Forest UCM Energy CurlFcons

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A force with a curl of zero is a conservative force.

Thus taking the curl of the force is an easier way to test for conservative forces rather than calculating the work and inspecting to see if it only depends on the endpoints of the motion.

Definition of curl

We have seen that the gradient operator is defined in cartesian coordinates as

[math]\vec \nabla = \frac{\partial}{\partial x} \hat i +\frac{\partial}{\partial y} \hat j +\frac{\partial}{\partial z} \hat k[/math]

can be used to find the functional form of a conservative force given its potential energy

Stokes Theorem

closed line integral

A closed integral is a mathematical expressionwhich may be used to calculate the work done by a force when an object moves to some distant point and then returns to its point of origin

[math]\oint \vec F \cdot d \vec r = \int_{r_1}^{r_2} \vec F \cdot d \vec r + \int_{r_2}^{r_1} \vec F \cdot d \vec r[/math]
[math] = W + (-W) =0 [/math]

The above is true if you have a conservative force where the work done only depends on the endpoints.

Stokes theorem

Stokes theorem relates the line integral of a vector field over its closed boundary

[math]\oint \vec F \cdot d \vec r[/math]
(the circle around the integral indicates a closed path, you go to some point and then back)

to the surface integral of the curl of the vector field over a surface

[math]\iint (\vec \nabla \times \vec F) \cdot d \vec A[/math]

Stokes theorem equates the two integrals

[math]\oint \vec F \cdot d \vec r= \iint (\vec \nabla \times \vec F) \cdot d \vec A[/math]

Thus if you have a conservative force then

[math]\oint \vec F \cdot d \vec r= 0 = \iint (\vec \nabla \times \vec F) \cdot d \vec A[/math]


[math] \iint (\vec \nabla \times \vec F) \cdot d \vec A=0[/math]

then one way for this integral (sum) to be zero is if you add up something that is zero everywhere

[math] (\vec \nabla \times \vec F) =0[/math]
A second way to get zero is

If we have a conservative force such that a potential may be defined where

[math]F = -\vec \nabla U[/math]
[math]\vec \nabla \times \vec F = \vec \nabla \times (-\vec \nabla U) [/math]
[math]\vec \nabla \times \vec \nabla = 0 [/math] The cross product of the same vector is zero since it is parallel to itself.

A test for path independence of a force

We now have a test to determine if the work done by a force is path independent ( ie it is a conservative force)


[math] \vec \nabla \times \vec F=0[/math]

Then [math]\vec F[/math] is a conservative force

or if you are given a potential for the force such that

[math] F = -\vec \nabla U[/math]

Then you can be confident that the force is conservative.

conservative Force test examples

Force that depends on r

Consider the coulomb force

[math]\vec F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^3} \vec r \equiv \frac{C}{r^3} \vec r[/math]

Test if

[math]\vec \nabla \times \vec F =0[/math]

to determine if the force is conservative.

[math]\vec \nabla \times \vec F = \left ( \begin{matrix} \hat i & \hat j & \hat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{C}{r^3} x &\frac{C}{r^3} y &\frac{C}{r^3} z \end{matrix} \right )[/math]

Working on the [math]\hat i[/math] component

[math]\left . \left ( \vec \nabla \times \vec F \right ) \right |_x = \left ( \frac{\partial}{\partial y} \frac{C}{r^3} z- \frac{\partial}{\partial z} \frac{C}{r^3} y\right )[/math]
[math]= C \left ( z \frac{\partial}{\partial y} r^{-3}- y \frac{\partial}{\partial z} r^{-3} \right )[/math]

[math]\frac{\partial}{\partial y} r^{-3} = \frac{\partial}{\partial y} (x^2 + y^2 + z^2)^{-3/2}[/math]
[math]= \frac{-3}{2}(x^2 + y^2 + z^2)^{-5/2} (2y) = -\frac{3y}{r^5} [/math]
[math]\left . \left ( \vec \nabla \times \vec F \right ) \right |_x = -3C \left ( -\frac{y}{r^5} z- -\frac{z}{r^5} y\right ) =0 [/math]

You can show the same for the other components thereby proving that for the coulomb force

[math]\vec \nabla \times \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^3} \vec r =0[/math]

The potential

The potential energy for this force may be defined according to the work integral

[math]U(r) = - W = - \int \vec F \cdot d \vec r[/math]

Aside on uniform electric field

The first part of the Lorentz force indicates that

[math]\vec F = q \vec E[/math]

if q is near a point charge then

[math]\vec E = \frac{1}{4 \pi \epsilon_0} \frac{ q_2}{r^3} \vec r \equiv \frac{C}{r^3} \vec r[/math]

and we can use the coulomb force argument to show that the force is conservative.

If you construct a capacitor with a uniform electric field between its plates, the Force is constant as a result of the summation of all the Forces between individual charges but the fundamental force responsible for the uniform electric field is itself a function of position.


Maxwells equations have

[math]\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}[/math]

If the magnetic field is not time dependent then you clearly have a curl of zero for the force (electrostatics)