Deriving the spherical volume element

  • #1
I’m trying to derive the infinitesimal volume element in spherical coordinates. Obviously there are several ways to do this. The way I was attempting it was to start with the cartesian volume element, dxdydz, and transform it using

$$dxdydz = \left (\frac{\partial x}{\partial r}dr + \frac{\partial x}{\partial \theta }d\theta + \frac{\partial x}{\partial \phi }d\phi \right )\left ( \frac{\partial y}{\partial r}dr + \frac{\partial y}{\partial \theta }d\theta + \frac{\partial y }{\partial \phi}d\phi \right )\left ( \frac{\partial z}{\partial r}dr + \frac{\partial z}{\partial \theta }d\theta + \frac{\partial z}{\partial \phi}d\phi \right )$$

Unfortunately, I can’t see how I will arrive at the correct expression, ##r^{2}sin\theta drd\theta d\phi ##.

For one reason, when completely expanded, I get terms with repeated differentials like ##dr^{3} ## that don’t cancel.

Why is my method of derivation invalid?
 

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  • #2
Orodruin
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You are not looking at the volumes actually spanned by the basis vectors corresponding to the spherical coordinates. The volume element spanned by three vectors ##\vec a_i## is the triple product ##\vec a_1 \cdot (\vec a_2 \times \vec a_3)##. Hence, if you look at the volume of a small parallelepiped spanned by the coordinate lines with coordinate differences ##dy^i##, then this is given by
$$
dV = \frac{\partial \vec x}{\partial y^1} \cdot \left(\frac{\partial \vec x}{\partial y^2} \times \frac{\partial \vec x}{\partial y^3}\right) dy^1 dy^2 dy^3.
$$

Alternatively, the volume element can be seen as the wedge product between the differentials. You must then remember that the wedge product is completely anti-symmetric and therefore anything like ##dr\wedge dr = 0##.
 
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  • #4
Thanks for the help! Using a wedge between the terms in my original expression got me to the answer!
 
Last edited:

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