- #1
cr7einstein
- 87
- 2
Dear all,
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:
$$|\frac{\partial y^i}{\partial x^j}| |\frac{\partial x^\alpha}{\partial y^\beta}|= 1$$, ........(1)
With $ |\frac{\partial y^i}{\partial x^j}|$ being the jacobian for transformation $$y^i=y^i(x^1...x^n)$$, and $$|\frac{\partial x^\alpha}{\partial y^\beta}|$$ 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. Thanks in advance!
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:
$$|\frac{\partial y^i}{\partial x^j}| |\frac{\partial x^\alpha}{\partial y^\beta}|= 1$$, ........(1)
With $ |\frac{\partial y^i}{\partial x^j}|$ being the jacobian for transformation $$y^i=y^i(x^1...x^n)$$, and $$|\frac{\partial x^\alpha}{\partial y^\beta}|$$ 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. Thanks in advance!