# Question #a65c6

Sep 11, 2016

The Divergence and Curl of a vector field $\vec{A}$ are defined as $\vec{\nabla} \cdot \vec{A}$ and$\vec{\nabla} X \vec{A}$

#### Explanation:

So far so good.

We shall now try to understand a simple physical interpretation of the concepts defined above.

Considering the divergence, let us have a vector field $\vec{A}$

The flux of $\vec{A}$ over an arbitrary surface $S$ is defined as,

${\oint}_{S} \vec{A} \cdot \mathrm{dv} e c \sigma$, where $\mathrm{dv} e c \sigma$ is an elemental area vector on the surface $S$. The direction of the area vector element is the direction normal on it.

Now, by Gauss' divergence theorem, let us convert the surface integral into a volume integral,

${\oint}_{S} \vec{A} \cdot \mathrm{dv} e c \sigma = {\int}_{V} \vec{\nabla} \cdot \vec{A} d \tau$

This, is how we related divergence of a vector field with the flux.

For a special understanding, consider the Magnetic field $\vec{B}$

We know that the flux of a magnetic field over a closed surface is $0$.

Thus,

${\oint}_{S} \vec{B} \cdot \mathrm{dv} e c \sigma = {\int}_{V} \vec{\nabla} \cdot \vec{B} d \tau = 0$

Since, $d \tau$ is completely arbitrary, $\vec{\nabla} \cdot \vec{B} = 0$

This result is simply stated as, the magnetic field vector has no sources or sinks inside the surface concerned.

This can be generalized to any vector field, if $\vec{A}$ has no sources or sinks in the concerned volume, it's net flux over the surface enclosing the volume must be zero.

Consider now, the electric field, $\vec{E}$

We know by Gauss' law, that ${\oint}_{S} \vec{E} \cdot \mathrm{dv} e c \sigma = {Q}_{i} / {\epsilon}_{0}$ where ${Q}_{i}$ is the charge enclosed.

Now, if the volume charge density in the region be $\rho$, then $\rho d \tau = {Q}_{i}$

Now, similarly we have

${\oint}_{S} \vec{E} \cdot \mathrm{dv} e c \sigma = {\int}_{V} \vec{\nabla} \cdot \vec{E} d \tau$

$\implies \vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon} _ 0$

Now for example, the area contains a net positive charge. $\rho > 0$ which implies that $\vec{\nabla} \cdot \vec{E} > 0$.

This is simply stated as the vector field $\vec{E}$ has a source inside the surface. This is quite obvious from our observations. The net positive charges generate the field, They are placed inside the surface and thus, the volume encloses a source of the field.

if the net charge is negative, $\rho < 0$ which implies $\vec{\nabla} \cdot \vec{E} < 0$

This result is simply stated as the vector field $\vec{E}$ has a sink inside the volume. This is quite obvious from the common sense. We know that electric field terminates at the negative charges and thus, the electric field sinks inside the surface i.e. it enters the surface from outside and terminates. (Even though the negative charge is the one producing the field, but, here we consider the directional property of the field. Here, the field appears as if it enters the surface and terminates at the negative charges).

For $\rho = 0$ the Div of $\vec{E}$ is zero and as with the case of magnetic field, we say that it has no sources and sinks inside. This is quite obvious as well from our experience.

Thus, for any vector field $\vec{A}$ if

$\vec{\nabla} \cdot \vec{A} = 0$, $\vec{A}$ has no sources or sinks and is called solenoidal.

$\vec{\nabla} \cdot \vec{A} > 0$, $\vec{A}$ has a source.

$\vec{\nabla} \cdot \vec{A} < 0$, $\vec{A}$ has a sink.

This is the physical interpretation of the divergence.

Another physical interpretation may be obtained from continuity equation is fluid dynamics,

$\frac{\partial \rho}{\mathrm{dt}} + \vec{\nabla} \cdot \left(\rho \vec{v}\right) = 0$

Consult some standard text on vector calculus for a more detailed and intuitive approach.

Now, the Curl of the vector field is associated with rotational properties of it.

I would although like to give you a simple physical representation of he Curl of a conservative force.

From definition of conservative force $\vec{F}$, the work done by it in a closed path must be zero, thus, the line integral given below vanishes.

${\oint}_{C} \vec{F} \cdot \mathrm{dv} e c R = 0$ where $C$ is any contour and $\mathrm{dv} e c R$ is the line element along it.

By Stokes' theorem,

${\oint}_{C} \vec{F} \cdot \mathrm{dv} e c R = {\int}_{S} C u r l \vec{F} \cdot \mathrm{dv} e c \sigma$ where $S$ is surface enclosed by $C$

Since, the line integral vanishes and $\mathrm{dv} e c \sigma$ being completely an arbitrary area vector element, the curl must vanish.

Thus, $C u r l \vec{F} = 0$

In general, curls of all conservative forces and fields are zero.

Such fields are also called curl less or irrotational fields.

In a region, any irrotational field must have the path independence property.

I hope it helped.