# Help with notation.

1. May 8, 2009

### J Hill

I've recently begun to study continuum mechanics, and am having difficulty with tensor derivatives, primarilly because of notation... so can anyone explain what these equations mean, as far as notation goes:
1. $$\frac{\partial f}{\partial\mathbf{v}}\cdot\mathbf{u}=Df(\mathbf{v})[\mathbf{u}]$$
2. $$\frac{\partial \mathbf{f}}{\partial \mathbf{v}}\cdot\mathbf{u}=D\mathbf{f}(\mathbf{v})[\mathbf{u}]$$

2. May 9, 2009

### BobbyBear

Hi J,

$$\frac{\partial \mathbf{f}}{\partial \mathbf{v}}$$

is a second order tensor: it is the gradient of the vector field f, which in cartesian coordinates would be represented by the Jacobian matrix of the function f as a vector valued function of the vairables vi ,
f=(f1(v1,v2,...,vn),f2(v1,v2,...,vn),...,fn(v1,v2,...,vn))
(though please note that

$$\frac{\partial \mathbf{f}}{\partial \mathbf{v}}$$

is a tensor, not a matrix). I prefer to write it as:

$\nabla \mathbf{f}$

as this way it's coordinate independent.
The dot product is seen as a contraction, so that

$\nabla \mathbf{f} \cdot\mathbf{u}$

yields a vector (or first order tensor), which is basically the directional derivative of the field f in the direction of u (or proportional to it if u has modulus different to 1).

I don't like the notation
$$D\mathbf{f}(\mathbf{v})[\mathbf{u}]$$
because this suggests that
$$D\mathbf{f}(\mathbf{v})$$
and
$$[\mathbf{u}]$$
are matrices rather than tensors (a square matrix and a column matrix respectively), and all you have to do is the matrix multiplication. Oh well as long as you don't forget that the matrices just represent the components of the tensors in a certain basis :p

My favourite way of writing

$\nabla \mathbf{f}$

though, is using indicial notation, because it leaves no room for ambiguity. Thus,

$\nabla \mathbf{f}$
is written as:
$\mathbf{e_i}\frac{\partial\mathbf{f}}{\partial v_i} = \mathbf{e_i}\frac{\partial f^j \mathbf{e_j}}{\partial v^i}$
so that

$\nabla \mathbf{f} \cdot\mathbf{u} = \mathbf{e_i}\frac{\partial\mathbf{f}}{\partial v_i} \cdot\mathbf{u} = \mathbf{e_i}\frac{\partial f^j \mathbf{e_j}}{\partial v^i} \cdot u^k \mathbf{e_k}$

In a Cartesian coordinate system this would equal:
$\nabla \mathbf{f} \cdot\mathbf{u} = = \mathbf{e_i}\frac{\partial f_j \mathbf{e_j}}{\partial v^i} \cdot u_k \mathbf{e_k} = \mathbf{e_i}\frac{\partial f_j }{\partial v_i} \mathbf{e_j} \cdot u_k \mathbf{e_k} = \mathbf{e_i}\frac{\partial f_j }{\partial v_i} u_k (\mathbf{e_j} \cdot \mathbf{e_k}) = \mathbf{e_i}\frac{\partial f_j }{\partial v_i} u_k \delta_j_k = \mathbf{e_i}\frac{\partial f_j }{\partial v_i} u_j$

Hope this helps! :)

3. May 9, 2009

### BobbyBear

Oops, I showed you what 2. is only, 1. is the same except that f is a scalar field, ie. same as the second case but with just one component, so that grad(f) is a vector and the result of your operation is a scalar:
$\nabla f \cdot\mathbf{u} = \mathbf{e_i}\frac{\partial f }{\partial v^i} \cdot u_k \mathbf{e_k} = \frac{\partial f}{\partial v_i} u_k (\mathbf{e_i} \cdot \mathbf{e_k}) = \frac{\partial f}{\partial v_i} u_k \delta_i_k = \frac{\partial f}{\partial v_i} u_i$

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook