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Integrals along a Curve

Consider the type of integral that everyone learns initially:

$$E(b) - E(a) = \int_a^b f(x) dx$$ (14-1)

The equation implies that $f$ is integrable and

$$dE = f dx = \frac{dE}{dx} dx$$ (14-2)

so that the integral can be written in the following way:

$$E(b) - E(a) = \int_{a}^{b} dE$$ (14-3)

where $a$ and $b$ represent ``points'' on some line where $E$ is to be evaluated.

Of course, there is no reason to restrict integration to a straight line--the generalization is the integration along a curve (or a path) $\vec{x}(t) = (x_1(t) , x_2(t), \ldots , x_n(t))$ .

$$E(b) - E(a) = \int_{\vec{x}(a)}^{\vec{x}(b)} \vec{f}(\vec{x}) \cd... ...) dt = \int_a^b \ensuremath{\nabla}E \cdot \frac{d\vec{x}}{dt} dt = \int_a^b dE$$ (14-4)

This last set of equations assumes that the gradient exists-i.e., there is some function $E$ that has the gradient $\nabla E = \vec{f}$ .

Path-Independence and Path-Integration

If the function being integrated along a simply-connected path (Eq. 14-4) is a gradient of some scalar potential, then the path between two integration points does not need to be specified: the integral is independent of path. It also follows that for closed paths, the integral of the gradient of a scalar potential is zero.¹ A simply-connected path is one that does not self-intersect or can be shrunk to a point without leaving its domain.

There are familiar examples from classical thermodynamics of simple one-component fluids that satisfy this property:

$$\oint dU = \oint \ensuremath{\nabla}_{\vec{\mathcal S}} U \cdot d \vec{\mathcal S} = 0 \oint dS = \oint \ensuremath{\nabla}_{\vec{\mathcal S}} S \cdot d \vec{\mathcal S} = 0 \oint dG = \oint \ensuremath{\nabla}_{\vec{\mathcal S}} G \cdot d \vec{\mathcal S} = 0$$ (14-5)

$$\oint dP = \oint \ensuremath{\nabla}_{\vec{\mathcal S}} P \cdot d \vec{\mathcal S} = 0 \oint dT = \oint \ensuremath{\nabla}_{\vec{\mathcal S}} T \cdot d \vec{\mathcal S} = 0 \oint dV = \oint \ensuremath{\nabla}_{\vec{\mathcal S}} V \cdot d \vec{\mathcal S} = 0$$ (14-6)

Where $\vec{\mathcal S}$ is any other set of variables that sufficiently describe the equilibrium state of the system (i.e, $U(S,V)$ , $U(S,P)$ , $U(T,V)$ , $U(T,P)$ for $U$ describing a simple one-component fluid).

The relation curl    grad $f = \nabla \times \nabla f = 0$ provides method for testing whether some general $\vec{F}(\vec{x})$ is independent of path. If

$$\vec{0} = \nabla \times \vec{F}$$ (14-7)

or equivalently,

$$0 = \ensuremath{\frac{\partial{F_j}}{\partial{x_i}}} - \ensuremath{\frac{\partial{F_i}}{\partial{x_j}}}$$ (14-8)

for all variable pairs $x_i$ , $x_j$ , then $\vec{F}(\vec{x})$ is independent of path. These are the Maxwell relations of classical thermodynamics.

MATHEMATICA$^{\text{\scriptsize {\textregistered }}}$ Example

(notebook Lecture-14)

(html Lecture-14)

(xml+mathml Lecture-14)

Path Dependence, Curl, and Curl=0 subspaces This example will show that the choice of path matters for a vector-valued function that does not have vanishing curl and that it doesn't matter when integrating a function with vanishing curl.

Path-dependent/Non-conserving Field

1. Verify that the function $\vec{v}(\vec{x}) = x y z ( \hat{i} + \hat{k} + \hat{z} )$ does not have vanishing curl.
2. Integrate $\vec{v}$ along a path that is wrapped around a cylinder of radius $R$ , (e.g., $(x(t), y(t), z(t)) = (R \cos t , R \sin t, A P_{2 \pi}(t))$ , where $P_{2 \pi}(t=0) = P_{2 \pi}(t=2 \pi)$ )
3. Calculate the integral specifically for $P_{2 \pi}(t) = \cos t$ , $P_{2 \pi}(t) = \sin t$ , $P_{2 \pi}(t) = t (t - 2 \pi)$ , and $P_{2 \pi}(t) = \cos N t$ .

Path-independent/Conservative Field

1. Verify that, for the function $\vec{w}(\vec{x}) = e^{x y z} ( y z \hat{i} + z x \hat{k} + x y \hat{z} )$ , $\nabla \times \vec{w} = 0$ . In fact, $\vec{w} = \nabla e^{x y z}$ .
2. Integrate $\vec{w}$ along the same cylindrical-type path as above and see that the integral always vanishes--it is path-independent.

Path independent on a Subspace

1. The vector function $\vec{v}(\vec{x}) = (x^2 + y^2 - R^2)\hat{z}$ only vanishes on the cylinder or radius $R$ .
2. It is easy to find $\vec{w}$ such that $\vec{w} = \nabla \times v$ :
   $$\vec{w} = \frac{1}{2} \left( y R^2 \left[1 - x^2 - \frac{y^2}{3}\right] \hat{x}+ -x R^2 \left[1 - y^2 - \frac{x^2}{3}\right] \hat{y} \right)$$
   In fact, because we could add any vector function that has vanishing curl to $\vec{w}$ there are an infinite number of $\vec{w}$ such that $\vec{w} = \nabla \times v$ .
3. Therefore, if we integrate $\vec{w}$ along a path that is restricted to the cylinder it should be path independent.
4. Using the same methods as above, we find that the integral on the cylinder will be independent of $P$ --the vector function $\vec{w}$ is independent of path as long as the path remains on the cylinder.