# Flux; Green’s (in plane)



My notes of the excellent lecture 23 of “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Flux is the amount of something (water, wind, electric field, magnetic field) passing through a surface.

## Flux in a plane

Let $$C$$ be a plane curve, and $$\vec F$$ be a vector field in that plane.

The flux of $$\vec F$$ across $$C$$ is defined as $$\shaded{ \rm{flux}=\int_C\vec F\cdot\hat n\;ds } \nonumber$$

Where $$\hat n$$ is the unit normal vector to curve $$C$$ pointing 90° clockwise from $$\hat T$$. So, it is to the right of the curve, as you travel along the curve. Vector field $$\vec F$$ through $$C$$

### Compared to work

Flux compared to the line integral for work

Flux versus line integral for work
Flux Work
$$\lim_{\Delta s\to 0}\sum\vec F\cdot\hat n\;\Delta s \nonumber$$ $$\lim_{\Delta s\to 0}\sum\vec F\cdot\hat T\;\Delta s \nonumber$$
Sums the tangent component of $$\vec F$$. Sums the normal component of $$\vec F$$.
Measures how much the curve goes with $$\vec F$$. Measures how much $$\vec F$$ goes across the curve.
$$\vec F$$ is a velocity field. $$\vec F$$ is a force field.

### Physics Interpretation

Let $$\vec F$$ be a velocity field of water, and $$C$$ being a curve, then flux measures how much fluid passes through $$C$$ per unit of time.

If you imagine that you have a river and you are building a damn with holes, then flux measures how much water passes through your membrane per unit time. Velocity $$\vec F$$ passing through $$C$$

Zoomed in enough, the flow will be the same everywhere. In respect to the flow, the curve $$C$$ moves left. The parallelogram represents the fluid that passes through $$C$$. Fluid that passes through $$C$$

The size of the parallelogram is base $$\times$$ height. The height is the normal component of $$\vec F$$ \begin{align*} \rm{area} &= \rm{base}\;.\rm{height} \\ &= \Delta s\,(\vec F\cdot\hat n) \nonumber \end{align*}

The flux is the summation \begin{align*} \rm{flux} &= \lim_{\Delta s\to 0}\sum\vec F\cdot\hat n\;\Delta s \\ \Rightarrow \rm{flux} &= \int_S \vec F\cdot\hat n\,ds \end{align*}

Note

• what flows across a small segment of $$C$$ from left-to-right is counted positively, while
• what flows left-to-right is counted negatively.

### Examples using geometry

#### One

\begin{align*} C &: \rm{circle\ of\ radius\ } a {\ at\ }(0,0),\rm{\ counterclockwise} \\ \vec F &= x\hat\imath+y\hat\jmath \end{align*} Velocity field $$\vec F$$ and curve $$C$$

Along $$C$$ \require{cancel} \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \vec F\;&\parallelsum\;\hat n \\ \Rightarrow\ \vec F\cdot\hat n &=\left|\vec F\right|\;\bcancel{\left|\hat n\right|}_1=\left|\vec F\right| \\ &=\sqrt{x^2+y^2}=a \\ \Rightarrow \int_C\vec F\cdot\hat n\;ds &=\int_C a\;ds =a\;.\rm{length}(C)=2\pi a^2 \end{align*} There would be water added uniformly. Because as you go from the origin, it is going faster and faster.

#### Two

\begin{align*} C &: \rm{circle\ of\ radius\ } a {\ at\ }(0,0),\rm{\ counterclockwise} \\ \vec F &= \left\langle -y,x\right\rangle \end{align*} Velocity field $$\vec F$$ and curve $$C$$

When $$\vec F$$ is tangent to $$C$$ \require{cancel} \begin{align*} \vec F&\perp\hat n \\ \Rightarrow\ \vec F\cdot\hat n &=0 \\ \Rightarrow \int_C\vec F\cdot\hat n\;ds &=\int_C 0\;ds = 0 \end{align*}

### Compute using components

Let $$\vec F$$ be a vector field with components $$M(x,y)$$, and $$N(x,y)$$, and let $$\Delta s$$ be a tiny length of curve $$C$$.

Remember: for work we sum how much the vector field goes along the curve $$\Longrightarrow$$ it uses the dot-product between $$\vec F$$ and the tangent vector of the curve $$\hat T$$

$$\left. \begin{array}{l} d\vec r=\underline{\hat T\;ds} = \left\langle dx ,dy\right\rangle \\ \vec F = \left\langle M, N\right\rangle \end{array} \right\} \Rightarrow \vec F\cdot \hat T\,ds = Mdx+Ndy \nonumber$$ Curve $$C, \Delta\vec r$$ and $$\hat T$$

For flux, we sum how much the vector field goes across the curve (=along the normal vector) $$\Longrightarrow$$ the dot-product between $$\vec F$$ and the normal vector of the curve $$\hat n$$. Where $$\hat n$$ equals the tangent vector $$\hat T$$ rotated by 90° clockwise. $$\hat n\,\Delta s$$

Rotating $$\hat T\,ds=\left\langle dx ,dy\right\rangle$$ by 90&deg clockwise, makes the $$x$$-component $$y$$, and the $$y$$-component $$-x$$ $$\left.\begin{array}{l} \underline{\hat n\;ds}=\left\langle dy,-dx \right\rangle \\ \vec F=\left\langle M,N\right\rangle \end{array}\right\} \Rightarrow \vec F\cdot\hat n\;ds = -N\,dx + M\,dy \nonumber$$

The flux integral can then be written as the differential $$\shaded{ \int_C \vec F\cdot\hat n\;ds = \int_C -N\,dx + M\,dy } \nonumber$$

### Green’s theorem in normal form

For work, Green’s theorem, replaces a line integral along a closed curve with a double integral.

If $$C$$ is a closed curve, enclosing region $$R$$, counterclockwise, and vector field $$\vec F$$ is defined and differentiable everywhere in $$R$$, then $$\shaded{ \oint_C\vec F\cdot d\vec r = \iint_R\mathrm{curl}(\vec F)\;dA } \nonumber$$

For flux, it is similar: Green’s theorem for flux

If $$C$$ is a closed curve, enclosing region $$R$$, counterclockwise, and vector field $$\vec F$$ is defined and differentiable everywhere in $$R$$, then $$\oint_C\vec F\cdot\hat n\,ds = \iint_R\mathrm{div}(\vec F)\,dA \nonumber$$

Recall: divergence for vector field $$\vec F=\left\langle P,Q\right\rangle$$, where $$P(x,y)$$ and $$Q(x,y)$$

$$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} \rm{div}\left\langle P,Q\right\rangle = \pdv{}{x}P + \pdv{}{y}Q = P_x + Q_y \nonumber$$

Green’s theorem in coordinates $$\shaded{ \oint_C -Q\,dx + P\,dy = \iint_R(P_x +Q _y)\;dA } \label{eq:greens}$$

Interpretation of divergence:

1. Measures how much the flow is “expanding”. E.g. how much a gas expands.
2. “Source rate”. Amount of fluid added to the system per unit time and area.

#### Proof

In equation $$\eqref{eq:greens}$$, substitute $$M=-Q$$ and $$N=P$$ $$\oint_C \underbrace{-Q}_{M}\,dx + \underbrace{P}_{N}\,dy = \iint_R(\underbrace{P_x}_{N} +\underbrace{Q _y}_{-M_y})\;dA \nonumber$$

$$\oint_C M\,dx+N\,dy = \iint_R (N_x-M_y)\,dA \nonumber$$

By renaming the components, we go from tangential form to the normal form. Therefore it is really the same theorem.

#### Examples

##### One

Look at the earlier example \begin{align*} C&:\rm{circle\ of\ radius\ } a {\ at\ }(0,0),\rm{\ counterclockwise} \\ \vec F&= x\hat\imath+y\hat\jmath = \left\langle x,y\right\rangle \end{align*}

Picture Velocity field $$\vec F$$ and curve $$C$$

Start with $$\rm{div}(\vec F)$$ $$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} \rm{div}(\vec F) = \pdv{}{x}x+\pdv{}{y}y=1+1=\underline{2} \nonumber$$

Apply Green’s theorem \begin{align*} \oint_C\vec F\cdot\hat n\,dx &= \iint_R\underline{2}\,dA = 2\overbrace{\iint_R\,dA}^{\text{Area}(R)} \\ &= 2\pi a^2 \end{align*} \nonumber

#### Two

Determine the flux for \begin{align*} C&:\rm{circle\ of\ radius\ } a {\ NOT\ at\ }(0,0),\rm{\ counterclockwise} \\ \vec F&= x\hat\imath+y\hat\jmath \end{align*} This will be the same, because we never used the location of the circle. We only used its area. The flux is the same everywhere.

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