Regardless of the form of a vector field F , one always has div(curl F ) = 0
The Correct Answer and Explanation is:
Yes, the statement is correct: the divergence of the curl of any vector field is always zero, mathematically written as:div(∇×F)=0\text{div} (\nabla \times \mathbf{F}) = 0div(∇×F)=0
To understand why this is true, let’s break it down step by step.
- Curl of a vector field: The curl of a vector field F=(F1,F2,F3)\mathbf{F} = (F_1, F_2, F_3)F=(F1,F2,F3) represents the rotational tendency of the field at each point. The curl of F\mathbf{F}F is defined as:
∇×F=(∂F3∂y−∂F2∂z,∂F1∂z−∂F3∂x,∂F2∂x−∂F1∂y)\nabla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} – \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} – \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} – \frac{\partial F_1}{\partial y} \right)∇×F=(∂y∂F3−∂z∂F2,∂z∂F1−∂x∂F3,∂x∂F2−∂y∂F1)
- Divergence of a vector field: The divergence of a vector field A=(A1,A2,A3)\mathbf{A} = (A_1, A_2, A_3)A=(A1,A2,A3) represents the net rate of change of the field’s strength at a point. The divergence of A\mathbf{A}A is defined as:
div(A)=∂A1∂x+∂A2∂y+∂A3∂z\text{div} (\mathbf{A}) = \frac{\partial A_1}{\partial x} + \frac{\partial A_2}{\partial y} + \frac{\partial A_3}{\partial z}div(A)=∂x∂A1+∂y∂A2+∂z∂A3
- Divergence of the curl: Now, when we take the divergence of the curl of F\mathbf{F}F, we apply the divergence operator to each component of the curl. That is:
div(∇×F)=∂∂x(∂F3∂y−∂F2∂z)+∂∂y(∂F1∂z−∂F3∂x)+∂∂z(∂F2∂x−∂F1∂y)\text{div} (\nabla \times \mathbf{F}) = \frac{\partial}{\partial x} \left( \frac{\partial F_3}{\partial y} – \frac{\partial F_2}{\partial z} \right) + \frac{\partial}{\partial y} \left( \frac{\partial F_1}{\partial z} – \frac{\partial F_3}{\partial x} \right) + \frac{\partial}{\partial z} \left( \frac{\partial F_2}{\partial x} – \frac{\partial F_1}{\partial y} \right)div(∇×F)=∂x∂(∂y∂F3−∂z∂F2)+∂y∂(∂z∂F1−∂x∂F3)+∂z∂(∂x∂F2−∂y∂F1)
- Simplifying the expression: When we simplify this expression using the mixed partial derivatives (the fact that the order of partial derivatives does not matter for sufficiently smooth functions), we see that each term cancels out. Specifically, terms like ∂2F1∂x∂y\frac{\partial^2 F_1}{\partial x \partial y}∂x∂y∂2F1 will cancel with ∂2F1∂y∂x\frac{\partial^2 F_1}{\partial y \partial x}∂y∂x∂2F1, and similarly for the other components.
Thus, the result is:div(∇×F)=0\text{div} (\nabla \times \mathbf{F}) = 0div(∇×F)=0
This is true for any vector field F\mathbf{F}F, and it’s a fundamental identity in vector calculus. The reason is that the curl measures the local rotation of the field, and the divergence measures the net “source” or “sink” of the field. Since rotation cannot have a net source or sink, the divergence of the curl is zero everywhere.
