My notes of the excellent lecture 5 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

The connection between electric potential $$V$$, and electric field, $$\vec E$$.

## Gradient field, $$\vec E=-\nabla V$$

Earlier [1,2,3], we found the following equations \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} \label{eq:gauss} \\[1pt] V_P &= \int_P^\infty \vec E\cdot d\vec r \label{eq:electpotential} \\[1pt] V_A – V_B &= \int_A^B \vec E\cdot d\vec l \label{eq:electpotdiff} \end{align}

### Work over closed line

If the electric field is a static electric field (no moving charges) $$\Longrightarrow$$ the forces are conservative forces.

If I take a charge $$q$$ from point $$A$$, all around the conservative field, and back to point $$A$$, then the work based on equation $$\eqref{eq:electpotdiff}$$ is $$V_A-V_A=\int_A^A \vec E\cdot d\vec l = 0 \nonumber$$

We mark the integral with a circle to indicate that it is a closed path. So moving through static electric field, over a closed line, the work is $$0$$ $$\shaded{ \oint \vec E\cdot d\vec l = 0 }$$

### From potentials to electric field

If you look at equation $$\eqref{eq:electpotential}$$ and $$\eqref{eq:electpotdiff}$$, you see that the potential is the integral of the electric field $$\Longrightarrow$$ the electric field must be the derivative of the potential.

Let there be a charge $$Q$$ a distance $$r$$ from point $$P$$

Recall: we derived the electric potential at that location, the scalar

$$V = \frac{Q}{4\pi\varepsilon_0 r} \nonumber$$

The derivative of this potential $$\newcommand{dv}{\frac{d #1}{d #2}} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \dv{V}{r} = \dv{}{r} \frac{Q}{4\pi\varepsilon_0 r} = -\underline{\frac{Q}{4\pi\varepsilon_0 r^2}} \label{eq:pdvV}$$

Recall: the electric field is defined as the vector

$$\vec E = \underline{\frac{Q\,\hat r}{4\pi\varepsilon_0 r^2}} \nonumber$$

Substituting $$\eqref{eq:pdvV}$$ in this, shows that if you know the potential everywhere in space, you can retrieve the electric field $$\newcommand{dv}{\frac{d #1}{d #2}} \shaded{ \vec E = -\dv{V}{r}\,\hat r } \nonumber$$

That means that if you release a charge at $$0$$ speed, it will always start perpendicular to an equipotential surface, because it always starts to move in the direction of a field line.

#### Going deeper

1. Imagine that I am somewhere in space at position $$P$$. At that position the potential is $$V_P$$. At that position there is an electric field.
2. Now I make a tiny step in the $$x$$-direction. If I measure
• no change in potential, over that little step, it means that the component of the electric field in the direction of $$x$$ is $$0$$,
• a change in potential, it means that the magnitude of the $$x$$-component of the electric field $$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} |E_x| = \left|\pdv{V}{x}\right| \nonumber$$ The partial derivative $$\frac{\partial}{\partial x}$$, means you keep the other variables $$y,z$$ constant.
3. Equally for the $$y$$ and $$z$$-directions $$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} |E_y| = \left|\pdv{V}{y}\right|,\quad |E_z| = \left|\pdv{V}{z}\right| \nonumber$$

Until now we used the unit $$\tfrac{\rm N}{\rm C}$$ for the electric field, but from now on we will use $$\tfrac{\rm V}{\rm m}$$ because it gives a bit more insight. You move a bit in meters, and see how much the potential changes in volts.

Now we can write the relation between electric field and potential in Cartesian coordinates (instead of just a function of distance $$r$$) \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} \vec E &= \left(E_x\hat x+ E_y\hat y+ E_z\hat z\right) \\ &= – \left( \pdv{V}{x}\hat x + \pdv{V}{y}\hat y + \pdv{V}{z}\hat z \right) \\ &= -\left\langle \pdv{V}{x}, \pdv{V}{y},\pdv{V}{z} \right\rangle \end{align*} \nonumber

Using the symbolic $$\nabla$$-operator, this is written as $$\shaded{ \vec E = -\rm{grad}\,V = -\nabla\,V } \tag{gradient} \label{eq:gradient}$$

Note that here in physics the sign is opposite as used in the math notation for a gradient field.

#### Example: given potentials

We have a potential field $$V(x)=10^5 x$$ for $$0\lt x\lt 10^{-2}\rm m$$. What is the electric field in space?

The electric field $$\vec E$$ follows from equation $$\eqref{eq:gradient}$$ \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} \vec E &= -\nabla\,V = -\left\langle\pdv{10^5 x}{x},\pdv{10^5 x}{y},\pdv{10^5 x}{z}\right\rangle \\ &= -\left\langle 10^5,0,0\right\rangle = \underline{-10^5\,\hat x} \end{align*}

This is constant and opposite to $$\hat x$$.

#### Example: parallel plates

Continue the charged plates example. If the plates are close together $$d\ll \sqrt{A}$$ where $$A$$ is the area of a plate, we can approximate the field between these plates as those coming from a pair of infinite planes. We saw that the electric field inside is independent of the distance.

Assume this electric field is the $$\vec E=-10^5\hat x$$ from the previous example.

The potential difference follows from equation $$\eqref{eq:electpotdiff}$$, where going from $$A$$ to $$B$$ $$\Rightarrow$$ $$d\vec l=d\vec x$$ \begin{align*} V_A – V_B &= \int_A^B \vec E\cdot d\vec x = \int_A^B \underline{-10^5\,\hat x}\cdot d\vec x \\ &= -10^5 \int_A^B \hat x\cdot\vec x \end{align*}

The unit vector $$\hat x$$ is in the same direction as $$d\vec x$$, so the dot-product becomes a multiplication of scalars $$|\hat x|\,|d\vec x|\,\cos 0=$$$$1.dx.1$$ \begin{align*} V_A – V_B &= -10^5 \int_A^B dx = -10^5\Big[x\Big]_A^B \\ &= -10^5 (x_B – x_A) \end{align*}

If plates $$A$$ and $$B$$ are $$10\,\rm{cm}$$ apart $$V_A – V_B = -10^5\,10^{-2} = -1000\,\rm V \nonumber$$

Note

• $$A$$ is $$1000\,\rm V$$ lower than $$B$$.
• From $$A$$ to $$B$$ the potential increases linearly.

Instead of choosing the $$0$$-potential at $$\infty$$, we could choose it at plate $$A$$. Then plate $$B$$ would be 1000 V, and the potential $$V$$ and the electric field $$\vec E$$ become \shaded{ \left\{ \begin{align*} V &= 10^5 x \\ \vec E &= -10^5\,\hat x \end{align*} \right. } \nonumber

## Electrostatic Shielding

As long as there is no charge moving, and we’re dealing with solid conductors $$\Longrightarrow$$ we have static electric fields (the charges are not heavily moving) $$\Longrightarrow$$ the field inside the conductor is always $$0$$. Because conductors have free electrons, and if these free electrons see an electric field inside, then they start to move until they no longer experience a force $$\Longrightarrow$$ they kill the electric field inside.

So the charge in a conductor always rearranges itself so that the electric field becomes zero.

### Charging a solid conductor

Suppose we bring a plus charge near a solid conductor.

For a very short moment, there will be an electric field inside the conductor. This field will move the free electrons close to the plus charge (induction), leaving net positive charge behind. Once these charges cancel out the electric field, there is no longer an electric field to move them, so the electrons stay still. [MIT]

So, after this very brief moment, the electric field inside is zero. This implies

• This inside of the solid contains no charge. (Assume a small Gaussian surface inside the conductor, then equation $$\eqref{eq:gauss}$$ tell us that this inside surface contains no charge.)
• The entire solid is an equipotential. (Based on equation $$\eqref{eq:electpotdiff}$$.)

The charge has to be on the surface. It however is not uniformly distributed with this odd shape (as we see later). With the solid being an equipotential, the field lines are perpendicular to the surface of the solid.

External electric field lines are perpendicular to the surface. Because if a component of the $$\vec E$$-field was tangent to the surface, charges would flow along the surface until there was no longer any tangential component.

### Charging a hollow conductor

The same shape, but hollow. Once more, we charge it.

Again, for a very short moment, there will be an electric field inside the that moves the free electrons so they cancel out the “external” electric field. After that, there is no longer an electric field to move them.

So, after this very brief moment, the electric field inside is zero. This implies

• This inside of the shell and cavity contains no charge. (Assume a small Gaussian surface $$S$$ just under the surface of the conductor, then equation $$\eqref{eq:gauss}$$ tell us that it encloses no charge.)
• The whole shell including this cavity is an equipotential. (Based on equation $$\eqref{eq:electpotdiff}$$.)

There is never any electric field anywhere inside the shell or cavity. The only electric field is on the outside of the shell. So, again, the charge has to be on the outside.

Pretty amazing: one side becomes negative, the other side becomes positive; the whole things becomes an equipotential; no charge inside.

Earlier we saw that a sphere has no electric field inside. Here, we demonstrated that it doesn’t have to be a sphere (and also can be solid). Any conductor shape will give you an electric field of $$0$$ inside.

An external electric field (such as from a van der Graaff generator), can cause polarization on the outside of the conductor, but inside, even the inside wall will have no charge (no electric field). After a very brief moment, the free moving electrons rearranged themselves, in such a way that the electric field is zero everywhere inside the conductor. Metal box with imaginary opening, to show inside $$E=0$$

The inside is electrically shielded from the outside world. You would not notice a strong electric field outside. This is called electrostatic shielding. The object will be called a Faraday cage. Named after the English scientist Michael Faraday, who invented them in 1836

### A charge inside a conductor

A hollow metallic sphere is initially uncharged. Now imagine that a positive point charge $$q$$ is placed somewhere (not in the middle) in the sphere without touching the walls.

Now there is positive charge inside, so there has be an electric field inside. For a very short moment the free electrons in the conducting wall will move so that the electric field in the wall becomes zero.

There must be charge on the inside surface, because if we take a Gaussian surface $$S$$ halfway the shell, the electric field is zero at that surface, so the enclosed charge is zero. That means that negative charge must have accumulated on the inside wall of the conductor, to make the net charge inside surface $$S$$ is zero.

Originally the conductor had no charge. On the outside surface we now see charge $$q$$, because the minus charge on the inside surface came from electrons that moved from the outside surface.

The plus charge inside, creates an electric field inside, that creates a negative charge on the inside, and a plus charge on the outside.

Because of the spherical symmetry, the charge on the outside will be uniformly distributed. That is the only way nature can obey the laws of physics. (The conductor must become an equipotential; there can be no electric field inside the conductor; the electric field lines have to perpendicular to the surface; the $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \oiint_S \vec E \cdot d\vec l$$ must be zero everywhere.)

The uniform outside charge is independent of the position of the charge $$q$$ inside. The outside world will not notice if you move the charge $$q$$ around. It has no way of knowing what happens on the inside. We call that electrostatic shielding, the effect of the Faraday cage.



My notes of the excellent lecture 30 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

The fundamental theorem of multivariable calculus tell us (using the symbolic $$\nabla$$-operator)

If you take the line integral of the gradient of a function, what you get back is the function.
$$\shaded{ \int_C\nabla f\cdot d\vec r = f(P_1) – f(P_0) } \label{eq:fundthm}$$

Similar to the plane, in space

When vector field $$\vec F$$ is a gradient of function $$f(x,y)$$, it is called a gradient field $$\newcommand{pdv}{\tfrac{\partial}{\partial #1}} \shaded{ \vec F = \nabla f = \left\langle \pdv{x}f, \pdv{y}f, \pdv{z}f \right\rangle } \nonumber$$ where $$f(x,y,z)$$ is called the potential.

## When is a vector field a gradient field?

In the plane, to check if a vector field is a gradient field we only had to check the condition

$$\newcommand{dv}{\frac{d #1}{d #2}} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}{\frac{\partial^2 #1}{\partial #2\partial #3}} \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \newcommand{ppdv}{\frac{\partial^2 #1}{\partial #2^2}} \pdv{M}{y} = \pdv{N}{x} \nonumber$$

In space, we want to know if $$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} \vec F = \left\langle P,Q,R\right\rangle \stackrel{?}{=} \nabla f = \left\langle \pdv{}{x}f, \pdv{}{y}f, \pdv{}{z}f \right\rangle \nonumber$$ where $$\vec F=\left\langle P,Q,R\right\rangle$$ be defined in a simply connected region.

If $$\vec F$$ is a gradient field, $$\vec F=\nabla f$$, then \newcommand{dv}{\frac{d #1}{d #2}} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \left\{ \begin{align*} P &= \pdv{}{x}f = f_x \\ Q &= \pdv{}{y}f = f_y \\ R &= \pdv{}{z}f = f_z \end{align*} \right.

Take the partial derivatives of $$P$$, $$Q$$ and $$R$$ \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}{\frac{\partial^2 #1}{\partial #2\partial #3}} \left\{ \begin{align*} P_y=\pdv{P}{y}=\ppdv{}{x}{y}f = f_{xy} \\ P_z=\pdv{P}{z}=\ppdv{}{x}{z}f = f_{xz} \\ Q_x=\pdv{Q}{x}=\ppdv{}{y}{x}f = f_{yx} \\ Q_z=\pdv{Q}{z}=\ppdv{}{y}{z}f = f_{yz} \\ R_x=\pdv{R}{x}=\ppdv{}{z}{x}f = f_{zx} \\ R_z=\pdv{R}{y}=\ppdv{}{z}{y}f = f_{zy} \end{align*} \right.

Recall:

The mixed second derivatives are the same, no matter what order you take them $$\newcommand{dv}{\frac{d #1}{d #2}} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}{\frac{\partial^2 #1}{\partial #2\partial #3}} \pdv{}{\color{red}x}\left(\pdv{f}{\color{blue}y}\right) =\ppdv{f}{\color{red}x}{\color{blue}y} =\ppdv{f}{\color{blue}y}{\color{red}x} =\pdv{}{\color{blue}y}\left(\pdv{f}{\color{red}x}\right) \nonumber$$

Based on this second partial derivative rule \begin{align*} \underline{P_y} = f_{xy} &= f_{yx} = \underline{Q_x} \\ \underline{P_z} = f_{xz} &= f_{zx} = \underline{R_x} \\ \underline{Q_z} = f_{yz} &= f_{zy} = \underline{R_y} \end{align*} \nonumber

Therefore, if $$\vec F=\left\langle P,Q,R\right\rangle$$, defined in a simply connected region is a gradient field, when \shaded{ \left\{ \begin{align*} P_y &= Q_x \\ P_z &= R_x \\ Q_z &= R_y \end{align*} \right. } \nonumber

### Exact differential

We can also think of it in terms of differentials.

If we have a differential of the form $$P\,dx + Q\,dy + R\,dz \nonumber$$ is an exact differential. That means it is going to equal to $$df$$ for some function $$F$$ exactly and of the same conditions. Just another way of saying it.

#### Example

For which $$a$$ and $$b$$ is the differential below exact? $$\underbrace{axy}_P\,dx + (\underbrace{x^2+z^3}_Q)dy + (\underbrace{byz^2+4z^3}_R)dz \nonumber$$

Or, for which $$a$$ and $$b$$ is $$\vec F$$ a gradient field? $$\vec F=\left\langle axy,x^2+z^3, byz^2-4z^3\right\rangle \nonumber$$

Compare $$\left. \begin{array}{lllll} P_y &= ax &= 2x &= Q_x &\Rightarrow a = 2 \\ P_z &= 0 &= 0 &= R_x \\ Q_z &= 3z^2 &= bz^2 &= R_y &\Rightarrow b=3 \end{array} \right\} \nonumber$$

For those values of $$a$$ and $$b$$ we can look for a potential. For any other values we would have to set up the line integral.

## Potential of a gradient field

Recall: for the plane

When the field is a gradient, and you know the function $$f$$, you can simplify the evaluation of the line integral for work. $$\shaded{ \int_C\nabla f\cdot d\vec r=f(P_1)-f(P_0) } \nonumber$$ where $$f(x,y)$$ is called the potential

Once again, you have two methods: computing line integrals, or using antiderivatives.

### Method 1: Computing line integrals

Similar to in the plane, apply the fundamental theorem, equation $$\eqref{eq:fundthm}$$, to find an expression for the potential at $$(x_1,y_1,z_1)$$ \begin{align} &\int_C\vec F\cdot d\vec r=f(x_1,y_1,z_1)-f(0,0,0) \nonumber \\ \Rightarrow & f(x_1,y_1,z_1) = \underbrace{\int_C\vec F\cdot d\vec r}_\text{work} + \underbrace{f(0,0,0)}_{\mathrm{constant}} \label{eq:method1} \end{align}

The work in a gradient field is path independent $$\Longrightarrow$$ find the easiest path

Apply the work differential, to find the work along $$C$$ in gradient field $$\vec F$$

Then add the line integrals together, to get the total work. $$\underline{\int_C\vec F\cdot d\vec r} = \int_{C_1}\ldots + \int_{C_2}\ldots + \int_{C_3}\ldots \nonumber$$

Substitute this back in equation $$\eqref{eq:method1}$$ and drop the subscripts $$f(x,y,z) = \underline{\int_C\vec F\cdot d\vec r} + \rm{c} \nonumber$$ If you would take the gradient of $$f(x,y,z)$$, you should get $$\vec F$$ back.

### Method 2: Using Antiderivatives

Similar to in the plane. No integrals, but you have to follow the procedure very carefully.

Continue with the earlier example, with $$a=2$$ and $$b=3$$. Solve \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align} \pdv{f}{x} &= f_x = 2xy \label{eq:potential1} \\ \pdv{f}{y} &= f_y = x^2 + z^3 \label{eq:potential2} \\ \pdv{f}{z} &= f_z = 3yz^2 – 4z^3 \label{eq:potential3} \end{align}

Integrate $$\eqref{eq:potential1}$$ in respect to $$x$$. The integration constant might depend on $$y$$ and $$z$$, so call it $$g(y,z)$$ $$\newcommand{pdv}{\tfrac{\partial #1}{\partial #2}} f_x = 2xy \xrightarrow{\int dx} \underline{f = x^2y + g(y,z)} \nonumber$$

To get information on $$g(y,z)$$ we look at the other partials. Take the derivative of $$f$$ in respect to $$y$$ and compare to $$\eqref{eq:potential2}$$. The integration constant might depend on $$y$$, so call it $$h(z)$$ \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} x^2 + \pdv{g}{y} &= x^2 + z^3 \\ \Rightarrow \pdv{g}{y} &= z^3 \xrightarrow{\int dy} g = \underline{yz^3 + h(z)} \\ \end{align*}

Substituting $$g$$ in $$f$$ $$f = x^2y + \underline{yz^3 + h(z)} \nonumber$$

To get information on $$h(z)$$ we look at the other partials. Take the derivative of $$f$$ in respect to $$z$$ and compare to $$\eqref{eq:potential3}$$. The integration constant $$\rm{c}$$ is a true constant \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} 3yz^2 + \pdv{h}{z} &= 3yz^2 – 4z^3 \\ \Rightarrow \pdv{h}{z} &= – 4z^3 \xrightarrow{\int dz} h = \underline{-z^4 + \rm{c}} \\ \end{align*}

Substituting $$h$$ in $$f$$ $$f = x^2y + yz^3 \underline{-z^4\,(+ \rm{c})} \nonumber$$

If you want to find one potential, you can just forget about the constant. If you want to find all the potentials: they differ by this constant.



My notes of the excellent lectures 20 and 21 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

## Potential

When vector field $$\vec F$$ is a gradient of function (written using the symbolic $$\nabla$$-operator) $$f(x,y)$$, it is called a gradient field $$\newcommand{pdv}{\tfrac{\partial}{\partial #1}} \shaded{ \vec F = \nabla f = \left\langle \pdv{x}f, \pdv{y}f \right\rangle = \left\langle f_x, f_y \right\rangle } \nonumber$$

Where $$f(x,y)$$ is called the potential.

### Fundamental theorem

Recall the fundamental theorem of calculus

If you integrate the derivative, you get back the function. $$\int_a^b \frac{df(t)}{dt}\,dt=f(b)-f(a) \label{eq:fndcalc}$$

In multivariable calculus, it is the same

If you take the line integral of the gradient of a function, what you get back is the function. $$\shaded{ \int_C\nabla f\cdot d\vec r = f(P_1) – f(P_0) } \label{eq:fundthm}$$ where $$f(x,y)$$ is called the potential.

Only when the field is a gradient, and you know the function $$f$$, you can simplify the evaluation of the line integral for work. $$\shaded{ \int_C\nabla f\cdot d\vec r=f(P_1)-f(P_0) } \nonumber$$

### Proof

In coordinates, the gradient field $$\nabla f$$ is expressed as $$\newcommand{pdv}{\tfrac{\partial}{\partial #1}} \nabla f =\left\langle \pdv{x}f, \pdv{y}f \right\rangle =\left\langle M, N \right\rangle \nonumber$$

Recall: the work integral in differential form
$$\int_C\vec F\cdot d\vec r = \int_C\left( M\,dx + N\,dy \right) \nonumber$$

Substituting $$M$$ and $$N$$ from the gradient field into the work integral \renewcommand{dv}{\frac{d #1}{d #2}} \renewcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align} \int_C \nabla f\cdot d\vec r &= \int_C \pdv{f}{x}dx+\pdv{f}{y}dy \nonumber \\ &= \int_C \underline{ \left(\pdv{f}{\color{red} x}\dv{\color{red}x}{t} + \pdv{f}{\color{blue}y}\dv{\color{blue}y}{t}\right)}\,dt \label{eq:subgrad} \end{align}

Recall: the multivariable calculus chain rule

$$\renewcommand{dv}{\frac{d #1}{d #2}} \renewcommand{pdv}{\frac{\partial #1}{\partial #2}} \dv{}{t}\,f\left(\,\color{red}x(t),\,\color{blue}y(t)\,\right) = \pdv{f}{\color{red}x}\frac{d\color{red}x}{dt}\,+\, \pdv{f}{\color{blue}y}\frac{d\color{blue}y}{dt} \nonumber$$

Substitute the reverse chain rule to equation $$\eqref{eq:subgrad}$$, and integrate the differential of a function \renewcommand{dv}{\frac{d #1}{d #2}} \renewcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} \int_C \nabla f\cdot d\vec r &= \int_{t_0}^{t_1} \dv{}{t} f\color{grey}{\big(\,x(t),\,y(t)\,\big)}\,dt = \int_{t_0}^{t_1} f\color{grey}{\big(\,x(t),\,y(t)\,\big)} \end{align*}

By the fundamental theorem of calculus $$\eqref{eq:fndcalc}$$ \renewcommand{dv}{\frac{d #1}{d #2}} \renewcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} \int_C \nabla f\cdot d\vec r &= f\color{grey}{\big(\,x(t_1),\,y(t_1)\,\big)} – f\color{grey}{\big(\,x(t_0),\,y(t_0)\,\big)} \end{align*}

With the points \left\{ \begin{align*} P_0 &= \big(\,x(t_0),\,x(t_0)\,\big) \\ P_1 &= \big(\,x(t_1),\,x(t_1)\,\big) \end{align*} \right.

So, the work done in gradient field $$f$$ can be expressed as the difference in potential $$\renewcommand{dv}{\frac{d #1}{d #2}} \renewcommand{pdv}{\frac{\partial #1}{\partial #2}} \shaded{ \int_C \nabla f\cdot d\vec r = f(P_1)-f(P_0) }$$

### Physics (using math notation)

A lot of forces are gradients of potentials such as the electric force and the gravitational force. However, magnetic fields are not gradients.

The work done by the electrical (or gravitational) force, is given by the change of the potential energy from the starting point to the ending point.

Note that: physics potentials are the opposite of mathematical potentials. The force $$\vec F$$ will be negative the gradient. So in physics, it would be expressed as $$\vec F=-\nabla f \nonumber$$

### Properties

Equivalent properties of the work in a gradient field $$\rm W = \int_C\nabla f\,d\vec r \nonumber$$

1. Path-independent: the work only depends on at the start and end points, $$f(P_0)$$ and $$f(P_1)$$.
2. Conservative: the work is $$0$$ along all closed curves. This means a closed loop in a gradient field does not provide energy. Conservativeness means no energy can be extracted from the field for free. The total energy is conserved.
3. $$Mdx+Ndy$$ is an exact differential. That means it can be put in the form $$df$$.

### Examples

#### One

Let’s look at the earlier example again: Curve $$C$$ starting and ending at $$(0,0)$$ through vector field $$\vec F$$ $$\begin{array}{l} \vec F = \left\langle y,x \right\rangle \\ C_1: (0,0)\ \mathrm{to}\ (1,0) \\ C_2: \mathrm{unit\ circle\ from}\ (1,0)\ \mathrm{to\ the\ diagonal} \\ C_3: \mathrm{from\ the\ diagonal\ to}\ (0,0) \end{array} \nonumber$$

##### Try

Try function $$f=xy$$. The gradient is \renewcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} \nabla f &= \left\langle \pdv{}{x}xy, \pdv{}{y}xy \right\rangle = \left\langle y,x \right\rangle \end{align*} That means the line integral can just be evaluated by finding the values of $$f$$ at the endpoints.

##### Visualize

Visualize using a contour plot of $$f=xy$$ through gradient field $$\vec F$$

Along the segments

• On $$C_1$$ the potential stays $$0$$.
• On $$C_2$$ \begin{align*} \int_{C_2}\vec F\cdot d\vec r &= f\left(\frac{1}{\sqrt 2},\frac{1}{\sqrt 2}\right) – f(1,0) \\ &= \frac{1}{2}-0=\frac{1}{2} \end{align*}
• On $$C_3$$ it decreases back to $$0$$.
The sum of the work therefore is $$0$$.

## When is a vector field a gradient field?

Let vector field $$\vec F=\left\langle M,N\right\rangle$$ where $$M$$ and $$N$$ are functions of $$x$$ and $$y$$.

When is this a gradient field? $$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} \vec F = \left\langle M, N \right\rangle \stackrel{?}{=} \nabla f = \left\langle \pdv{}{x}f, \pdv{}{y}f \right\rangle \nonumber$$

If $$\vec F$$ is a gradient field, $$\vec F=\nabla f$$, then \newcommand{dv}{\frac{d #1}{d #2}} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \left\{ \begin{align*} M=\pdv{}{x}f = f_x \\ N=\pdv{}{y}f = f_y \end{align*} \right.

Take the partial derivatives of $$M$$ and $$N$$ \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}{\frac{\partial^2 #1}{\partial #2\partial #3}} \begin{align} M_y=\pdv{M}{y}=\ppdv{}{\color{red}x}{\color{blue}y}f = f_{xy} \label{eq:proof1} \\ N_x=\pdv{M}{x}=\ppdv{}{\color{blue}y}{\color{red}x}f = f_{yx} \label{eq:proof2} \end{align}

Recall: the second partial derivative of function $$f$$

$$\newcommand{dv}{\frac{d #1}{d #2}} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}{\frac{\partial^2 #1}{\partial #2\partial #3}} \pdv{}{\color{red}x}\left(\pdv{f}{\color{blue}y}\right) =\ppdv{f}{\color{red}x}{\color{blue}y} =\ppdv{f}{\color{blue}y}{\color{red}x} =\pdv{}{\color{blue}y}\left(\pdv{f}{\color{red}x}\right) \nonumber$$

Based on the second partial derivative rule, equations $$\eqref{eq:proof1}$$ and $$\eqref{eq:proof2}$$ are the same. That implies that a gradient field should have the property \newcommand{dv}{\frac{d #1}{d #2}} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}{\frac{\partial^2 #1}{\partial #2\partial #3}} \left. \begin{align*} M_y = \ppdv{}{\color{red}x}{\color{blue}y}f = f_{xy} \\ N_x = \ppdv{}{\color{blue}y}{\color{red}x}f = f_{yx} \end{align*} \right\} \Rightarrow M_y=N_x

Therefore, $$\vec F=\left\langle M,N \right\rangle$$, defined and differentiable everywhere, is a gradient field, when $$\shaded{ M_y = N_x } \nonumber$$ (Also see the Definition of Curl.)

So, if $$\vec F=\left\langle M,N\right\rangle$$ is a gradient field in a region of the plane.

• $$\Leftrightarrow$$ Conservative if $$\int_C \vec F\cdot d\vec r=0$$ for any closed curve. To note it is along a closed curve, we note it as $$\oint_C$$ $$\oint_C \vec F\cdot d\vec r=0 \nonumber$$
• $$\Rightarrow$$ $$N_x=M_y$$ at every point.
• $$\Leftarrow$$ $$N_x=M_y$$ at every point, if $$\vec F$$ is defined in the entire plane (or, in a simply connected region). (see later)

### Example

#### One

Is $$\vec F$$ a gradient field? $$\vec F=\underbrace{-y}_{M}\hat\imath+\underbrace{x}_N\hat\jmath=\left\langle -y,x \right\rangle \nonumber$$

$$\vec F$$ is not a gradient field, because \newcommand{dv}{\frac{d #1}{d #2}} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}{\frac{\partial^2 #1}{\partial #2\partial #3}} \left. \begin{align*} \pdv{M}{y}&=\pdv{}{y}(-y)=-1 \\ \pdv{N}{x}&=\pdv{}{x}x=1 \end{align*} \right\} \Rightarrow \pdv{M}{y}\neq \pdv{N}{x}

#### Two

For what value of $$a$$ is $$\vec F$$ a gradient field? $$\vec F = \underbrace{(4x^2+axy)}_{M}\hat\imath+\underbrace{(3y^2+4x^2)}_N\hat\jmath = \left\langle 4x^2+axy, 3y^2+4x^2\right\rangle \nonumber$$

\newcommand{dv}{\frac{d #1}{d #2}} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}{\frac{\partial^2 #1}{\partial #2\partial #3}} \left. \begin{align*} \pdv{M}{y}&=\pdv{}{y}(4x^2+axy)=ax \\ \pdv{N}{x}&=\pdv{}{x}(3y^2+4x^2)=8x \end{align*} \right\} \Rightarrow a=8 Note that $$x=0$$ is not an answer everywhere.

## Finding the potential

Recall: from earlier

When the field is a gradient, and you know the function $$f$$, you can simplify the evaluation of the line integral for work. $$\shaded{ \int_C\nabla f\cdot d\vec r=f(P_1)-f(P_0) } \nonumber$$ where $$f(x,y)$$ is called the potential

To show the two methods, we will find the potential of the gradient field $$\vec F$$ $$\vec F = \left\langle \underbrace{4x^2+axy}_{=M}, \underbrace{3y^2+4x^2}_{=N} \right\rangle \nonumber$$

### Compute line integrals

Apply the fundamental theorem, equation $$\eqref{eq:fundthm}$$, to find an expression for the potential at $$(x_1,y_1)$$ \begin{align} &\int_C\vec F\cdot d\vec r=f(x_1,y_1)-f(0,0) \nonumber \\ \Rightarrow & f(x_1,y_1) = \underbrace{\int_C\vec F\cdot d\vec r}_{\rm{work}} + \underbrace{f(0,0)}_{\mathrm{constant}} \label{eq:method1} \end{align}

Apply the work differential, to find the work along $$C$$ in gradient field $$\vec F$$ \begin{align*} \underline{\int_C\vec F\cdot d\vec r} &= \int_C M\,dx+N\,dy \\ &= \int_C\left(4x^2+8xy\right)dx+\left(3y^2+4x^2\right)dy \end{align*}

The work in a gradient is path independent $$\Longrightarrow$$ find the easiest path Paths $$C, C_1, C_2$$

The easiest path is $$\begin{array}{lll} C_1: & x\ \mathrm{from}\ 0\ \mathrm{to}\ x_1 & y=0 &\Rightarrow dy=0 \\ C_2: & x=x_1 & y\ \mathrm{from}\ 0\ \mathrm{to}\ y_1 &\Rightarrow dx=0 \end{array} \nonumber$$

Work along the curves

• Along $$C_1$$ \begin{align*} \int_{C_1}\vec F\cdot d\vec r &= \int_0^{x_1}(4x^2+0)\,dx + 0 \\ &= \left[\frac{4}{3}x^3\right]_0^{x_1} = \frac{4}{3}{x_1}^3 \end{align*}
• Along $$C_2$$ \begin{align*} \int_{C_2}\vec F\cdot d\vec r &= \int_0^{y_1}0+(3y^2+4{x_1}^2)\,dy \\ &= \left[y^3+4{x_1}^2y \right]_0^{y_1} = {y_1}^3+4{x_1}^2y_1 \end{align*}

The total work $$\int_C\vec F\cdot d\vec r = \int_{C_1}\ldots + \int_{C_2}\ldots = \frac{4}{3}{x_1}^3 + {y_1}^3+4{x_1}^2y_1 \nonumber$$

Substitute $$\int_{C_1}, \int_{C_2}$$ back in $$\eqref{eq:method1}$$ \begin{align*} f(x_1,y_1) &= \int_C\vec F\cdot d\vec r + \rm{c} \\ &= \frac{4}{3}{x_1}^3 + {y_1}^3+4{x_1}^2y_1 + \rm{c} \end{align*}

Drop the subscripts $$f(x,y) = \frac{4}{3}x^3 + 4x^2y_1 + y^3\, (+ \rm{c}) \nonumber$$ If you would take the gradient, you should get $$\vec F$$ back.

### Compute using antiderivatives

No integrals, but you have to follow the procedure very carefully. A common pitfall, is to treat the second equation, like the first one.

For the example, we want to solve \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \left. \begin{align} \pdv{f}{x}&=f_x=4x^2+8xy \label{eq:anti1} \\ \pdv{f}{y}&=f_y=3y^2+4x^2 \label{eq:anti2} \end{align} \right.

Integrate equation $$\eqref{eq:anti1}$$ in respect to $$x$$. The integration constant might depend on $$y$$, so we call it $$g(y)$$ $$\newcommand{pdv}{\frac{\partial #1}{\partial #2}} \pdv{f}{x} = 4x^2+8xy \xrightarrow{\int dx} f = \underline{\frac{4}{3}x^3 + 4x^2y + g(y)} \label{eq:anti}$$

To get information of $$g(y)$$, we look at the other partial. Take the derivative of $$f$$ in respect to $$y$$ and compare to $$\eqref{eq:anti}$$ \require{cancel} \newcommand{pdv}{\frac{\partial #1}{\partial #2}} \begin{align*} \pdv{}{y}\left(\frac{4}{3}x^3 + 4x^2y + g(y)\right) &= 3y^2+\bcancel{4x^2} \\ 0 + \bcancel{4x^2}+\pdv{}{y}g(y) &= 3y^2+\bcancel{4x^2} \Rightarrow \pdv{}{y}g(y) = 3y^2\\ \xrightarrow{\int dy} g(y) &= \int \pdv{}{y}g(y)\,dy = \underline{y^3 + c} \end{align*} $$g(y)$$ only depends on $$y$$, so $$c$$ is a true constant.

Plug this back into equation $$\eqref{eq:anti}$$, gives the potential $$f(x,y)$$ $$f = \frac{4}{3}x^3 + 4x^2y + \underline{y^3\ (+ \rm{c})} \nonumber$$