Clarification about Divergence

Question

So I know $\mathrm{Div}(F)=∇·F$ and $$ ∇=\frac{\partial \:}{\partial \:x}i+\frac{\partial \:}{\partial \:y}j+\frac{\partial \:}{\partial \:z}k$$ So for example, if $$ F=(2x-\cos z)i+(x^2+z)j+(x\sin z+y)k $$

My intuition was to find the gradient: $$ ∇=2i+0j+(x\cos z)k $$

and then do the dot product of this with F so my final answer would be: $$ \mathrm{Div}(F)=(2x\cos z)(2)+(x^2+z)(0)+(x\sin z+y)(x\cos z) $$

Now I know this is wrong and that the real answer is actually just $$ \mathrm{Div}(F)=2+0+(x\cos z) $$

but I am confused as to why because isn't that just the gradient? and If that is the answer then what is the gradient? or are they the same?

Answer 1

The Divergence operator applied to a vector field $F = (F_x,F_y,F_z)$, is defined by

$$ Div(F) := \partial_x F_x + \partial_y F_y + \partial_z F_z. $$

The notation $\nabla\cdot F$ is just a way of remembering the definition. (Strictly speaking, the dot product of an operator and a vector is undefined.)

How did you take the gradient of a vector and get another vector?

Answer 2

Gradient is for functions that map from 2 or 3 dimensional space to 1-dimensional space. And gradient is a vector.

Divergence is a scalar and is for functions that map from 3 space to 3 space (or 2-space to 2-space.)

Answer 3

I thik that your confusion is in the interpretation of how $\nabla$ operator work. This operator explicity is $\nabla=(\partial_x,\partial_y,\partial_z)$ (in the three dimensional case) and it can be ussually writen with canonical coordinates as $\nabla=\partial_x i+\partial_y j+\partial_z k$. The notation $\nabla\cdot F$ indicate that you must calculate the correspondig partial derivaty of each component of $F$ (that is $\nabla\cdot F=(\partial_x F_x,\partial_y Fy,\partial_zF_z)$ where $F=(F_x,F_y,F_z)$).

Answer 4

As stated in the other answers, the gradient operator is usually just defined for functions, i.e. maps $f\colon\mathbb{R}^n \to \mathbb{R}$. What you where looking for is a generalisation of the gradient to vector fields, i.e. maps $F\colon\mathbb{R}^n \to \mathbb{R}^m$. As you suspected such a generalisation does exist, but its not the map you thought it was. The generalisation is the following.

Let $F\colon\mathbb{R}^n \to \mathbb{R}^m$ be a vectorfield and write $F=(F^1, \dots, F^m)^T$, i.e. $F^i (x)$ is the $i$-th coordinate of the vector $F(x)$ for all $x\in \mathbb{R}^n$. Then we can define the gradient $\nabla F$ of $F$ as the transpose of the Jacobian of F: $$\nabla F = (\frac{\partial F^i}{\partial x_j})_{i,j} = \begin{pmatrix} \frac{\partial F^1}{\partial x_1} & \dots & \frac{\partial F^m}{\partial x_1} \\ \vdots & & \vdots \\ \frac{\partial F^1}{\partial x_n} & \dots & \frac{\partial F^m}{\partial x_n} \\ \end{pmatrix}$$

(See e.g. the wikipedia page on the gradient.) In the case $n=m$ we can then obtain the divergence by taking the trace, thus $Div(F)=Tr(\nabla F)$.

Note that in this procedure we use $F$ only once, namely when calculating all the derivatives of $F$. The point where it went wrong in your naive calculation of the divergence wasn't as much in your definition of $\nabla F$, but in the fact that you then again used $F$ to try to find $Div(F)$. You thus had two applications of $F$ in your calculation of the divergence. Such aproaches will never work, since F is only used once in the calculation of $Div(F)$. You probably missed this since you wrote just $\nabla$ for what was your variant of $\nabla F$.