# Schwarzchild spacetime singularity

1. Oct 18, 2009

### alle.fabbri

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 Schwarzchild 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?

2. Oct 18, 2009

### haushofer

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.

3. Oct 18, 2009

### alle.fabbri

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?

4. Jan 13, 2010

### utku

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 doesnt say anything.observer doesnt see the any body cross the event horizon.

5. Jan 13, 2010

### bcrowell

Staff Emeritus
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.

6. Jan 14, 2010

### yuiop

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.