How a Multivariable Function Changes with Respect to a Radial Coordinate

How a Multivariable Function Changes with Respect to a Radial Coordinate post thumbnail image

First focus on the Cartesian coordinate $x$, which depends on the spherical coordinates $r,\theta,$ and $\phi$.

For a function $f(x)$, the chain rule yields

$ \frac{\partial f(x)}{\partial r} = ( \frac{\partial f}{\partial x} ) ( \frac{\partial x}{\partial r} )$.

Similarly for functions $g(y)$ and $h(z)$:

$ \frac{\partial g(y)}{\partial r}  = (\frac{\partial g}{\partial y}) (\frac{\partial y}{\partial r} )$.

$ \frac{\partial h(z)}{\partial r}  = (\frac{\partial h}{\partial z})( \frac{\partial z}{\partial r} )$.

Suppose $F(x,y,z) \equiv f(x) g(y) h(z)$. This is a significant assumption.

Then, using the product rule twice,

$ \frac{\partial F(x)}{\partial r} = \frac{\partial}{\partial r} [f(x) g(y) h(z)] = g(y) h(z) \frac{\partial f(x)}{\partial r} + f(x) [ h(z) \frac{\partial  g(y)  }{\partial r} + g(y) \frac{\partial  h(z) }{\partial r} ] $.

Simplifying,

$ \frac{\partial F(x,y,z) }{\partial r} = g(y) h(z) \frac{\partial f(x)}{\partial r} + f(x)  h(z) \frac{\partial  g(y)  }{\partial r} + f(x) g(y) \frac{\partial  h(z) }{\partial r} $.

Substituting,

$ \frac{\partial F(x,y,z) }{\partial r} = g(y) h(z) \frac{\partial f}{\partial x} \frac{\partial x}{\partial r}  + f(x)  h(z) \frac{\partial g}{\partial y} \frac{\partial y}{\partial r} + f(x) g(y) \frac{\partial h}{\partial z} \frac{\partial z}{\partial r} $.

Rearranging,

$ \frac{\partial F(x,y,z)}{\partial r} =  \frac{\partial x}{\partial r} (\frac{\partial f}{\partial x}) ( g(y) h(z) )  +  \frac{\partial y}{\partial r} ( \frac{\partial g}{\partial y} )  (f(x)  h(z) )  +  \frac{\partial z}{\partial r} ( \frac{\partial h}{\partial z}) ( f(x) g(y) )$.

Each partial derivative only depends on one variable, so it is possible to write:

$ \frac{\partial F(x,y,z) }{\partial r} =  \frac{\partial x}{\partial r} (\frac{\partial f(x) g(y) h(z)}{\partial x})   +  \frac{\partial y}{\partial r} ( \frac{\partial f(x)  h(z)  g(y)}{\partial y} )   +  \frac{\partial z}{\partial r} ( \frac{\partial f(x) g(y) h(z) }{\partial z}) $.

Since $F(x,y,z) \equiv f(x) g(y) h(z)$,

$ \frac{\partial F(x,y,z) }{\partial r} = ( \frac{\partial x}{\partial r} ) \frac{\partial F(x,y,z) }{\partial x}  + ( \frac{\partial y}{\partial r} ) \frac{\partial F(x,y,z) }{\partial y}    +  ( \frac{\partial z}{\partial r} )\frac{\partial F(x,y,z)  }{\partial z} $.

Rewriting,

$ \frac{\partial F(x,y,z)}{\partial r}  = ( \frac{\partial x}{\partial r} ) \frac{\partial  }{\partial x} F(x,y,z)  +  ( \frac{\partial y}{\partial r} ) \frac{\partial }{\partial y} F(x,y,z)   +  ( \frac{\partial z}{\partial r} )  \frac{\partial  }{\partial z} F(x,y,z)  $.

Rewriting again,

$ \frac{\partial }{\partial r} F(x,y,z) = \bigg[ ( \frac{\partial x}{\partial r} ) \frac{\partial  }{\partial x}   +  ( \frac{\partial y}{\partial r} ) \frac{\partial }{\partial y}    +  ( \frac{\partial z}{\partial r} )  \frac{\partial  }{\partial z} \bigg] F(x,y,z)  $.

Removing the $F(x,y,z)$ from the right,

$ \frac{\partial }{\partial r}  = ( \frac{\partial x}{\partial r} ) \frac{\partial  }{\partial x}   +  ( \frac{\partial y}{\partial r} ) \frac{\partial }{\partial y}    +  ( \frac{\partial z}{\partial r} )  \frac{\partial  }{\partial z}   $.

The same procedure can be done for $\frac{\partial }{\partial \theta} $ and $\frac{\partial }{\partial \phi} $.


Alternatively, consider the result:

$ \frac{\partial }{\partial r}  = ( \frac{\partial x}{\partial r} ) \frac{\partial  }{\partial x}   +  ( \frac{\partial y}{\partial r} ) \frac{\partial }{\partial y}    +  ( \frac{\partial z}{\partial r} )  \frac{\partial  }{\partial z}   $.

By analogy, it is possible to write, by replacing $r$ with the spherical coordinate of interest:

$ \frac{\partial }{\partial \theta}  = ( \frac{\partial x}{\partial \theta} ) \frac{\partial  }{\partial x}   +  ( \frac{\partial y}{\partial \theta} ) \frac{\partial }{\partial y}    +  ( \frac{\partial z}{\partial \theta} )  \frac{\partial  }{\partial z}   $.

and

$ \frac{\partial }{\partial \phi}  = ( \frac{\partial x}{\partial \phi} ) \frac{\partial  }{\partial x}   +  ( \frac{\partial y}{\partial \phi} ) \frac{\partial }{\partial y}    +  ( \frac{\partial z}{\partial \phi} )  \frac{\partial  }{\partial z}   $.

References

[1] http://www.thphys.nuim.ie/Notes/MP469/Laplace.pdf

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