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For a velocity field, curl measures the rotation component of the motion.

Curl also measures how far the vector field is from being conservative.

Definition

Let \(\vec F=\left\langle M,N\right\rangle\) be a vector field with components \(M(x,y)\), and \(N(x,y)\).

Curl is defined as (written using the symbolic \(\nabla\)-operator) $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \shaded{ \mathrm{curl}\left(\vec F\right) = \nabla\times\vec F = \pdv{N}{x}-\pdv{M}{y} = N_x-M_y } \label{eq:curldef} $$

It measures how far the vector field is from being conservative.

A conservative vector field is a vector field that is the gradient of some function \(f(x,y)\). The line integral of a conservative vector field is path independent; the choice of any path between two points does not change the value of the line integral.

Recall: a conservative vector field

If \(\vec F\) is defined and differentiable everywhere, and \(M_y=M_x\), then \(\vec F\) is a gradient field, a conservative vector field.

Some expressions

1. If the curl is \(0\) and defined everywhere, then the field is conservative \(\Longrightarrow\) a line integral on a closed curve will be \(0\). $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \oint_C\vec F\cdot d\vec r = \iint_R\underbrace{\mathrm{curl}(\vec F)}_{=0}\;dA = 0 \nonumber $$
2. For a gradient field, \(N_x=M_y \Longrightarrow\mathrm{curl}(\vec F)=0\). $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \rm{curl}\left(\vec F\right) = \pdv{N}{x}-\pdv{M}{y} = 0 \nonumber $$
3. For \(\vec F=\left\langle a,b\right\rangle\), where \(a\) and \(b\) are constants $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \rm{curl}\left(\vec F\right) = \pdv{b}{x}-\pdv{a}{y} = 0 \nonumber $$
4. In a radial vector field, \(\vec F=\left\langle x,y\right\rangle\), there is no rotation $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \mathrm{curl}\left(\vec F\right)=\pdv{}{x}y-\pdv{}{y}x=0 \nonumber $$
5. In a rotation vector field \(\vec F=\left\langle -y,x\right\rangle\), $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \mathrm{curl}\left(\vec F\right)=\pdv{}{x}x-\pdv{}{y}(-y)=1+1=2 \nonumber $$

Topological considerations (Lec. 33)

Curl for a velocity field

In a velocity field, curl measures (twice) the angular velocity (\(\omega\)) of the rotation component of the field. If you would put something light in a fluid, the curl tells you how fast it will spin at a given time (\(\times 2\)).

The curl measures how strong the rotation effect of a velocity field is at a given point. For instance, if you’re looking at wind vector fields, the regions with high curl trend to be tornedos or hurricanes.

Curl singles out the rotation component of motion (while divergence singles out the stretching component).

Curl for a force field

We will continue to use \(\vec F\) for the force field, and will use \(\vec f\) for force.

Recall: Newton’s Second Law and its angular analogue (see moment of inertia)

Linear motion			Circular motion
Mass	\(=m\)		Moment of inertia	\(=I=m\,r^2\)
Force	\(=\vec F\)		Torque	\(=\vec\tau = \vec r\times\vec f\)
Velocity	\(=\vec v\)		Angular velocity	\(=\vec\omega=\displaystyle\frac{d}{dt}\theta\)
Acceleration	\(=\vec a = \displaystyle\frac{\vec F}{m} = \frac{d}{dt}\vec v\)		Angular acceleration	\(=\vec\alpha = \displaystyle\frac{\vec\tau}{I} = \displaystyle\frac{d}{dt}\vec\omega\)

Let point mass \(m\) be in a force field (or rather acceleration field) \(\vec F\). The field is the force per unit mass that would be exerted on a small mass at that point. So, the force \(\vec f\) exerted on \(m\) is $$ \vec f = \vec F\,m \nonumber $$

Let \(m\) be connected with a light rod to the origin. The \(\vec f\) exerts a torque \(\vec\tau\) at the origin $$ \vec\tau = \vec r \times \vec f = \vec r \times \vec F\,m \nonumber $$

Translation vs. rotation movement

Applying Newton’s 2nd law, to both translation movement and rotation moment (ignoring the factor \(2\)) $$ \begin{matrix} \frac{\text{Force}}{\rm{Mass}} &=& \text{Acceleration} &=& \frac{d}{dt}(\text{Velocity}) \\ \downarrow\rlap{\color{blue}{\text{curl}}} && \downarrow\rlap{\color{blue}{\text{curl}}} && \downarrow\rlap{\color{blue}{\text{curl}}} \\ \frac{\text{Torque}}{\text{Moment of inertia}} &=& \text{Angular acceleration} &=& \frac{d}{dt} (\text{Angular velocity}) \end{matrix} \nonumber $$

Curl connects the translation and rotation moments. Thinking of curl as an operation

• We know that the curl of a velocity field, gives the angular velocity.
• Then, the curl of an acceleration field, gives the angular acceleration (in the rotation part of the acceleration effect).
• Therefore, the curl of a force field measures the torque per unit moment of inertia.

So, it measures how much torque the force field exerts on a small solid. It will measure how much it starts spinning if you leave it in this force field. In particular, a force field with no curl does not generate rotation movement. It may accelerate in some direction, but it will not start spinning.

Just as the curl of a velocity field measures the angular velocity of its rotation, the curl of a force field measures the torque it exerts on a mass per unit moment of inertia.

The curl of a force field tells you how quickly the angular velocity (\(\omega\)) is going to increase or decrease.

Consequence

If force field \(\vec F\) derives from a potential (=conservative), then its curl \(\nabla\times\vec F = 0\) $$ \nabla\times\vec F = \nabla\times(\nabla f) = 0 \nonumber $$

\(\Longrightarrow\) \(\vec F\) does not induce any rotational motion.

That means that the earth by itself doesn’t spin because of gravitational attraction. (It spins because it was formed spinning.). It didn’t start spinning because of gravitational effects.

(This is however not strictly true, because actually the Earth is deformable. Friction and tidal effects due to Earth’s gravitational attraction explain why the Moon’s rotation and revolution around Earth are synchronous. This is why we always see the same side of the moon).

Green’s Theorem in tangential form

Green’s theorem:

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

Substitute the work integral in differential form, and equation \(\eqref{eq:curldef}\) for curl $$ \shaded{ \oint_C Mdx+Ndy = \iint_R(N_x-M_y)\;dA } \nonumber $$

Comparing the left- and right-hand side

\(\displaystyle\oint_C Mdx+Ndy\)		\(\displaystyle\iint_R(N_x-M_y)\;dA\)
Lives only on the curve \(C\)		Lives everywhere inside region \(R\)
\((x,y)\) are related through the curve \(C\).		\((x,y)\) are independent within boundary \(R\).

FYI: A interesting practical application is the Planimeter.

Example

Let curve \(C\) be a circle of radius \(1\) centered at \((2,0)\), counterclockwise. Let \(\vec F\) be the vector field $$ \vec F= \left\langle \underbrace{\rm ye^{-x}}_M, \underbrace{\tfrac{1}{2}x^2-\rm e^{-x}}_N \right\rangle \nonumber $$

Compute the work

Using the contour integral

$$ \oint_C\vec F\cdot d\vec r =\oint_C M\,dx+N\,dy = \oint_C y\rm e^{-x}\,dx+\left(\tfrac{1}{2}x^2-\rm e^{-x}\right)dy \nonumber $$

Picture

To do it directly, you would parametrize the curve. $$ \left\{ \begin{align*} x=2+\cos\theta &\Rightarrow dx=-\sin\theta\;d\theta \\ y=\sin\theta &\Rightarrow dy=\cos\theta\;d\theta \\ 0 \leq \theta\leq 2\pi \end{align*} \right. \nonumber $$ Instead, we will use Green’s theorem

Using Green’s theorem

$$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \oint_C\vec F\cdot d\vec r &= \iint_R \left( N_x-M_y \right)\;dA \\ &= \iint_R\left( \pdv{}{x} \left( \tfrac{1}{2}x^2-\rm e^{-x} \right) – \pdv{}{y}ye^{-x} \right)\;dA \\ &= \iint_R\left( \left( x+\rm e^{-x} \right) – e^{-x} \right)\;dA \\ &= \iint_R x\;dA \end{align*} \nonumber $$

Proof

Let \(\vec F=\left\langle M,N\right\rangle\) be a vector field with components \(M(x,y)\), and \(N(x,y)\); and \(C\) be closed curve around region \(R\).

We want to proof

$$ \oint_C Mdx+Ndy = \iint_R(N_x-M_y)\;dA \nonumber $$

Observations

1. Proof the special case, where \(N=0\). $$ \oint_C Mdx = \iint_R-M_y\;dA \nonumber $$ A similar argument will show $$ \oint_C Ndy = \iint_R N_x\;dA \nonumber $$ Then summing, we get Green’s theorem
2. We can decompose \(R\) into simpler regions,

   

   if we can proof $$ \left\{ \begin{align*} \oint_{C_1} Mdx &= \iint_{R_1}-M_y\;dA \\ \oint_{C_2} Mdx &= \iint_{R_2}-M_y\;dA \end{align*} \right. \nonumber $$ then because the boundary between \(R_1\) and \(R_2\) is traversed twice, but in opposite directions. When you sum, them these pieces will cancel. $$ \oint_C Mdx = \oint_{C_1} + \oint_{C_2} = \iint_{R_1} + \iint_{R_2} = \iint_R -M_ydA \nonumber $$

Proof the special case

Let’s proof $$ \shaded{ \oint_C Mdx = \iint_R-M_y\;dA } \nonumber $$

Start with the left hand side. Cut \(R\) into “vertically simple” regions.

$$ a\lt x\lt b, f_1(x)\lt y\lt f_2(x) \nonumber $$

If \(R\) is vertically simple, and \(C\) is the boundary of \(R\) counterclockwise,

then \(\oint_{C_i}M(x,y)\,dx\) for the curve segments \(C_i\) $$ \left. \begin{array}{ll} C_1:& y=f_1(x), x\ \rm{from}\ a\ \rm{to}\ b \\ C_2:& x=b \Rightarrow dx=0 \\ C_3:& y=f_2(x), x\ \rm{from}\ b\ \rm{to}\ a \\ C_4:& x=a \Rightarrow dx=0 \end{array} \right\} \Rightarrow \ \left\{ \begin{align*} \oint_{C_1} M(x,y)\;dx &= \int_a^b M\left(x,f_1(x)\right) dx \\ \oint_{C_2} M(x,y)\;dx &= 0 \\ \oint_{C_3} M(x,y)\;dx &= \int_b^a M\left(x,f_2(x)\right) dx \\ &= -\int_a^b M\left(x,f_2(x)\right) dx \\ \oint_{C_4} M(x,y)\;dx &= 0 \end{align*} \right. $$

Combining the curve segments \(C_{1,2,3,4}\) $$ \begin{align} \oint_C M\;dx &= \int_a^b M\left(x,f_1(x)\right) dx – \int_a^b M\left(x,f_2(x)\right) dx \nonumber \\ &= \int_a^b \Big(M\left(x,f_1(x)\right) – M\left(x,f_2(x)\right)\Big)\ dx \label{eq:greenspr1} \end{align} $$

Now, evaluate the right hand side $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \iint_R -M_y\;dA &= -\int_a^b\underbrace{\int_{f_1(x)}^{f_2(x)} \frac{\partial M}{\bcancel{\partial y}}\;\bcancel{dy}}_{\rm{}inner\ integral}\;dx \\ &= -\int_a^b\underbrace{\Big[ M\Big]_{f_1(x)}^{f_2(x)}}\ dx \\ &= -\int_a^b\Big(M(x,f_2(x)) – M(x,f_1(x))\Big)\ dx \\ &= \int_a^b \Big(M(x,f_1(x)) – M(x,f_2(x))\Big)\ dx \end{align*} \nonumber $$

This matches equation \(\eqref{eq:greenspr1}\). For this special case, when we have only an \(x\)-component and a vertically simple region, things work.

Generalize

Now, we can remove the assumption that things are vertically simple using this second observation. We can just glue the various pieces together, and prove it for any region. Then, we do same thing with the \(y\)-component. That’s the first observation. When we add things together, we get Green’s theorem in its full generality.

More about validity (Lec. 27)

Let \(\vec F\) be vector field $$ \vec F(x,y) = \frac{-y\hat\imath+x\hat\jmath}{x^2+y^2} = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle \nonumber $$

The vector field is not defined at origin, so we can’t use Green’s theorem directly.

$$ \int_C\vec F\cdot d\vec r=\iint_R\rm{curl}\left(\vec F\right)\,dA \nonumber $$

Everywhere, except at the origin, its curl is \(0\) $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \rm{Curl}\left(\vec F\right) &= \pdv{}{x}\left(\frac{x}{x^2+y^2}\right) – \pdv{}{y}\left(\frac{-y}{x^2+y^2}\right) \\ &= \frac{ (x^2+y^2)\pdv{x}{x} – x\pdv{x^2+y^2}{x}}{\left(x^2+y^2\right)^2} – \frac{ (x^2+y^2)\pdv{(-y)}{y} – (-y)\pdv{x^2+y^2}{y}} {\left(x^2+y^2\right)^2} \\ &= \frac{(x^2+y^2)-2x^2}{\left(x^2+y^2\right)^2} – \frac{-(x^2+y^2)+2y^2}{\left(x^2+y^2\right)^2} = \frac{0}{\left(x^2+y^2\right)^2} \end{align*} $$

If the curve wouldn’t include the origin, we could use Green’s theorem. So let’s cut out a smaller region \(C”\) clockwise around the origin.

In this region with a hole at the origin, the curl is \(0\). This tells us that the line integrals are equal to each other, but we don’t know the values. $$ \int_{C’}\vec F\,dr-\int_{C”}\vec F\,dr = \iint_R\rm{curl}\left(\vec F\right)\,dA = 0 \nonumber $$

To make a single closed curve, introduce a slit to connect the outer and inner curves. Now, Green’s theorem is well defined.

Total line integral, with \(C”\) now going clockwise, and the two horizontal segments cancelling out $$ \int_{C’}\vec F\,dr – \int_{C”}\vec F\,dr = \iint_R\rm{curl}\left(\vec F\right)\,dA \nonumber $$

This same argument applies to Green’s theorem for flux.

Simply connected region

Definition

A connected region \(R\) in the plane is simply connected if the interior of any closed curve in \(R\) is also contained in \(R\).

In other words: the region \(R\) doesn’t have any holes inside it.

In the example above, the domain of definition is the plane minus the origin. The region is not simply connected, because there is a hole around the origin.

Condition for Green’s theorem:

If domain where \(\vec F\) is defined (and differentiable) is simply connected, then we can always apply Green’s theorem.

If a vector field is defined in a simply connected region, and its curl is \(0\), then the vector field is conservative and is a gradient field. Thus we need to update:

If \(\rm{curl}\,\vec F=0\) and domain where \(\vec F\) defined is simply connected, then \(\vec F\) is conservative (=is a gradient field).