How to caculate the inverse metric tensor

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To calculate the inverse metric tensor g^{\mu \nu} from a given metric tensor g_{\mu \nu}, one must use matrix inversion methods rather than simply inverting individual entries, especially since the matrix is not diagonal. The original metric tensor provided was incorrectly assumed to be symmetric, leading to confusion in the calculations. The correct approach involves recognizing the symmetry of the metric tensor, which simplifies the inversion process. Users are encouraged to verify their calculations and ensure they understand the properties of the metric tensor. Understanding these concepts is crucial for accurate results in tensor calculations.
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Given a metric tensor gmn, how to calculate the inverse of it, gmn. For example, the metric
<br /> g_{\mu \nu }= <br /> \left[ \begin{array}{cccc} f &amp; 0 &amp; 0 &amp; -w \\ 0 &amp; -e^m &amp; 0 &amp;0 \\0 &amp; 0 &amp; -e^m &amp;0\\0 &amp; 0 &amp; 0 &amp; -l \end{array} \right]<br />

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

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

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

Thanks
 
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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.
 
Matterwave said:
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.

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
<br /> g^{\mu \nu }= <br /> \left[ \begin{array}{cccc} \frac{1}{f} &amp; 0 &amp; 0 &amp; -\frac{w}{fl} \\ 0 &amp; -e^{-m} &amp; 0 &amp;0 \\0 &amp; 0 &amp; -e^{-m} &amp;0\\0 &amp; 0 &amp; 0 &amp; -\frac{1}{l} \end{array} \right]<br />

That's why I wonder how the author get that result..
 
Last edited:
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!
 
Bill_K said:
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!
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 don't think it is written wrongly in the paper.
 
I checked it many times.
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.
 
Ich said:
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.
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.
 
haha, I can't believe I didn't even notice the non-symmetry...>.>
 

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