Mathematics refresher

Thermal physics uses a small, specific set of mathematical tools, and it uses them constantly. This page is the toolkit — not a course in calculus, but the particular corners of it that this subject leans on, with the notation this course will use for them.

It is a reference, not a narrative. Read the section the readiness map sent you to, work its examples with a pen, and come back when a later module makes you unsure.

Dimensional analysis and units¶

Every physical quantity carries dimensions, built from the SI base quantities. Four of the seven matter for us:

Everything else is a product of powers of these. Energy is $\mathsf{M}\mathsf{L}^2\mathsf{T}^{-2}$ (the joule); pressure is force per area, $\mathsf{M}\mathsf{L}^{-1}\mathsf{T}^{-2}$ (the pascal).

Reading the units of an unfamiliar constant. You never need to memorise them. From $PV = N \kB T$, with $N$ a pure count,

so $\kB$ is an energy per unit temperature — which is the whole content of the statement “temperature is a measure of energy per degree of freedom”.

Checking a candidate formula. Suppose you cannot remember whether the root-mean-square speed of a gas molecule is $\sqrt{3\kB T/m}$ or $\sqrt{3\kB T/m^2}$ or $\sqrt{3\kB T m}$. Only one of them is a speed:

The other two fail. This costs ten seconds and catches a large fraction of all algebra errors.

When two different quantities share a unit. Pressure is a force per area and energy density is an energy per volume, and they are the same thing dimensionally:

That is not a coincidence — it is the hint that pressure is going to turn out to be about energy — but it is also not an identity. A monatomic ideal gas has $P = \tfrac{2}{3}u$ and isotropic radiation has $P = \tfrac{1}{3}u$, where $u$ is the energy density. Sharing a dimension tells you two quantities can be related; it never tells you by what factor. (This is prediction 4 on the orientation page; the physics behind the factors arrives in modules 4 and 16.)

Comparing numbers honestly. Two quantities agree when their fractional difference is small:

Absolute differences mean nothing without a scale to compare them to: the energies in this course are routinely around $10^{-18}\,\mathrm{J}$, so a test asking whether two of them differ by less than $10^{-12}\,\mathrm{J}$ will answer “no difference” whatever the physics did.

Partial derivatives, and saying what is held fixed¶

For a function of several variables, the partial derivative with respect to one of them is the ordinary derivative taken with the others held constant. For $f(x,y) = x^2 y$:

The subscript is not decoration. In ordinary calculus the notation $\partial f/\partial x$ is unambiguous because the list of variables is fixed by how $f$ was written. In thermodynamics it is not: $P$, $V$, $T$ and $N$ are tied together by an equation of state, so any of them can be eliminated in favour of the others, and “the derivative with respect to $T$” means different things depending on what you chose to hold.

Mixed partials commute. For any function with continuous second derivatives,

For $f = x^2y$ both routes give $2x$. This innocuous fact turns into a machine for generating thermodynamic identities — the Maxwell relations of module 10 are nothing more than this statement applied to the thermodynamic potentials — and it is also the exactness criterion below.

The total differential¶

The total differential of $f(x,y)$ collects both first-order responses:

For $f = x^2 y$ this reads $\mathrm{d}f = 2xy\,\mathrm{d}x + x^2\,\mathrm{d}y$. Read practically: it says how much $f$ changes when you nudge $x$ and $y$ by small amounts, to first order in the nudges.

Along a path, the partial and the total derivative differ. Suppose we move along the line $y = x$. Substituting first gives $f = x^3$, hence $\mathrm{d}f/\mathrm{d}x = 3x^2$, which is 3 at $x=1$. But $(\partial f/\partial x)_y$ at $(1,1)$ is $2xy = 2$. The two numbers are different and both are correct — they answer different questions. Dividing the total differential by $\mathrm{d}x$ shows exactly where the gap comes from:

The partial derivative ignores the fact that $y$ is changing too. The total derivative does not. If prediction 5 on the module page caught you, this is the identity that resolves it.

Line integrals of a differential form¶

An expression $\omega = M(x,y)\,\mathrm{d}x + N(x,y)\,\mathrm{d}y$ is called a differential form. It is not a number; it becomes one only once you supply a path $C$ to integrate along:

To evaluate it, parametrise the path as $\big(x(t), y(t)\big)$ for $t$ from $t_0$ to $t_1$ and substitute:

Worked example. Take $\omega = y\,\mathrm{d}x$ and go from $(0,0)$ to $(1,1)$ three ways.

• Along the diagonal $y = x$: put $x = t$, $y = t$, so $\int_0^1 t \,\mathrm{d}t = \tfrac{1}{2}$.

• Across, then up: on the first leg $y = 0$, so the integrand vanishes; on the second leg $x$ is constant so $\mathrm{d}x = 0$. Total: 0.

• Up, then across: on the first leg $\mathrm{d}x = 0$; on the second $y = 1$ and $x$ runs from 0 to 1, giving $\int_0^1 1 \,\mathrm{d}x = 1$. Total: 1.

Same start, same finish, three different answers: 0, $\tfrac{1}{2}$, 1. The integral of a differential form is in general a property of the route, not of the endpoints.

Exact and inexact forms¶

Sometimes a form is the total differential of some function, and then the route stops mattering.

Apply the test to the two forms we have been using:

• $\omega_1 = y\,\mathrm{d}x + x\,\mathrm{d}y$: here $\partial M/\partial y = 1$ and $\partial N/\partial x = 1$. Equal, so $\omega_1$ is exact — and indeed $\omega_1 = \mathrm{d}(xy)$. Every route from $(0,0)$ to $(1,1)$ gives $1 \cdot 1 - 0 = 1$.

• $\omega_2 = y\,\mathrm{d}x$: here $\partial M/\partial y = 1$ but $\partial N/\partial x = 0$. Unequal, so $\omega_2$ is inexact — which is why the three routes above disagreed. There is no function whose differential it is, and writing “$\mathrm{d}(\text{something})$” for it would be a mistake.

This distinction is the reason this page exists. In module 5 the plane becomes the $P$–$V$ plane, and one of the forms you will integrate is exact while another is not. The exact one integrates to a change in internal energy, which depends only on the endpoints; the inexact one integrates to heat or to work, which depend on the route. Nothing physical is needed to see the difference — it is already here, in $y\,\mathrm{d}x$.

Quick self-test¶

If you can do these without notes, this page has nothing further for you.

1. Give the SI units of $\kB$, and of $\kB T / m$.

2. For $f(x,y) = x/y$, write $(\partial f/\partial x)_y$, $(\partial f/\partial y)_x$ and $\mathrm{d}f$.

3. Is $x\,\mathrm{d}x + y\,\mathrm{d}y$ exact? If so, give its potential.

4. Is $x\,\mathrm{d}y - y\,\mathrm{d}x$ exact? Integrate it from $(1,0)$ to $(0,1)$ along the straight line, and then along the unit circle, and compare.

5. A formula gives an answer of $2.7 \times 10^{-21}\,\mathrm{J}$ where the expected value is $2.6 \times 10^{-21}\,\mathrm{J}$. Do they agree?