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Section 6.1 Conservative Vector Fields

In Section 5.1, I talked about two motivating examples: fluid flows and fields of force. In this section, I'll focus particularly on fields of force. I'm interested in describing force (per unit mass) of gravitational attration due to a massive particle and force (per unit charge) of electronmatic attraction or repulsion due to a charged particle. I'm going to dip briefly into some physics to motivate the definitions of this section.

There are many ways to describe the force or gravity or the electromagnetic force. One of them is to describe changes in potential energy. Potential energy is a scalar that describes the state of an object with relation to another electromagnetic or massive object. State in terms of potential energy, force is understood at the tendency for object to minimize their potential energy as quickly as possible. That is, an object will move in the direction of greatest decrease. Thankfully, we already have a tool to describe the direction of greatest increase/decrease for a scalar field: the gradient. So, the fields of force I want to discuss are fields of force that can be described as gradients of some potential energy. There is a name for these vector fields.

Definition 6.1.1.

Let \(f\) be a scalar field. Then field \(\nabla f\) created from a potential energy field is called a conservative vector field. If \(F\) is a vector field with \(F = \nabla f\) then \(f\) is a scalar potential for \(F\text{.}\)

Given a general vector field \(F\text{,}\) I would like to check whether or not it is conservative. From Proposition 5.3.10, for any scalar field \(f\text{,}\) \(\nabla \times \nabla f=0\text{.}\) Therefore, all conservative fields are irrotation. Is the converse true? If it were, then checking the curl would be a very convenient way to find conservative vector fields. It turns out that the converse is often true, but I need to define a technical definition to describe the conditions when this work.

Definition 6.1.2.

Let \(U\) be an open subset of \(\RR^n\text{.}\) \(U\) is called path connected if there is a parametric curve connecting any two points \(a, b \in U\text{.}\) (Recall in the definition in Calculus III that parametric curves are always continuous.) A parametric curve is falled closed if its start and end poitns are the same. If \(U\) is path connected, it is also called simply connected if any closed path can be contracted down to a point. (Think of the path as a loop in a rope, and contraction as pulling the rope so that the loop disappears. This needs to happen in such a way that all the intermediate steps are still paths in the set \(U\)).

Naively, a simply connected set has no holes in it. A solid cylinder is simply connected. However, the hollow cylinder is not, since a loop around the cylinder can never be contracted. A disc is simply connected; a ring is not. Now I can state the test for conservative vector fields.

If \(F\) is conservative, then there exists at least one (in fact, infinitely many!) scalar potential \(f\) with \(F = \nabla f\text{.}\) Given a conservative vector fields, I'd like to be able to calculate this scalar potent. How is it done? I'll start by expanding the equation \(F = \nabla f\) in its components. I'll work just in \(\RR^3\text{,}\) though potentials can happy in any dimension.

\begin{align*} F_1 \amp = \frac{\del f}{\del x} \amp \amp F_2 \amp = \frac{\del f}{\del y} \amp \amp F_3 \amp = \frac{\del f}{\del z} \end{align*}

I can integrate each of these equations. However, I have to be careful with the ‘constants’: in each integral, since all varaible other than the variable of integration are being treated as constant, the constants of integration can be functions of both of the remaining variables.

\begin{align*} f \amp = \int F_1 dx + g_1(y,z)\\ f \amp = \int F_2 dy + g_2(x,z)\\ f \amp = \int F_3 dz + g_3(x,y) \end{align*}

Finding \(f\) amounts to finding a scalar field that fits this system. For reasonable \(F\text{,}\) this isn't too terrible. Often, the information from the other equations will help determine the functions \(g_i\text{.}\) Since there is all this choice, however, it is reasonable to wonder how many scalar potentials a conservative vector field actually has.

This situation is pretty good; from all the possibility variability in the system of differential equations, the difference in potential basically comes down to a constant (an actual, real constant!) of integration. Let me do a couple exmaples.

If \(F = (y \cos (xy), x \cos (xy), 2z)\) then \(F\) is defined on \(\RR^3\text{,}\) which is simply connected.

\begin{align*} \nabla \times F \amp = \left( \frac{\del}{\del y} 2z - \frac{\del}{\del z} x \cos (xy), \frac{\del}{\del z} y \cos (xy) - \frac{\del}{\del x} 2z, \right. \\ \amp \left. \frac{\del}{\del x} x \cos (xy) - \frac{\del}{\del y} y \cos (xy) \right)\\ \amp = (0, 0, \cos xy - xy \sin xy - \cos xy + xy \sin xy) = (0,0,0) \end{align*}

Therefore, \(F\) is a conservative field. I can try calculate its potential.

\begin{align*} f \amp = \int y \cos (xy) dx + g_1(y,z) = \sin (xy) + g_1(y,z)\\ f \amp = \int x \cos (xy) dy + g_2(x,z) = \sin (xy) + g_2(x,z)\\ f \amp = \int 2z dz + g_3(x,y) = z^2 + g_3(x,y)\\ f \amp = \sin (xy) + z^2 + c \end{align*}

I can check \(\nabla f\) to see that we recover the original field \(F\text{.}\)

If \(F = (3y^2z - 8xz^2 + 3x^2 y, 6xyz + x^3 + 2yz^2, 3xy^2 - 8x^2z + 2y^2z)\text{,}\) then \(F\) is defined on all \(\RR^3\text{.}\)

\begin{align*} \amp \nabla \times F \\ \amp = (6xy+4yz-6xy-4yz, 3y^2 -16z -3y^2 + 16z, 6yz + 3x^2 - 3yz - 3x^2) \\ \amp = (0,0,0) \end{align*}

Therefore, \(F\) is conservative. I will try to calculate its potential.

\begin{align*} f \amp = \int F_1 dx = 3xy^2 z - 4x^2z^2 + x^3 y + g_1(y,z)\\ f \amp = \int F_2 dy = 3xy^2 z + x^3 y + y^2 z^2 + g_2(x,z)\\ f \amp = \int F_3 dz = 3xy^2 z - 4x^2 z^2 + y^2 z^2 + g_3 (x,y)\\ f \amp = 3xy^2 z - 4x^2 z^2 + x^3 y + y^2 z^2 + c \end{align*}

Returning to the physics that motivated this whole section: what does this constant of integration mean? If a force is described by potential energey, then why does potential energy have an unknown or arbitrary constant?

The constant of integration reflects the fact that energy levels are arbitrarily set. Energy is relative to some arbitrary set base level. On the surface of the earth, you can set potential energy to be zero when something is on the ground, so that it can gain potential energey as it moves up. In planetary systems with the gravitional source at the origin, the convention is to set potential energy to \(-\infty\) (in the limit) at the origin, increasing to \(0\) as the distance from the origin goes to \(\infty\text{.}\) These are convenient conventions for the calculations physicists want to do with energy, but they are just convention.

This leads me to an interesting and important point: energy is a fiction. In the mathematical models of Newtonian physics (leaving the subtleties of relativity and quantumn mechanics aside for now), things like mass, charge, distance and time are real, fixed quantities. They mean something about the physical universe. Energy is not like these. Energy is fiction we made up and set to an arbitrary reference level, because it leads to very nice explanations of movement (i.e. objects accelerate to minimize their potential energy). Whether energy is positive or negative, large or small in magnitude, doesn't matter to the system. There is no intrinsic energy. What is real is difference in energey, how much potential energy changes with position, how much kinetic energey changes with velocity (relatively, of course, to some fixed reference frame).

In the initial discussion on potential energy, you may have noticed some small confusion about increase and decrease. Gradient points in the direction of greatest increase. However, in the interpretations of phyics for forces, objects move in the direction of greatest decrease. There is a sign issue here. The terminology in this section is correct: if \(F\) is a conservative with with \(F = \nabla f\text{,}\) then \(f\) is its scalar potential. However, for the actual conservative forces in physics (gravitational and electromagnetic), to account for this sign difference, the force will be the negative gradient: \(F = -\nabla f\)

Let me end with one final observation about the interpretation of conservative forces and their potential energy scalar fields. If \(-f\) is the (sign-corrected) scalar potential for \(F\text{,}\) then \(F\) points in the direction of greatest decrease of \(f\text{.}\) The integral curves of \(F\) are parametric curves whose tangents are all these direction of greatest decrease. They are the direction of movement of an object starting at rest under the action of the force. In this sense, a potential energy \(f\) cause movement (on object starting at rest) along the integral curves of its gradient.