# Basic metric diagonalization questions

1. May 11, 2006

### lonelyphysicist

I understand it is always possible to diagonalize a metric to the form

diag$[1,-1,\dots,-1]$

at any given point in spacetime because the metric is symmetric and we can always re-scale our eigenvectors.

But is this achievable via a coordinate transformation? That is, would the basis vectors in such a diagonalized metric always be coordinate vectors $\{ \partial / \partial x^i \}$? More explicitly, if we start with the coordinates $\{ y^i \}$, can we always find $\{ x^i \}$ such that

$diag[1,-1,\dots,-1]_{ij} = \frac{\partial y^{a'}}{\partial x^i} \frac{\partial y^{b'}}{\partial x^j} g_{a'b'}$

Last edited: May 11, 2006
2. May 11, 2006

### pmb_phy

Yes. You can diagonalize the metric to diag(1,-1,-1,-1) from something else by a coordinate transformation.

Pete

3. May 11, 2006

### pervect

Staff Emeritus
The answer is basically yes. If you have an orthonormal basis of vectors at some point, $x^i$ you also have an orthonormal basis of one forms at that point, $x_i = g_{ij} x^j$

The orthonormal basis of one-forms defines a local coordinate system. You may or may not have problems with the behavior of the coordinate system over all of space-time, but "close to" the point in question, you can use the one-forms to define a coordinate basis. You need to map points nearby the point of origin into vectors via some arbitrary mapping (one point -> one vector). Then the one-forms give you the diagonalized coordinates directly from the vectors, because a one-form maps a vector into a scalar. Your 4 basis one-forms give you 4 coordinates.

If this is too abstract, "Fermi normal" coordinates are a good non-abstract example of a coordinate system with the desired properties near the point in question.

4. May 11, 2006

### clj4

There is a theorem in linear algebra that states that any linear transform that is represented by a symmetric matrix in one basis can be represented by a diagonal matrix through a change of basis. The diagonal matrix has the eigenvalues on the diagonal.

5. Nov 27, 2006

### Chris Hillman

Diagonalizing the metric tensor?

Hi, lonelyphysicist,

I am glad you said "at any one event" because this is NOT true (for D=4) without this crucial qualification. All spacetime models admit infinitel many coordinate charts, but many have none at all which diagonalize the metric tensor.

Ah, yes, but eigenthings work differently in $$E^{1,n-1}$$ than they do in $$E^n$$. This is discussed in the book by Barrett O'Neill, Semi-Riemannian Geometry: with Applications to Relativity, Academic Press, 1983.

Here you are probably both thinking algebraically, at the level of tangent spaces. A better way to achieve much of what you probably want, lonelyphysicist, is the notion of a coframe.

Here is a specific example: the coframe read off the usual expression for the line element of Minkowski spacetime in a cylindrical coordinate chart is $$-dt, \; dz, \; dr, \; d\phi$$. A simple coframe for Minkowski spacetime would be:
$\sigma^0 = -dt$
$\sigma^1 = dz$
$\sigma^3 = dr$
$\sigma^4 = r \, \d\phi$
The point is that the line element can be expressed as:
$ds^2 = -\sigma^0 \otimes \sigma^0 + \sigma^1 \otimes \sigma^1 + \sigma^2 \otimes \sigma^2 + \sigma^3 \otimes \sigma^3$
(Note: tensor product, not exterior product!)

The dual frame (four orthonormal vector fields) is:
$\vec{e}_0 = \partial_t$
$\vec{e}_1 = \partial_z$
$\vec{e}_2 = \partial_r$
$\vec{e}_3 = \frac{1}{r} \, \partial_\phi$
Here, the vector fields $$\vec{e}_2, \; \vec{e}_3$$ do not commute. The minus sign on dt in the coframe ensures that its dual vector is forward pointing.

A more interesting coframe, also written in cylindrical coordinates, is:
$\sigma^0 = -dt + a \, r^2 \, d\phi$
$\sigma^1 = \exp(-a^2 \,r^2/2) \, dz$
$\sigma^3 = \exp(-a^2 \,r^2/2) \, dr$
$\sigma^4 = r \, \dphi$
The corresponding line element is
$ds^2 = -dt^2 + 2 \, a \, r^2 dt \, d\phi + \exp(-a^2 \, r^2) \, \left( dz^2 + dr^2 \right) + r^2 \, d\phi^2 = -\sigma^0 \otimes \sigma^0 + \sigma^1 \otimes \sigma^1 + \sigma^2 \otimes \sigma^2 + \sigma^3 \otimes \sigma^3$
which happens to give an exact dust solution in gtr, the van Stockum dust (1937).

The dual frame field consists of four orthonormal vector fields (one timelike and three spacelike):
$\vec{e}_0 = \partial_t$
$\vec{e}_1 = \exp(a^2 \, r^2/2) \, \partial_z$
$\vec{e}_2 = \exp(a^2 \, r^2/2) \, \partial_r$
$\vec{e}_3 = a \, r \, d\phi + \frac{1}{r} \, \partial_\phi$
Again, not all of these commute.

In this example, the metric tensor is not diagonal (on some neighbhorhood) in any coordinate chart, but

See the article "Frame fields in general relativity" archived at http://en.wikipedia.org/wiki/User:H...ry:Mathematical_methods_in_general_relativity
and see the excellent book by Flanders, Differential Forms with Applications to the Physical Sciences or the book by Frankel, The Geometry of Physics, for more applications of differential forms to Riemannian geometry.

No. We can always construct infinitely many frames (the four vector fields dual to the four covector fields of our coframe), but these will generally not be commuting vector fields, hence the term "anholonomic basis".

Pete omitted to add: "but, in general, this can be achieved only at a single event".

Last edited: Nov 27, 2006