Using Vector Operations In Mathematica

Question

I would like to calculate

$$ \mathbf T = (\mathbf B \bullet \nabla) \mathbf B$$

Where nabla (the upside down triangle) is the grad operator $(\partial/\partial x,\partial/\partial y,\partial/\partial z)$ and the dot is the divergence operator. This is the magnetic tension force, for those who are interested.

I have a vector

$$ \mathbf B = (B_x, B_y, B_z ) $$

At the moment I break this down into single-dimensional operations, do that for each dimension, then create the $\mathbf T$ vector, but I wanted to know if I could do it using the vector operations that Mathematica offers.

The struggle is that I am doing an operation on an operator, then the latter operator carries out its operation. So, the divergence operation will give us a scaler:

$$ (\mathbf B \bullet \nabla) = (B_x * \partial/\partial x + B_y * \partial/\partial y + B_z * \partial/\partial z)$$

Then for instance in the x dimension:

$$T_x = (B_x * \partial/\partial x + B_y * \partial/\partial y + B_z * \partial/\partial z) B_x = B_x \frac{\partial B_x}{\partial x} + B_y \frac{\partial B_x}{\partial y} + B_z \frac{\partial B_x}{\partial z}$$

So the full vector would be:

$$\mathbf T = (B_x \frac{\partial B_x}{\partial x} + B_y \frac{\partial B_x}{\partial y} + B_z \frac{\partial B_x}{\partial z}, B_x \frac{\partial B_y}{\partial x} + B_y \frac{\partial B_y}{\partial y} + B_z \frac{\partial B_y}{\partial z} , B_x \frac{\partial B_z}{\partial x} + B_y \frac{\partial B_z}{\partial y} + B_z \frac{\partial B_z}{\partial z}) $$

Mathematica offers the following functions Grad[] and Div[]; however, I can't work out how to get them to work in the required fashion.

Answer 1

Here is a way to do the manipulation. First I define a convenient way to set up vector fields, then I do the dot product and gradient. The main thing is that the application to the vector argument on the right is done element-wise, so that the natural Mathematica operation is Map (/@).

To make the order of operations clearer, I also use two different vector fields:

vec[a_Symbol, r_: {x, y, z}] := 
 Through[Array[Subscript[SymbolName[a], Map[SymbolName, r][[#]]] &, 3] @@ r]

vec[a].Grad[#, {x, y, z}] & /@ vec[b]

$$\left\{\text{a}_{\text{x }}(x,y,z) \text{b}_{\text{x}}{}^{(1,0,0)}(x,y,z)+\text{a}_{\text{y}}(x,y,z) \text{b}_{\text{x}}{}^{(0,1,0)}(x,y,z)+\text{a}_{\text{z }}(x,y,z) \text{b}_{\text{x}}{}^{(0,0,1)}(x,y,z),\text{a}_{\text{x }}(x,y,z) \text{b}_{\text{y}}{}^{(1,0,0)}(x,y,z)+\text{a}_{\text{y }}(x,y,z) \text{b}_{\text{y}}{}^{(0,1,0)}(x,y,z)+\text{a}_{\text{z }}(x,y,z) \text{b}_{\text{y}}{}^{(0,0,1)}(x,y,z),\text{a}_{\text{x }}(x,y,z) \text{b}_{\text{z}}{}^{(1,0,0)}(x,y,z)+\text{a}_{\text{y }}(x,y,z) \text{b}_{\text{z}}{}^{(0,1,0)}(x,y,z)+\text{a}_{\text{z }}(x,y,z) \text{b}_{\text{z}}{}^{(0,0,1)}(x,y,z)\right\}$$

With the function vec, you can specify the set of independent variables as an optional argument. It returns an array in which each entry is labeled by the first argument and by one of the coordinate names in the second argument (whose default is {x,y,z}).

By using Map, we're able to preserve the order of operations exactly the way it was written in the question.