# Tensor infinities

1. Jul 18, 2008

### snoopies622

If in a particular coordinate system every component of a tensor is zero, then they are also zero in any other coordinate system. Is there an analogous relationship with infinities? If a tensor has at least one component equal to infinity in one coordinate system, must this be the case in all coordinate systems?

2. Jul 18, 2008

### Hurkyl

Staff Emeritus
By definition, the components of the coordinate representation of a tensor are real numbers, and therefore cannot be infinite.

3. Jul 18, 2008

### snoopies622

Perhaps I should put it this way. Suppose I have a tensor field with singularities -- that is -- points at which at least one of the tensor components becomes undefined. Can a coordinate transformation of the field eliminate the singularities? My guess is "yes", but this would mean that one would not be able to find the components of the tensor field at a (former) singularity in the new coordinate system by using the tensor transformation law that is appropriate for the rest of the field, and yet they would still be the components of a tensor. Is this correct?

4. Jul 18, 2008

### Mentz114

One can transform away the singularity at the horizon of a Schwarzschild space-time but not the one a r=0.

In the FLRW space-time it's possible to transform away the big bang singularity at t=0 ( I heard this from Prof. Steven Weinberg himself).

5. Jul 18, 2008

### Hurkyl

Staff Emeritus
If the field doesn't exist at some point, then you can't change that fact by choosing a coordinate chart, and then looking at the coordinate representation of that field.

Coordinate changes don't change the field -- they only change the coordinate representation of the field.

6. Jul 18, 2008

### snoopies622

Does this include the curvature tensor singularities? It looks like six of it's 256 components become undefined there. If I understand what Hurkyl is saying, this tensor does not exist at $$r=r_s$$ in any coordinate system.

7. Jul 19, 2008

### haushofer

No. You have to distinguish between two singularities: coordinate singularities and physical singularities. One can do that by computing scalar curvature quantities like $$R^{\mu\nu\rho\sigma}R_{\mu\nu\rho\sigma}$$ for a particular solution of Einstein's equations ( I believe it's called the Kretsch scalar ). If this quantity blows up, it will blow up in every coordinate system. For the Schwarzschild radius this quantity is perfectly finite, so this indicates that the Schwarzschild radius exhibits a coordinate singularity, and that it can be removed by maximal analytic extension.

Compare this with someone who flies into a black hole. An observer from the outside observes that this person will stand still at the Schwarzschild radius and never enters the black hole; the coordinate time of this process is infinite. The person itself however just goes by this radius; the eigentime to pass the Schwarzschild radius is finite.

We say that we can extend geodesics from outside the Schwarzschild radius to the interior of the black hole. However, this is not possible for the r=0 singularity; here the geodesics terminate.

8. Jul 19, 2008

### snoopies622

So at the Schwarzschild radius, the metric tensor can be recovered with a coordinate transformation, but the Riemann tensor cannot be?

9. Jul 21, 2008

### haushofer

At the Schwarzschild radius the metric appears to be singular in spherical coordinates, but you can transform this away with a proper coordinate transformation. Everything behaves nicely there, and because the Riemann tensor is a function of the metric and its first two derivatives, also the Riemann tensor will behave nicely. The curvature is not "infinite" at the Schwarzschild radius, so there is no reason for the Riemann curvature tensor to behave badly there.

A frequently maded comparison is the r=0 singularity if you use polar coordinates in flat 2dimensional space. The Jacobian dissapears at r=0, but ofcourse this is just our choice of coordinates ( or, our choice of the origin ). If we would calculate the Riemann tensor in polar coordinates in 2 dimensions it would be zero because the connection terms are zero.

That r=0 is a problematic point can be seen by the fact that for r=0 but an arbitrary angle we remain at the same point; $$(r, \phi) = (0,\phi)$$ for arbitrary $$\phi$$ is one single point on the manifold. That's no good, because we would like that every different coordinate indicates one unique point. Often we need more than one chart to cover a manifold.

10. Jul 21, 2008

### Hurkyl

Staff Emeritus
I should explicitly point out that a coordinate singularity lies outside of the domain of a coordinate chart. Since the coordinates at such a point are not well-defined, all bets are off when you try to represent a tensor relative to those coordinates.