The problem is asking to prove the identity:

$\text{curl} (\nabla \phi) = \mathbf{0},$

where $\phi$ is a scalar field (denoted here as $r_m$).

Proof:

1. Understanding Curl and Gradient:

   • The curl of a vector field $\mathbf{F}$ is denoted as $\nabla \times \mathbf{F}$.
   • The gradient of a scalar field $\phi$ is denoted as $\nabla \phi$, which results in a vector field.

2. Expression for Curl of a Gradient:

   • Let $\phi$ be a scalar field. The gradient is $\nabla \phi$.
   • The curl of this gradient is: $\nabla \times (\nabla \phi).$

3. Curl of Gradient is Zero:

   • The curl of the gradient of any scalar field $\phi$ is always zero. This is a property of vector calculus, derived from the fact that mixed partial derivatives commute (i.e., $\frac{\partial^2 \phi}{\partial x \partial y} = \frac{\partial^2 \phi}{\partial y \partial x}$).

   Mathematically:

   \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z} \end{vmatrix}.$$ Expanding this determinant, you find that all terms cancel out due to the commutativity of partial derivatives, leaving: $$\nabla \times (\nabla \phi) = \mathbf{0}.$$

4. Conclusion:

   • Therefore, $\text{curl} (\nabla \phi) = 0$, which proves the identity.

Let me know if you'd like a more detailed explanation of any step or have further questions!

Related Questions:

1. What is the geometric interpretation of the gradient and curl in vector calculus?
2. How does the curl of a vector field relate to circulation and rotation?
3. What are the conditions under which $\nabla \cdot (\nabla \times \mathbf{F}) = 0$?
4. Can the divergence of a gradient ever be zero? Why or why not?
5. How does this property relate to conservative fields in physics?

Tip:

Always keep in mind that vector identities, such as $\nabla \times (\nabla \phi) = 0$, rely heavily on the smoothness (differentiability) of the scalar field $\phi$.