# Vector Calculus Derivation

## Main Question or Discussion Point

I was reading a paper and came across this equation:

Fmagnetic0(M<dot>)H

Is this the correct expansion below? (I'm not too experienced with vectors operating on the gradient operator)

Fmagnetic0[(MxH/∂x)i + (MyH/∂y)j + (MzH/∂z)k]
_____________

chiro
Hey hjel0743 and welcome to the forums.

Just for clarification, is M a constant vector and H some kind of function?

Thanks for the reply chiro! I imagine I'll be here a few more times before my thesis is done... M is a function of H actually, and H is a function of the vector r, representing the radius.

The equation as initially written, describes the force on a particle by a magnetic field. H is the "H-field" and M is the magnetization.

$$({\bf M} \cdot \nabla) {\bf H} = ( ({\bf M} \cdot \nabla H_x) \widehat{i} + ({\bf M} \cdot \nabla H_y) \widehat{j} + ({\bf M} \cdot \nabla H_z) \widehat{k}).$$

Where, for example,
$${\bf M} \cdot \nabla H_x = M_x \frac{\partial H_x}{\partial x} + M_y \frac{\partial H_x}{\partial y} + M_z \frac{\partial H_x}{\partial z}.$$

chiro
Assuming your M is a function of x,y,z (in vector form you have M = (Mx,My,Mz) where Mx,My,Mz map R^3 to R for each component) then

del(M) = (d/dx . Mx + d/dy . My + d/dz . Mz)H (I'm assuming everything is Cartesian not a general tensor)
= (dMx/dx + dMy/dy + dMz/dz) H.

Now this will give you the product of two functions but if H is a vector (like M with each component have some transformation from R^3 -> R) then this means you use the scalar form of a*v = (a*v1,a*v2,a*v3) which means if H = (Hx,Hy,Hz) then the whole thing is equal to

(dMx/dx + dMy/dy + dMz/dz) * <Hx,Hy,Hz>

Now M . grad(Hx) = <Mx,My,Mz> . <dHx/dx,dHx/dy,dHx/dz>
= Mx*dHx/dx + My.dHx/dy + Mz.dHx/dz

So they both look the same when they are expanded out, so I imagine you are right in your assertion. (It's been a while since I've done this kind of thing myself).