# Vector calculus identities proof

## Main Question or Discussion Point

Hello.

How can I prove something like
$$\nabla\cdot(\mathbf fv)=(\nabla v)\cdot\mathbf f+v(\nabla\cdot \mathbf f)$$

using only the definition of divergence
$$\text{div}\mathbf V=\lim_{\Delta v\rightarrow0}\frac{\oint_S\mathbf V\cdot d\mathbf s}{\Delta v},$$
i.e. without referring to any particular coordinate system? I have yet to see a book that does not assume cartesian coordinates.

HallsofIvy
Homework Helper
Where did you get that defition? I don't seee how it makes any sense if you don't say what roll "$\Delta v$" plays in
$$\frac{\oint_S V\cdot ds}{\Delta v}$$
and what is the contour S? Is it a circle around the point with radius $\Delta v$?

Oh sorry, I thought it was a common definition. Here it is at http://mathworld.wolfram.com/Divergence.html" [Broken]. Maybe it is more of a physicist's definition? But I know that you can use it to derive the expressions for the divergence in various coordinate systems.

In words, the divergence is the outward flux of a vector field per unit volume as the volume tends to zero. So it is a closed surface integral on some surface. I think this surface can have pretty much any shape. $\Delta v$ is the volume enclosed by the surface.

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If you don't use the definition I have shown, how does one then derive the expressions for $\nabla$ in various coordinate systems? To me it seems that the definition I have shown here is fundamental.

Let me try and ask in another way: When I see an expression such as
$$\nabla\cdot(\mathbf fv)=(\nabla v)\cdot\mathbf f+v(\nabla\cdot \mathbf f)$$

which contains no reference to any particular coordinate system, how do I know when it is valid? Is it valid for any kind of weird non-orthogonal coordinate system?

HallsofIvy
The transformation from one coordinate system to another is alway "homogenous". That, is the "0" vector in one coordinate system is the "0" vector in any coordinate system. If a vector equation says $\vec{v}= \vec{u}$ (where $\vec{u}$ and $\vec{v}$ can be very complicated formulas, but the result is a vector) then $\vec{v}- \vec{u}= \vec{0}$ in any coordinate system.
Ah okay, but do we not need to take into account that the $\nabla$-operator changes when the coordinate system changes? It sounds like you are basically saying is that if we can prove an identity in one coordinate system, then it holds in all coordinate systems.