Schwarzchild spacetime singularity

alle.fabbri
Messages
31
Reaction score
0
Hi all! I'm studying black holes and there's a point that I cannot understand. The book I'm reading is Modeling black hole evaporation, by Fabbri and Navarro Salas. The path is the following.
After introducing the Schwarzschild metric
ds^2 = \left(1 - \frac{2M}{r} \right) \ dt^2 - \left(1 - \frac{2M}{r} \right)^{-1} \ dr^2 - r^2 d \Omega^2
they get the radial null geodesic equation
dt^2 = \frac{dr^2}{\left(1 - \frac{2M}{r} \right)^2 }
that once solved gives, implicitly, the law of motion for a ray of light radially falling
t = r - 2M \ln \frac{|r-2M|}{2M}
Inspired by this one we can introduce the ingoing Eddington-Finkelstein coordinate by means of
v = t + r - 2M \ln \frac{|r-2M|}{2M}
and then switch to another system of coordinate, in order to remove the singularity in r=2M which is not physical. And here problems begin. One can make two choices for the coordinate system:
- the set (v,r,\Omega) for which the metric becomes
ds_r^2 = - \left(1 - \frac{2M}{r} \right) \ dv^2 - 2 dr dv - r^2 d \Omega^2
or
- the set (v,t,\Omega) for which
ds_t^2 = - \left(1 - \frac{2M}{r} \right) \left( dv^2 - 2 dt dv \right) - r^2 d \Omega^2
This is a straightforward calculation, so no probs. Then they say

" It is clear that only in the first case we can analytically continue the metric to all possible values of the radial coordinate r>0. In the second case we still have a singularity at r=2M. The coordinates (v,r,\Omega) are called the ingoing (or advanced) coordinates and because of the cross term drdv the metric is not singular at r=2M."


And I really don't understand the meaning of this. Any insight?
 
Physics news on Phys.org
Well, the essential problem I think is that at r=2M the determinant of the metric is 0 in the second case (just write down the matrix representing the metric). That's a problem, because you would like the vector spaces and their duals to be isomorphic. In the first case you don't have this problem, and you can do an analytic extension beyond this sphere r=2M. Soe everything is well-defined here.
 
So the point is that when the metric has zero determinant is not invertible so we can't use it to define covariant vectors (crudely speaking: you can't lower tensor indices?), i.e. the elements of the dual space?
 
Hi friends.
In schwarzchild vacuum solition spacetime metric singularity exist when you put the r=2gm (also called schwarzhild radius),when we use Eddington-Finkelstein coordinat we can see this singularity is depend the coordinat.So it is not real singularity.
When we transform our coordinat to Kruskal's coordinats so solition is extended all region of black holes so there exist white hole and worm hole.
All the transformations of Sch metri exhibite that the singularity of r=2gm is depend our coordinat.
example,
when we fall in black hole we can croos event horizon(regardless of tidal fores) but a observer at infinity doesn't say anything.observer doesn't see the anybody cross the event horizon.
 
alle.fabbri said:
So the point is that when the metric has zero determinant is not invertible so we can't use it to define covariant vectors (crudely speaking: you can't lower tensor indices?), i.e. the elements of the dual space?

Right. Another way of saying it is that GR is completely independent of the coordinates you express it in, except for certain restrictions. The restrictions are that the functions describing a change of coordinate have to be smooth and one-to-one. If they weren't one-to-one, then you could have spacetime events that were identical in one coordinate system, but different in the other coordinate system. That would mean that one observer would say that particles A and B collided, but another observer would say that A and B didn't collide.

As an example, suppose you transform from Cartesian coordinates to polar coordinates in the Euclidean plane. The mapping isn't one-to-one, because for r=0, all the different values of theta describe the same point. This expresses itself as a coordinate singularity at the origin. The coordinate singularity has no physical significance, because space is still flat and Euclidean everywhere. At the origin, g_{\theta\theta} vanishes, and the metric is not invertible. This is simply because the change of coordinates wasn't invertible at that point.
 
alle.fabbri said:
...
they get the radial null geodesic equation
dt^2 = \frac{dr^2}{\left(1 - \frac{2M}{r} \right)^2 }
that once solved gives, implicitly, the law of motion for a ray of light radially falling
t = r - 2M \ln \frac{|r-2M|}{2M}

The last equation has to be treated with caution. If you carry out indefinite integration of 1/(1-2*m/r) with respect to r the result is:

t = r + 2M\ ln(r-2M) +k

and not t = r - 2M \ln \frac{|r-2M|}{2M} or t = r - 2M \ln {|r-2M|}

as can be confirmed by copying and pasting 1/(1-2*m/r) into this online integrator http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=calculus&s2=integrate&s3=advanced .

I have added in the constant of integration k.
For values of r>2M the constant of integration is zero for real values of t.
For values of r<2M the constant of integration is -\pi i for real values of t.

The requirement for two different constants of integration implies the function is not smooth and continuous across r=2M.

If you use the online integrator to carry our definite integration using variables and limits of r,1/2m,m or r,2*m,4*m the values for t are real, but if you try to integrate across r=2m using for example r,1/2*m,2*m the result is complex and again implies the path is smooth and continuous across r=2m.

(The missing denominator of 2M inside the log function is not a concern as the end result is the same)

Now by using two different constants of integration, which is what they are effectively doing by inserting an arbitary absolute function within the log function, the continuity of the light path across r=2m is artificially and arbitarily built in at an early stage of the EF derivation.
 
OK, so this has bugged me for a while about the equivalence principle and the black hole information paradox. If black holes "evaporate" via Hawking radiation, then they cannot exist forever. So, from my external perspective, watching the person fall in, they slow down, freeze, and redshift to "nothing," but never cross the event horizon. Does the equivalence principle say my perspective is valid? If it does, is it possible that that person really never crossed the event horizon? The...
In this video I can see a person walking around lines of curvature on a sphere with an arrow strapped to his waist. His task is to keep the arrow pointed in the same direction How does he do this ? Does he use a reference point like the stars? (that only move very slowly) If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in? So ,although one refers to intrinsic curvature...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
Back
Top