Potential Functions

Remember that if $\vec{F}$ is a conservative vector field then there is some scalar function f such that $\vec{F} = \mbox{grad}f$. This function is called the scalar potential function. Can we reconstruct it simply from knowledge of the vector field?

If we write $\vec{F}(x,y,z) = F_1(x,y,z) \hat{i} + F_2(x,y,z)\hat{j} + F_3(x,y,z)\hat{k}$ then the statement that $\vec{F} = \mbox{grad}f$translates to

$$F_1 \hat{i} + F_2 \hat{j} + F_3 \hat{k} = \frac{\partial f}{... ... f}{\partial y}\hat{j} + \frac{\partial f}{\partial z} \hat{k}.$$ (6)

This is really the three equations

$$F_1 = \frac{\partial f}{\partial x}, \qquad F_2 = \frac{\partial f}{\partial y}, \qquad F_3 = \frac{\partial f}{\partial z}.$$ (7)

We can integrate the first of these equations with respect o x to get

$$f(x,y,z) = \int F_1(x,y,z) dx + g(y,z)$$ (8)

where the arbitrary function g(y,z) takes the place of the constant of integration since we have three variables and are only integrating with respect to one of them.

We can now differentiate this expression for f(x,y,z) with respect to y and compare this with F₂ to get

$$F_2 = \frac{\partial f}{\partial y} = \frac{\partial}{\parti... ...left( \int F_1(x,y,z)dx\right) + \frac{\partial g}{\partial y}.$$ (9)

Rearranging and integrating will let us find an expression for g(y,z) which will contain an arbitrary function of z, h(z). Repeating and using the third condition above, we can pin down h(z) except for a constant factor.

An example will serve to illustrate this more clearly. What is the potential function for $\vec{F}(x,y) = \frac{x}{x^2 + y^2} \hat{i} + \frac{y}{x^2 + y^2} \hat{j}$?

$$F_1 = \frac{x}{x^2 +y^2} = \frac{\partial f}{\partial x} \Ri... ...2 + y^2} dx + g(y) = \frac{1}{2}\ln \vert x^2 + y^2\vert + g(y)$$ (10)

Now we differentiate with respect to y and set this equal to F₂.

$$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\... ...ght) + \frac{\partial g}{\partial y} = F_2 = \frac{y}{x^2+ y^2}$$ (11)

A little calculus reveals that this is the same as

$$\frac{y}{x^2 + y^2} + \frac{\partial g}{\partial y} = \frac{y}{x^2 + y^2} \Rightarrow \frac{\partial g}{\partial y} = 0$$ (12)

Thus, g(y) is a constant and we can write the scalar potential for $\vec{F}$ as

$$f(x,y) = \frac{1}{2} \ln \vert x^2 + y^2\vert + c.$$ (13)

Note that sometimes it might be easier to integrate one of the other equations first and then begin taking derivatives. Only experience and pages of work will tell you. Also, it's an excellent idea to check that the gradient of the potential you find is actually the same as the vector field you started with.