The book Goreux, Introduction to Tensor Calculus (p.21) gives the following explanation of this rule (which I'll call the delta relationship): $$ \frac{\partial x^i}{\partial \hat{x}^j} \frac{\partial \hat{x}^j}{\partial x^k} = \delta_k^i $$

The equation involves original coordinates $x^i$ and new (transformed) coordinates $\hat{x}^i$.

The explanation starts by expressing the original coordinates in terms of the new (eq1), $$ x^i = x^i( \hat{x}^1, \hat{x}^2, \cdots, \hat{x}^n ) $$ and vice-versa (eq2) $$ \hat{x}^i = \hat{x}^i( x^1, x^2, \cdots, x^n ) $$ which is possible if the transform between the coordinates is not singular at the point being considered.

Then, the book suggests substituting the second equation into the first, i.e. expressing the $\hat{x}^i$ on the right hand side of the first equation in terms of the $x^i$ using the second equation. The book says "Since these coordinates are independent, this expression of $x^i$ must reduce to $x^i$ itself". So far, this is obvious to me.

The next sentence is the puzzle: "We deduce that $$ \frac{\partial x^i}{\partial \hat{x}^j} \frac{\partial \hat{x}^j}{\partial x^k} = \delta_k^i $$ "

I don't see how this follows.


1 Answer 1


To elaborate on my comment: for each $i$, you have an equation of the form: $$ x^i = x^i(\hat x^1(x^1,\dots,x^n),\hat x^2(x^1,\dots,x^n),\dots,\hat x^n(x^1,\dots,x^n)) $$ For any $k$, take the derivative with respect to $x^k$ of each side of the above. The left-hand side is $\delta^i_k$, and for the right-hand side, we use the chain rule with each $\hat x^j$ as an intermediate variable. Thus $$ \delta^i_k = \frac{\partial x^i}{\partial \hat x^j}\frac{\partial \hat x^j}{\partial x^k} $$ (I assume you know about the Einstein summation convention—the repeated $j$ index means that index is summed over.)

  • $\begingroup$ that answers it for me. Thank you. $\endgroup$
    – Bull
    Jan 3, 2017 at 4:31

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