Proof that divergence of vector field is a scalar

Question

Let $\underline{x}$ be a vector with components $x_i$ in a given frame $\underline{e}_i$. The expression for components $x_i'$ in a new frame $\underline{e}_i'$ is $$x_i'=l_{ij}x_j$$ where $l_{ij}=\underline{e}_i'·\underline{e}_j$ is the passive transformation. How can I show that $\frac{\partial x_j}{\partial x_i'}=l_{ij}$? Using that result I need to show that if $\underline{u}(\underline{x})$ is a vector field then $\underline{\nabla}·\underline{u} $ is a scalar field. How do I go about this?

Answer 1

You've already got the vector transformation law $x_i ' = l_{ij} x_j$ (it pains me, but I'll write everything with a lower index since we're working with a euclidean metric anyway. If you continue your study of differential geometry, you'll learn to distinguish between raised and lowered indices, but for now it's not important)

To obtain the expression for $l_{ij}$, just calculate $$ \frac{\partial x_i '}{\partial x_k} = \frac{\partial}{\partial x_k} (l_{ij} x_j) = l_{ij} \frac{\partial x_k}{\partial x_j} = l_{ij} \delta_{kj} = l_{ik} $$ Note that I have assumed that $l_{ij}$ does not depend on $x$. This is a necessary assumption, otherwise the result is not true in general. Again, more on this when you study differential geometry.

Anyway, we calculate $$ \begin{align} \nabla ' \cdot u' &= \frac{\partial}{\partial x_i '} u_i ' = \frac{\partial}{\partial x_i '} \left( l_{ij} u_j \right) \\&= l_{ij} \frac{\partial u_j}{\partial x_i '} \\&= l_{ij} \frac{\partial x_k}{\partial x_i '} \frac{\partial u_j}{\partial x_k} \\&= \frac{\partial x_i '}{\partial x_j}\frac{\partial x_k}{\partial x_i '} \frac{\partial u_j}{\partial x_k} \\&= \delta_{jk} \frac{\partial u_j}{\partial x_k} \\&= \frac{\partial u_k}{\partial x_k} = \nabla \cdot u \end{align} $$ where we have used the multi-variable chain rule and the well-known fact that $ \frac{\partial x_i '}{\partial x_j}\frac{\partial x_k}{\partial x_i '} = \delta_{jk}$.

You should know that this calculation only makes sense for cartesian coordinates and changes of variables that do not depend on $x$ (a pretty heavy restriction). In any case, the calculation demonstrates that divergence is a scalar for this restricted class of coordinate transformations.

The more general fact essentially ends up being true by definition. But you'll have to learn some differential geometry before you can understand how divergence is even defined in the more general setting.

Answer 2

The generalized coordinates notation used by your question is highly abstract, unfortunately, but the concept is straightforward.

Let $\vec F$ be a vector field, for instance

$$\vec F = \begin{pmatrix} F_x \\ F_y \\ F_z \end{pmatrix}$$

The by the definition for divergence,

$$div(\vec F) = \nabla \cdot \vec F = \frac{dF_z}{dx} + \frac{dF_y}{dy} + \frac{dF_z}{dz}$$

This computed quantity will by definition be a scalar number.

An numerical example is always helpful. For instance,

$$\vec F = \begin{pmatrix} x \\ x-z \\ yz \end{pmatrix}$$

Then

$$ div(\vec F) = \nabla \cdot \vec F$$ $$= \frac{d}{dx}(x) + \frac{d}{dy}(x-z) + \frac{d}{dz}(yz)$$

$$=1 + 0 + y$$ $$ = y+1$$

is a scalar field. Hope this helps.