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Math 344: Calculus III

16.3 Fundamental Theorem for Path Integrals

16.3.1 Statement of the Theorem

A gradient vector field $\nabla f$ of a function of several variables $f$ is in some sense a derivative of the function $f$, since it is composed of partial derivatives.

$$ \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \newcommand{\pzi}[2][]{\dfrac{\partial #1}{\partial #2}} \begin{align*} \nabla f(x,y) &= f_x(x,y)\,\ihat + f_y(x,y)\,\jhat \\ \\ \nabla f(x,y,z) &= f_x(x,y,z)\,\ihat + f_y(x,y,z)\,\jhat + f_z(x,y,z)\,\khat \end{align*} $$
With this in mind, we can consider a path integral through a gradient field and this theorem as the analogue to the Fundamental Theorem for path integrals.

Theorem

Fundamental Theorem for Path Integrals
Let $C$ be a smooth curve defined by the function $\mathbf{r}(t)$, $t\in [a,b]$. Let $f$ be a differentiable function of several variables whose gradient vector $\nabla f$ is continuous on $C$. Then

$$ \int_C \nabla f\cdot d\mathbf{r} = f(\mathbf{r}(b)) - f(\mathbf{r}(a)) $$

Proof:

Recall the formula

$$ \begin{align*} \int_c \nabla f\cdot d\mathbf{r} &= \int_a^b \nabla f(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt \\ \\ &= \int_a^b \left(f_x(x,y,z)\, x'(t) + f_y(x,y,z)\, y'(t) + f_z(x,y,z)\, z'(t)\right)\, dt \\ \\ &= \int_a^b \dfrac{d}{dt} f(\mathbf{r}(t))\,dt \\ \\ &= f(\mathbf{r}(b)) - f(\mathbf{r}(a)) \end{align*} $$
∎

Note that while the theorem here was proven for a smooth curve $C$, if $C$ is piecewise smooth the result can be shown component-wise. Therefore, the result holds for piecewise smooth $C$ as well.

This theorem gives a result reminiscent of one part of the Fundamental Theorem of Calculus for single-variable functions, also known at the Net Change Theorem:

$$ \int_a^b f'(x)\,dx = f(b) - f(a) $$
Which essentially states that to find the value of the integral over the derivative $f'$ on the interval $[a,b]$, it is only necessary to evaluate the function $f$ at the end points (the boundary) of the interval. The result for path integrals is much more impressive, since we get the same statement of only needing to evaluate the function at the end points of the path $C$, but nothing about $C$ was specified outside of it being a smooth curve.

A example of the smooth curve C. The curve is not self intersecting and has arrow marks to indicate the direction of travel.

Example 16.3.1

Work in physics is easily computed by using this theorem, so let us determine the work done by the gravitational field

$$ \mathbf{F}(\mathbf{x}) = -\dfrac{mMG}{|\mathbf{x}|^3}\mathbf{x} $$
on a particle of mass $m$ from the point $(2,3,6)$ to the point $(3,0,3)$ along a piecewise smooth curve $C$.

Solution

We know that $\mathbf{F}$ is a conservative vector field obeying $\mathbf{F} = \nabla f$ with potential function

$$ f(x,y,z) = \dfrac{mMG}{\sqrt{x^2 + y^2 + z^2}} $$ The work done is given by

$$ \begin{align*} W &= \int_C \mathbf{F}\cdot d\mathbf{r} \\ \\ &= \int_C \nabla f\cdot d\mathbf{r} \\ \\ &= f(3,0,3) - f(2,3,6) \\ \\ &= \dfrac{mMG}{\sqrt{3^2 + 3^2}} - \dfrac{mMG}{\sqrt{2^2 + 3^2 + 6^2}} \\ \\ &= mMG\left(\dfrac{1}{3\sqrt{2}} - \dfrac{1}{7}\right) \end{align*} $$

Path Independence

In Example 16.3.1, we did not need to know any details about the path $C$ the particle moved along, so long as the path was piecewise smooth. This property is known as path independence , and it is a special property of path integrals for conservative vector fields.

A path integral of a conservative vector field is path independent.

We value this highly, and prefer to work with conservative vector fields when possible due to the ease with which path integrals may be computed.

16.3.2 Types of Paths

Another important consideration for working with path integrals over conservative vector fields is whether or not the path in question is closed . A closed path is one where the initial point of the path coincides with the terminal point.

A pair of simple (not self-intersecting) smooth closed curves.

Closed curves may evaluated using any point on the curve as its initial and terminal point, or broken up arbitrarily into smaller curves by dividing it into sections. Taking one of the above curves, we can choose two points $a$ and $b$ to "cut" the curve into smaller segments $C_1$ and $C_2$, where $C = C_1\cup C_2$.

A curve from the previous example with two points a and b selected to break the curve into two segments C1 and C2.

The specific choices of $a$ and $b$ here do not matter, but they do serve to show an important property of conservative vector fields. Since $\int_C \mathbf{F}\cdot d\mathbf{r}$ is independent of path for conservative $\mathbf{F}$, we may write

$$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_{C_1} \mathbf{F}\cdot d\mathbf{r} + \int_{C_2} \mathbf{F}\cdot d\mathbf{r} = \int_{C_1} \mathbf{F}\cdot d\mathbf{r} - \int_{-C_2} \mathbf{F}\cdot d\mathbf{r} = 0 $$
since $C_1$ and $-C_2$ share $a$ and $b$ as the initial and terminal points and traveling in the negative orientation on the path introduces a minus sign to the integral (just like switching the limits of integration does in the single-dimensional case).

The curves shown here are both simple . That is, the curves do not self-intersect or cross themselves. When discussing paths for path integrals, we are concerned with whether or not the path has one or both of these properties.

An example of a simple, not closed curve and a not simple, not closed curve side by side.

Both of the curves above are not closed, but the first features no self-intersections while the second does.

An example of a simple, closed curve and a not simple, closed curve side by side.

Here both curves are closed, but the second features a self-intersection. In most situations, we must work with simple curves to ensure that the mathematical environment we are working in is suitably well-behaved.

Historically, this has been issue. An extremely important theorem on this subject, the Jordan curve theorem , is notoriously difficult to prove.

Types of Regions

Another important classification when considering path integrals is the type of region . When a domain $D$ is considered, we want to know if that region is simply-connected . A simply connected region is one that is contiguous with no holes.

A region bounded a simple, closed curve with no holes.

A region may fail to be simply-connected if it has holes or is separated into more than one piece.

Several examples of regions that fail to be simply connectedd, by either having holes or being composed to multiple pieces.

In practice, we tend to avoid regions that are not simply-connected, as they interfere with our path integral theorems.

16.3.3 Conservative Vector Fields

We say that a vector field is conservative if it obeys the equation

$$ \mathbf{F} = \nabla f $$
If we suppose that $\mathbf{F} = P\,\ihat + Q\,\jhat$ in a two-dimensional case, then there must exist some scale function $f$ where

$$ \mathbf{F} = P\,\ihat + Q\,\jhat = \pzi[f]{x}\,\ihat + \pzi[f]{y}\,\jhat = \nabla f $$
This means that we must have

$$ P = \pzi[f]{x} \qquad\qquad\text{and}\qquad\qquad Q = \pzi[f]{y} $$
which by Clairaut's Theorem implies

$$ \pzi[P]{y} = \dfrac{\partial^2\! f}{\partial y\,\partial x} = \dfrac{\partial^2\! f}{\partial x\,\partial y} = \pzi[Q]{x} $$
yielding the theorem

Theorem

If $\,\mathbf{F}(x,y) = P(x,y)\,\ihat + Q(x,y)\,\jhat$ is a conservative vector field, where $P$ and $Q$ have continuous first-order partial derivatives on some domain $D$, then in all of $D$

$$ \pzi[P]{y} = \pzi[Q]{x} $$

In order to have the converse of this statement, we must specify that $D$ is an open, simply connected region. A general domain will not do.

Theorem

If $\,\mathbf{F}(x,y) = P(x,y)\,\ihat + Q(x,y)\,\jhat$ is a vector field on an open, simply connected domain $D$ and the functions $P$ and $Q$ have continuous first-order partial derivatives with

$$ \pzi[P]{y} = \pzi[Q]{x} $$
in all of $D$, then $\mathbf{F}$ is a conservative vector field.

Example 16.3.2

Determine if the vector field

$$ \mathbf{F}(x,y) = (2x + y)\,\ihat + (x - y^2)\,\jhat $$
is conservative.

Solution

Set $P(x,y) = 2x - y$ and $Q(x,y) = x - y^2$, then

$$ \pzi[P]{y} = -1 \qquad\qquad\qquad \pzi[Q]{x} = 1 $$
We see that $\pzi[P]{y}\neq\pzi[Q]{x}$, so $\mathbf{F}$ is not conservative.

Example 16.3.3

Determine if the vector field

$$ \mathbf{F}(x,y) = (y^2 - 2x)\,\ihat + 2xy\,\jhat $$
is conservative. If it is, find an $f$ such that $\mathbf{F} = \nabla f$.

Solution

Set $P(x,y) = y^2 - 2x$ and $Q(x,y) = 2xy$, then

$$ \pzi[P]{y} = 2y = \pzi[Q]{x} $$
which satisfies the condition of the theorem, and the domain of $\mathbf{F}$ is all of $\mathbb{R}^2$, an open and simply-connected region, so $\mathbf{F}$ is conservative.

To find $f$, we note that $f_x = y^2 - 2x$ and $f_y = 2xy$. We select one of these functions and integrate with respect to the appropriate variable to get an expression for $f$.

$$ \int f_x\,dx = \int y^2 - 2x\,dx = xy^2 - x^2 + g(y) $$
where $g(y)$ is some function that depends solely on $y$ (this takes the place of the constant of integration since we have multiple independent variables). If we differentiate this expression with respect to $y$, it must be equal to $f_y$, so

$$ \pzi{y} \left(xy^2 - x^2 + g(y)\right) = 2xy + g'(y) = f_y = 2xy $$
This implies that $g'(y) = 0$, so $g(y) = k$, a constant. Therefore, the function $f$ whose gradient is $\mathbf{F}$ is

$$ f(x,y) = xy^2 - x^2 + k $$

Example 16.3.4

Find a function $f$ such that $\nabla f = \mathbf{F}$ for

$$ \mathbf{F}(x,y,z) = \left(y^2 z+2xz^2\right)\,\ihat + 2xyz\,\jhat + \left(xy^2 + 2x^2 z\right)\,\khat $$

Solution

Such a function $f$ would have first-order partials of the form

$$ \begin{align*} f_x(x,y,z) &= y^2 z+2xz^2 \\ \\ f_y(x,y,z) &= 2xyz \\ \\ f_z(x,y,z) &= xy^2 + 2x^2 z \end{align*} $$
Integrating $f_x$ with respect to $x$, we have

$$ f(x,y,z) = xy^2 z+x^2 z^2 + g(y,z) $$
where $g(y,z)$ replaces the constant of integration in this multivariate case. If we differentiate with respect to $y$, this expression must match $f_y$, so

$$ \pzi{y}\left(xy^2 z+x^2 z^2 + g(y,z)\right) = 2xyz + g_y(y,z) = 2xyz = f_y $$
Therefore, $g(y,z)$ is solely a function of $z$ since $g_y = 0$. Set $g(y,z) = h(z)$ due to this. Now looking at $f$ and differentiating with respect to $z$, we have

$$ \pzi{z}\left(xy^2 z+x^2 z^2 + h(z)\right) = xy^2 + 2x^2 z + h'(z) = xy^2 + 2x^2 z = f_z $$
This implies that $h'(z) = 0$ and so $h(z) = k$, some constant. Therefore, we can say that our function $f$ is

$$ f(x,y,z) = xy^2 z+x^2 z^2 + k $$

16.3.4 Conservation of Energy

Conservative vector fields are named as such because they conserve energy .

Example 16.3.5

Consider a conservative vector field, where the force applied by that field is at some point along a curve $C$ is given by

$$ \mathbf{F}(\mathbf{r}(t)) = m\mathbf{r}''(t) $$
where $m$ is the mass of an object and $\mathbf{r}''(t)$ is the acceleration along the path $C$ parametrized by $\mathbf{r}(t)$. The work done by the force on this object is

$$ \begin{align*} W &= \int_c \mathbf{F}\cdot d\mathbf{r} \\ \\ &= \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt \\ \\ &= \int_a^b m\mathbf{r}''(t)\cdot\mathbf{r}'(t)\,dt \\ \\ &= \dfrac{m}{2}\int_a^b \dfrac{d}{dt}\left[\mathbf{r}'(t)\cdot\mathbf{r}'(t)\right]\,dt \\ \\ &= \dfrac{m}{2}\int_a^b \dfrac{d}{dt}\left|\mathbf{r}'(t)\right|^2\,dt \\ \\ &= \dfrac{m}{2}\left[\dfrac{d}{dt}\left|\mathbf{r}'(t)\right|^2\right]_{\,a}^{\,b} \\ \\ &= \dfrac{m}{2}\left[\left|\mathbf{r}'(b)\right|^2 - \left|\mathbf{r}'(a)\right|^2\right]\ \end{align*} $$
Since $\mathbf{r}' = \mathbf{v}$, the velocity, we have that

$$ W = \frac{1}{2}m\left[\left|\mathbf{v}(b)\right|^2 - \left|\mathbf{v}(a)\right|^2\right] $$
We know from physics that $\frac{1}{2}m\left|\mathbf{v}(t)\right|^2$ is the kinetic energy of the body, and using the shorthand $K(\mathbf{r}(t)) = \frac{1}{2}m\left|\mathbf{v}(t)\right|^2$ we can represent the kinetic energy as a function of position along the curve $C$ at time $t$.

Since our vector field is conservative, we know that $\mathbf{F} = \nabla f$ for some $f$. The potential energy of an object at $(x,y,z)$ is defined as $P(x,y,z) = -f(x,y,z)$ and hence $\mathbf{F} = -\nabla P$. Using the Fundamental Theorem for Path Integrals, the work done is

$$ W = \int_c \mathbf{F}\cdot d\mathbf{r} = -\int_c \nabla P\cdot d\mathbf{r} = -\left[P(\mathbf{r}(b)) - P(\mathbf{r}(a))\right] = P(\mathbf{r}(a)) - P(\mathbf{r}(b)) $$
Both expressions for $W$ must match, so we have that

$$ W = K(\mathbf{r}(b)) - K(\mathbf{r}(a)) = P(\mathbf{r}(a)) - P(\mathbf{r}(b)) $$
and so

$$ P(\mathbf{r}(a)) + K(\mathbf{r}(a)) = P(\mathbf{r}(b)) + K(\mathbf{r}(b)) $$
Since the sum of the potential and kinetic energies remain constant while moving due to a conservative vector field, the total energy is constant and we say that energy is conserved .

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