Q51P - Introduction to Electrodynamics

Question

For Theorem 1, show that $\mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{\Rightarrow }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{,}\mathbf{ }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{\Rightarrow }\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{,}\mathbf{ }\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{\Rightarrow }\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{,}\mathbf{ }\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{\Rightarrow }\mathbf{\left(}\mathbf{c}\mathbf{\right)}$ and  $\left(c\right)\mathbf{\Rightarrow }\left(a\right)$

Step-by-Step Solution

Answer

1. The statement  $\left(a\right)\Rightarrow \left(c\right)$ has been shown.
2. The statements $\left(c\right)\Rightarrow \left(b\right)$ and  $\left(b\right)\Rightarrow \left(c\right)$ has been shown.
3. The statement  $\left(c\right)\Rightarrow \left(a\right)$ has been shown.

Step 1: Describe the given information

The theorem (1), describing the line integral of a vector along a closed path, shows that for a closed path $\left(d\right)\Rightarrow \left(a\right), \left(a\right)\Rightarrow \left(c\right), \left(c\right)\Rightarrow \left(b\right), \left(b\right)\Rightarrow \left(c\right)$ and $\left(c\right)\Rightarrow \left(a\right)$ .

Step 2: Define the line integral

The line integral of a vector F  along a route  dl  is defined as $\mathbf{\int }\mathit{F}\mathbf{·}\mathit{d}\mathit{l}$ . If the line integral is independent of the path, then the line integral of function F must be zero along that closed path.

Step 3: Prove ( a ) ⇒ ( c )

Let F be a gradient of a scalar function as $F=-\nabla v$ . Now find the curl of function F  as,

$$\begin{gathered} \nabla \times F=\nabla \times \left[-\left(\nabla v\right)\right] \\ =-\nabla \times \left[\left(\nabla v\right)\right] \\ =0 \end{gathered}$$ 

 

The result is zero because the curl of gradient of a vector is always zero. Thus we can say that $d\Rightarrow a$ .

 

According to strokes theorem,

$\oint F·dl=\int \left(\nabla \times F\right)·da$

 

Since the curl of vector function F is given as zero, that is $\nabla \times F=0$ . So, above equation becomes, 

 

$$\begin{gathered} \oint F·dl=\int 0·da \\ =0 \end{gathered}$$ 

 

Thus, we can say, $a\Rightarrow c$ .

Step 4: Prove ( c ) ⇒ ( b ) and ( b ) ⇒ ( c )

The theorem (1) says that the path  dl  from  a  to b is an independent path. As discussed, $F=-\nabla v$ , then using strokes theorem , we can write the following line integral,

 

${\int }_a^{b}F·dl=-{\int }_a^b\left(\nabla v\right)·dl$ 

 

Thus, the path from a  to  b  is independent according to the gradient theorem (1). Hence, from $a\Rightarrow b$.

 

As the line integral ${\int }_a^{b}F·dl$ is independent of path from a  to b , then for any closed loop, this line integral gives same value as the final and initial points in any closed loop coincide with each other. Thus, we can write, $b\Rightarrow c$ .

Step 4: Prove ( c ) ⇒ ( a )

As discussed, if the line integral is path independent , then for any closed loop, the value of line integral $\oint F·dl$ is always zero, that is,  $\oint F·dl=0$

 

According to strokes theorem,

$\oint F·dl=\int \left(\nabla \times F\right)·da$

 

Since the line integral of vector function F is zero, that is, $\oint F·dl=0$ . So, above equation becomes, 

 

$$\begin{gathered} 0=\int \left(\nabla \times F\right)·da \\ \nabla \times F=0 \end{gathered}$$ 

 

Thus, we can say, $c\Rightarrow a$ .