Divergence & Curl
Divergence
If \(\vec{F} \longrightarrow \) Vector Point Function + Continuous + Differentiable then divergence of \(\vec{F}\) is \[ div\vec{F} = \nabla .\vec{F} \] i.e. dot product between \(\nabla\) and \(\vec{F}\) \[ = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right).(F_1\hat{i} + F_2\hat{j} + F_3\hat{k}) \]
\[ \large\boxed{div\vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} }\]
Note : The function is vector but its divergence is scalar.
Solenoidal Vector :
A vector is said to be Solenoidal if \(div\vec{F} = 0 \) \[ \large\boxed{\nabla .\vec{F} = 0 }\]
Sample Problem 1 :
Evaluate \( div(3x^2\hat{i} + 5xy^2\hat{j} + xyz^3\hat{k}) \) at point (1,2,3).
Solution
Let \( f = 3x^2\hat{i} + 5xy^2\hat{j} + xyz^3\hat{k} \) \[ div(f) = \nabla .f \] \[ = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right).(3x^2\hat{i} + 5xy^2\hat{j} + xyz^3\hat{k}) \] \[ = 6x + 10xy + 3xy^2 \] At point (1,2,3) \[ div(f) = 6(1) + 10(1)(2) + 3(1)(2)(3)^2 \] \[ = \boxed{80} \]
Curl of a Vector
If \(\vec{F} \longrightarrow \) Vector Point Function + Continuous + Differentiable
then \( curl(\vec{F}) = \nabla \times \vec{F} = \)
\[ \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)\times (F_1\hat{i} + F_2\hat{j} + F_3\hat{k}) \]
\[ =\large \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\\ \Large\frac{\partial}{\partial x} & \Large\frac{\partial}{\partial y} & \Large\frac{\partial}{\partial z} \\\\ F_1 & F_2 & F_3 \end{vmatrix} \]
Note : The function is vector and its curl is also a vector.
Irrotational Vector Field :
A vector field is said to be irrotational if \( curl(\vec{F}) = 0 \) \[ \boxed{ \nabla \times \vec{F} = 0 } \]
Sample Problem 2 :
Find curl of \( \vec{F} = xy\hat{i} + y^2\hat{j} + zx\hat{k} \) at point (-2,4,1).
Solution :
\[ curl(\vec{F}) = \nabla \times \vec{F} \] \[ = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right)\times (xy\hat{i} + y^2\hat{j} + zx\hat{k}) \] \[ =\large \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\\ \Large\frac{\partial}{\partial x} & \Large\frac{\partial}{\partial y} & \Large\frac{\partial}{\partial z} \\\\ xy & y^2 & zx \end{vmatrix} \] \[ = \hat{i}\left[\frac{\partial}{\partial y}(zx) - \frac{\partial}{\partial z}(y2) \right] \] \[ - \hat{j}\left[\frac{\partial}{\partial x}(zx) - \frac{\partial}{\partial z}(xy) \right] \] \[ + \hat{k}\left[\frac{\partial}{\partial x}(y^2) - \frac{\partial}{\partial y}(xy) \right] \] \[ = \hat{i}[0 - 0] - \hat{j}[z - 0] + \hat{k}[0 - x] \] \[ = 0\hat{i} - z\hat{j} - x\hat{k} \] At point (-2,4,1) \[ curl(\vec{F}) = -(1)\hat{j} - (-2)\hat{k} \] \[ \Rightarrow \boxed{ curl(\vec{F}) = -\hat{j} + 2\hat{k} } \]
Sample Problem 3 :
Find \( div(\vec{F})\) & \( curl(\vec{F}) \) where \( \vec{F} = grad(x^3 + y^3 + z^3 - 3xyz) \)
Solution :
Here we haven't been given the vector function directly, instead we have to first find gradient and then evaluate divergence and curl. \[ \vec{F} = grad(x^3 + y^3 + z^3 - 3xyz) \] \[ = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right)(x^3 + y^3 + z^3 - 3xyz) \] \[ \Rightarrow \vec{F} = (3x^2 - 3yz)\hat{i} + (3y^2 - 3xz)\hat{j} + (3z^2 - 3xy)\hat{k} \] \[ \text{Now } div(\vec{F}) = \nabla .\vec{F} \] \[ = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right)\left[(3x^2 - 3yz)\hat{i} + (3y^2 - 3xz)\hat{j} + (3z^2 - 3xy)\hat{k} \right] \] \[ \Rightarrow div(\vec{F}) = 6x\hat{i} + 6y\hat{j} + 6z\hat{k} \] \[ \text{Now } curl(\vec{F}) = \nabla \times \vec{F} \] \[ = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right)\times \left[(3x^2 - 3yz)\hat{i} + (3y^2 - 3xz)\hat{j} + (3z^2 - 3xy)\hat{k} \right] \] \[=\large \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\\ \Large\frac{\partial}{\partial x} & \Large\frac{\partial}{\partial y} & \Large\frac{\partial}{\partial z} \\\\ 3x^2 - 3yz & 3y^2 - 3xz & 3z^2 - 3xy \end{vmatrix}\] \[ = \hat{i}\left[\frac{\partial}{\partial y}(3z^2 - 3xy) - \frac{\partial}{\partial z}(3y^2 - 3xz) \right] \] \[ - \hat{j}\left[\frac{\partial}{\partial x}(3z^2 - 3xy) - \frac{\partial}{\partial z}(3x^2 - 3yz) \right] \] \[ + \hat{k}\left[\frac{\partial}{\partial x}(3y^2 - 3xz) - \frac{\partial}{\partial y}(3x^2 - 3yz) \right] \] \[ = \hat{i}\left[(0 - 3x) - (0 - 3x) \right] - \hat{j}\left[(0 - 3y) - (0 - 3y) \right] + \hat{k}\left[(0 - 3z) - (0 - 3z) \right] \] \[ \Rightarrow curl(\vec{F}) = 0\hat{i} + 0\hat{j} + 0\hat{k} = 0 \]
Sample Problem 4
If \( \hspace{7pt} \vec{R} = x\hat{i} + y\hat{j} + z\hat{k} \hspace{7pt}\) then prove that \(\hspace{7pt} div(r^n\vec{R}) = (n + 3)r^n \hspace{7pt} \) and find the value of \(n\) if \( \hspace{6pt} r^n\vec{R} \hspace{6pt}\) is a solenoidal vector.
Solution
\[ \vec{R} = x\hat{i} + y\hat{j} + z\hat{k} \]
\[ r = |\vec{R}| = \sqrt{x^2 + y^2 + z^2} \]
L.H.S. = \( div(r^n\vec{R}) = \nabla r^n\vec{R} \)
\[ = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)\left[r^n(x\hat{i} + y\hat{j} + z\hat{k}) \right] \]
\[ = \frac{\partial}{\partial x}(xr^n) + \frac{\partial}{\partial y}(yr^n) + \frac{\partial}{\partial z}(zr^n) \]
\[ = \left(r^n + nxr^{n-1}\frac{\partial r}{\partial x}\right) + \left(r^n + nyr^{n-1}\frac{\partial r}{\partial y}\right) + \left(r^n + nzr^{n-1}\frac{\partial r}{\partial z} \right) \]
\[ \text{Now } \frac{\partial r}{\partial x} = \frac{2x}{2\sqrt{x^2 + y^2 + z^2}} = \frac{x}{r} \]
\[ \frac{\partial r}{\partial y} = \frac{y}{r}, \frac{\partial r}{\partial z} = \frac{z}{r} \]
Using these values
\[ \nabla r^n\vec{R} = 3r^n + nxr^{n-1}\times \frac{x}{r} + nyr^{n-1}\times \frac{y}{r} + nzr^{n-1}\times \frac{z}{r} \]
\[ = 3r^n + nr^{n-2}(x^2 + y^2 + z^2) \]
\[ = 3r^n + nr^{n-2}(r^2) \]
\[ = (n + 3)r^n \]
= R.H.S., Hence proved :)
Since \(r^n\vec{R} \) is a solenoidal vector
\[ \Rightarrow div(r^n\vec{R}) = 0 \]
\[\Rightarrow (n + 3)r^n = 0 \]
\[ \text{As } r^n \neq 0 \]
\[ \Rightarrow n + 3 = 0 \]
\[ \Rightarrow \boxed{n = -3} \]
Sample Problem 5
If \( \hspace{7pt} \vec{R} = x\hat{i} + y\hat{j} + z\hat{k} \hspace{7pt}\) then prove that \(\hspace{6pt} curl(r^n\vec{R}) = 0 \) and find value of \(n\) if \(r^n\vec{R}\) is irrotational vector.
Solution
\[ \vec{R} = x\hat{i} + y\hat{j} + z\hat{k} \]
\[ r = |\vec{R}| = \sqrt{x^2 + y^2 + z^2} \]
L.H.S. = \( curl(r^n\vec{R}) = \nabla r^n\vec{R} \)
\[ = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)\times \left[r^n(x\hat{i} + y\hat{j} + z\hat{k}) \right] \]
\[ =
\large
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\\\
\Large\frac{\partial}{\partial x} & \Large\frac{\partial}{\partial y} & \Large\frac{\partial}{\partial z} \\\\
xr^n & yr^n & zr^n
\end{vmatrix}
\]
\[ = \hat{i}\left[\frac{\partial}{\partial y}(zr^n) - \frac{\partial}{\partial x}(yr^n) \right] \]
\[ - \hat{j}\left[\frac{\partial}{\partial x}(zr^n) - \frac{\partial}{\partial z}(xr^n) \right] \]
\[ + \hat{k}\left[\frac{\partial}{\partial x}(yr^n) - \frac{\partial}{\partial y}(xr^n) \right] \]
\[ = \hat{i}\left[znr^{n-1}\frac{\partial r}{\partial y} - ynr^{n-1}\frac{\partial r}{\partial z} \right] \]
\[ - \hat{j}\left[znr^{n-1}\frac{\partial r}{\partial x} - xnr^{n-1}\frac{\partial r}{\partial z} \right]\]
\[+ \hat{k}\left[ynr^{n-1}\frac{\partial r}{\partial x} - xnr^{n-1}\frac{\partial r}{\partial y} \right] \]
As we had calculated in previous problems
\[\frac{\partial r}{\partial x} = \frac{x}{r}, \frac{\partial r}{\partial y} = \frac{y}{r}, \frac{\partial r}{\partial z} = \frac{z}{r} \]
\[ = \hat{i}\left[\cancel{znr^{n-1}\times \frac{y}{r}} - \cancel{ynr^{n-1}\times \frac{y}{r}} \right]\]
\[ - \hat{j}\left[\cancel{znr^{n-1}\times \frac{x}{r}} - \cancel{xnr^{n-1}\times \frac{z}{r}} \right] \]
\[ + \hat{k}\left[\cancel{ynr^{n-1}\times \frac{x}{r}} - \cancel{xnr^{n-1}\times \frac{y}{r}} \right] \]
\[ \Rightarrow curl(r^n\vec{R}) = 0 \]
Hence Proved :)
Since \(r^n\vec{R} \) is irrotational
\[ \Rightarrow curl(r^n\vec{R}) = 0 \]
which we already proved true, so \(n\) can have any value.
