# Metric tensors

1. Jan 31, 2008

### nolanp2

in my fields course we are using the metric tensor g=diagonal(1,-1,-1,-1), off diagonal(0)
i'm looking for an explanation of why the time coordinate has to be orientated oppositely to the spacial coordinates. can anyone give me an explanation of this?

i'm also lost with upper and lower indices. i dont understand why multiplying a lower index by the metric tensor gives an upper index.

any help appreciated, thanks

2. Jan 31, 2008

### Andy Resnick

The simple answer is that by defining the metric tensor that way, there are "null distances" of length zero which correspond to path that light rays travel. It does not *have* to be defined that way, but in the General Theory of Relativity, it can be defined that way.

The upper and lower indices are more subtle- they are not, in general, movable. Upper indices correspond to vectors, lower indices correspond to forms. If a metric tensor can be defined on a generic geometry, then it is possible to move them up and down.

Misner, Thorne, and Wheeler's book "Gravitation" is an excellent way to get started understanding this stuff.

3. Jan 31, 2008

### nolanp2

so the upper and lower indices are like dual spaces? what advantage is there to raising indices, why is it necessary that they be used in Classical field theory?

4. Jan 31, 2008

### pam

The basic reason for the "indefinite metric", with time having the opposite sign from space, is that the combination t^2-x^2-y^2-z^2 is an invariant under Lorentz transformation.
The two types of indices, upper and lower, is a relatively simple way to incorporate this.

5. Jan 31, 2008

### jcsd

Yes the upper indexed components are the components of some vector in some basis. The lower indexed components are the components of it's dual vector in some dual basis.

The metric tensor maps a vector to it's dual.

Index gymnastics is just a way in that the relationship between tensors refelcts the relationship between their components in different bases.