Show that the metric tensor is independent of coordinate choice

[PrecPoint]

Homework Statement

Let $\overline{x^j}, \overline{\overline{x^j}}$ denote the coordinates of an arbitrary point P of $E_n$ referred to two distinct rectangular coordinate systems. An arbitrary curvlinear system in $E_n$ is related to the two rectangular systems according to:

Relevant Equations

$\overline{x^j}=\overline{x^j}(x^h),\qquad \overline{\overline{x^j}}=\overline{\overline{x^j}}(x^h)$

Show that:

$$\frac{\partial{\overline{x^j}}}{\partial{x^h}} \frac{\partial{\overline{x^j}}}{\partial{x^k}}=\frac{\partial{\overline{\overline{x^j}}}}{\partial{x^h}} \frac{\partial{\overline{\overline{x^j}}}}{\partial{x^k}}$$

I need to use some property of the relalation between the coordinate systems to prove that g_{hk} is independent of the choice of the underlying rectangular coordinate system.

I will try to borrow an idea from basic linear algebra. I expect any transformation between the rectangular systems to be orthogonal and hence should be able to use orthogonality. To illustrate my idea, I expect something along these lines (in linear algebra pseudocode):

\overline{\overline{x}}=A\overline{x},\quad A^TA=I,\quad (AJ)^T(AJ)=J^TA^TAJ=J^TJ

Lets try:

\overline{x}^j=\frac{\partial{\overline{x^j}}}{\partial{\overline{\overline{x^m}}}}\overline{\overline{x^m}}

Now the part where I get stuck

\frac{\partial{\overline{x^j}}}{\partial{x^h}}=\frac{\partial{^2\overline{x^j}}}{\partial{x^h}\partial{\overline{\overline{x^m}}}}\overline{\overline{x^m}}+\frac{\partial{\overline{x^j}}}{\partial{\overline{\overline{x^m}}}}\frac{\partial{\overline{\overline{x^m}}}}{\partial{x^h}}

And likewise for the other factor. But a) I am very unsure of the derivates (this is my first tensor problem) and b) there is no easy identity- or delta quantity to be found. I suspect I am on the wrong track :(

Edit: btw, this is not a homework problem, but posted here anyway since there were no other suitable place to be found. The problem is from the book on Tensors by Lovelock and Rund

 

[TSny]

Welcome to PF!

Suggestions:

Use $\overline{\overline{x}}=A\overline{x}$ to show that $\large \frac{\partial{\overline{\overline{x^j}}}}{\partial{\overline{x^r}}}$ is a particular matrix element of $A$.

Use the chain rule to express $\large \frac{\partial{\overline{\overline{x^j}}}}{\partial{x^h}}$ in terms of partial derivatives of $\overline{\overline{x^j}}$ with respect to the $\overline{x^r}$'s and the partial derivatives of the $\overline{x^r}$'s with repsect to $x^h$

 

[PrecPoint]

Thank you very much TSny!

Using your suggestions I first thought about "matrix A" which should be:

A_{jr}=\frac{\partial\overline{\overline{x^j}}}{\partial{\overline{x^r}}}

Now, using my analogy A^TA=I (the transformation is orthogonal) we note that:

\frac{\partial\overline{\overline{x^j}}}{\partial{\overline{x^r}}}\frac{\partial\overline{\overline{x^j}}}{\partial{\overline{x^t}}}=\delta_{rt}

\frac{\partial\overline{\overline{x^j}}}{\partial{x^h}}=<br /> \frac{\partial\overline{\overline{x^j}}}{\partial{\overline{x^r}}}\frac{\partial\overline{x^r}}{\partial{{x^h}}},\quad \frac{\partial\overline{\overline{x^j}}}{\partial{x^k}}=<br /> \frac{\partial\overline{\overline{x^j}}}{\partial{\overline{x^t}}}\frac{\partial\overline{x^t}}{\partial{{x^k}}}<br />

Applying this to the RHS of the original statement:

\frac{\partial\overline{\overline{x^j}}}{\partial{x^h}}\frac{\partial\overline{\overline{x^j}}}{\partial{x^k}}=\frac{\partial\overline{\overline{x^j}}}{\partial{\overline{x^r}}}\frac{\partial\overline{x^r}}{\partial{x^h}}\frac{\partial\overline{\overline{x^j}}}{\partial{\overline{x^t}}}\frac{\partial\overline{x^t}}{\partial{x^k}}=\delta_{rt}\frac{\partial\overline{x^r}}{\partial{x^h}}\frac{\partial\overline{x^t}}{\partial{x^k}}=\frac{\partial\overline{x^t}}{\partial{x^h}}\frac{\partial\overline{x^t}}{\partial{x^k}}

Which is what we wanted.

 

[TSny]

Looks good!