Three Dimensional Transformation (Help Needed ASAP Please)

**AA23** · August 9th 2012, 05:39 AM

I have been given the following problem and a unsure if I have established the correct answer or not? See attachment

Thank You in advance for any help

**JohnDMalcolm** · August 12th 2012, 06:23 AM

I think the problem is asking you to choose a particular alpha and beta, and show that the expression on the left equals the Kronecker delta. Recall,

${\delta^{\alpha}}_{\beta} = \left\{ \begin{array}{cc} 1, & \mbox{ if } \alpha = \beta \\ 0, & \mbox{ if } \alpha \neq \beta \end{array} \right.$

Let's do the example $\alpha = \beta = r$ , where $r = \sqrt{x^2 + y^2 + z^2}$

$\frac{\partial z^{\alpha}}{\partial x^{\mu}} \frac{\partial x^{\mu}}{\partial z^{\beta}} = \frac{\partial r}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial r}{\partial y} \frac{\partial y}{\partial r} + \frac{\partial r}{\partial z} \frac{\partial z}{\partial r}$

$= \frac{x \sin{\theta} \cos{\phi}}{\sqrt{x^2 + y^2 + z^2}} + \frac{y \sin{\theta} \sin{\phi}}{\sqrt{x^2 + y^2 + z^2}} + \frac{z \cos{\theta}}{\sqrt{x^2 + y^2 + z^2}}$

$= \frac{x^2 + y^2 + z^2}{r^2}$

$= 1$

See if you can do it for other choices of alpha and beta.

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