Flux; Green’s (in plane)

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.

Compared to work

Flux compared to the line integral for work

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.

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 \(C\).

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

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

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

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

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 \(\hat T\)

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

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.

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 \(Q(x,y)\)

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

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 \(N=P\) $$ \oint_C \underbrace{-Q}_{M}\,dx + \underbrace{P}_{N}\,dy = \iint_R(\underbrace{P_x}_{N} +\underbrace{Q _y}_{-M_y})\;dA \nonumber $$

Compare to Green’s theorem (for work) in tangential form

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

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

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

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.