# Metric tensor with diagonal components equal to zero

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1. Apr 9, 2015

### kkz23691

Hello,

Let $g_{jk}$ be a metric tensor; is it possible for some $i$ that $g_{ii}=0$, i.e. one or more diagonal elements are equal to zero? What would be the geometrical/ topological meaning of this?

2. Apr 9, 2015

### Staff: Mentor

It is certainly possible for some diagonal elements to be zero. For example, consider the metric on Minkowski spacetime in null coordinates, i.e., $u = t + x$, $v = t - x$ (for simplicity I leave out the other two spatial dimensions, they will be unchanged); we have $ds^2 = - du dv$ instead of $ds^2 = - dt^2 + dx^2$.

As for the meaning, it depends on the metric; you can't really make any general statements. In the above case, the geometry is the same as the one described by the usual Minkowski metric, so there's no real "geometric meaning" to it at all; it's just a different set of coordinates.

3. Apr 9, 2015

### Orodruin

Staff Emeritus
What are your thoughts on the matter? Do you have any ideas and have you tried checking what would be implied?

4. Apr 9, 2015

### kkz23691

If there are some nonzero off-diagonal components of $g_{jk}$ then the coordinate system is not orthogonal. Thinking along these lines, probably there is a meaning behind zero diagonal elements

5. Apr 9, 2015

### Orodruin

Staff Emeritus
So how do you reach the conclusion that non-zero off-diagonal elements imply non-orthogonal basis vectors? By comparison to this argumentation, what would it imply if a diagonal element is zero?

6. Apr 9, 2015

### kkz23691

http://en.wikipedia.org/wiki/Skew_coordinates

That is the question indeed... when you think about it, a skew-symmetric matrix has zero diagonal elements because $a_{ij} = -a_{ji}$
would this mean the metric tensor cannot have zero diagonal elements as $g$ is symmetric?

Last edited: Apr 9, 2015
7. Apr 9, 2015

### Matterwave

Peter just gave you an example of a metric with 0 diagonal elements in his post #2. A matrix can be symmetric and have 0 diagonal elements no problem. A skew-symmetric matrix must have 0 diagonal elements but that tells us nothing about diagonal elements of symmetric matrices.

A metric with 0 diagonal elements means that the basis vectors are null, that's about it. I think Peter told you this back in post #2, but I just wanted to reiterate since you didn't seem to pick it up.

8. Apr 9, 2015

### Orodruin

Staff Emeritus
But the metric is symmetric and not skew symmetric. The reason I ask about the implication for off diagonal elements of the metric is not that I am wondering, it is in order to gauge your understanding so that we can start from there.

9. Apr 9, 2015

### Staff: Mentor

You appear to believe that "skew coordinates" and a "skew-symmetric matrix" for the metric are the same thing. They are not. "Skew coordinates" just means the basis vectors aren't all orthogonal to each other; that can be true for a metric that is a symmetric matrix, and of course the metric is always symmetric. I gave you an example of "skew coordinates"--basis vectors not all orthogonal, as can be easily verified for my example--with a symmetric metric.

10. Apr 10, 2015

### kkz23691

11. Apr 12, 2015

### kkz23691

There is this one last question - the above discussion was about a metric tensor with all diagonal elements equal to zero, which corresponds to null coordinates. What would be the meaning of a metric tensor where one (or some) of the diagonal elements are zero. This seems to suggest that the corresponding coordinate cannot contribute to the measurement of distance, expect if an off-diagonal element in the same row is nonzero.

Then, if a metric tensor is diagonal, what would it mean if only one (or a few) diagonal elements are zero?

12. Apr 12, 2015

### Staff: Mentor

More precisely, two diagonal elements were equal to zero. If we put back in the other two coordinates, they would remain spacelike, so the full line element would be $ds^2 = - du dv + dy^2 + dz^2$, so there would still be two nonzero diagonal elements.

(That's not to say it's impossible to have a metric tensor with all four diagonal elements equal to zero. I've never seen one, but I've never really looked for one, either.)

Obviously, for a coordinate differential to contribute to the line element, it has to appear somewhere. I don't see how it would be possible for a coordinate differential to not appear at all in the line element; that would imply that the "coordinate" wasn't really a valid coordinate.

In the line element I wrote down, obviously $du$ and $dv$ can contribute to a nonzero distance--as long as they both are nonzero. So any line element with either $du = 0$ or $dv = 0$ gets no contribution from those two coordinates (it still might from $dy$ or $dz$ if those are nonzero).

13. Apr 12, 2015

### kkz23691

This is very interesting as such discussions cannot be found in textbooks :) Then, in any metric tensor, at least one element per row must be nonzero?

14. Apr 12, 2015

### Staff: Mentor

Yes. Textbooks probably don't mention this because it's considered too obvious to need mentioning. (Although if you dig into textbooks on differential geometry, which is where all of this stuff is made mathematically rigorous, there is probably something in the full definition of a valid coordinate chart that is equivalent to the above statement.)

15. Apr 13, 2015

### Orodruin

Staff Emeritus
A metric with a full row equal to zero would have a non-trivial kernel and thus be degenerate. This violates the definition of the merric on a pseudo-Riemannian manifold.

16. Apr 13, 2015

### Ben Niehoff

If you want your coordinates to be real numbers, and your metric to have signature (-+++), then this is the best you can do. If you allow complex coordinates, then you could write

$$ds^2 = - du \, dv + dw \, d\bar w$$