# Divergence of a vector field is a scalar field?

merrypark3
Hello.

How can I show the Divergence of a vector field is a scalar field(in $$E^{3}$$) ?
Should I show that Div is invariant under rotation?

$$x^{i'}=a^{ij}x^{j},V^{'}_{i}(\stackrel{\rightarrow}{x})=a_{ij}v_{j}(\stackrel{\rightarrow}{x})$$

then

$$\frac{\partial V^{'}_{i}(\stackrel{\rightarrow}{x})}{\partial x^{'i}}=\frac{\partial(a_{ij}V_{j} (\stackrel{\rightarrow}{x})) }{\partial(a^{ij} x^{j} ) } = \frac{\partial V_{i} (\stackrel{\rightarrow}{x})}{\partial x^{i}}$$

How can I prove the last equality?

## Answers and Replies

chrisk
Start with the dot product of the del operator with the vector field; this is the expression for divergence.

Staff Emeritus
Homework Helper
Hello.

How can I show the Divergence of a vector field is a scalar field(in $$E^{3}$$) ?
Should I show that Div is invariant under rotation?

$$x^{i'}=a^{ij}x^{j},V^{'}_{i}(\stackrel{\rightarrow}{x})=a_{ij}v_{j}(\stackrel{\rightarrow}{x})$$

then

$$\frac{\partial V^{'}_{i}(\stackrel{\rightarrow}{x})}{\partial x^{'i}}=\frac{\partial(a_{ij}V_{j} (\stackrel{\rightarrow}{x})) }{\partial(a^{ij} x^{j} ) } = \frac{\partial V_{i} (\stackrel{\rightarrow}{x})}{\partial x^{i}}$$

How can I prove the last equality?
Use the chain rule to express the derivative with respect to x'i in terms of the derivatives with respect to the unprimed coordinates.