Fernando Garcia
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PHYS405 - Electricity & Magnetism I
Spring 2024
Problem 1.53
Recall from problem 1.15 the vectors:
\begin{align*} \mathbf{v}_a &= x^2 \hat{\mathbf{x}}+3xz^2 \hat{\mathbf{y}}-2xz\hat{\mathbf{z}} \\ \mathbf{v}_b &= xy \hat{\mathbf{x}}+2yz \hat{\mathbf{y}}+3zx\hat{\mathbf{z}} \\ \mathbf{v}_c &= y^2 \hat{\mathbf{x}}+(2xy+z^2 ) \hat{\mathbf{y}}+2yz\hat{\mathbf{z}} \end{align*}a) Which of them can be expressed as the gradient of a scalar? Find a scalar function that does the job.
b) Which can be expressed as the curl of a vector? Find such a vector.
To gain geometrical intuition as to when a vector field is the gradient of scalar or the curl of another vector field, it is relevant to see what these fields look like: (In order $\mathbf{v}_a,\mathbf{v}_b,\mathbf{v}_c $)
a) Which of them can be expressed as the gradient of a scalar? Find a scalar function that does the job.
Recall that a vector field has a scalar potential if and only if the vector field is curl-less (also known as an irrotational field):
\[\nabla \times \mathbf{F}=\mathbf{0}\;\Leftrightarrow \; \mathbf{F}=-\nabla V \tag{Eq. 1.103, page 51} \]By recalling that the curl is given by:
\[\nabla \times \mathbf{v}= \left( \frac{\partial v_z}{\partial y}-\frac{\partial v_y}{\partial z} \right) \hat{\mathbf{x}} + \left( \frac{\partial v_x}{\partial z}-\frac{\partial v_z}{\partial x} \right) \hat{\mathbf{y}} + \left( \frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y} \right) \hat{\mathbf{z}} \tag{As seen on the cover of your book}\]It is straightforward to see that
\begin{align*} \nabla \times \mathbf{v}_a &= -6xz\hat{\mathbf{x}} +2z \hat{\mathbf{y}} +3z^2 \hat{\mathbf{z}} \\ \nabla \times\mathbf{v}_b &= -2y\hat{\mathbf{x}} -3z \hat{\mathbf{y}} -x \hat{\mathbf{z}} \\ \nabla \times\mathbf{v}_c &= \mathbf{0} \end{align*}Thus, $\mathbf{v}_c $ is the only vector field out of those 3 which can be written as the gradient of a scalar field.
b) Which can be expressed as the curl of a vector? Find such a vector.
Recall that a vector field has a vector potential if and only if the vector field is divergence-less (also known as an solenoidal field):
\[\nabla \cdot \mathbf{F}=0\;\Leftrightarrow \; \mathbf{F}=\nabla \times \mathbf{A} \tag{Eq. 1.104, page 52} \]By recalling that the divergence is given by:
\[\nabla \cdot \mathbf{v}=\frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} +\frac{\partial v_z}{\partial z} \tag{As seen on the cover of your book}\]It is straightforward to see that
\begin{align*} \nabla \cdot \mathbf{v}_a &= 0 \\ \nabla \cdot\mathbf{v}_b &= 3 x + y + 2 z \\ \nabla \cdot\mathbf{v}_c &= 2 x + 2 y \end{align*}Thus, $\mathbf{v}_a $ is the only vector field out of those 3 which can be written as the gradient of a scalar field.