For a vector field $\vec{A}$, which one of the following is FALSE?

Vector Field Fundamentals

A vector field is a function that assigns a vector to each point in space. These fields are fundamental in physics and engineering, representing quantities like fluid flow, gravitational forces, or electromagnetic forces. Understanding their properties, such as being solenoidal or irrotational, is crucial for various applications.

The question asks us to identify the statement that is FALSE regarding a vector field $\vec{A}$. Let's analyze each option provided.

Solenoidal Vector Field Definition

A vector field $\vec{A}$ is defined as solenoidal if its divergence is zero. The divergence of a vector field measures the magnitude of its source or sink at a given point.

- Mathematical Expression: A vector field $\vec{A}$ is solenoidal if $\nabla \cdot \vec{A}=0$.
- Interpretation: This means that at any point within the field, there is no net outflow or inflow of the "fluid" or quantity represented by the vector field. Examples include the magnetic field, which is always solenoidal, implying no magnetic monopoles.
- Option 1 Analysis: The statement "$\vec{A}$ is solenoidal if $\nabla \cdot \vec{A}=0$ " directly aligns with the definition of a solenoidal field. Therefore, this statement is TRUE.

Curl of a Vector Field

The curl of a vector field ($\nabla \times \vec{A}$) is another vector field that represents the infinitesimal rotation or "curliness" of the field at a given point. It indicates the tendency of the field to rotate a small object placed within it.

- Mathematical Expression: If $\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$, then its curl is given by: $$\nabla \times \vec{A}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ A_x& A_y& A_z\end{array}\right|=\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)\hat{i}+\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)\hat{j}+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)\hat{k}$$
- Interpretation: As seen from the formula, the result of the curl operation on a vector field is a new vector field. Each component of the resulting vector is a scalar, but together they form a vector.
- Option 2 Analysis: The statement "$\nabla \times \vec{A}$ is another vector field" correctly describes the nature of the curl operation. Therefore, this statement is TRUE.

Irrotational Vector Field Condition

A vector field $\vec{A}$ is defined as irrotational (or conservative) if its curl is zero. This means there is no rotational tendency within the field.

- Mathematical Expression: A vector field $\vec{A}$ is irrotational if $\nabla \times \vec{A}=\vec{0}$.
- Relationship with Scalar Potential: An irrotational vector field can always be expressed as the gradient of a scalar potential function, i.e., $\vec{A}=\nabla \varphi$.
- **Laplacian of a Vector Field (${\nabla }^2\vec{A}$):** The expression ${\nabla }^2\vec{A}=0$ means that each component of the vector field satisfies Laplace's equation. A vector field whose Laplacian is zero is called a harmonic vector field. While some irrotational fields might also be harmonic, ${\nabla }^2\vec{A}=0$ is not the defining condition for an irrotational field.
- Counterexample: Consider the vector field $\vec{A}=xy\hat{i}$. - Let's check if it's irrotational: $$\nabla \times \vec{A}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ xy& 0& 0\end{array}\right|=\hat{i}\left(\frac{\partial (0)}{\partial y}-\frac{\partial (0)}{\partial z}\right)-\hat{j}\left(\frac{\partial (0)}{\partial x}-\frac{\partial (xy)}{\partial z}\right)+\hat{k}\left(\frac{\partial (0)}{\partial x}-\frac{\partial (xy)}{\partial y}\right)$$ $$=\hat{i}(0-0)-\hat{j}(0-0)+\hat{k}(0-x)=-x\hat{k}$$ Since $\nabla \times \vec{A}=-x\hat{k}\ne \vec{0}$, the field $\vec{A}=xy\hat{i}$ is not irrotational. - Now let's check its Laplacian: $${\nabla }^2\vec{A}=\left(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}\right)(xy\hat{i})=\left(\frac{{\partial }^2(xy)}{\partial x^2}+\frac{{\partial }^2(xy)}{\partial y^2}+\frac{{\partial }^2(xy)}{\partial z^2}\right)\hat{i}$$ $$=(0+0+0)\hat{i}=\vec{0}$$ So, for $\vec{A}=xy\hat{i}$, we have ${\nabla }^2\vec{A}=\vec{0}$, but it is not irrotational. This clearly shows that ${\nabla }^2\vec{A}=0$ does not imply that $\vec{A}$ is irrotational.
- Option 3 Analysis: The statement "$\vec{A}$ is irrotational if ${\nabla }^2\vec{A}=0$ " is incorrect. The correct condition for an irrotational field is $\nabla \times \vec{A}=\vec{0}$. Therefore, this statement is FALSE.

Vector Identity

There are several fundamental vector identities that are widely used in vector calculus and physics. One such identity relates the curl of a curl to the gradient of a divergence and the Laplacian of the vector field.

- Mathematical Expression: The identity is given by: $$\nabla \times \left(\nabla \times \vec{A}\right)=\nabla \left(\nabla \cdot \vec{A}\right)-{\nabla }^2\vec{A}$$
- Interpretation: This identity is crucial for simplifying complex vector expressions and is often used in electromagnetic theory and fluid dynamics. It connects the rotational and divergence properties of a vector field with its Laplacian.
- Option 4 Analysis: The statement "$\nabla \times \left(\nabla \times \vec{A}\right)=\nabla \left(\nabla \cdot \vec{A}\right)-{\nabla }^2\vec{A}$ " is a standard and widely accepted vector identity. Therefore, this statement is TRUE.

Conclusion

Based on our analysis of each statement:

- Option 1: TRUE
- Option 2: TRUE
- Option 3: FALSE (The correct condition for irrotational is $\nabla \times \vec{A}=\vec{0}$, not ${\nabla }^2\vec{A}=0$.)
- Option 4: TRUE

The question asks for the statement that is FALSE. Thus, the false statement is Option 3.