Path Integrals

A path integral in $\real^2$ is the integral of a scalar function $f(x, y)$ along a path $\vec{\sigma}$ in the x-y-plane.

Represent the curve in parametrized form: $\sigma(t) = (x(t), y(t))$ , or $x = x(t)$ , $y = y(t)$ for $a \le t \le b$ . Then

$$\int_{\vec{\sigma}} f\,ds = \int_a^b f(\vec{\sigma}(t)) \|\vec{\sigma}\,'(t)\|\,dt.$$

Path integrals in higher dimensions are defined in similar fashion.

Heuristically, the curve is divided into little pieces. A small piece of the curve has length $ds$ , where

$$ds = \sqrt{dx^2 + dy^2} = \sqrt{\left(\der x t\right)^2 + \left(\der y t \right)^2}\,dt.$$

Above the small piece of the curve, I build a rectangle using f to obtain the height (for example, by plugging a point on the small piece of the curve into f). A careful definition would use Riemann sums, as usual.

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It is like building a rectangle sum along a curve, rather than the x-axis as you do with ordinary single-variable integrals.

Example. Compute $\int_C (3 x + 2 y)\,ds$ , where C is the segment from $(1, 3)$ to $(2, -1)$ .

The segment from $(1, 3)$ to $(2, -1)$ is

$$(x, y) = (1 - t) \cdot (1, 3) + t \cdot (2, -1) = (1 + t, 3 - 4 t).$$

Thus,

$$x = 1 + t, \quad y = 3 - 4 t.$$

Hence,

$$ds = \sqrt{x'(t)^2 + y'(t)^2}\,dt = \sqrt{1 + 16}\,dt = \sqrt{17}\,dt.$$

In addition,

$$3 x + 2 y = 3 (1 + t) + 2 (3 - 4 t) = 9 - 5 t.$$

Hence,

$$\int_C (3 x + 2 y)\,ds = \int_0^1 (9 - 5 t) \sqrt{17}\,dt = \sqrt{17} \left[9 t - \dfrac{5}{2} t^2\right]_0^1 = \dfrac{13 \sqrt{17}}{2} = 26.80018 \ldots.\quad\halmos$$

Example. Compute $\displaystyle \int_{\vec{\sigma}} f\,ds$ , where $f(x, y) = x^2 - y^2$ and $\vec{\sigma}(t) = (\cos t, \sin t)$ , $0 \le t \le \dfrac{\pi}{4}$ .

First, I'll find $ds$ :

$$\vec{\sigma}\,'(t) = (-\sin t, \cos t), \quad\hbox{so}\quad |\vec{\sigma}\,'(t)| = \sqrt{(\sin t)^2 + (\cos t)^2} = 1.$$

Therefore, $ds = 1 \cdot dt = dt$ .

Next, I convert f to a function of t:

$$f(t) = (\cos t)^2 - (\sin t)^2 = \cos 2 t.$$

Therefore,

$$\int_{\vec{\sigma}} f\,ds = \int_0^{\pi / 4} \cos 2 t\,dt = \left[\dfrac{1}{2} \sin 2 t\right]_0^{\pi/4} = \dfrac{1}{2}.\quad\halmos$$

Path integrals work in similar fashion in $\real^3$ .

Example. Compute $\displaystyle \int_C (x^2 + y + 2 z)\,ds$ , where C is the segment from $(1, 2, 1)$ to $(2, 0, 1)$ .

The segment from $(1, 2, 1)$ to $(2, 0, 1)$ is

$$(x, y, z) = (1 - t) \cdot (1, 2, 1) + t \cdot (2, 0, 1) = (1 + t, 2 - 2 t, 1).$$

Hence,

$$ds = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}\,dt = \sqrt{1 + 4 + 0}\,dt = \sqrt{5}\,dt.$$

In addition,

$$x^2 + y + 2 z = (1 + t)^2 + (2 - 2 t) + 2 = t^2 + 5.$$

Therefore,

$$\int_C (x^2 + y + 2 z)\,ds = \int_0^1 (t^2 + 5) \cdot \sqrt{5}\,dt = \sqrt{5} \left[\dfrac{1}{3} t^3 + 5 t\right]_0^1 = \dfrac{16 \sqrt{5}}{3} = 11.92569 \ldots.\quad\halmos$$

Example. A wire is bent into the shape of the helix

$$\vec{\sigma}(t) = (\cos t, \sin t, t), \quad 0 \le t \le 4\pi.$$

The density is proportional to the square of the distance from the origin. Find the mass of the wire.

In this case, the curve is 3-dimensional, so I can't picture it as a "fence" as I did with the 2-dimensional curves. However, the computation is essentially the same.

The density is $\delta = k(x^2 + y^2 + z^2)$ , where k is a constant. The mass is just $\displaystyle \int_{\vec{\sigma}} \delta\,ds$ .

The velocity is

$$\vec{\sigma}\,'(t) = (-\sin t, \cos t, 1), \quad\hbox{so}\quad |\vec{\sigma}\,'(t)| = \sqrt{(\sin t)^2 + (\cos t)^2 + 1} = \sqrt{2}.$$

Hence, $ds = \sqrt{2}\,dt$ .

Write $\delta$ in terms of t:

$$\delta = k\left[(\cos t)^2 + (\sin t)^2 + 1\right] = k(1 + t^2).$$

The mass is

$$\int_0^{4 \pi} k (1 + t^2) \cdot \sqrt{2}\,dt = k \sqrt{2} \left[t + \dfrac{1}{3} t^3\right]_0^{4 \pi} = k \sqrt{2} \left(4 \pi + \dfrac{64}{3} \pi^3\right).\quad\halmos$$

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