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The two-state paramagnet from counting to measurement: the tanh law for energy and magnetization, Curie's law at high temperature, saturation, and the Schottky heat capacity peak.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find the magnetization and energy of a paramagnet from $\mu B/kT$, use Curie's law to predict changes, tell the Curie regime from saturation, and derive the tanh law from the entropy.
You can count the microstates of a two-state paramagnet, $\binom{N}{N_\uparrow}$, you know temperature as $1/T = \partial S/\partial U$, and you know that the paramagnet's energy is $U = \mu B(N_{\downarrow} - N_{\uparrow})$. This lesson combines them into predictions a magnetometer can test, and meets the hyperbolic tangent, $\tanh x = (e^{x} - e^{-x})/(e^{x} + e^{-x})$.
| Term | What it means |
|---|---|
| Paramagnet | A material whose magnetic dipoles align with an applied field and do not interact with each other. |
| Magnetization | $M = \mu(N_{\uparrow} - N_{\downarrow})$, the net magnetic moment; per unit volume, in A/m. |
| Bohr magneton | $\mu_B = 9.27 \times 10^{-24}$ J/T $= 5.788 \times 10^{-5}$ eV/T, the magnetic moment of an electron's spin. |
| Curie's law | At high temperature, $M \propto B/T$. |
| Saturation | The low-temperature limit in which nearly every dipole is aligned, $M \to N\mu$. |
| Hyperbolic tangent | $\tanh x = \dfrac{e^{x} - e^{-x}}{e^{x} + e^{-x}}$: rises from $0$, is close to $x$ for small $x$, and levels off at $1$. |
| Schottky anomaly | The peak in a two-state system's heat capacity near $kT \approx \mu B$. |
Take $N$ dipoles, each with energy $-\mu B$ along the field or $+\mu B$ against it. Its entropy, from $\Omega = \binom{N}{N_{\uparrow}}$ with Stirling's approximation, and its energy $U = \mu B(N - 2N_{\uparrow})$ give, through $1/T = \partial S/\partial U$,
$$U = -N\mu B\tanh\frac{\mu B}{kT}, \qquad M = N\mu\tanh\frac{\mu B}{kT}.$$
Equivalently, each dipole is along the field with probability $e^{x}/(e^{x} + e^{-x})$, where
$$x = \frac{\mu B}{kT}.$$
Everything depends on that one ratio: the magnetic energy that pulls a dipole into line, against the thermal energy that knocks it out.
$$M \approx \frac{N\mu^2B}{kT}.$$
The magnetization is proportional to the field and inversely proportional to the temperature. This is Curie's law, found experimentally by Pierre Curie in 1895. - Low temperature, $x \gg 1$. Then $\tanh x \to 1$ and $M \to N\mu$: every dipole aligned, the magnetization saturated.
For electron spins, $\mu = \mu_B = 5.788 \times 10^{-5}$ eV/T. At room temperature $kT = 0.0259$ eV, so even in a $1$ T field $x = 0.0022$: the paramagnet is deep in the Curie regime, and only about two spins in a thousand point along the field on balance. Saturation needs temperatures of a few kelvin or fields of many tesla.
Another way: picture
Picture a field of wheat on a breezy day with a steady wind from one side. The steady wind is the magnetic field, bending every stalk the same way; the gusts are thermal motion, whipping stalks in every direction. In gusty weather — high temperature — the field of wheat leans only slightly with the wind, and the lean is proportional to the steady wind's strength. On a still day — low temperature — every stalk bends fully with the wind.
Another way: steps
Checks. The net alignment must lie between $0$ and $1$. Raising the field or lowering the temperature must raise it. In the Curie regime the answer should equal $x$ itself to within a few percent. The energy must be negative for positive temperature, since more dipoles point along the field than against it. And the fraction along the field must be at least one half: at any positive temperature the lower-energy state is the more populated.
With $N_{\uparrow}$ dipoles along the field, Stirling's approximation gives the entropy
$$\frac{S}{k} = N\ln N - N_{\uparrow}\ln N_{\uparrow} - N_{\downarrow}\ln N_{\downarrow}.$$
Since $U = \mu B(N - 2N_{\uparrow})$, adding energy $dU$ means $dN_{\uparrow} = -dU/(2\mu B)$, and
$$\frac{1}{T} = \frac{\partial S}{\partial U} = \frac{k}{2\mu B}\ln\frac{N_{\uparrow}}{N_{\downarrow}}.$$
Solve for the ratio: $N_{\uparrow}/N_{\downarrow} = e^{2\mu B/kT}$. So the populations are in the ratio of $e^{x}$ to $e^{-x}$, and with $N_{\uparrow} + N_{\downarrow} = N$,
$$N_{\uparrow} - N_{\downarrow} = N\frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} = N\tanh x.$$
Notice the ratio of populations, $e^{2\mu B/kT}$, is the ratio of $e^{-E/kT}$ for the two states. That is the Boltzmann factor, which the fifth unit derives in general: here it has emerged from nothing but counting and the definition of temperature.
Differentiating $U = -N\mu B\tanh(\mu B/kT)$ with respect to $T$ gives the heat capacity
$$C = Nk\,\frac{x^2}{\cosh^2 x}, \qquad x = \frac{\mu B}{kT}.$$
It is small at both ends. At low temperature the dipoles are all aligned and a little extra heat cannot flip any of them, because a flip costs $2\mu B$, far more than $kT$. At high temperature they are already half and half, and extra heat hardly changes that. In between, near $kT \approx 0.8\mu B$, the heat capacity peaks at about $0.44Nk$.
This bump, called a Schottky anomaly, appears in any system with two closely spaced energy levels, and physicists use its position to measure the spacing. Paramagnetic salts cooled toward a kelvin show it clearly, sitting on top of the much smaller lattice heat capacity, and it is the reason these salts are used to reach temperatures below a kelvin by adiabatic demagnetization.
The model in this lesson treats every dipole as independent: each one feels only the applied field and the thermal jostling of its surroundings. That is a good description of a dilute paramagnetic salt, such as a compound of iron or gadolinium ions spread out among many nonmagnetic atoms, where neighboring dipoles are too far apart to influence each other. It is why Curie's law was first confirmed in salts like these.
In iron metal itself the dipoles are close and interact strongly, each tending to align its neighbors. Below a critical temperature, $1043$ K for iron, that interaction wins over thermal jostling even with no applied field, and the material magnetizes itself: it is a ferromagnet. Above that temperature it behaves as a paramagnet, but with the Curie-Weiss law, $M \propto B/(T - T_c)$, which diverges as the temperature falls toward $T_c$. The Ising model, taken up at the end of the course, is the simplest model of that interaction, and the two-state paramagnet of this lesson is its non-interacting limit. Everything in this lesson is what that richer model reduces to when the dipoles are far enough apart to ignore each other.
The protons in body tissue are a two-state paramagnet, with magnetic moment $\mu = 1.41 \times 10^{-26}$ J/T, about $660$ times smaller than an electron's. In a $1.5$ T clinical scanner at body temperature, $310$ K,
$$x = \frac{\mu B}{kT} = \frac{1.41 \times 10^{-26} \times 1.5}{1.38 \times 10^{-23} \times 310} = 4.9 \times 10^{-6}.$$
So the net alignment is about five protons per million. That seems hopeless, but a cubic millimeter of tissue holds about $7 \times 10^{19}$ protons, and five millionths of that is $3.5 \times 10^{14}$ excess aligned protons — enough to produce a measurable signal when a radio pulse tips them and they precess.
Because the alignment is in the Curie regime, it is proportional to $B$. A $3$ T scanner doubles the imbalance and the signal, and the $7$ T research scanners approved by the FDA in 2017 give nearly five times the signal of a $1.5$ T machine. That is the whole reason hospitals invest in larger, more expensive magnets: higher field buys signal, and signal buys resolution or shorter scans.
A paramagnetic salt can be used as a refrigerator to reach temperatures far below one kelvin. Start with the salt in contact with a helium bath at about $1$ K and apply a strong field, say $1$ T. The spins align ($x \approx 0.67$ for electron spins, and more for the larger moments of rare-earth ions), lowering the salt's entropy; the heat released flows into the bath.
Now isolate the salt and slowly reduce the field. With no heat exchange the process is isentropic, and the paramagnet's entropy depends only on $x = \mu B/kT$. Keeping the entropy fixed means keeping $B/T$ fixed: reduce the field a hundredfold, to $0.01$ T, and the temperature falls a hundredfold, from $1$ K to $0.01$ K. The limit comes from the small internal fields the spins exert on each other, which act like a residual field that cannot be switched off.
Adiabatic demagnetization, proposed by Debye and Giauque in 1926, took physicists below one kelvin for the first time. Its descendant, nuclear demagnetization using the much smaller moments of copper nuclei, has reached temperatures of microkelvin and below, and space telescopes such as the Hitomi X-ray observatory carried demagnetization refrigerators to cool their detectors to $50$ millikelvin.
It is natural to picture a magnet pulling every dipole in a paramagnetic sample into line. At room temperature that picture is badly wrong. The magnetic energy of an electron spin in a $1$ T field — a strong field, twenty thousand times Earth's — is $5.8 \times 10^{-5}$ eV, while the thermal energy is $0.026$ eV. The thermal jostling wins by a factor of about four hundred and fifty, and the net alignment is a couple of spins per thousand.
The opposite confusion is to expect the magnetization to keep growing in proportion to the field forever. Curie's law is the small-$x$ form of the tanh; as the field rises or the temperature falls, the magnetization bends over and approaches a ceiling, $N\mu$, which no field can exceed because there are no more dipoles to align. A paramagnet is also not a ferromagnet: its dipoles do not align each other, so remove the field and the magnetization vanishes at once.
Electron spins in a $1.0$ T field at $300$ K. Find the magnetic energy of one spin.
$\mu_B B = 5.788 \times 10^{-5} \times 1.0 = 5.79 \times 10^{-5}\ \text{eV}$
The Bohr magneton in electronvolts per tesla.
Find the thermal energy.
$kT = 8.617 \times 10^{-5} \times 300 = 0.0259\ \text{eV}$
Boltzmann's constant in electronvolts per kelvin.
Form the ratio.
$x = \dfrac{5.79 \times 10^{-5}}{0.0259} = 0.00224$
Tiny: the thermal energy is nearly five hundred times the magnetic.
Find the net alignment.
$\tanh 0.00224 \approx 0.00224$
For small $x$ the tanh equals its argument.
Interpret the result.
$\dfrac{N_{\uparrow} - N_{\downarrow}}{N} \approx 2.2 \times 10^{-3}$
About two spins in a thousand, on balance: a room-temperature paramagnet is almost unmagnetized.
Cool the salt to $1.0$ K in the same $1.0$ T field. Find the thermal energy.
$kT = 8.617 \times 10^{-5}\ \text{eV}$
Three hundred times smaller than at room temperature.
Form the ratio.
$x = \dfrac{5.788 \times 10^{-5}}{8.617 \times 10^{-5}} = 0.672$
Now the magnetic and thermal energies are comparable.
Evaluate the tanh.
$\tanh 0.672 = \dfrac{e^{0.672} - e^{-0.672}}{e^{0.672} + e^{-0.672}} = \dfrac{1.958 - 0.511}{1.958 + 0.511} = 0.586$
Evaluate the exponentials first.
Find the fraction along the field.
$\dfrac{N_{\uparrow}}{N} = \dfrac{1 + 0.586}{2} = 0.793$
Four spins in five point along the field.
Compare with the Curie-law estimate.
$x = 0.672 \quad \text{vs} \quad \tanh x = 0.586$
Curie's law would overestimate the alignment by about fifteen percent here, because the tanh is starting to level off.
Find the energy per spin.
$\dfrac{U}{N} = -\mu_B B\tanh x = -5.788 \times 10^{-5} \times 0.586 = -3.39 \times 10^{-5}\ \text{eV}$
Negative: more spins point along the field.
Write the magnetization.
$M = N\mu\tanh x, \qquad x = \dfrac{\mu B}{kT}$
Derived from counting and $1/T = \partial S/\partial U$.
Expand the tanh for small $x$.
$\tanh x = x - \dfrac{x^3}{3} + \cdots \approx x$
The Taylor series; the cubic term is negligible when $x \ll 1$.
Substitute the small-x form.
$M \approx N\mu \cdot \dfrac{\mu B}{kT} = \dfrac{N\mu^2B}{kT}$
The leading term.
Name the result.
$M = C_{\text{Curie}}\,\dfrac{B}{T}, \qquad C_{\text{Curie}} = \dfrac{N\mu^2}{k}$
Curie's law, with a constant that depends only on the number and moment of the dipoles.
Use it to predict a change.
$M_1 = 4.0\ \text{A/m at } 1.0\ \text{T}, 300\ \text{K} \quad\Rightarrow\quad M_2 = 4.0 \times \dfrac{2.0}{1.0} \times \dfrac{300}{150} = 16\ \text{A/m at } 2.0\ \text{T}, 150\ \text{K}$
Double the field and halve the temperature: four times the magnetization.
Check that the regime still holds.
$x = \dfrac{5.788 \times 10^{-5} \times 2.0}{8.617 \times 10^{-5} \times 150} = 0.009 \ll 1$
Still far into the Curie regime, so the prediction is sound.
Say where it would fail.
$x \gtrsim 0.3$
Once $\mu B$ is a sizable fraction of $kT$ the cubic term matters and the magnetization falls below Curie's law, heading for saturation.
Write the fraction along the field.
$\dfrac{N_{\uparrow}}{N} = \dfrac{1 + \tanh x}{2}$
From $N_{\uparrow} - N_{\downarrow} = N\tanh x$.
Evaluate the tanh.
$\tanh 1 = \dfrac{e - e^{-1}}{e + e^{-1}} = \dfrac{2.718 - 0.368}{2.718 + 0.368} = 0.762$
Exponentials first.
Find the fraction.
A paramagnetic salt at room temperature is well within the Curie regime. Its absolute temperature is divided by $3$ and the field is multiplied by $2$, and it stays in the Curie regime. By what factor does its magnetization change?
Complete the worked solution: electron spins ($\mu_B = 5.788 \times 10^{-5}$ eV/T) sit in $3$ T at $2$ K. What fraction of full magnetization do they reach? Use $k = 8.617 \times 10^{-5}$ eV/K.
Multiply the moment by the field.
$\mu_B B = 5.788 \times 3 \times 10^{-5} =$ a $\times 10^{-5}$ eV
The magnetic energy that pulls a spin into line.
Multiply Boltzmann's constant by the temperature.
$kT = 8.617 \times 2 \times 10^{-5} =$ c $\times 10^{-5}$ eV
The thermal energy that jostles it out of line.
Take the tanh of their ratio.
$\dfrac{M}{N\mu} = \tanh\dfrac{\mu_B B}{kT} =$ h
The paramagnet's magnetization depends on this ratio alone.
A paramagnet of $N$ spins sits in a field of $5$ T. Match each regime to its behavior.
| a small magnetization proportional to $B/T$ | nearly every dipole along the field | half the dipoles each way, zero energy | more dipoles against the field than along it | |
|---|---|---|---|---|
| $kT$ much larger than $\mu B$ | ||||
| $kT$ much smaller than $\mu B$ | ||||
| infinite temperature | ||||
| negative temperature |
For a paramagnet in a $4$ T field, fill in the net alignment $\tanh x$ and the fraction of dipoles along the field for each value of $x = \mu B/kT$, to four decimal places.
| $\tanh x$ | fraction along the field | |
|---|---|---|
| $x = 0.1$ | ||
| $x = 0.5$ | ||
| $x = 1$ | ||
| $x = 2$ |
A sample of a paramagnetic salt has magnetization $2$ A/m in a field of $2$ T at $300$ K. Assuming Curie's law holds, what is its magnetization in a field of $4$ T at $150$ K?
Answer: A/m
A dilute paramagnetic salt has electron spins with magnetic moment $\mu_B = 5.788 \times 10^{-5}$ eV/T. It sits in a field of $5$ T at $10$ K. What fraction of the maximum magnetization does it reach? Use $k = 8.617 \times 10^{-5}$ eV/K.
Answer:
An MRI scanner images the hydrogen nuclei in tissue at $310$ K. A proton's magnetic moment is $\mu = 1.41 \times 10^{-26}$ J/T. In a $1.5$ T scanner, what is the net excess of protons aligned with the field, per million protons? Use $k = 1.38 \times 10^{-23}$ J/K.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For a paramagnet in a $2$ T field, fill in the net alignment $\tanh x$ and the fraction of dipoles along the field for each value of $x = \mu B/kT$, to four decimal places.
| $\tanh x$ | fraction along the field | |
|---|---|---|
| $x = 0.1$ | ||
| $x = 0.5$ | ||
| $x = 1$ | ||
| $x = 2$ |
You can predict how strongly a paramagnet magnetizes at any field and temperature. Explain to someone why a one-tesla magnet barely magnetizes a paramagnetic salt at room temperature.
15. Your turn: a paramagnet is at $x = \mu B/kT = 1$. What fraction of its dipoles point along the field?, step 3
$\dfrac{1 + 0.762}{2} = 0.881$
Nearly nine in ten.