Curl Rf 0 Where F is a Scalar Function With Continuous Secondorder Partial Derivatives

Gradient

The Gradient operation is performed on a scalar function to get the slope of the function at that point in space,for a \(f(x,y,z), \nabla f\) can be defined as:
$$ \nabla f = \left(\hat{x}\frac{\partial}{\partial x} + \hat{y}\frac{\partial}{\partial y} + \hat{z}\frac{\partial}{\partial z} \right)f $$
The del operator represented by the symbol \(\nabla\) can be defined as:
$$ \nabla = \hat{x}\frac{\partial}{\partial x} + \hat{y}\frac{\partial}{\partial y} + \hat{z}\frac{\partial}{\partial z}$$
Essentially we can say that the del when acted upon(multiplied to a scalar function) gives a vector in terms of the coordinates giving information about the slope of the multiplied function.

We will look into some related questions in later sections of this post.

Divergence

The physical meaning of divergence can be understood as a measure of spreading out(diverging) of a vector at any point(space coordinates).
Mathematically the divergence of a vector can be computed by taking a dot product of the vector with del(\(\nabla \))
So if \(\overrightarrow{V} = V_x\hat{x} + V_y\hat{y} + V_z\hat{z}\) then the divergence of \(\overrightarrow{V}\) at any point (x,y,z) can be computed as:
$$ \nabla\cdot V = \left(\hat{x}\frac{\partial}{\partial x} + \hat{y}\frac{\partial}{\partial y} + \hat{z}\frac{\partial}{\partial z} \right)\cdot\left( V_x\hat{x} + V_y\hat{y} + V_z\hat{z}\right) $$
$$ = \frac{\partial V_x}{\partial x} + \frac{\partial V_y}{\partial y} + \frac{\partial V_z}{\partial z} $$

Fig. 1(a) shows vector with high divergence while 1(b) and 1(c) has zero divergence

Curl

The physical meaning of curl can be understood as a measure of how much a vector curls around the point of interest(space coordinates).
Mathematically, the curl of a vector can be computed by taking the cross product of del operator with the vector.
So if \(\overrightarrow{V} = V_x\hat{x} + V_y\hat{y} + V_z\hat{z}\) then the curl of \(\overrightarrow{V}\) at any point (x,y,z) can be computed as:
$$ \nabla\times\ V = \left|
\begin{array}{ccc}
\hat{x} & \hat{y} & \hat{z} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
V_x & V_y & V_z \\
\end{array}
\right| $$

$$ = \hat{x}\left(\frac{\partial V_z}{\partial y} – \frac{\partial V_y}{\partial z}\right) + \hat{y}\left(\frac{\partial V_x}{\partial z} – \frac{\partial V_z}{\partial x}\right) +
\hat{z}\left(\frac{\partial V_y}{\partial x} – \frac{\partial V_x}{\partial y}\right) $$

Fig. 2(a) shows vector of high curl while figure 2(b) shows vector of zero curl

Product Rules

Addition of two functions \(f\) and \(g\):
$$ \nabla (f+g) = \nabla f + \nabla g $$
$$ \nabla\cdot (A+B) = (\nabla\cdot A) + (\nabla\cdot B)$$
$$ \nabla\times(A+B) = (\nabla\times A) + (\nabla\times B)$$
Multiplication with scalar:
$$ \nabla (kf) = k\nabla f$$
$$ \nabla\cdot(kA) = k(\nabla\cdot A),$$
$$ \nabla\times (kA) = k(\nabla\times A)$$
Product of two function:
$$ \nabla (fg) = f\nabla g + g\nabla f $$
$$ \nabla(A\cdot B) = A\times(\nabla\times B) + B\times(\nabla\times A) + (A\cdot\nabla)B + (B\cdot\nabla)A $$
$$ \nabla\cdot(fA) = f(\nabla\cdot A) + A\cdot(\nabla f)$$
$$ \nabla\cdot(A\times B) = B\cdot (\nabla\times A) – A\cdot (\nabla\times B)$$
$$ \nabla\times(fA) = f(\nabla\times A) – A\times(\nabla f)$$
$$ \nabla\times(A\times B) = (B\cdot \nabla)A – (A\cdot\nabla)B + A(\nabla\cdot B) – B(\nabla\cdot A)$$

Solved Questions

Q. Find the divergence of a vector \(\overrightarrow{A} = x^2\hat{x} + 6y^2\hat{y} + z^3\hat{z} \) at point P(2,4,1).
Sol. $$ \nabla\cdot \bar{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}$$
$$ = 2x + 12y + 3z^2$$
$$ \nabla\cdot \bar{A}|_{(2,4,1)} = 2(2) + 12(4) +3(1)^2 = 55$$

Q. A potential field is given by \(V = 3x^2y – yz\). Find electric field at (1, 0, -1).
Sol. As we know E = \(-\nabla V\)
$$ E = -\left[ \frac{\partial V}{\partial x}\hat{x} + \frac{\partial V}{\partial y}\hat{y} + \frac{\partial V}{\partial z}\hat{z}\right]$$
$$ = -\left[6xy\hat{x} + (3x^2 – z)\hat{y} – y\hat{z} \right]$$
$$ \bar{E}_{(1, 0, -1)} = -\left[6(1)(0)\hat{x} + \left[3(1)^2 – (-1)\right]\hat{y} – 0\hat{z} \right] = -4\hat{y}$$

Q. Given \(\overrightarrow{F} = (y^2 – x^2 + y)\hat{x} + x(2y + 1)\hat{y}\) check and comment about it's divergence and curl.
Sol. Given, $$ \overrightarrow{F} = (y^2 – x^2 + y)\hat{x} + x(2y + 1)\hat{y}$$
$$ \nabla\cdot\overrightarrow{F} = \frac{\partial V}{\partial x} + \frac{\partial V}{\partial y} + \frac{\partial V}{\partial z} = -2x + 2x = 0$$
Therefore, the given vector is divergent free
$$ \nabla\times\overrightarrow{F} = \left|
\begin{array}{ccc}
\hat{x} & \hat{y} & \hat{z} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y^2 – x^2 + y & x(2y + 1) & 0 \\
\end{array}
\right|
$$
$$ [0]\hat{x} – [0]\hat{y} + [(2y+1) – (2y+1)]\hat{z} = 0$$
Therefore, \(\overrightarrow{F}\) is irrotational

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Source: https://prialogue.com/gradient-divergence-and-curl/

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