Solved thermodynamics exercise

Integrating a differential form along a path

Exercise 6 · Lesson 4 — The First Law of Thermodynamics

differential form
exact differential
Schwarz criterion
line integral

Draft version — this exercise is still under review.

Statement

Consider the two differential forms

\omega_1 = y\,\mathrm{d}x + x\,\mathrm{d}y, \qquad \omega_2 = y\,\mathrm{d}x,

and three paths joining the origin $O(0,0)$ to the point $M(1,1)$ : path $\gamma_1$ , consisting of the horizontal segment $O\to(1,0)$ followed by the vertical segment $(1,0)\to M$ ; path $\gamma_2$ , consisting of the vertical segment $O\to(0,1)$ followed by the horizontal segment $(0,1)\to M$ ; and path $\gamma_3$ , the straight-line segment joining $O$ directly to $M$ .

Show that $\omega_1$ is exact by finding a function $f$ such that $\omega_1 = \mathrm{d}f$ .
Calculate $\int_{\gamma_1}\omega_1$ and $\int_{\gamma_2}\omega_1$ by parametrizing each segment. Recover the result without calculation.
Apply Schwarz's criterion to $\omega_2$ . What can be concluded?
Calculate $\int_{\gamma_1}\omega_2$ , $\int_{\gamma_2}\omega_2$ , and $\int_{\gamma_3}\omega_2$ . Comment on the results.

Hint

On a horizontal segment

\mathrm{d}y = 0

, and on a vertical segment

\mathrm{d}x = 0

: each integral reduces to an ordinary single integral.

Detailed solution

Solution

Question 1. The function

f(x,y) = xy

works, since

\partial f/\partial x = y

and

\partial f/\partial y = x

Question 2. For a curve parametrized by $t\mapsto\bigl(x(t),y(t)\bigr)$ , we have

\int_\gamma\omega_1 = \int \left[y(t)x'(t)+x(t)y'(t)\right]\,\mathrm{d}t.

Path $\gamma_1$ is the union of two segments. On the first, $x(t)=t$ and $y(t)=0$ , with $t\in[0,1]$ ; hence $\mathrm{d}x=\mathrm{d}t$ and $\mathrm{d}y=0$ . On the second, $x(t)=1$ and $y(t)=t$ , again with $t\in[0,1]$ ; hence $\mathrm{d}x=0$ and $\mathrm{d}y=\mathrm{d}t$ . Therefore,

\begin{aligned} I_1 = \int_{\gamma_1}\omega_1 &= \int_0^1\left(0\times 1+t\times 0\right)\,\mathrm{d}t + \int_0^1\left(t\times 0+1\times 1\right)\,\mathrm{d}t \\ &= 0+\int_0^1\mathrm{d}t = 1. \end{aligned}

For $\gamma_2$ , the first segment is parametrized by $x(t)=0$ , $y(t)=t$ , and the second by $x(t)=t$ , $y(t)=1$ , with $t\in[0,1]$ in both cases. We therefore obtain

\begin{aligned} I_2 = \int_{\gamma_2}\omega_1 &= \int_0^1\left(t\times 0+0\times 1\right)\,\mathrm{d}t + \int_0^1\left(1\times 1+t\times 0\right)\,\mathrm{d}t \\ &= 0+\int_0^1\mathrm{d}t = 1. \end{aligned}

The two values agree. This could have been predicted without calculation: because $\omega_1=\mathrm{d}f$ with $f(x,y)=xy$ , its integral depends only on the endpoints,

\int_\gamma\omega_1=f(M)-f(O)=f(1,1)-f(0,0)=1.

Question 3. Write $\omega_2=A(x,y)\,\mathrm{d}x+B(x,y)\,\mathrm{d}y$ . Here $A(x,y)=y$ and $B(x,y)=0$ . If the form were exact, Schwarz's criterion would give

\frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}.

However, $\partial A/\partial y=1$ , whereas $\partial B/\partial x=0$ . The mixed derivatives differ, so $\omega_2$ is not exact. Its integral may therefore depend on the path followed between $O$ and $M$ , as the following calculation confirms.

Question 4. Since $\omega_2=y\,\mathrm{d}x$ , only a change in $x$ at a nonzero value of $y$ can contribute to the integral. Using the previous parametrizations, on $\gamma_1$ we find

\begin{aligned} J_1 =\int_{\gamma_1}\omega_2 &=\int_0^1 0\times 1\,\mathrm{d}t +\int_0^1 t\times 0\,\mathrm{d}t \\ &=0. \end{aligned}

On $\gamma_2$ , the first segment satisfies $x(t)=0$ , so $\mathrm{d}x=0$ ; on the second, $x(t)=t$ , $y(t)=1$ , and $\mathrm{d}x=\mathrm{d}t$ . Thus

\begin{aligned} J_2 =\int_{\gamma_2}\omega_2 &=\int_0^1 t\times 0\,\mathrm{d}t +\int_0^1 1\times 1\,\mathrm{d}t \\ &=1. \end{aligned}

Finally, the diagonal segment $\gamma_3$ is explicitly parametrized by

x(t)=t,\qquad y(t)=t,\qquad t\in[0,1],

so $\mathrm{d}x=\mathrm{d}t$ and $\mathrm{d}y=\mathrm{d}t$ . Hence

J_3 =\int_{\gamma_3}\omega_2 =\int_0^1 y(t)x'(t)\,\mathrm{d}t =\int_0^1 t\,\mathrm{d}t =\frac12.

The three paths have the same endpoints but yield three different values: $J_1=0$ , $J_2=1$ , and $J_3=1/2$ . This is precisely the path dependence characteristic of a non-exact differential form.