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Suppose I want to transform a partial derivative operator from spherical to Cartesian coordinates. I have found the following relation based on the chain rule here:

$$ \frac{\partial }{\partial \theta } = \frac{\partial x}{\partial \theta} \frac{\partial}{\partial x} + \frac{\partial y}{\partial \theta} \frac{\partial}{\partial t} + \dots $$

As I know from calculus, the chain rule is commonly defined, for example when want to take derivative of some 'functions' wrt to some variables.

So, how can the chain rule be defined for the derivative operators? I don't exactly understand the chain rule in this context for the system transformation.

Thank you.

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Suppose we have $f = f(r,s)$ where $r=r(x,y)$ and $s=s(x,y)$. Then, by the usual chain rule for functions you mentioned: $$\frac{\partial}{\partial x}f(r(x,y),s(x,y)) = \frac{\partial r}{\partial x}\frac{\partial f(r,s)}{\partial r} + \frac{\partial s}{\partial x}\frac{\partial f(r,s)}{\partial x}$$

Thus, if $\frac{\partial}{\partial x}$ is viewed as an operator acting on $f$, you can see $\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial s}{\partial x}\frac{\partial}{\partial x}$ also as an operator acting on $f$. This is what we mean by: $$\frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial}{\partial r} + \frac{\partial s}{\partial x}\frac{\partial}{\partial x}$$ which can be interpreted as a chain rule for operators.

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  • $\begingroup$ Thanks. It means that the relation i wrote does not hold in general, and It is correct only if we have a function? I mean, don't we define chain rule for operators(without any function) ? $\endgroup$ – Rob Aug 1 at 18:49
  • $\begingroup$ I guess a better way to express it would be: the "chain rule for operators" is nothing but an operational rule which comes from the actual chain rule for functions. $\endgroup$ – IamWill Aug 1 at 19:31
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Transforming spherical coordinates to Cartesian coordinates is just a mapping:

\begin{align} x(r, \theta, \varphi) & = r \sin \theta \cos \varphi \\ y(r, \theta, \varphi) & = r \sin \theta \sin \varphi \\ z(r, \theta, \varphi) & = r \cos \theta. \end{align}

Given $f\equiv g\circ(x,y,z)$, it follows from the chain rule (assuming the appropriate differentiability) that $$ \frac{\partial f}{\partial\theta}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial\theta}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial\theta}+\frac{\partial g}{\partial z}\frac{\partial z}{\partial\theta}. $$

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    $\begingroup$ In your final formula, did you mean to use $f$ rather than $g$ on the left? $\endgroup$ – littleO Aug 1 at 19:05
  • $\begingroup$ @littleO: Thanks, fixed now. $\endgroup$ – parsiad Aug 2 at 22:55

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