Intuition of definition of divergence

Question

Intution : The divergence of a three-dimensional vector field is the extent to which the vector field flow behaves like a source at a given point.

But if my vector field is $F=\langle P,Q,R\rangle$ then formula is for divergence is given as $P_x+Q_y+R_z$.

I want to know how this formula capute that intutitve idea.

I studied using MIT OCW. But professor didn't talk about how we get formula from the idea. So,I see videos from Khan academy, if we consider $i$ component i.e. $P$ , the divergence is positive ( Intution) if $P_x>0$ and vice versa.similarly for $j,k$ component. But a vector field need not be oriented along axis .he tells something like for an arbitrary vector field we can decompose vectors into components.but I didn't get that thing

Could someone please explain how we get this formula from intution of divergence.

Thanks!

Answer 1

Basically the intuition comes from the divergence theorem, which says that the integral of the divergence of a field $F$ over a volume $V$ is equal to the outward flux of $F$ through the boundary of $V$. By taking the volume small you recover the differential interpretation.

The divergence theorem itself is more or less a generalization of the fundamental theorem of calculus. Indeed in the 1D situation, the contributions to the outward flux from $[a,b]$ at $a,b$ are $-f(a),f(b)$ respectively (the minus sign coming because the outward normal at $a$ points to the left), so the flux is $f(b)-f(a)$ which is of course $\int_a^b f'(x) dx$.

Answer 2

You can use the divergence theorem as the motivation.

We are calculating the flux $\Phi_V$ of the vector field $\vec{F}$ through a closed surface $\partial V$ enclosing the volume $V$. Notice that if we divide $V$ into smaller volumes $V_1, \ldots, V_n$ then $$\Phi_V = \Phi_{V_1} + \cdots + \Phi_{V_n}$$ because the flux on the dividing surfaces cancels out.

Hence letting $n\to\infty$ and $V_i \to 0$ and using the divergence theorem we get $$\int_{V} (\operatorname{div} \vec{F})\,dV = \Phi_V = \sum_{i=1}^n\Phi_{V_i}= \sum_{i=1}^n \frac{\Phi_{V_i}}{V_i} V_i\xrightarrow{V_i \to 0}\int_V \left(\lim_{V_i \to 0} \frac{\Phi_{V_i}}{V_i}\right)\,dV$$

Since $V$ was arbitrary, we conclude $\operatorname{div} \vec{F} = \lim_{V_i \to 0} \frac{\Phi_{V_i}}{V_i}$.

To calculate $\operatorname{div} \vec{F}$, consider an infinitesimal box $V = [x, x+dx]\times[ y,y+dy] \times [z,z+dz]$.

Let's calculate the flux through the bottom and the top sides $\Phi_{\text{bottom and top}}$. Since the box is small, we can assume that $\vec{F}$ is constant on each side of the box, and equal to $\vec{F}(x,y,z)$ and $\vec{F}(x,y,z+dz)$ respectively. Hence the flux through the bottom and the top is \begin{align}\Phi_{\text{bottom and top}} &= \text{area of top}\cdot \vec{F}(x,y,z+dz)\cdot(0,0,1) + \text{area of bottom}\cdot \vec{F}(x,y,z)\cdot (0,0,-1) \\ &= \big(F_z(x,y,z+dz) - F_z(x,y,z)\big)\,dx\,dy \\ &= \left(\frac{\partial F_z}{\partial z}(x,y,z) \,dz\right)\,dx\,dy \end{align}

By symmetry, the entire flux is \begin{align} \Phi_V &= \Phi_{\text{bottom and top}} + \Phi_{\text{left and right}} + \Phi_{\text{front and back}} \\ &= \left(\frac{\partial F_x}{\partial x}(x,y,z) + \frac{\partial F_y}{\partial y}(x,y,z)+\frac{\partial F_z}{\partial z}(x,y,z) \right)\,dx\,dy\,dz \end{align}

Since the volume of the box is $dV = dx\,dy\,dz$, we have $$\operatorname{div} \vec{F} = \frac{\Phi_V}{dV} = \frac{\partial F_x}{\partial x}(x,y,z) + \frac{\partial F_y}{\partial y}(x,y,z)+\frac{\partial F_z}{\partial z}(x,y,z)$$

Now, any volume $V$ can be divided into very small boxes, this result holds for any $V$.

Answer 3

Let's start with a simple and intuitive idea.

Imagine you have a closed, three-dimensional region, which we'll call the volume V, and it's bounded by a closed surface, which we'll call S. Now, within this volume, there's some quantity of interest – this quantity could be mass, fluid flow, electric charge, or anything else you can think of.

This quantity can flow through the surface S. In some regions of the surface, the quantity flows out, and in other regions, it flows in. This flow in and out of the surface is what we call "flux." Essentially, flux quantifies the net transfer of this quantity through the closed surface.

So, if you were to measure the quantity and find that it has a positive net flux, this means that more of the quantity is leaving the volume than entering it. Consequently, the net quantity within the volume should decrease because it's losing more than it's gaining. In other words, the volume is losing some of this quantity.

An example

Certainly, let's expand on the concept with an example using mass as the quantity of interest. If we observe that more mass is entering the closed volume than leaving through its boundary surface, then this implies that the total mass inside the volume should be increasing (Thus, a negative flux).

Mathematical Definition

So, defining $\vec{F}$ as the vector field of this quantity, then what goes through the surface should be the sum of the quantities that goes through all the infinitestimal areas. In integral form it is defined as

$$\iint _{S} (\vec{F}\cdot \vec{n}) dS$$

Now, we need to address the amount of this quantity that has been created or destroyed within the volume. To understand this, we must examine how this quantity has varied within the volume. Considering that we have defined this quantity as $\vec{F}$, then it's variations is defined as $\nabla \cdot \vec{F}$. So, if we sum all the internal variations of this quantities inside the volume it should be expressed as

$$\iiint _{V} (\nabla \cdot \vec{F})dV$$

A brief note before we proceed: the definition of $\nabla \cdot \vec{F}$ is also called the divergence of $\vec{F}$ or $div(\vec{F})$, defined as $\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} + \frac{\partial F}{\partial z}$.

And what does this sum signify? It essentially quantifies the total amount of the quantity that has been generated in all spatial directions in one specific point, hence why we sum it up. For instance, imagine that 5 kilograms of mass have been generated in the x-direction ($\frac{\partial m}{\partial x}$), and an additional 7 kilograms of mass have been generated in the y-direction ($\frac{\partial m}{\partial y}$) within a certain volume element $dV$. Consequently, the overall mass generated amounts to $5kg + 7kg$, or expressed as $\frac{\partial m}{\partial x} + \frac{\partial m}{\partial y}$.

Assembling the Concepts

Thus, Gauss Divergence Theorem states that:

$$\iint _{S} (\vec{F}\cdot \vec{n}) dS = \iiint _{V} (\nabla \cdot \vec{F})dV$$

Or, in layman terms, whatever the net flux through the surface is, that quantity should have been created or destroyed inside the volume.