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For any transformation from Rectangular (Cartesian) to spherical coordinates in three dimensions, we use the transformation equation

$x = r\cos(\theta)\sin(\phi)$

among others, where $\theta$ is the azimuth and $\phi$ is the polar angle.

Now,

$\cfrac{\partial x}{\partial r} = \cos(\theta)\sin(\phi) = \cfrac{x}{r} \tag{1}$.

However, $r = \sqrt{x^2 + y^2 + z^2}$ so

$\cfrac{\partial r}{\partial x} = \cfrac{x}{\sqrt{x^2 + y^2 + z^2}} = \cfrac{x}{r} \tag{2}$.

What's wrong with the above?

1 Answers 1

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Absolutely nothing, except that it is recommended not to mix different coordinate systems in the same expression.

I suspect you are concerned that this doesn't match up with $ \frac{dy}{dx} = \left( \frac{dx}{dy} \right)^{-1} $. This rule only works for total derivatives: for partial derivatives, the relationship between them depends on what is being held constant when you take the derivative.

Consider, for example, the change of coordinates $x = u$, $y=u+v$, so $u=x$ and $v=y-x$. You might think that $\partial f/\partial u$ is the same as $\partial f/\partial x$, but it isn't: we find $$ \frac{\partial f}{\partial u} = \frac{\partial x}{\partial u}\frac{\partial f}{\partial x} + \frac{\partial y}{\partial u} \frac{\partial f}{\partial y} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}, $$ so even though $u$ and $x$ look the same, holding the other coordinate constant in the differentiation gives different results (draw the lines $y=\text{const.}$ and $v=\text{const.}$ to see this clearly). We have in this case $$ \frac{\partial u}{\partial x} = 1, \quad \frac{\partial v}{\partial x} = -1 \\ \frac{\partial u}{\partial y} = 0, \quad \frac{\partial v}{\partial y} = 1, $$ but $$ \frac{\partial x}{\partial u} = 1, \quad \frac{\partial y}{\partial u} = 1 \\ \frac{\partial x}{\partial v} = 0, \quad \frac{\partial y}{\partial v} = 1, $$ so we see that we have a different reciprocal relation: the matrices of partial derivatives are inverses, or if we write $x=x_1$, $x=x_2$, $u=u_1$, $v=u_2$ $$ \sum_{j=1}^2 \frac{\partial x_i}{\partial u_j} \frac{\partial u_j}{\partial x_k} = \delta_{ik}. $$ This relation tells us very little about the individual partial derivatives.

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    Thank you! In the derivation of the Laplacian, I have seen people use $\cfrac{\partial r}{\partial x} = \cos(\theta)\sin(\phi)$ (eqn 2 corollary) over eqn 1. Why is this? – 2017-02-14
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    It's a matter of what you already know: you start with a (Cartesian) expression with $x,y,z$ derivatives, and you want to get to one with $r,\theta,\phi$ derivatives; the most direct way to do that is use $\frac{\partial f}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial f}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial f}{\partial \theta} + \frac{\partial \phi}{\partial x} \frac{\partial f}{\partial \phi} $ and so on. If you start with $\frac{\partial f}{\partial r}= \dotsb$, &c., you would have to invert to find a formula for the Laplacian, which is awful. – 2017-02-14