**NOTES ON STATISTICS, PROBABILITY and
MATHEMATICS**

### The Metric Tensor:

The **length or magnitude** of the vector \(dS\) **in Cartesian
coordinates** is given by:

\[dS^2=\left(dX^1\right)^2 +
\left(dX^2\right)^2 + \left(dX^3\right)^2 + \cdots = \large \delta_{mn}
\;\color{blue}{dX^m}\, dX^n\tag 1\]

Notice that the dummy indexes are set up to sum over. We start with
\(m=1\), and we see that all terms are
going to be zero unless when \(n=1\),
in which case it will be \(dX^1\, dX^1 =
(dX^1)^2.\)

But how to write the same for the curvilinear system?

We have to refer back to the following expression of a change of
coordinate system at the beginning of the notes on
tensors:

\[dy^n = \frac{\partial y^n}{\partial
x^\color{blue}{m}} dx^{\color{blue}{m}} \tag{Ref.1}\]

In this case we have that the part in blue in Eq. 1 can be expressed
as:

\[\color{blue}{dX^m}= \frac{\partial
x^m}{\partial y^r} dy^r \tag2\]

Hence, Eq. 1 becomes,

\[dS^2 = \large \delta_{mn} \;dX^m\, dX^n=
\large \underset{\text{METRIC TENSOR }\Large
g^{(y)}_{\color{red}{rs}}}{\underbrace{\delta_{mn}\frac{\partial
x^m}{\partial y^\color{red}{r}}\frac{\partial x^n}{\partial
y^\color{red}{s}}}} \;dy^r\; dy^s\]

\[\Large \bbox[10px, border:2px solid
aqua]{g^{(y)}_{\color{red}{rs}}=\delta_{mn}\frac{\partial x^m}{\partial
y^\color{red}{r}}\frac{\partial x^n}{\partial y^\color{red}{s}}}\tag
3\]

Here is another
derivation: