# How to caculate the inverse metric tensor

1. Apr 26, 2012

### ngkamsengpeter

Given a metric tensor gmn, how to calculate the inverse of it, gmn. For example, the metric
$$g_{\mu \nu }= \left[ \begin{array}{cccc} f & 0 & 0 & -w \\ 0 & -e^m & 0 &0 \\0 & 0 & -e^m &0\\0 & 0 & 0 & -l \end{array} \right]$$

From basic understanding, I would think of divided it, that is
$$g^{\mu \nu }= \left[ \begin{array}{cccc} 1/f & 0 & 0 & -1/w \\ 0 & -e^{-m} & 0 &0 \\0 & 0 & -e^{-m} &0\\0 & 0 & 0 & -1/l \end{array} \right]$$

But the author gave some different answer, that is
$$g^{\mu \nu }= \left[ \begin{array}{cccc} \frac{l}{fl+w^2} & 0 & 0 & -\frac{w}{fl+w^2} \\ 0 & -e^{-m} & 0 &0 \\0 & 0 & -e^{-m} &0\\0 & 0 & 0 & -\frac{f}{fl+w^2}\end{array} \right]$$

So how should I calculate the inverse metric tensor as shown by the author?

Thanks

2. Apr 26, 2012

### Matterwave

The inverse metric is, like the name suggests, just the inverse matrix.

You should have learned how to calculate inverse matrices in Linear algebra, there are many methods.

The way you are trying, where you just invert the entries only works if the matrix is diagonal, which this one is not.

3. Apr 26, 2012

### ngkamsengpeter

Well, I thought the same way at first but I have tried find the inverse matrix but the result is different from the given one.
Mathematica give the me the inverse matrix as
$$g^{\mu \nu }= \left[ \begin{array}{cccc} \frac{1}{f} & 0 & 0 & -\frac{w}{fl} \\ 0 & -e^{-m} & 0 &0 \\0 & 0 & -e^{-m} &0\\0 & 0 & 0 & -\frac{1}{l} \end{array} \right]$$

That's why I wonder how the author get that result..

Last edited: Apr 26, 2012
4. Apr 26, 2012

### Bill_K

Well in the first place you have written the metric tensor incorrectly. It should be symmetric. Also you have either entered it into Mathematica incorrectly, or incorrectly copied down the result. Really, you should be able to invert a simple matrix like this all by yourself!

5. Apr 26, 2012

### ngkamsengpeter

Nope. It is correct. I copied the metric tensor and its inverse directly from the Mallett paper. I checked it many times. I copied it correctly. It is from Mallett paper on circulating beam. I dont think it is written wrongly in the paper.

6. Apr 26, 2012

### Ich

That doesn't help if you lack the background. The metric tensor is always symmetric, that's why it is enough that Mallet provides g_03. g_30 = g_03 is implicit.

7. Apr 26, 2012

### ngkamsengpeter

I see. I think I know my mistake already. Thanks. for the help. I am new to this, always forget about the symmetric. Many thanks to all the help.

8. Apr 26, 2012

### Matterwave

haha, I can't believe I didn't even notice the non-symmetry...>.>