What are the operations on vector field A and how do I simplify the results?

[gcooke]

Hello. I am stuck trying to find an understandable answer to this online:

Carry out the following operations on the vector field A reducing the results to their simplest forms:

a. (d/dx i + d/dy j + d/dz k) . (Ax i + Ay j + Ax k)
b. (d/dx i + d/dy j + d/dz k) x (Ax i + Ay j + Ax k)

I know this is a dot product and cross product thing, and I think that A should become "integral Anda" (div) but I'm not sure what to do with the curl for b.

I can't find this worked out online. Can anyone direct me to a source?

Thank you!
Graham.

 

[Thaakisfox]

The best to do when you get problems as this one, is to use the abstract index notation. your first expression is the divergence of a vectorfield, and the second one is the curl, of the same field. So I will show the problem, for an arbitrary \vec v vector field, multiplied by some A scalar field. I will denote the differential operator as: \frac{\partial}{\partial x_j} \equiv \partial_j. And we use the einstein summation convention. That is we sum automaticaly on double indices. So:

\text{div}(A\vec{v}=\partial_k(Av_k)=v_k\partial_k A + A\partial_k v_k =(\vec{v}\nabla)A + A \text{div}\vec{v}
As we see the first part is the "substantive" part of the total time derivative, we use in hydrodynamics. The operator explicitly in cartesian coordinates:
\vec{v}\nabla = v_x\frac{\partial}{\partial x}+v_y\frac{\partial}{\partial y}+v_z\frac{\partial}{\partial z}
Now in your problem: \vec{v}=(x,y,x). The A scalar field is not given explicitly. So we have for the divergence according to the previous:
\text{div}(A\vec{v})=x\frac{\partial A}{\partial x}+y\frac{\partial A}{\partial y}+x\frac{\partial A}{\partial z}+A\left(\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial x}{\partial z}\right)=x\frac{\partial A}{\partial x}+y\frac{\partial A}{\partial y}+x\frac{\partial A}{\partial z}+2A

For the second part we have to calculate the curl. Using the same convention as above:

(\text{curl}(A\vec{v}))_k=\epsilon_{klm}\partial_lAv_m=A\epsilon_{klm}\partial_l v_m + \epsilon_{klm}(\partial_lA)v_m=

=(A\text{curl}\vec{v}+(\text{grad}A)\times\vec{v})_k<br />

Where \epsilon_{klm} is the three indice totaly antisymmetric tensor, the levi civita tensor.

So now using the given field:

\text{curl}(A\vec{v})=A(0\;,\;-1\;,\;0)+(\text{grad}A)\times\vec{v}

And we are done.

 

[gcooke]

Thank you very much Thaakisfox!

G.