Exact one-forms and conservative vector fields

Section 2.2 Exact one-forms and conservative vector fields

We introduce a class of one-forms that arise as differentials of a function; the associated vector field is the gradient of the function. But not all one-forms are differentials of functions: we call such one-forms “exact”, and their corresponding vector fields “conservative”.

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Objectives

You should be able to:

Calculate the differential of a function, and intepret it as a one-form.
Translate between the differential of a function and its gradient.
Define exact one-forms.
Define conservative vector fields and their associated potentials.
Translate between exact one-forms and conservative vector fields.
Define closed one-forms using partial derivatives.
Show that exact one-forms are always closed.
Rephrase the statement as the "screening test" for conservative vector fields.
Use the screening test to show that a given vector field cannot be conservative.

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Subsection 2.2.1 Differential of a function

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One-forms may seem strange, but in fact there is a large class of one-forms that can be obtained directly from functions.

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Definition 2.2.1. Differential of a function.

Let $f$ be a smooth function on an open subset $U \subseteq R^{3} .$ Its differential (it will become known as “exterior derivative” shortly) is

d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y + \frac{\partial f}{\partial z} d z .

It is a one-form on $U .$ A similar definition of course holds for $R,$ $R^{2},$ and in fact $R^{n} .$

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Using the correspondence in Principle 2.1.3 between one-forms and vector fields, we can find the vector field associated to the differential of a function.

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Fact 2.2.2. Correspondence between the differential and the gradient of a function.

The vector field associated to the differential $d f$ of a function $f$ is the vector field with component functions

(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}),

which is nothing else but the gradient $\nabla f .$

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Recall from Calculus III that at a given point

(x_{0}, y_{0}, z_{0}) \in U,

the gradient vector field

\nabla f (x_{0}, y_{0}, z_{0})

gives the direction of fastest increase of

f .

In three dimensions, this direction is orthogonal to the level surfaces. In two dimensions, it is perpendicular to the level curves: think of a topographical map, the direction of fastest increase is the direction perpendicular to the contour lines.

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Example 2.2.3. The differential and gradient of a function.

Consider the function $f (x, y) = x^{2} e^{y} .$ Its differential is

d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y = 2 x e^{y} d x + x^{2} e^{y} d y,

which is a one-form on $R^{2} .$ Its associated vector field is the function $F : R^{2} \to R^{2}$ with

F (x, y) = (2 x e^{y}, x^{2} e^{y}) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) = \nabla f .

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Remark 2.2.4.

Using the definition of the differential of a function, we can somewhat make sense of the placeholders $d x,$ $d y,$ and $d z .$ Consider for instance the function $f (x, y, z) = x .$ Its differential becomes

d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y + \frac{\partial f}{\partial z} d z = d x .

In other words, the placeholders $d x,$ $d y,$ and $d z$ are simply the differentials of the component functions $x,$ $y,$ and $z$ on $R^{3} .$

Note that we will give a more satisfying algebraic meaning for these placeholders later on in Subsection 4.1.1.

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Subsection 2.2.2 Exact one-forms and conservative vector fields

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Differentials of functions are one-forms, but not all one-forms are differentials of functions, just like not all vector fields can be written as the gradient of a function. Such one-forms and vector fields are special, and hence have their own name.

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Definition 2.2.5. Exact one-forms.

We say that a one-form $ω$ on $U \subseteq R^{3}$ is exact if $ω = d f$ for some function $f$ on $U .$

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We can define a similar concept for vector fields.

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Definition 2.2.6. Conservative vector fields.

A vector field $F$ on $U \subseteq R^{3}$ is conservative if $F = \nabla V$ for some function $V$ on $U .$ We call $V$ a potential for $F .$ ¹

Note that in physics, the potential of a conservative vector field is usually defined as $F = - \nabla V,$ with an extra minus sign. The difference is purely conventional; the minus sign is introduced so that when $F$ is a force field, then $V$ becomes the potential energy physically.

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Clearly, by Fact 2.2.2, we see that if a one-form is exact, its associated vector field is conservative, and vice-versa.

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Example 2.2.7. An exact one-form and its associated conservative vector field.

Consider the one-form $ω = \cos x d x - \sin y d y$ on $R^{2} .$ $ω$ is an exact one-form, since it can be written as $ω = d f$ for the function $f (x, y) = \sin x + \cos y .$ In the language of vector fields, this is the statement that the vector field $F : R^{2} \to R^{2}$ associated to $ω,$ which is given by $F (x, y) = (\cos x, - \sin y),$ is conservative, since it can be written as $F = \nabla f$ for the function $f$ above. The function $f$ is called the potential of the vector field $F .$

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Example 2.2.8. The gravitational force field is conservative.

The gravitational force field $F : R^{3} ∖ {(0, 0, 0)} \to R^{3}$ that a mass $M$ at the origin exerts on a mass $m$ at position $(x, y, z)$ is given by

F (x, y, z) = - \frac{G M m}{r^{3}} (x, y, z),

where $r = \sqrt{x^{2} + y^{2} + z^{2}}$ and $G$ is the gravitional constant. The corresponding one-form is

ω = - \frac{G M m}{r^{3}} (x d x + y d y + z d z) .

It is easy to see that the gravitational force field is conservative, or equivalently that the one-form $ω$ is exact, since $ω = d φ$ with the function $φ$ given by

φ (x, y, z) = \frac{G M m}{r} .

$φ$ is the potential function, which from physics you may recognize as minus the gravitational potential energy.

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Subsection 2.2.3 Closed one-forms in $R^{2}$

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Since not all one-forms can be written as differentials of functions, i.e. not all one-forms are exact, a natural question arises: can we determine, looking at a one-form, whether it is exact or not? Similarly, can we easily determine whether a vector field is conservative? Unfortunately we will not be able to fully answer this question at the moment, we will come back to it in Section 3.6. For the time being, we will be able to find a necessary condition for a one-form to be exact, which in the context of vector calculus is sometimes called a "screening test" for conservative vector fields.

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Let us focus first on one-forms and vector fields on

U \subseteq R^{2} .

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Definition 2.2.9. Closed one-forms in $R^{2}$ .

We say that a one-form $ω = f d x + g d y$ on $U \subseteq R^{2}$ is closed if

\frac{\partial f}{\partial y} = \frac{\partial g}{\partial x} .

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This definition may seem ad hoc, but it is important because of the following lemma.

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Lemma 2.2.10. Exact one-forms in $R^{2}$ are closed.

If a one-form $ω$ on $U \subseteq R^{2}$ is exact, then it is closed.

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Proof.

Suppose that $ω$ is exact: then it can be written as

ω = f d x + g d y = d F = \frac{\partial F}{\partial x} d x + \frac{\partial F}{\partial y} d y

for some smooth function $F$ on $U \subseteq R^{2} .$ It will then be closed if

\frac{\partial f}{\partial y} = \frac{\partial^{2} F}{\partial y \partial x}

is equal to

\frac{\partial g}{\partial x} = \frac{\partial^{2} F}{\partial x \partial y} .

Equality of the two expressions follows from the Clairaut-Schwarz theorem, which states that partial derivatives commute, as long as they are continuous. But continuity of the partial derivatives is guaranteed by the fact that all partial derivatives of $F$ exist and are continuous, since by definition (see Definition 2.1.1) one-forms are assumed to have smooth component functions. Therefore the one-form is closed.

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Note however that the converse statement is not necessarily true: not all closed one-forms on

U \subseteq R^{2}

are exact. In fact, the question of when closed one-forms are exact is an important one; the result is known as Poincare's lemma. We will come back to this in Section 3.6. But what Lemma 2.2.10 tells us is that one-forms that are not closed cannot be exact.

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There is of course an analogous statement for conservative vector fields. The only minor difference is that we need to impose a condition on the component functions of vector fields, since vector fields are not always smooth by definition (see Definition Definition 2.1.2.

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Lemma 2.2.11. Screening test for conservative vector fields in $R^{2}$ .

If a vector field $F = (f_{1}, f_{2})$ on $U \subseteq R^{2}$ is conservative and has component functions that are continuously differentiable, then it passes the screening test:

\frac{\partial f_{1}}{\partial y} = \frac{\partial f_{2}}{\partial x} .

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Proof.

Same as for Lemma 2.2.10, but for the associated vector fields.

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In the context of vector fields, it is called a screening test, because it is a quick test to determine whether a vector field has a chance at all to be conservative. In other words, if a vector field does not pass the screening test, then it is certainly not conservative. However, if it passes the screening test, at this stage we cannot conclude anything. Just like closed one-forms are not necessarily exact.

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Example 2.2.12. Exact one-forms are closed.

Consider the exact one-form on $R^{2}$ from Example 2.2.7, $ω = \cos x d x - \sin y d y .$ We show that it is closed, according to Definition 2.2.9. The partial derivatives are easily calculated:

\frac{\partial}{\partial y} \cos x = 0, \frac{\partial}{\partial x} (- \sin y) = 0.

Thus Definition 2.2.9 is satisfied, and $ω$ is closed.

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Example 2.2.13. Closed one-forms are not necessarily exact.

If a vector field is conservative, then it passes the screening test. Correspondingly, if a one-form is exact, then it is closed. But the converse statement is not necessarily true (we will revisit it later in Section 3.6). Consider for instance the one-form

ω = - \frac{y}{x^{2} + y^{2}} d x + \frac{x}{x^{2} + y^{2}} d y .

Calculating the partial derivatives for Definition 2.2.9, we get that

\frac{\partial}{\partial y} (- \frac{y}{x^{2} + y^{2}}) = \frac{y^{2} - x^{2}}{x^{2} + y^{2}}, \frac{\partial}{\partial x} (\frac{x}{x^{2} + y^{2}}) = \frac{y^{2} - x^{2}}{x^{2} + y^{2}} .

The two expressions are equal, and thus $ω$ is closed. However, one can show that $ω$ is not exact: there does not exist a function $f$ such that $ω = d f$ (see Exercise 3.4.3.2).

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Subsection 2.2.4 Closed one-forms in $R^{3}$

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We focused on

R^{2}

for simplicity, but similar results hold for one-forms and vector fields in

R^{3} .

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Definition 2.2.14. Closed one-forms in $R^{3}$ .

We say that a one-form $ω = f d x + g d y + h d z$ on $U \subseteq R^{3}$ is closed if

\frac{\partial f}{\partial y} = \frac{\partial g}{\partial x}, \frac{\partial f}{\partial z} = \frac{\partial h}{\partial x}, \frac{\partial g}{\partial z} = \frac{\partial h}{\partial y} .

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We then have the lemma:

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Lemma 2.2.15. Exact one-forms in $R^{3}$ are closed.

If a one-form $ω$ on $U \subseteq R^{3}$ is exact, then it is closed.

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Proof.

Suppose that $ω$ is exact: then it can be written as

ω = f d x + g d y + h d z = d F = \frac{\partial F}{\partial x} d x + \frac{\partial F}{\partial y} d y + \frac{\partial F}{\partial z} d z

for some smooth function $F$ on $U \subseteq R^{3} .$ It will then be closed if

\frac{\partial^{2} F}{\partial y \partial x} = \frac{\partial^{2} F}{\partial x \partial y}, \frac{\partial^{2} F}{\partial z \partial x} = \frac{\partial^{2} F}{\partial x \partial z}, \frac{\partial^{2} F}{\partial z \partial y} = \frac{\partial^{2} F}{\partial y \partial z} .

As before, these equalities follow from the Clairaut-Schwarz theorem, which states that partial derivatives commute as long as they are continuous.

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The analogous statement for vector fields goes as follows:

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Lemma 2.2.16. Screening test for conservative vector fields in $R^{3}$ .

If a vector field $F = (f_{1}, f_{2}, f_{3})$ on $U \subseteq R^{3}$ is conservative and has component functions that are continuously differentiable, then it passes the screening test:

\frac{\partial f_{1}}{\partial y} = \frac{\partial f_{2}}{\partial x}, \frac{\partial f_{1}}{\partial z} = \frac{\partial f_{3}}{\partial x}, \frac{\partial f_{2}}{\partial z} = \frac{\partial f_{3}}{\partial y} .

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Proof.

Same as for Lemma 2.2.15, but for the associated vector fields.

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Remark 2.2.17.

At this stage the definition of closeness for one-forms and the associated screening test for vector fields appear to be rather ad hoc. Sure, they are necessary conditions for a one-form to be exact and a vector field to be conservative, but is that it? No, not really. In fact, those conditions will come out very naturally when we go beyond one-forms and introduce the theory of $k$ -forms in general. We will see how we can "differentiate" forms - this is the notion of exterior derivative. Then the definition of closed one-forms in Definition 2.2.9 will be simply that a one-form $ω$ is closed if its exterior derivative $d ω$ vanishes. Furthermore, if $F$ is the vector field associated to a one-form $ω,$ then the vector field associated to its derivative $d ω$ will be called the "curl" of $F,$ and denoted by $\nabla \times F .$ The screening test for vector fields will then be that $F$ is "curl-free", that is $\nabla \times F = 0 .$ All fun stuff, but it will have to wait for a bit! Coming in Chapter 4.

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Exercises 2.2.5 Exercises

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1.

Consider the function $f (x, y, z) = \sin (y) + z x .$ Find its differential $d f .$

Solution.

$d f$ is given by:

\begin{aligned} d f = & \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y + \frac{\partial f}{\partial z} d z \\ = & z d x + \cos (y) d y + x d z . \end{aligned}

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2.

Consider the one-form $ω = x d x - y d y$ on $R^{2} .$ Show that it is exact, and find a function $f$ such that $ω = d f .$

Solution.

Suppose that there exists a function $f (x, y)$ such that $d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y = x d x - y d y .$ Then we must have

\frac{\partial f}{\partial x} = x, \frac{\partial f}{\partial y} = - y .

Integrating the first equation (recalling that this is a partial derivative, so the “constant of integration” is any function of $y$ alone), we get:

f (x, y) = \frac{x^{2}}{2} + g (y)

for some function $g (y) .$ Substituting into the second equation, we get

\frac{\partial f}{\partial y} = g^{'} (y) = - y,,

which can be integrated to

g (y) = - \frac{y^{2}}{2} + C

for any constant $C .$ Since we are only interested in finding one $f$ such that $d f = ω,$ we can choose $C = 0 .$ We conclude that the function

f (x, y) = \frac{x^{2}}{2} - \frac{y^{2}}{2}

is such that $d f = ω,$ and hence that $ω$ is an exact one-form.

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3.

True or False. If $F$ and $G$ are both conservative, then $F + G$ is also conservative.

Solution.

True. If $F$ and $G$ are both conservative, then $F = \nabla f$ and $G = \nabla g$ for some functions $f$ and $g .$ But then $F + G = \nabla (f + g),$ and hence $F + G$ is also conservative.

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4.

Consider the vector field $F (x, y, z) = (y z e^{x y}, x z e^{x y}, e^{x y}) .$ Show that it is conservative and find a potential.

Solution.

$F$ is conservative if there exists a function $f$ such that $F = \nabla f .$ So we need to solve the equations

\frac{\partial f}{\partial x} = y z e^{x y}, \frac{\partial f}{\partial y} = x z e^{x y}, \frac{\partial f}{\partial z} = e^{x y} .

Let us start by integrating the last one. We get:

f (x, y, z) = z e^{x y} + g (x, y)

for some function $g (x, y) .$ Substituting back into the second one, we get:

\frac{\partial f}{\partial y} = x z e^{x y} + \frac{\partial g}{\partial y} = x z e^{x y},

from which we conclude that $\frac{\partial g}{\partial y} = 0,$ that is, $g (x, y) = h (x)$ for some function $h (x) .$ Finally, substituting back into the first equation, we get:

\frac{\partial f}{\partial x} = y z e^{x y} + \frac{d h}{d x} = y z e^{x y},

from which we get $h^{'} (x) = 0,$ that is $h (x) = C$ for some constant $C .$ We choose $C = 0,$ and we conclude that $F$ is conservative, with potential

f (x, y, z) = z e^{x y} .

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5.

Show that the one-form $ω = f^{'} (x) d x + g^{'} (y) d y$ on $R^{2}$ is exact, for any smooth functions $f, g : R \to R .$

Solution.

Consider the function $F (x, y) = f (x) + g (y) .$ Its differential is

d F = \frac{\partial F}{\partial x} d x + \frac{\partial F}{\partial y} d y = f^{'} (x) d x + g^{'} (y) d y .

Therefore $ω$ is exact, and it is a well defined one-form on $R^{2}$ as $f$ and $g$ are smooth.

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6.

Show that the one-form $ω = \cos (y) d x - x \sin (y) d y$ is closed. Is it exact?

Solution.

Let us write $ω = f d x + g d y$ for $f (x, y) = \cos (y)$ and $g (x, y) = - x \sin (y) .$ To show that it is closed, we calculate:

\frac{\partial f}{\partial y} = - \sin (y), \frac{\partial g}{\partial x} = - \sin (y) .

As the two partial derivatives are equal, we conclude that $ω$ is a closed one-form.

Is it exact? We are looking for a function $F (x, y)$ such that $d F = ω,$ that is,

\frac{\partial F}{\partial x} = \cos (y), \frac{\partial F}{\partial y} = - x \sin (y) .

Integrating the first equation, we get $F (x, y) = x \cos (y) + h (y)$ for some function $h (y) .$ Substituting in the second equation, we get

\frac{\partial F}{\partial y} = - x \sin (y) + h^{'} (y) = - x \sin (y),

from which we conclude that $h^{'} (y) = 0,$ that is $h (y) = C .$ We pick $C = 0 .$ We conclude that $ω = d F$ with $F (x, y) = x \cos (y),$ and hence it is exact.

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7.

Show that the vector field $F (x, y, z) = (y + z, x + z, x + y)$ passes the screening test.

Solution.

We simply need to calculate partial derivatives. If we denote the component functions by $(f_{1}, f_{2}, f_{3}),$ we get:

\begin{matrix} \frac{\partial f_{1}}{\partial y} = 1, \frac{\partial f_{1}}{\partial z} = 1, \frac{\partial f_{2}}{\partial z} = 1 \\ \frac{\partial f_{2}}{\partial x} = 1, \frac{\partial f_{3}}{\partial x} = 1, \frac{\partial f_{3}}{\partial y} = 1. \end{matrix}

It follows that $F$ passes the screening test. In fact, it is easy to show that it is conservative, with a potential given by $f (x, y, z) = x y + x z + y z .$

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8.

Determine whether the one-form $ω = x d x + x d y + z d z$ is exact. If it is, find a function $f$ such that $ω = d f .$

Solution.

If we write $ω = f d x + g d y + h d z,$ we see right away that

\frac{\partial f}{\partial y} = 0,

while

\frac{\partial g}{\partial x} = 1.

Thus $ω$ is not closed, and hence it cannot exact.

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9.

Consider the vector field $F (x, y) = (x y^{n}, 2 x^{2} y^{3})$ for some positive integer $n .$ Find the value of $n$ for which $F$ is conservative on $R^{2},$ and find a potential for this value of $n .$

Solution.

For $F$ to be conservative, it must pass the screening test. We write $F = (f_{1}, f_{2}),$ and calculate:

\frac{\partial f_{1}}{\partial y} = n x y^{n - 1}, \frac{\partial f_{2}}{\partial x} = 4 x y^{3} .

We see that the two partial derivatives are equal for all $(x, y)$ if and only if $n = 4,$ in which case the vector field becomes $F = (x y^{4}, 2 x^{2} y^{3}) .$ To show that it is conservative and find a potential function, we are looking for a function $f (x, y)$ such that $\nabla f = (x y^{4}, 2 x^{2} y^{3}) .$ So we must have

\frac{\partial f}{\partial x} = x y^{4}, \frac{\partial f}{\partial y} = 2 x^{2} y^{3} .

Integrating the first equation, we get

f (x, y) = \frac{1}{2} x^{2} y^{4} + g (y)

for some function $g (y) .$ Substituting in the second equation, we get

\frac{\partial f}{\partial y} = 2 x^{2} y^{3} + g^{'} (y) = 2 x^{2} y^{3},

from which we get $g^{'} (y) = 0,$ that is $g (y) = C .$ We pick $C = 0 .$ Thus $F$ is conservative for $n = 4,$ and a potential function is given by

f (x, y) = \frac{1}{2} x^{2} y^{4} .

Section 2.2 Exact one-forms and conservative vector fields

Objectives

Subsection 2.2.1 Differential of a function

Definition 2.2.1. Differential of a function.

Fact 2.2.2. Correspondence between the differential and the gradient of a function.

Example 2.2.3. The differential and gradient of a function.

Remark 2.2.4.

Subsection 2.2.2 Exact one-forms and conservative vector fields

Definition 2.2.5. Exact one-forms.

Definition 2.2.6. Conservative vector fields.

Example 2.2.7. An exact one-form and its associated conservative vector field.

Example 2.2.8. The gravitational force field is conservative.

Subsection 2.2.3 Closed one-forms in R2

Definition 2.2.9. Closed one-forms in R2.

Lemma 2.2.10. Exact one-forms in R2 are closed.

Proof.

Lemma 2.2.11. Screening test for conservative vector fields in R2.

Proof.

Example 2.2.12. Exact one-forms are closed.

Example 2.2.13. Closed one-forms are not necessarily exact.

Subsection 2.2.4 Closed one-forms in R3

Definition 2.2.14. Closed one-forms in R3.

Lemma 2.2.15. Exact one-forms in R3 are closed.

Proof.

Lemma 2.2.16. Screening test for conservative vector fields in R3.

Proof.

Remark 2.2.17.

Exercises 2.2.5 Exercises

1.

2.

3.

4.

5.

6.

7.

8.

9.

Subsection 2.2.3 Closed one-forms in $R^{2}$

Definition 2.2.9. Closed one-forms in $R^{2}$ .

Lemma 2.2.10. Exact one-forms in $R^{2}$ are closed.

Lemma 2.2.11. Screening test for conservative vector fields in $R^{2}$ .

Subsection 2.2.4 Closed one-forms in $R^{3}$

Definition 2.2.14. Closed one-forms in $R^{3}$ .

Lemma 2.2.15. Exact one-forms in $R^{3}$ are closed.

Lemma 2.2.16. Screening test for conservative vector fields in $R^{3}$ .