If \(\vec{F}\) is irrotational (i.e., \(\nabla \times \vec{F} = \vec{0}\)), then there exists a scalar point function \(\phi\) such that:
\[
\vec{F} = \nabla \phi
\]
Then \(\phi\) is called the scalar potential of \(\vec{F}\).
Note:
- A conservative vector field is also known as an irrotational vector field.
-
\[ \text{div}(\text{grad}\,\phi) = \nabla \cdot \nabla \phi = \nabla^{2}\phi = \frac{\partial^{2}\phi}{\partial x^{2}} + \frac{\partial^{2}\phi}{\partial y^{2}} + \frac{\partial^{2}\phi}{\partial z^{2}} \]
where \(\nabla^{2} = \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}}\) is called the Laplacian operator, and \(\nabla^{2}\phi = 0\) is called Laplace's equation.
Properties of Scalar Potential:
- The scalar potential \(\phi\) is determined up to an additive constant
- For a conservative field, the line integral between two points is path-independent and equals the difference in potential:
\[ \int_{A}^{B} \vec{F} \cdot d\vec{r} = \phi(B) - \phi(A) \]
- All gradient fields are irrotational (\(\nabla \times \nabla \phi = \vec{0}\))
Example 1: Show that the vector field defined by \(\vec{F}=(y\sin z-\sin x)\hat{i}+(x\sin z+2yz)\hat{j}+(xy\cos z+y^{2})\hat{k}\) is irrotational and find its scalar potential.
Given: \(\vec{F}=(y\sin z-\sin x)\hat{i}+(x\sin z+2yz)\hat{j}+(xy\cos z+y^{2})\hat{k}\)
\[
\text{curl}\,\vec{F} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y\sin z-\sin x & x\sin z+2yz & xy\cos z+y^{2}
\end{vmatrix}
\]
\[
= \left(\frac{\partial}{\partial y}(xy\cos z+y^{2}) - \frac{\partial}{\partial z}(x\sin z+2yz)\right)\hat{i}
\]
\[
- \left(\frac{\partial}{\partial x}(xy\cos z+y^{2}) - \frac{\partial}{\partial z}(y\sin z-\sin x)\right)\hat{j}
\]
\[
+ \left(\frac{\partial}{\partial x}(x\sin z+2yz) - \frac{\partial}{\partial y}(y\sin z-\sin x)\right)\hat{k}
\]
\[
= (x\cos z + 2y - x\cos z - 2y)\hat{i} - (y\cos z - y\cos z)\hat{j} + (\sin z - \sin z)\hat{k} = \vec{0}
\]
\(\therefore \vec{F}\) is irrotational
\[
\vec{F} = \nabla \phi \Rightarrow
\begin{cases}
\frac{\partial \phi}{\partial x} = y\sin z - \sin x \\
\frac{\partial \phi}{\partial y} = x\sin z + 2yz \\
\frac{\partial \phi}{\partial z} = xy\cos z + y^{2}
\end{cases}
\]
We know that:
\[
d\phi = \frac{\partial \phi}{\partial x}dx + \frac{\partial \phi}{\partial y}dy + \frac{\partial \phi}{\partial z}dz
\]
\[
= (y\sin z - \sin x)dx + (x\sin z + 2yz)dy + (xy\cos z + y^{2})dz
\]
\[
= (y\sin z\,dx + x\sin z\,dy + xy\cos z\,dz) + (2yz\,dy + y^{2}\,dz) - \sin x\,dx
\]
\[
= d(xy\sin z) + d(y^{2}z) - d(\cos x)
\]
\[
\phi = xy\sin z + y^{2}z + \cos x + c \quad \text{(where } c \text{ is a constant)}
\]
Example 2: Fluid motion is given by \(\vec{V}=ax\hat{i}+ay\hat{j}-2az\hat{k}\)
i) Is it possible to find the velocity potential? If so, find it.
ii) Is the motion possible for an incompressible fluid?
i)
First, check if the flow is irrotational (curl \(\vec{V} = \vec{0}\)):
\[
\text{curl}\,\vec{V} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
ax & ay & -2az
\end{vmatrix}
\]
\[
= \left(\frac{\partial}{\partial y}(-2az) - \frac{\partial}{\partial z}(ay)\right)\hat{i}
- \left(\frac{\partial}{\partial x}(-2az) - \frac{\partial}{\partial z}(ax)\right)\hat{j}
+ \left(\frac{\partial}{\partial x}(ay) - \frac{\partial}{\partial y}(ax)\right)\hat{k}
\]
\[
= (0 - 0)\hat{i} - (0 - 0)\hat{j} + (0 - 0)\hat{k} = \vec{0}
\]
The flow is irrotational, so a velocity potential \(\phi\) exists.
Find \(\phi\) such that \(\vec{V} = \nabla \phi\):
\[
\frac{\partial \phi}{\partial x} = ax \quad \Rightarrow \quad \phi = \frac{1}{2}ax^2 + f(y,z)
\]
\[
\frac{\partial \phi}{\partial y} = ay \quad \Rightarrow \quad \frac{\partial f}{\partial y} = ay \quad \Rightarrow \quad f(y,z) = \frac{1}{2}ay^2 + g(z)
\]
\[
\frac{\partial \phi}{\partial z} = -2az \quad \Rightarrow \quad \frac{dg}{dz} = -2az \quad \Rightarrow \quad g(z) = -az^2 + C
\]
\[
\phi = \frac{1}{2}ax^2 + \frac{1}{2}ay^2 - az^2 + C
\]
where \(C\) is an arbitrary constant.
ii)
Check if the flow is incompressible (div \(\vec{V} = 0\)):
\[
\text{div}\,\vec{V} = \frac{\partial}{\partial x}(ax) + \frac{\partial}{\partial y}(ay) + \frac{\partial}{\partial z}(-2az) = a + a - 2a = 0
\]
The flow is incompressible (div \(\vec{V} = 0\)).
Model problems on Scalar Potential
-
Show that \(\vec{F}=(y^{2}\cos x+z^{3})\hat{i}+(2y\sin x-4)\hat{j}+(3xz^{2}-2)\hat{k}\) is a conservative vector field and find its scalar potential.
Answer: \(y^{2}\sin x + z^{3}x - 4y + 2z + C\) -
Show that the vector field \(\vec{F}=(2xy^{2}+yz)\hat{i}+(2x^{2}y+xz+2yz^{2})\hat{j}+(2y^{2}z+xy)\hat{k}\) is irrotational and find its scalar potential.
Answer: \(x^{2}y^{2} + y^{2}z^{2} + xyz + C\) -
Show that the vector field \(\vec{F}=(x^{2}-y^{3})\hat{i}+(y^{2}-3x)\hat{j}+(z^{2}-xy)\hat{k}\) is irrotational and find its scalar potential.
Answer: Not irrotational -
Show that the vector field \(\vec{F}=(x^{2}-y^{2}+x)\hat{i}-(2xy+y)\hat{j}\) is irrotational and find its scalar potential.
Answer: \(\frac{1}{3}x^{3} - xy^{2} + \frac{1}{2}x^{2} - \frac{1}{2}y^{2} + C\) -
A fluid motion is given by \(\vec{V}=(y+z)\hat{i}+(z+x)\hat{j}+(x+y)\hat{k}\)
i) Is the motion irrotational? If so, find the velocity potential.
ii) Is the motion possible for an incompressible fluid?
Answer:
i) Yes, \(\phi = xy + yz + zx + C\)
ii) Yes, motion is possible for incompressible fluid -
Show that the vector field \(\vec{F}=(x^{2}+xy^{2})\hat{i}+(y^{2}+x^{2}y)\hat{j}\) is irrotational and find its scalar potential.
Answer: \(\frac{1}{3}x^{3} + \frac{1}{3}y^{3} + \frac{1}{2}x^{2}y^{2} + C\) -
Show that the vector field \(\vec{A}=(3x^{2}y)\hat{i}+(x^{3}-2yz^{2})\hat{j}+(3z^{2}-2y^{2}z)\hat{k}\) is irrotational but not solenoidal. Also find \(\phi(x,y,z)\) such that \(\nabla \phi = \vec{A}\).
Answer: \(\phi = x^{3}y - y^{2}z^{2} + z^{3} + C\)