0
$\begingroup$

In Schutz, Geometrical Methods of Mathematical Physics, is written a Jacobian coordinate transform $\Lambda$, $$ \Lambda^i_j = \frac{\partial x^i}{\partial y^j} $$ The inverse matrix is written $$ \Lambda^k_j = \frac{\partial y^k}{\partial x^j} $$ Schutz then says (the fact that it is an inverse) "is easily proved using the chain rule for partial derivatives": $$ \frac{\partial x^i}{\partial y^j} \frac{\partial y^j}{\partial x^k} = \frac{\partial x^i}{\partial x^k} = \delta^i_k $$ Question 1: is it correct to regard this as a tensor contraction over "$\partial y^j$" ?

Next, Koks, Explorations in Mathematical Physics, has a similar example, but written out in matrix notation: $$ \left[ \begin{array}{cc} \frac{\partial r}{\partial x} &\quad \frac{\partial r}{\partial y} \\ \frac{\partial\theta}{\partial x} &\quad \frac{\partial\theta}{\partial y} \\ \end{array} \right] \cdot \left[ \begin{array}{cc} \frac{\partial x}{\partial r} &\quad \frac{\partial x}{\partial\theta} \\ \frac{\partial y}{\partial r} &\quad \frac{\partial y}{\partial\theta} \\ \end{array} \right] = \left[\begin{array}{cc} 1 &\quad 0 \\ 0 &\quad 1 \end{array}\right] $$

Rewrite this using $\partial x^1 \equiv \partial r$, $\partial x^2 \equiv \partial\theta$, $\partial y^1 \equiv \partial x$, $\partial y^2 \equiv \partial y$ so that it matches the Shutz notation: $$ \left[ \begin{array}{cc} \frac{\partial x^1}{\partial y^1} &\quad \frac{\partial x^1}{\partial y^2} \\ \frac{\partial x^2}{\partial y^1} &\quad \frac{\partial x^2}{\partial y^2} \\ \end{array} \right] \cdot \left[ \begin{array}{cc} \frac{\partial y^1}{\partial x^1} &\quad \frac{\partial y^1}{\partial x^2} \\ \frac{\partial y^2}{\partial x^1} &\quad \frac{\partial y^2}{\partial x^2} \\ \end{array} \right] = \left[\begin{array}{cc} 1 &\quad 0 \\ 0 &\quad 1 \end{array}\right] $$ Now look at an individual element of the matrix product. Applying the chain rule from Schutz, $$ \frac{\partial x^i}{\partial y^j} \frac{\partial y^j}{\partial x^k} = \frac{\partial x^i}{\partial x^k} = \delta^i_k $$ however if we write this out explicitly for the 1,1 element: $$ \frac{\partial x^1}{\partial y^1}\frac{\partial y^1}{\partial x^1} + \frac{\partial x^1}{\partial y^2}\frac{\partial y^2}{\partial x^1} = 2 \frac{\partial x^1}{\partial x^1} = 2 $$ Question 2: this is incorrect...

$\endgroup$
3
  • 1
    $\begingroup$ You can't "cancel" $\partial y^1$ in $\frac{\partial x^1}{\partial y^1}\frac{\partial y^1}{\partial x^1}$, that's not how the chain rule works in several variables. $\endgroup$ May 3 '14 at 7:58
  • $\begingroup$ $\frac{\partial y}{\partial x}$ is not a quotient @beginner $\endgroup$
    – janmarqz
    May 4 '14 at 4:08
  • $\begingroup$ Is this not a duplicate of math.stackexchange.com/questions/778149/… ? I think that this answer is more satisfying, since it basically tells you that your method is not wrong but there is just an error in the application. $\endgroup$
    – xpnerd
    Apr 4 '17 at 7:02
2
$\begingroup$

There is an implicit sum over all values of $j$ in Schutz's notation; his statement is not that $$\frac{\partial x^1}{\partial y^1} \frac{\partial y^1}{\partial x^1} = 1$$ but rather that $$\frac{\partial x^1}{\partial y^1} \frac{\partial y^1}{\partial x^1} +\frac{\partial x^1}{\partial y^2} \frac{\partial y^2}{\partial x^1}= 1$$ i.e., $$\sum_j \frac{\partial x^1}{\partial y^j} \frac{\partial y^j}{\partial x^1} = 1$$ and his statement that $$\frac{\partial x^i}{\partial y^j} \frac{\partial y^j}{\partial x^k} = \delta^i_k$$ has an implicit summation: $$\sum_j \frac{\partial x^i}{\partial y^j} \frac{\partial y^j}{\partial x^k} = \delta^i_k$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.