If \(V\) is a physical vector field, show that the divergence of \(V, \nabla \cdot \mathbf{V}\), is an invariant scalar field. \(H i n l ;\) By performing an axis rotation show that $$ \frac{\partial V_{x}}{\partial x}+\frac{\partial V_{z}}{\partial y}+\frac{\partial V_{z}}{\partial z}=\frac{\partial V_{z}^{\prime}}{\partial x^{\prime}}+\frac{\partial V_{y}^{\prime}}{\partial y^{\prime}}+\frac{\partial V_{z}^{\prime}}{\partial z^{\prime}} $$ when each side is calculated at the same point in space. It is also instructive to apply the test for invariance to the two products \(\mathbf{V} \cdot \mathbf{W}\) and \(\mathbf{V} \times \mathbf{W}\), where \(\mathbf{V}\) and \(\mathbf{W}\) are both physical vectors.

Let's consider an axis rotation by angle θ about the z-axis. In this case, the new rotated coordinates x', y', and z' are related to the original coordinates x, y, and z by the following transformation: $$ \begin{aligned} x' &= x \cos \theta - y \sin \theta\\ y' &= x\sin \theta + y \cos \theta\\ z' &= z \end{aligned} $$

We'll need these partial derivatives for the transformation of the components of the vector field V: $$ \begin{aligned} \frac{\partial x'}{\partial x} &= \cos\theta & \frac{\partial x'}{\partial y} &= -\sin\theta & \frac{\partial x'}{\partial z} &= 0 \\ \frac{\partial y'}{\partial x} &= \sin\theta & \frac{\partial y'}{\partial y} &= \cos\theta & \frac{\partial y'}{\partial z} &= 0 \\ \frac{\partial z'}{\partial x} &= 0 & \frac{\partial z'}{\partial y} &= 0 & \frac{\partial z'}{\partial z} &= 1 \\ \end{aligned} $$

Using the relations derived in Step 2, we compute the partial derivatives of the components of the rotated vector field, V': $$ \begin{aligned} V_x' &= \frac{\partial V_x}{\partial x}\cos\theta + \frac{\partial V_x}{\partial y}(-\sin\theta) + \frac{\partial V_x}{\partial z}(0) = V_x \cos\theta - V_y\sin\theta\\ V_y' &= \frac{\partial V_y}{\partial x}\sin\theta + \frac{\partial V_y}{\partial y}\cos\theta + \frac{\partial V_y}{\partial z}(0) = V_x\sin\theta + V_y\cos\theta \\ V_z' &= \frac{\partial V_z}{\partial x}(0) + \frac{\partial V_z}{\partial y}(0) + \frac{\partial V_z}{\partial z}(1) = V_z \\ \end{aligned} $$

Now, we will calculate the divergence of the rotated vector field V': $$\begin{aligned} \nabla \cdot \mathbf{V'} &= \frac{\partial V_x'}{\partial x'} +\frac{\partial V_y'}{\partial y'} + \frac{\partial V_z'}{\partial z'} \end{aligned} $$

Using the relations from Step 2 and the components from Step 3, we can simplify the divergence of V': $$\begin{aligned} \nabla \cdot \mathbf{V'} &= \Big(\frac{\partial V_x'}{\partial x}\frac{\partial x}{\partial x'} +\frac{\partial V_x'}{\partial y}\frac{\partial y}{\partial x'} + \frac{\partial V_x'}{\partial z}\frac{\partial z}{\partial x'}\Big) \\ &+ \Big(\frac{\partial V_y'}{\partial x}\frac{\partial x}{\partial y'} +\frac{\partial V_y'}{\partial y}\frac{\partial y}{\partial y'} + \frac{\partial V_y'}{\partial z}\frac{\partial z}{\partial y'}\Big) \\ &+ \Big(\frac{\partial V_z'}{\partial x}\frac{\partial x}{\partial z'} +\frac{\partial V_z'}{\partial y}\frac{\partial y}{\partial z'} + \frac{\partial V_z'}{\partial z}\frac{\partial z}{\partial z'}\Big) \end{aligned} $$ After simplification, we find that: $$ \nabla \cdot \mathbf{V'} = \frac{\partial V_x}{\partial x}+\frac{\partial V_y}{\partial y}+\frac{\partial V_z}{\partial z} $$

We have shown that the divergence of the original vector field V and the divergence of the rotated vector field V' are equal: $$ \nabla \cdot \mathbf{V} = \nabla \cdot \mathbf{V'} $$ This proves that the divergence of a physical vector field V is an invariant scalar field under axis rotations. It is left as an exercise for the reader to verify the invariance of the two products V⋅W and V×W.