# Vector calculus identities proof

1. Jul 18, 2009

### daudaudaudau

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.

2. Jul 18, 2009

### HallsofIvy

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 $\Delta v$?

3. Jul 18, 2009

### daudaudaudau

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.

Last edited by a moderator: May 4, 2017
4. Jul 21, 2009

### daudaudaudau

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.

5. Jul 26, 2009

### daudaudaudau

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?

6. Jul 26, 2009

### HallsofIvy

?? If a vector equation is valid in one coordinate system, then it is valid in any coordinate system.

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.

7. Jul 26, 2009

### daudaudaudau

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.

8. Jul 27, 2009

### daudaudaudau

I guess you are right. The gradient is just a particular vector. And what the identity tells us is that one vector equals another vector. When we change coordinates, the gradient stays the same even though the gradient operator changes.