum is not always +1

Below the horizon, [itex]t[/itex] is a spatial coordinate, so [itex]dt/d\tau[/itex] can be positive or negative, just as [itex]dr/d\tau[/itex] can be positive or negative above the horizon.
 
  • #6
Below the horizon r is time coordinate, isn't 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
So -r is (one of) the time coordinate below the horizon?
 
  • #8
Sorry for forgetting about this.
mersecske said:
Below the horizon r is time coordinate, isn't it?

Yes.
mersecske said:
In this case dr/dtau has to be positive, but this cannot be true.
Or -r is the right time coordinate?

r is a past-directed timelike coordinate, and -r is a future-directed timelike coordinate.
mersecske said:
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

Signature doesn't have an ordering. For two common conventions for signature, see
http://en.wikipedia.org/wiki/Metric_signature.
mersecske said:
So -r is (one of) the time coordinate below the horizon?

Yes.
 
  • #9
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,
this radius is R=mr^2/(2mg)

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
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?
 

Related to Can we use the Schwarzschild metric under the horizon?

1. What is a thin shell under the horizon?

A thin shell under the horizon refers to a hypothetical situation in which an object or structure is located beneath the surface of a black hole. This thin shell would exist at the event horizon, which is the point of no return for anything that enters a black hole.

2. How is a thin shell under the horizon possible?

A thin shell under the horizon is possible due to the extreme gravitational pull of a black hole. This pull causes space and time to become distorted, allowing for the existence of a thin shell at the event horizon.

3. What would happen to an object inside a thin shell under the horizon?

An object inside a thin shell under the horizon would experience extreme tidal forces due to the intense gravitational pull of the black hole. These forces would stretch and distort the object, eventually tearing it apart.

4. Can we observe a thin shell under the horizon?

Currently, there is no way to directly observe a thin shell under the horizon as it is located inside a black hole. However, scientists can study the effects of the event horizon on surrounding matter to gain insight into the behavior of objects inside a thin shell under the horizon.

5. What implications does a thin shell under the horizon have for our understanding of the universe?

The existence of a thin shell under the horizon has significant implications for our understanding of gravity and the behavior of matter in extreme environments. It also provides insight into the nature of black holes and their role in shaping the structure of the universe.

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