Skip to content
Math Analysis 2

Curves, line integrals and conservative fields

Parametric curves and length, the integral of a function and the work of a field, conservative fields and potential, irrotational fields, the wire counterexample and finding the potential.

4 min read 0 comments

A curve in space is the path of a moving point. Along a curve you can integrate two different things: a scalar function (to compute a length or a mass) or a vector field (to compute work). This lesson introduces both integrals and arrives at conservative fields and potentials, which in Physics 2 will become the electric potential.

Parametric curves

A curve is a function \mathbf r: [a, b]\to\mathbb R^3, \mathbf r(t) = (x(t), y(t), z(t)). The vector \mathbf r'(t) is tangent to the curve and, if t is time, it is the velocity. The curve is regular if \mathbf r' is continuous and never zero.

Examples. The circle \mathbf r(t) = (R\cos t, R\sin t), t\in[0, 2\pi]. The helix \mathbf r(t) = (\cos t, \sin t, ct), the shape of a spring and a screw thread. The graph of y = f(x) is the curve \mathbf r(t) = (t, f(t)).

The length of a curve

Approximating the curve with a polygonal line and taking the limit, the length is the integral of the speed:

    \[ L = \int_a^b |\mathbf r'(t)|\,dt \]

Example. One turn of the helix (\cos t, \sin t, ct): |\mathbf r'| = \sqrt{\sin^2 t + \cos^2 t + c^2} = \sqrt{1 + c^2}, so L = 2\pi\sqrt{1 + c^2}. It is the hypotenuse of a right triangle with legs 2\pi (the circumference) and 2\pi c (the pitch): unrolling the cylinder, the helix becomes a straight line.

The length does not depend on how the curve is parametrised: travelling faster or slower does not change the road.

Line integral of a function

For a scalar function f (for example the linear density of a wire):

    \[ \int_\gamma f\,ds = \int_a^b f(\mathbf r(t))\,|\mathbf r'(t)|\,dt \]

where ds = |\mathbf r'|\,dt is the element of length. With f = 1 you get the length back; with f a density you get the mass of the wire. It does not depend on the direction of travel.

Line integral of a vector field: work

If \mathbf F is a force, the work done along the curve \gamma is

    \[ \int_\gamma\mathbf F\cdot d\mathbf r = \int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt \]

Only the component of \mathbf F tangent to the curve counts. Unlike the previous case, it changes sign if the direction is reversed.

A curve from A to B in a vector field with the field vectors and the tangent vectors
The field \mathbf F (grey and red arrows) and the curve \gamma from A to B with tangent vectors d\mathbf r (green). The line integral adds up, point by point, the dot product of field and displacement: where they agree the contribution is positive, where they oppose it is negative.

Example. \mathbf F = (-y, x) along the unit circle anticlockwise: \mathbf r' = (-\sin t, \cos t), \mathbf F(\mathbf r) = (-\sin t, \cos t), dot product 1, integral 2\pi. The field “turns” the same way as the curve and does positive work all the way round.

Conservative fields

A field \mathbf F is conservative if there is a function U, the potential, such that \mathbf F = \nabla U. (Physics often uses the opposite sign, \mathbf F = -\nabla V, with V the potential energy.)

Theorem. If \mathbf F = \nabla U, then for every curve from A to B

    \[ \int_\gamma\mathbf F\cdot d\mathbf r = U(B) - U(A) \]

Proof

By the chain rule \frac{d}{dt}U(\mathbf r(t)) = \nabla U(\mathbf r(t))\cdot\mathbf r'(t). So the integral is \int_a^b\frac{d}{dt}U(\mathbf r(t))\,dt = U(\mathbf r(b)) - U(\mathbf r(a)), by the fundamental theorem of calculus.

Three equivalent properties follow (in a connected domain):

  1. \mathbf F has a potential;
  2. the integral between two points does not depend on the path;
  3. the integral around every closed curve is zero.

The field (-y, x) of the previous example is not conservative: around a closed loop it gave 2\pi, not zero.

How to recognise a conservative field

If \mathbf F = (P, Q) = \nabla U, then P = U_x, Q = U_y, and by Schwarz’s theorem P_y = U_{xy} = U_{yx} = Q_x. The condition

    \[ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \]

(an irrotational field; in three dimensions \nabla\times\mathbf F = \mathbf 0) is therefore necessary. It is also sufficient if the domain is simply connected, that is with no “holes” a closed curve could go around.

The classic counterexample. \mathbf F = \left(\frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}\right) on the plane without the origin is irrotational, but its integral round the unit circle is 2\pi. The hole at the origin breaks the equivalence. It is, up to constants, the magnetic field of a straight wire carrying a current: its integral round a closed curve counts how many times the curve goes round the wire, and that is Ampère’s law.

How to find the potential

Example. \mathbf F = (2xy + 1, x^2 + 3y^2). Check: P_y = 2x = Q_x, and the domain is the whole plane. Integrating P in x: U = x^2 y + x + h(y). Differentiating in y and comparing with Q: x^2 + h'(y) = x^2 + 3y^2, so h = y^3. The potential is U = x^2y + x + y^3, and the work from (0, 0) to (1, 2) is U(1, 2) - U(0, 0) = 2 + 1 + 8 = 11, whatever the path.

Related articles

Subscribe
Notify of
guest
0 Comments
Oldest
Newest Most Voted
0
Would love your thoughts, please comment.x
()
x