To prove the vector calculus identity:

$\text{curl}(\phi \mathbf{F}) = \phi \, \text{curl}(\mathbf{F}) + (\nabla \phi) \times \mathbf{F}$

where $\phi$ is a scalar field, and $\mathbf{F}$ is a vector field, we can proceed step-by-step by applying vector calculus operations and leveraging the properties of the curl and gradient.

Step 1: Definition of Curl

Recall that the curl of a vector field $\mathbf{F} = (F_1, F_2, F_3)$ in three-dimensional Cartesian coordinates is defined as:

$\text{curl}(\mathbf{F}) = \nabla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right)$

Step 2: Expanding $\text{curl}(\phi \mathbf{F})$

The curl of the product of a scalar field $\phi$ and a vector field $\mathbf{F}$ is given by:

$\nabla \times (\phi \mathbf{F}) = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \times \left( \phi F_1, \phi F_2, \phi F_3 \right)$

We need to compute the curl of this product. Using the vector calculus identity for the curl of a product of a scalar field and a vector field:

$\nabla \times (\phi \mathbf{F}) = \phi (\nabla \times \mathbf{F}) + (\nabla \phi) \times \mathbf{F}$

This identity is derived from the product rule of differentiation. Let's break it down:

Step 3: Applying the Product Rule

The vector identity can be verified by expanding the curl of the product $\phi \mathbf{F}$:

$\nabla \times (\phi \mathbf{F}) = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \times \left( \phi F_1, \phi F_2, \phi F_3 \right)$

When you apply the curl operator to the components, you need to apply the distributive property of the cross product, and take derivatives on both the scalar field $\phi$ and the vector components $F_1, F_2, F_3$. This results in:

$\nabla \times (\phi \mathbf{F}) = \left( \frac{\partial \phi F_3}{\partial y} - \frac{\partial \phi F_2}{\partial z}, \frac{\partial \phi F_1}{\partial z} - \frac{\partial \phi F_3}{\partial x}, \frac{\partial \phi F_2}{\partial x} - \frac{\partial \phi F_1}{\partial y} \right)$

Using the product rule for differentiation:

$\frac{\partial}{\partial x} (\phi F_1) = \phi \frac{\partial F_1}{\partial x} + F_1 \frac{\partial \phi}{\partial x}$

Thus, we get:

$\nabla \times (\phi \mathbf{F}) = \phi (\nabla \times \mathbf{F}) + (\nabla \phi) \times \mathbf{F}$

Step 4: Conclusion

The identity is proven:

$\nabla \times (\phi \mathbf{F}) = \phi \, \nabla \times \mathbf{F} + (\nabla \phi) \times \mathbf{F}$

This concludes the proof.

Do you need further details or have any questions about this identity?

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5 Related Questions:

1. What is the physical interpretation of the curl operator in vector calculus?
2. How does the product rule apply to vector fields and scalar fields?
3. Can this identity be generalized to higher dimensions?
4. What is the difference between divergence and curl of a vector field?
5. How does the gradient of a scalar field relate to the curl of a vector field?

Tip:

In vector calculus, it's helpful to practice applying product rules to different combinations of scalar and vector fields to get a feel for how operations like divergence, curl, and gradient interact.