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I was revising on a bit of tensor calculus, when I stumbled upon this:

$$\delta^i_j = \frac{\partial y^i}{\partial x^\alpha} \frac{\partial x^\alpha}{\partial y^j}$$

And the next statement reads,

this expression yields:

$$ \left|\frac{\partial y^i}{\partial x^j}\right| \left|\frac{\partial x^\alpha}{\partial y^\beta}\right|= 1 ,$$

With $ \left|\dfrac{\partial y^i}{\partial x^j}\right|$ being the Jacobian for transformation $y^i=y^i(x^1.....x^n)$, and $\left|\dfrac{\partial x^\alpha}{\partial y^\beta}\right|$ being the Jacobian of the inverse transformation.

My question is, how do you get eq. 1 from the Kronecker delta, as they are merely jacobians of coordinate transformatios, and being inverse of each other, are 1. But how do they follow from $\delta^i_j$'s expansion? (i.e. DERIVE eq. 1 using the expansion for kronecker delta)? I am most probably making a conceptual error, but this is the first time I have seen such a representation of the kronecker delta.

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  • $\begingroup$ The simple explanation is that (taking into accound tensor summation convention, i.e same repeated index is to be summed over): $\partial y^i$ is (functionaly) independent of $\partial y^j$ same as $\partial x^\alpha$ is functionaly independent of $\partial x^\beta$ $\endgroup$ – Nikos M. Aug 17 '14 at 12:52
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    $\begingroup$ The question (v2) seems to follow directly from inverse function theorem in several variables. $\endgroup$ – Qmechanic Aug 17 '14 at 13:17
  • $\begingroup$ The conceptual error is expecting this to come from an expansion of the Kronecker delta. This is really just a property of coordinate transforms (and their inverse) written down in components. $\endgroup$ – Sanya Feb 19 '17 at 9:47
  • $\begingroup$ "My question is, how do you get eq. 1..." There is no equation 1 in your question. $\endgroup$ – Brian Moths Mar 24 '17 at 14:37
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The first equation states that two matrices' product is an identity matrix. The desired result states that the product of these matrices' determinants is $1$. This follows from taking determinants of the original equation.

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This is an old question, but just for the record:

The relation in the question, $$ \frac{\partial y^i}{\partial x^\alpha} \frac{\partial x^\alpha}{\partial y^j}=\delta^i_j ,$$ is essentially the product of two matrices $\frac{\partial y^i}{\partial x^\alpha}$ and $\frac{\partial x^\alpha}{\partial y^j}$ giving the identity. The relationship the OP asks for follows directly by taking the determinant of that matrix equation: the right-hand side gives $1$, and the left-hand side is the determinant of a product, which splits to the product of the determinants.

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You transform $y \rightarrow x$ then back $x \rightarrow y$ so back to the same coordinate system but not sure you have transform a particular basis vector to the same when you transform it back, $i$ may $\neq j$.

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    $\begingroup$ Welcome to Physics SE :) I think your answer, as it stands, is not easy to understand - could you elaborate and extend it a bit? :) $\endgroup$ – Sanya Feb 19 '17 at 9:45

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