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Adiabatic demagnetization 2026-10-18

A cooling method that reduces an applied magnetic field while suppressing heat transfer to or from the material. For a suitable material exhibiting paramagnetism, an approximately reversible field reduction can lower its temperature at approximately constant entropy.

The preparation matters as much as the field reduction. First establish thermal equilibrium with a cold heat bath, then raise the magnetic field while maintaining the bath temperature: this is magnetization in an isothermal process. The bath accepts heat and the material loses magnetic entropy. Next open a heat switch, thermally isolating the working material, and lower the magnetic field. If the sweep maintains internal thermal equilibrium with negligible entropy production, this second leg is an isentropic process. An adiabatic process alone does not guarantee constant entropy.

The attraction is a controllable energy scale: changing the magnetic field changes the separation of magnetic energy levels, even across a thermally insulating gap. This lets an already cold solid become colder without needing a colder contacting heat bath. A practical refrigerator also needs a load, a period of absorbing its heat, and a reset that rejects accumulated entropy to the bath. A single field sweep has finite cooling capacity.

Figure 1.
Functional cold stage and heat switch sequence for adiabatic demagnetization
. Original functional schematic. The controlled thermal link closes during magnetization in an isothermal process and opens before the magnetic field falls. The sensing coil is a readout, not a thermal connection. This is a functional diagram rather than a reconstruction of historical geometry. **Source**. **Details**.

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Adiabatic expansion Created 2026-10-05 Updated 2026-10-14

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Adiabatic expansion increases the accessible volume without heat transfer. In a loaded piston expansion of the fixed-composition ideal gas considered in adiabatic process, positive outward work comes from internal energy, so the final temperature decreases. This provides a cooling mechanism without removing heat during the stroke. In the Carnot cycle, reversible adiabatic expansion connects two different temperatures between its isothermal processes.^[ref]

For the same constant-

γ

reversible ideal-gas adiabat, put

b = V_{2} / V_{1} > 1

. The scaling and first law of thermodynamics give

\frac{T _{2}}{T _{1}} = b^{1 - γ}, \frac{p _{2}}{p _{1}} = b^{- γ}, W_{out} = C_{V} (T_{1} - T_{2}) = \frac{p _{1} V _{1}}{γ - 1} (1 - b^{1 - γ}) > 0.

(1)

With the figure's stipulated

γ = 1.4

p_{1} V_{1} = 1 kJ

and

b = 2

, the final ratios are

T_{2} / T_{1} ≃ 0.75786

and

p_{2} / p_{1} ≃ 0.37893

. Outward work is

0.60535 kJ

and

Δ U = - 0.60535 kJ

. Reversing this particular path from its cooled endpoint would require the same work input; the figure's separate compression starts at the original state, so its input is different. Endpoint entropy supplies another independent check:

Δ S = C_{V} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}} = 0.

(2)

The two logarithm terms cancel using

C_{V} (γ - 1) = n R

, confirming the reversible path is an isentropic process. ^[ref]

PASCO's expansion demonstration first compresses trapped air and waits for thermal equilibrium before the rapid expansion. Its manual warns that friction-warmed cylinder regions contact the gas during expansion, compromising quantitative comparison. ^[ref] The hold separates cooling caused by expansion from simply undoing the earlier hot compression. As proposed controls, observe the temperature dip and subsequent warming, compare rapid and slow strokes from matched initial states, and assess wall heat transfer and probe lag before fitting a reversible model.

Cooling is not a universal consequence of adiabatic expansion: free expansion of an ideal gas has no outward work and returns to the same equilibrium temperature. Its entropy increases, so the reversible power laws do not describe it. Reuse free expansion, Joule's free-expansion experiment and temperature sensitivity of a free-expansion experiment for that apparatus, its energy budget and the real-gas qualifications. The useful distinction is the load: expansion cools this ideal gas when it must spend internal energy doing work.

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Adiabatic exponents of a monatomic gas-radiation mixture 2026-10-06

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For fixed-composition monatomic ideal gas plus equilibrium radiation pressure, specific internal energy is

u = 3 R T /2 + a_{r} T^{4} / ρ

. An isentropic process satisfies

d u = P d ρ / ρ^{2}

, giving

Γ_{3} - 1 = (8 - 6 β) / (24 - 21 β)

and

Γ_{2} = \frac{32 - 24 β - 3 β ^{2}}{24 - 18 β - 3 β ^{2}}, \nabla_{ad} = \frac{8 - 6 β}{32 - 24 β - 3 β ^{2}} .

(1)

The stellar gas-pressure fraction must be allowed to change along the adiabat. The pure-radiation and pure-gas limits give

Γ_{1} = Γ_{2} = 4/3

and

5/3

, respectively. Partial ionization or different gas heat capacities change these formulas.

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Adiabatic process Created 2026-09-24 Updated 2026-10-20

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An adiabatic process has no heat transfer across the chosen boundary throughout the process:

δ Q = 0

, rather than merely zero net heat after incoming and outgoing transfers cancel. It is a condition on energy transfer, whereas an isothermal process holds temperature constant. A sealed cylinder can approach an adiabatic process through insulation or through motion short compared with the time needed for appreciable heat transfer. Its temperature can nevertheless change because work crosses the boundary.^[ref]

Take the trapped gas as a closed system, with fixed amount of substance and composition and no net change in bulk kinetic energy or potential energy between the endpoints. Define

Q_{in} > 0

as heat entering and

W_{out} > 0

as work delivered to the surroundings. The first law of thermodynamics gives

Δ U = Q_{in} - W_{out}, Q_{in} = 0 ⟹ Δ U = - W_{out} .

(1)

Here

U

is internal energy. An adiabatic compression normally supplies work, so

W_{out} < 0

; an adiabatic expansion against a resisting load delivers work, so

W_{out} > 0

. An adiabatic process need not preserve internal energy and need not make the closed system an isolated system. ^[ref]

For a reversible thermodynamic process,

d S = δ Q_{rev} / T

at positive absolute temperature

T

. A reversible adiabatic process therefore has

d S = 0

: it is an isentropic process. Irreversible adiabatic processes can instead generate entropy; free expansion is the existing counterexample, with its calculation in entropy change in free expansion. Conversely, constant entropy alone does not certify an adiabatic process or reversibility: outward heat can export the entropy generated internally. For a closed system the macroscopic entropy balance is

Δ S = \int \frac{δ Q _{in}}{T _{b}} + S_{gen}, S_{gen} \geq 0,

(2)

where

T_{b}

is the absolute boundary temperature where each heat transfer occurs and

S_{gen}

is internal entropy production. With no heat transfer, this reduces to

Δ S = S_{gen}

; reversibility makes

S_{gen} = 0

. ^[ref]

A motivating experiment measures the gas temperature directly during compression and expansion, instead of inferring it from how warm a cylinder feels. PASCO's authored TD-8565 manual describes a sealed piston cylinder, a fine nickel-wire thermometer, an absolute pressure sensor and a position readout for volume. A computer records all three channels together. The wire's electrical resistance changes with temperature; its small size permits a rapid response. ^[ref]

Use ordinary trapped air and the apparatus's designed travel stops; mount the base securely, close the valve during a stroke, and record the actual volume limits. John Hart's university demonstration uses a large transparent cylinder with a low-heat capacity thermocouple, a pressure sensor, a travel gauge and a safety stop. It explicitly keeps the release valve closed during the demonstration. The basic observable is a rise in temperature on compression and a fall on loaded expansion, alongside the corresponding pressure and volume changes. Hart discusses wall thermal conduction, sensor response and leakage as reasons real records differ from the reversible model. ^[ref]

For a proposed rapid-versus-slow comparison, let the air and walls return to the same initial thermal equilibrium before each run, then compare the same starting and finishing volume values with different stroke durations. Read temperature, absolute pressure and position continuously, including the hold after motion stops. A rapid compression gives a temporary temperature rise; holding the smaller volume lets heat leave and the temperature approach the surroundings again. Slow compression in good thermal contact allows that heat transfer during the stroke and approaches an isothermal compression. Loaded expansion reverses the signs: the rapid stroke cools, while heat entering during slow motion reduces the cooling. PASCO explicitly supports rapid nearly adiabatic processes and slow isothermal processes; the matched-run controls here are a proposed extension, not a reported set of measurements. ^[ref]

Insulation suppresses thermal conduction; it does not prove that heat transfer vanishes. As proposed controls, repeat the matched strokes with and without a removable insulating jacket, record room and wall temperatures, measure fixed-position drift, and check valve closure and seal leakage. Calibrate the thermometer and position readout independently, including any connected dead volume; account for the heat capacity of the probe rather than treating it as a passive observer. Hart explicitly identifies the small air sample relative to the probe and walls, sensor inertia and seal leakage as limitations. A fast display cannot eliminate those physical effects. ^[ref]

There are two distinct timing requirements. Let

τ_{eq}

describe internal equilibration,

τ_{probe}

the thermometer response,

t_{move}

the stroke duration and

τ_{heat}

the effective heat transfer time constant. A useful sufficient window for a resolved, nearly reversible adiabatic process is

τ_{eq}, τ_{probe} ≪ t_{move} ≪ τ_{heat},

(3)

with negligible friction. The left inequality permits a quasistatic process and a usable readout; the right limits heat leakage. A violent stroke may satisfy only the right inequality. PASCO reports a roughly

30

–

50 ms

lag of its temperature channel behind the other channels, so apparent simultaneous samples require care. If no such window exists, analyze equilibrium endpoints with the first law of thermodynamics, and regard the path as approximate. ^[ref]

Figure 1.
Direct temperature readout and a heat transfer time constant in an adiabatic process experiment
. Original apparatus schematic and analytical fixed-volume relaxation model, not measured records. After a stroke and internal equilibration, the stipulated Newton's law of cooling model gives $[T (t) - T_{b}] /∣ T (0) - T_{b} ∣ = \pm e^{- t / τ_{heat}}$ . The positive and negative curves illustrate the holds after compression and loaded expansion. The drawing is a generic sensor arrangement, not a reproduction of a commercial device; it distinguishes the small probe from the walls and their possible heat transfer. **Details**.

The diagram's model is deliberately a hold after the stroke, when boundary work is zero. For constant whole-sample heat capacity at constant volume

C_{V}

and effective thermal conductance

K > 0

, it assumes

C_{V} \frac{d T}{d t} = - K (T - T_{b}), τ_{heat} = \frac{C _{V}}{K}, T (t) - T_{b} = [T (0) - T_{b}] e^{- t / τ_{heat}} .

(4)

This is Newton's law of cooling with one uniform gas temperature and a fixed surrounding temperature; actual wall and probe dynamics can require more than one time constant. At

t / τ_{heat} = 0.1

, the fraction of the initial temperature anomaly erased is

1 - e^{- 0.1} ≃ 0.0952

; at one time constant it is about

0.6321

. These are analytical fractions, not experimental tolerances or evidence that the preceding stroke was reversible. Measuring the hold helps estimate heat leakage independently of the motion.

The usefulness of an adiabatic process is that it separates mechanical work from heat transfer. Its reversible limit connects temperature changes, mechanical stiffness and entropy in one model: Rüchardt experiment and Clément-Desormes experiment already use that connection to measure heat capacity ratio. The contrast with free expansion explains the deeper point: suppressing heat does not determine how much work is exchanged or how much entropy is produced.

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Carnot cycle Created 2026-09-29 Updated 2026-10-14

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A Carnot cycle realizes the reversible bound in Carnot's theorem. Its four steps for an ideal gas working substance are:

Reversible isothermal expansion at $T_{h}$ , absorbing $Q_{h}$ and increasing entropy by $Δ S$ .
Reversible adiabatic expansion, doing work while the temperature falls from $T_{h}$ to $T_{c}$ and entropy stays fixed.
Reversible isothermal compression at $T_{c}$ , rejecting $Q_{c}$ and decreasing entropy by $Δ S$ .
Reversible adiabatic compression, requiring work to restore $T_{h}$ at fixed entropy.

The temperature versus entropy plot in the Carnot's theorem figure is therefore a rectangle traversed clockwise. Because

δ Q_{rev} = T d S

, its width is

Δ S = Q_{h} / T_{h}

. For the illustrated engine,

Δ S = 5/3 J K^{- 1}

: the upper step absorbs

600 \times 5/3 = 1000 J

, the lower step rejects

300 \times 5/3 = 500 J

, and the enclosed area is the

500 J

net work. Geometry displays the thermal efficiency without solving the gas motion. The two vertical steps are isentropic processes, not constant-temperature steps. ^[ref] ^[ref]

Reversible heat transfer is a limit of vanishing temperature differences; ordinary finite-rate transfer introduces entropy production. The Carnot cycle is a performance benchmark rather than a guarantee of useful power. Running it backward gives the ideal operation of a refrigerator or heat pump, with work supplied to move heat toward the hotter thermal reservoir. ^[ref]

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Compressibility 2026-10-14

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Compressibility measures how readily a sample's volume decreases when its pressure increases. For a simple compressible system with fixed particle number

N

and composition, specify the constraint

X

κ_{X} = - \frac{1}{V} (\frac{\partial V}{\partial P})_{X, N}, \frac{d V}{V} = - κ_{X} d P (X, N fixed) .

(1)

Here

P

denotes pressure, also written

p

elsewhere. The minus sign makes compressibility positive on a mechanically stable equilibrium branch. Its units are inverse pressure,

P a^{- 1}

; its reciprocal

K_{X} = 1/ κ_{X}

is the constrained bulk modulus, measured in pascal units. Holding temperature fixed gives isothermal compressibility; holding entropy fixed gives isentropic compressibility

κ_{S}

along an isentropic process. The latter describes a reversible adiabatic compression, since an irreversible adiabatic process can increase entropy. ^[ref]

For an ideal gas, the ideal gas law

P V = N k_{B} T

and the reversible ideal-gas adiabat give, at the same equilibrium state,

κ_{T} = \frac{1}{P}, κ_{S} = \frac{1}{γ P}, γ = \frac{C _{P}}{C _{V}} > 1.

(2)

Here

C_{P}

and

C_{V}

are the whole sample's heat capacity at constant pressure and heat capacity at constant volume, and

γ

is the heat capacity ratio. Constant

γ

is required for the finite-path formula

P V^{γ} = constant

. Reversible adiabatic compression raises the temperature, making the pressure rise more steeply than in an isothermal compression. Thus keeping entropy fixed makes the gas harder to compress locally. For a classical monatomic gas at

100 kPa

γ = 5/3

, so

κ_{T} = 1 0^{- 5} P a^{- 1}

and

κ_{S} = 6 \times 1 0^{- 6} P a^{- 1}

; the corresponding values of the bulk modulus are

100 kPa

and

166.67 kPa

For a smooth, stable single thermodynamic phase at positive temperature, Maxwell relations also imply

\frac{κ _{T}}{κ _{S}} = \frac{C _{P}}{C _{V}}, κ_{T} - κ_{S} = \frac{T V α ^{2}}{C _{P}},

(3)

where

α

is the coefficient of thermal expansion. The ideal gas result is one example of this broader link between mechanical and thermal responses. The equality

κ_{T} = κ_{S}

is possible when

α = 0

and the other responses remain finite. These formulas compare equilibrium partial derivatives, rather than arbitrary fast or dissipative compression histories. ^[ref]

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Entropy change in free expansion 2026-10-14

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For the insulated gas-only free expansion of an ideal gas, the equilibrium endpoints are

(T, V_{1})

and

(T, V_{2})

with

V_{2} > V_{1}

. Because entropy is a state function, calculate their difference along a hypothetical reversible thermodynamic process: an isothermal expansion using a frictionless piston and a thermal reservoir at

T

. This comparison process receives heat and delivers work, even though the actual free expansion does neither. Its reversible heat increment is

δ Q_{rev} = p d V = n RT d V / V

, so

Δ S = \int_{1}^{2} \frac{δ Q _{rev}}{T} = n R \int_{V_{1}}^{V_{2}} \frac{d V}{V} = n R ln \frac{V _{2}}{V _{1}} > 0.

(1)

Here

R

is the molar gas constant, and the logarithm acts on a dimensionless ratio. The reversible reference gives the correct endpoint entropy difference because entropy is a state function; putting the actual

Q = 0

into

Δ S = Q / T

would misuse a relation requiring reversible heat transfer. For the actual isolated gas, all this increase is entropy production. Thus an adiabatic process can increase entropy and need not be an isentropic process. ^[ref]

Figure 1.
The reversible reference calculates entropy without supplying a path for the actual free expansion
. Toy ideal gas example with $n = 1 mol$ , $T = 300 K$ , $V_{1} = 0.025 m^{3}$ and stipulated $R = 8.314 J mo l^{- 1} K^{- 1}$ . The plotted curve belongs only to the hypothetical reversible isothermal expansion; the actual release has the same equilibrium endpoints but no equilibrium curve between them. The table refers to the gas, with work positive outward. No points are experimental data. **Details**.

In the figure's volume doubling, the ideal gas law gives

p_{1} = 99.768 kPa

and

p_{2} = 49.884 kPa

, while

Δ S = R ln 2 ≃ 5.763 J K^{- 1}

. The reference process receives and delivers

n RT ln 2 ≃ 1728.85 J

as heat and work; its thermal reservoir loses the same entropy that the gas gains. The actual free expansion has

Q_{in} = W_{out} = 0

and produces that entropy internally. Reversibly compressing the gas back at

300 K

would require

1728.85 J

of work input and rejection of equal heat. Restoring the gas alone therefore changes the surroundings: the actual release is irreversible.

For a real gas the final temperature may differ, so the ideal-gas logarithm formula cannot simply be copied. Evaluate

S (T_{2}, V_{2}) - S (T_{1}, V_{1})

using that gas's equilibrium properties or a suitable reversible comparison path. If the walls and water exchange heat with it, the Second law of thermodynamics constrains the total entropy of the isolated apparatus; the gas alone is then not an isolated system.

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Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2017 / 3 / a / Solution 2026-10-18

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Work in the rest frame of a thin planar shock wave, with normal upstream and downstream velocities

u_{1}, u_{2} > 0

. Assume a steady flow, negligible gravitational change across its thickness, no radiative energy loss, and negligible viscous stress and heat flux at the two remote faces of a control volume. Dissipation within the shock is permitted and is essential to the physical discontinuity.

Integrating mass conservation through that volume gives equal mass flux per unit area:

ρ_{1} u_{1} = ρ_{2} u_{2} = j .

(1)

The dimensions of

j

are

kg m^{- 2} s^{- 1}

. The normal momentum entering minus that leaving must balance the pressure forces. Equivalently, the conserved normal momentum flux is

p + ρ u^{2}

, giving

p_{1} + ρ_{1} u_{1}^{2} = p_{2} + ρ_{2} u_{2}^{2} .

(2)

For conservation of energy, matter advects kinetic and internal energy while pressure performs work at rate

p u

per unit area. Writing

ϵ

for specific internal energy, the energy flux is

ρ u (u^{2} /2 + ϵ) + p u

. With specific enthalpy

h = ϵ + p / ρ

, its conservation gives

\frac{u _{1}^{2}}{2} + h_{1} = \frac{u _{2}^{2}}{2} + h_{2} .

(3)

The division by

j

is valid for a genuine through-flow shock. Together these are the three Rankine-Hugoniot conditions. The conserved energy flux has units

W m^{- 2}

, and both momentum fluxes have units of pressure.

For a calorically perfect ideal gas,

ϵ = p / [(γ - 1) ρ]

and

h = γ p / [(γ - 1) ρ]

. The three jump conditions can be combined explicitly. Put

r = ρ_{2} / ρ_{1}

, so

u_{2} = u_{1} / r

and

p_{2} = p_{1} + ρ_{1} u_{1}^{2} (1 - r^{- 1}) .

(4)

Substitute these into the enthalpy equation and use

M_{1}^{2} = ρ_{1} u_{1}^{2} / (γ p_{1})

, with the upstream adiabatic sound speed. Besides the trivial no-shock root

r = 1

, the physical compression root is

r = \frac{( γ + 1 ) M _{1}^{2}}{( γ - 1 ) M _{1}^{2} + 2}, \frac{p _{2}}{p _{1}} = \frac{2 γ M _{1}^{2} - ( γ - 1 )}{γ + 1} .

(5)

For unchanged composition,

T_{2} / T_{1} = (p_{2} / p_{1}) / r

. A compressive shock has

M_{1} > 1

and increases the specific entropy. Thus an adiabatic shock conserves total energy but is not an isentropic process.

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Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2017 / 4 / a / Solution 2026-10-18

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Assume an inviscid barotropic fluid acted on by a conservative body force

- \nabla ψ

. Define its barotropic enthalpy function by

H (ρ) = \int^{ρ} \frac{p ^{'} ( s )}{s} d s, \nabla H = \frac{\nabla p}{ρ} .

(1)

The Euler equations for an inviscid fluid are then

\partial_{t} v + (v \cdot \nabla) v = - \nabla (H + ψ)

. Using the vorticity

ω = \nabla \times v

and the vector identity

(v \cdot \nabla) v = \nabla (v^{2} /2) - v \times ω

(2)

gives

\partial_{t} v + \nabla B = v \times ω

, where

B = v^{2} /2 + H + ψ

For steady flow, dotting this equation with

v

yields

v \cdot \nabla B = 0

. Thus Bernoulli's equation is

\frac{v ^{2}}{2} + \int^{ρ} \frac{p ^{'} ( s )}{s} d s + ψ = constant along each streamline .

(3)

The constant need not agree between different streamlines. If the flow is also irrotational, the vector equation gives

\nabla B = 0

, so one common constant applies throughout a connected flow region.

For an isentropic process with

p = K ρ^{γ}

H = γ p / [(γ - 1) ρ]

up to an additive constant, equal to the thermodynamic specific enthalpy. For an externally maintained isothermal gas,

H = a^{2} ln (ρ / ρ_{ref})

; this pressure potential is not the constant thermodynamic enthalpy of a fixed-temperature ideal gas.

If steady flow is not assumed, a useful additional case is unsteady potential flow

v = \nabla ϕ

. The same vector equation integrates spatially to

\partial_{t} ϕ + \frac{v ^{2}}{2} + H + ψ = C (t) .

(4)

General unsteady barotropic flow does not possess the steady streamline constant without further conditions.

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Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 317 / 3 / Solution Created 2026-10-03 Updated 2026-10-06

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Let

S

denote specific entropy, reserving

s

from the first question for envelope thickness. The stellar adiabatic exponents are defined at fixed composition by

Γ_{1} = (\frac{\partial lo g P}{\partial lo g ρ})_{S}, \frac{Γ _{2} - 1}{Γ _{2}} = (\frac{\partial lo g T}{\partial lo g P})_{S} .

(1)

It is useful to introduce

Γ_{3} - 1 = (\partial lo g T / \partial lo g ρ)_{S}

, so

(Γ_{2} - 1) / Γ_{2} = (Γ_{3} - 1) / Γ_{1}

Take the stellar gas to be a fully ionized, nonrelativistic monatomic ideal gas, with fixed specific gas constant

R

. Its mixture with thermal radiation has

P = R ρT + \frac{1}{3} a_{r} T^{4}, u = \frac{3}{2} R T + \frac{a _{r} T ^{4}}{ρ} .

(2)

Here

u

is specific internal energy, and the stellar gas-pressure fraction is

β = R ρT / P

. The assumption of monatomic gas fixes its heat capacity; the perfect-gas pressure law alone would not determine it.

At fixed

T

and fixed

ρ

, respectively, the pressure derivatives are

χ_{ρ} = (\frac{\partial l o g P}{\partial l o g ρ})_{T} = β, χ_{T} = (\frac{\partial l o g P}{\partial l o g T})_{ρ} = 4 - 3 β .

(3)

For an isentropic process, the first law of thermodynamics gives

d u = P d ρ / ρ^{2}

. Differentiate the internal energy rather than artificially holding

β

fixed during the perturbation:

(\frac{3}{2} R + \frac{4 a _{r} T ^{3}}{ρ}) d T = \frac{P + a _{r} T ^{4}}{ρ ^{2}} d ρ .

(4)

The parenthesis is

(P / ρT) [(3/2) β + 12 (1 - β)]

, giving

Γ_{3} - 1 = \frac{4 - 3 β}{12 - ( 21/2 ) β} = \frac{8 - 6 β}{24 - 21 β} .

(5)

Since

d lo g P = χ_{ρ} d lo g ρ + χ_{T} d lo g T

Γ_{1} = β + (4 - 3 β) (Γ_{3} - 1) = \frac{32 - 24 β - 3 β ^{2}}{24 - 21 β} .

(6)

Using the relation between the exponents gives the adiabatic exponents of a monatomic gas-radiation mixture

Γ_{2} = \frac{32 - 24 β - 3 β ^{2}}{24 - 18 β - 3 β ^{2}}, \nabla_{ad} = \frac{Γ _{2} - 1}{Γ _{2}} = \frac{8 - 6 β}{32 - 24 β - 3 β ^{2}} .

(7)

The requested values, including the adiabatic temperature gradient, are

β 0 1/2 1 Γ_{1} 4/3 77/54 5/3 Γ_{2} 4/3 77/57 5/3 \nabla_{ad} 1/4 20/77 2/5 .

(8)

The pure-radiation and pure-gas rows are understood as limits of the mixture.

For uniform composition, the Schwarzschild criterion for stellar convective instability is

\nabla \equiv \frac{d lo g T}{d lo g P} > \nabla_{ad} = \frac{Γ _{2} - 1}{Γ _{2}} .

(9)

To test whether a radiative configuration becomes unstable, use its required stellar radiative temperature gradient

\nabla_{rad}

for

\nabla

. Since

d P / d r < 0

, the equivalent radial condition is

- \frac{d T}{d r} > \frac{Γ _{2} - 1}{Γ _{2}} \frac{T}{P} (- \frac{d P}{d r}) .

(10)

An outward-displaced parcel then cools less than its new surroundings, remains less dense at the same pressure, and is further accelerated outward.

The requested Eddington-model convective-core mass fraction follows from the usual global constant-

β

closure. At the outer radiative surface, put

m = M

L_{r} = L

, and

P_{rad} = (1 - β) P

. Dividing the radiation-pressure gradient by the stellar hydrostatic equation gives

\frac{d P _{rad}}{d P} = \frac{κ L _{r}}{4 π c G m}, L = \frac{4 π c GM}{κ} (1 - β) = (1 - β) L_{Edd} .

(11)

There is no energy generation outside the convective core, so use

L_{r} = L

there. The local radiative-gradient formula is

\nabla_{rad} (m) = \frac{3 κ L P}{16 π a _{r} c G m T ^{4}} = \frac{κ L}{16 π c G m ( 1 - β )} ≃ \frac{M}{4 m} .

(12)

The same representative

β

has been used in the global luminosity closure and at the boundary. Setting this gradient equal to

\nabla_{ad}

m = M_{c}

produces

\frac{M _{c}}{M} = \frac{1}{4 \nabla _{ad}} = \frac{Γ _{2}}{4 ( Γ _{2} - 1 )} = \frac{32 - 24 β - 3 β ^{2}}{4 ( 8 - 6 β )} .

(13)

This is

1

77/80

, and

5/8

for

β = 0, 1/2, 1

, respectively, with

β = 0

a marginal radiation-dominated limit.

There is a consistency qualification to this last model estimate. It cannot be an exact stellar solution if all the printed assumptions are enforced pointwise. An exactly uniform nonzero radiation fraction would give

d P_{rad} / d P = 1 - β

at every radiative point. The exact equations would then require

\frac{L _{r}}{m} = \frac{4 π c G}{κ} (1 - β) .

(14)

L_{r}

is constant throughout an envelope with positive mass density,

m

increases with radius, so this equality cannot hold there. Equivalently uniform

β

fixes the actual gradient to

1/4

, whereas

\nabla_{ad} > 1/4

for

β > 0

. The boxed core fraction is the intended Eddington closure and boundary estimate, which relaxes exact constancy of

β

in the detailed envelope. The constant-beta radiative-envelope obstruction identifies the missing approximation; it is not legitimate to silently assert exact compatibility.

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 346 / 1 / a / Solution Created 2026-10-03 Updated 2026-10-05

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Write

x

for a comoving coordinate,

u

for the peculiar velocity,

H = \overset{a}{˙} / a

for the Hubble parameter, and

δ = δ ρ / \overset{ρ}{ˉ}

for the density contrast. For a homogeneous self-gravitating fluid, the linearized continuity equation, Euler equations for an inviscid fluid, and cosmological Poisson equation are

\dot{δ} = - \frac{1}{a} \nabla \cdot u, \dot{u} + H u = - \frac{c _{s}^{2}}{a} \nabla δ - \frac{1}{a} \nabla ϕ, \nabla^{2} ϕ = 4 π G a^{2} \overset{ρ}{ˉ} δ .

(1)

Here

c_{s}

is the physical adiabatic sound speed, gradients refer to comoving coordinates, and the homogeneous background force has already been subtracted. Differentiating the first equation and eliminating

u

and

ϕ

gives

\ddot{δ} + 2 H \dot{δ} - \frac{c _{s}^{2}}{a ^{2}} \nabla^{2} δ - 4 π G \overset{ρ}{ˉ} δ = 0.

(2)

A Fourier mode with comoving wavenumber

k

therefore obeys

\ddot{δ}_{k} + 2 H \dot{δ}_{k} + (\frac{c _{s}^{2} k ^{2}}{a ^{2}} - 4 π G \overset{ρ}{ˉ}) δ_{k} = 0.

(3)

The Jeans wavenumber occurs where the coefficient in parentheses vanishes. Since a comoving wavelength is

2 π / k

, the result is

k_{J} = \frac{a 4 π G ρ ˉ}{c _{s}}, λ_{J}^{com} = \frac{c _{s}}{a} \frac{π}{G ρ ˉ} .

(4)

Below this comoving Jeans length, pressure restores a displaced fluid element: there are acoustic waves rather than growing Jeans instability. In a static background they oscillate; expansion changes their amplitude and frequency, and dissipation can damp them. Pressure support alone does not erase them. In the pre-recombination photon-baryon fluid, Silk damping supplies a separate erasure mechanism.

For collisionless matter, replace the sound speed by a characteristic velocity dispersion. For an isotropic Maxwell-Boltzmann velocity distribution, defining

σ^{2} = ⟨ u_{x}^{2} ⟩ = ⟨ ∣ u ∣^{2} ⟩ /3

, the static collisionless marginal-stability calculation gives

λ_{J, collisionless}^{com} = \frac{σ}{a} \frac{π}{G ρ ˉ} = \frac{v _{rms}}{a} \frac{π}{3 G ρ ˉ} .

(5)

The coefficient depends on the distribution and the convention for velocity dispersion. Small-scale suppression is now collisionless free streaming and phase mixing, rather than collisional acoustic waves. An instantaneous collisionless Jeans length must be distinguished from the accumulated comoving distance

\int v d t / a

traveled since decoupling.

For the requested thermal histories, an Einstein-de Sitter universe describes the matter-era limit: it cannot literally also have a radiation-dominated epoch. Interpret the question as a spatially flat universe with negligible cosmological constant, passing from radiation domination,

a \propto t^{1/2}

, to matter domination,

a \propto t^{2/3}

. For the usual schematic dynamical Jeans scale use

\overset{ρ}{ˉ}_{tot}

in the estimate, so

λ_{dyn}^{com} \propto {c_{s} a, c_{s} a^{1/2}, t ≪ t_{eq}, \overset{ρ}{ˉ}_{tot} \propto a^{- 4}, t ≫ t_{eq}, \overset{ρ}{ˉ}_{tot} \propto a^{- 3} .

(6)

Near the particle horizon a relativistic perturbation calculation replaces the Newtonian derivation; the scaling estimate still identifies the relevant sound-crossing or streaming scale.

Before cosmological recombination, adiabatic initial conditions keep the entropy per baryon fixed in the tightly coupled photon-baryon fluid. With

R_{b} = 3 \overset{ρ}{ˉ}_{b} / (4 \overset{ρ}{ˉ}_{γ}) \propto a

, its photon-baryon sound speed is

c_{s}^{2} = \frac{c ^{2}}{3 ( 1 + R _{b} )} .

(7)

For example, this follows from

δ ρ_{γ} / \overset{ρ}{ˉ}_{γ} = (4/3) δ ρ_{b} / \overset{ρ}{ˉ}_{b}

and

δ P = c^{2} δ ρ_{γ} /3

. In the radiation-dominated limit

R_{b} ≪ 1

c_{s} ≃ c / 3

; in the strongly baryon-loaded matter-era limit

R_{b} ≫ 1

c_{s} \propto a^{- 1/2}

. Thus the baryon Jeans length across recombination has the asymptotic history

λ_{J, b}^{com} \propto ⎩ ⎨ ⎧ a \propto t^{1/2}, approximately constant, a^{- 1/2} \propto t^{- 1/3}, t ≪ t_{eq}, t_{eq} ≪ t < t_{rec}, t > t_{rec} with adiabatic gas cooling .

(8)

The transition to the plateau is smooth: retaining

1 + R_{b}

gives

λ_{J, b}^{com} \propto a^{1/2} / 1 + R_{b}

in matter domination. If baryon loading is still small, this intermediate segment instead rises as

a^{1/2}

until loading becomes important.

At photon decoupling, radiation ceases to provide pressure support to the baryons, so the sound speed drops from the coupled-fluid value to that of a nonrelativistic ideal gas,

c_{s, b}^{2} = γ k_{B} T_{b} / (μ m_{p})

, with

γ = 5/3

. The drop in comoving Jeans length at fixed density is the same ratio of sound speeds, typically several orders of magnitude. Subsequently an isentropic process gives

T_{b} \propto \overset{ρ}{ˉ}_{b}^{γ - 1} \propto a^{- 2}

, so

c_{s, b} \propto a^{- 1}

and

λ_{J, b}^{com} \propto a^{- 1/2}

. Residual energy exchange through Compton scattering can initially keep

T_{b} \propto a^{- 1}

, producing a short approximately flat gas segment before the decreasing segment. The idealized sketch assumes immediate adiabatic cooling:

log(comoving baryon Jeans scale)
  ^
  |                   __________________
  |                 /                   |
  |               /                     |  loss of photon pressure
  |             /                       |
  |           /                         +---\
  |         /                                \
  +-------------------|-----------------|----------> log(time)
                    t_eq              t_rec
        slope 1/2          slope ~0        slope -1/3

For the cold-dark-matter kinetic Jeans scale, take the stated toy history to mean that collisions maintain a particle temperature equal to the photon temperature until kinetic decoupling. This is an assumption about thermal coupling; a negligible collision rate after decoupling makes the subsequent treatment collisionless. Assume the ordering

t_{NR} < t_{dec} < t_{eq} < t_{rec}

. While relativistic, the characteristic particle speed is constant; while nonrelativistic but thermally coupled,

σ \propto T \propto a^{- 1/2}

; after kinetic decoupling, particle momentum redshifts as

a^{- 1}

and

σ \propto a^{- 1}

. Combining these with the background dynamical density gives

λ_{dyn, CDM}^{com} \propto ⎩ ⎨ ⎧ a \propto t^{1/2}, a^{1/2} \propto t^{1/4}, constant, a^{- 1/2} \propto t^{- 1/3}, t < t_{NR}, t_{NR} < t < t_{dec}, t_{dec} < t < t_{eq}, t > t_{eq} .

(9)

There is no sudden change at cosmological recombination: cold dark matter has already decoupled. Its small initial velocity dispersion makes its suppression scale much smaller than the pre-recombination baryon scale.

log(comoving CDM dynamical Jeans scale)
  ^
  |                        _______________
  |                    ___/               \
  |                ___/                    \       no jump
  |              /                          \      at t_rec
  |            /                             \
  +-----------|------------|--------------|----|------> log(time)
            t_NR         t_dec          t_eq t_rec
   slope 1/2    slope 1/4       slope 0       slope -1/3

The density convention matters. If one instead defines a species' self-gravitating collisionless Jeans length using only its own nonrelativistic density

\overset{ρ}{ˉ}_{m} \propto a^{- 3}

, its thermally coupled segment is constant and its decoupled segment decreases as

a^{- 1/2}

even during radiation domination. The constant radiation-era segment in the second sketch is the usual instantaneous background dynamical streaming estimate, not an exact single-species self-gravity threshold. A complete multicomponent calculation uses the separate perturbation equations and their combined cosmological Poisson equation.

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 35D / a / Solution Created 2026-09-24 Updated 2026-09-29

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Macroscopically, an adiabatic process exchanges no heat with its surroundings:

δ Q = 0

. A reversible adiabatic process is therefore isentropic, since

d S = δ Q_{rev} / T = 0

. Microscopically, the system is thermally isolated, so changes in its energy arise from work done by changing external parameters rather than random energy exchange with a heat bath; in a sufficiently slow change, the populations of the corresponding microscopic states are carried along without producing entropy.

For repeated operation the working substance must execute a thermodynamic cycle. Reversible isothermal contact at

T_{2}

absorbs heat

Q_{2}

, a reversible adiabat changes the temperature to

T_{1}

, reversible isothermal contact at

T_{1}

rejects heat

Q_{1} > 0

, and a second adiabat returns to the initial state. This is the Carnot cycle. Reversibility and zero net entropy change give

\frac{Q _{2}}{T _{2}} - \frac{Q _{1}}{T _{1}} = 0, \frac{Q _{1}}{Q _{2}} = \frac{T _{1}}{T _{2}} .

(1)

The thermal efficiency is the net work output divided by the absorbed heat:

η = \frac{W}{Q _{2}} = \frac{Q _{2} - Q _{1}}{Q _{2}} .

(2)

Consequently

η_{m a x} = 1 - \frac{T _{1}}{T _{2}} .

(3)

The Carnot theorem shows that any irreversible cycle has smaller efficiency, so the two isotherms joined by two adiabats are optimal.

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 35D / Solution Created 2026-09-24 Updated 2026-10-03

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For one molecule whose translational motion separates from its internal degrees of freedom, the classical one-particle partition function is

z_{1} (T, V) = \frac{V}{λ ^{3}} q_{int} (T), λ = \frac{2 π ℏ ^{2}}{m k _{B} T},

(1)

where

q_{int}

contains rotational, vibrational, and other internal states. For

N

indistinguishable noninteracting molecules, the classical ideal-gas partition function is

Z_{N} = z_{1}^{N} / N!

. The thermodynamic pressure therefore obeys

P = k_{B} T (\frac{\partial l o g Z _{N}}{\partial V})_{T, N} = \frac{N k _{B} T}{V},

(2)

P V = N k_{B} T .

(3)

The internal structure changes the energy, entropy, and heat capacity through

q_{int} (T)

, but it does not change the ideal gas law while the internal partition function is independent of volume and the molecules have negligible interactions.

For a dilute monatomic gas

q_{int} = 1

. Using the Stirling formula in

Z_{N} = (V / λ^{3})^{N} / N!

and differentiating the Helmholtz free energy

F = - k_{B} T lo g Z_{N}

gives the Sackur-Tetrode equation

S = N k_{B} [lo g (\frac{V}{N} (\frac{2 πm k _{B} T}{h ^{2}})^{3/2}) + \frac{5}{2}] .

(4)

With

N

fixed, an isentropic process therefore satisfies

V T^{3/2} = constant

, or

T V^{2/3} = constant

. Eliminating

T

with the ideal-gas law yields the monatomic reversible ideal-gas adiabat

P V^{5/3} = constant .

(5)

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / Section B / 5 / b / Solution 2026-10-20

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Holding

y

fixed, the chain rule gives

(\frac{\partial v}{\partial u})_{y} (\frac{\partial u}{\partial x})_{y} = (\frac{\partial v}{\partial x})_{y},

(1)

where the local change of variable from

x

u

is valid.

For the internal energy

U (S, V)

, its exact differential identifies

T = U_{S}

and

p = - U_{V}

. The Maxwell relation from equality of mixed partial derivatives is

T_{V} ∣_{S} = - p_{S} ∣_{V}

. At fixed volume, the constant-volume heat capacity satisfies

C_{V} = (\frac{\partial U}{\partial T})_{V} = T (\frac{\partial S}{\partial T})_{V},

(2)

T_{S} ∣_{V} = T / C_{V}

, assuming nonzero

C_{V}

and a locally invertible relation. Applying the chain rule to

p (T, V)

gives the isentropic temperature-volume derivative

(\frac{\partial T}{\partial V})_{S} = - (\frac{\partial p}{\partial T})_{V} (\frac{\partial T}{\partial S})_{V} = - \frac{T}{C _{V}} (\frac{\partial p}{\partial T})_{V} .

(3)

For the stated one-mole ideal gas, the equation of state gives

p_{T} ∣_{V} = R / V

and

C_{V} = 3 R /2

, so the right side is

- 2 T / (3 V)

. Along an isentropic process,

p \propto V^{- 5/3}

and

T = p V / R \propto V^{- 2/3}

, giving the same left side:

(\frac{\partial T}{\partial V})_{S} = - \frac{2 T}{3 V} .

(4)

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Past exam of the physics B course of the University of Cambridge / 2019 / Paper 1 / Section A / A5 / Solution 2026-10-18

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For a reversible adiabatic process, an ideal gas obeys

T V^{γ - 1} = constant

. A monatomic gas has heat capacity ratio

γ = 5/3

. Taking the stated expansion factor to mean

V_{f} = 9 V_{i}

T_{f} = 320 (\frac{1}{9})^{2/3} K ≃ 74.0 K .

(1)

This assumes the usual quasistatic isentropic process. Heat insulation alone does not determine the final temperature: an ideal-gas free expansion is also adiabatic but leaves the temperature unchanged. The numerical answer uses nine times the original volume, the conventional intended meaning of the source's “nine times larger”.

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Past exam of the physics B course of the University of Cambridge / 2021 / Paper 1 / Section A / A5 / Solution 2026-10-18

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For the intended reversible adiabatic process, an ideal gas obeys

T V^{γ - 1} = constant

. A monatomic gas has heat capacity ratio

γ = 5/3

, so

T_{f} = T_{i} (\frac{V _{i}}{V _{f}})^{γ - 1} = 400 (\frac{1}{8})^{2/3}, T_{f} = 100 K .

(1)

The required extra assumption is that “adiabatic” denotes the usual quasistatic isentropic process. Absence of heat transfer alone would not fix the final temperature: an ideal gas undergoing free expansion into vacuum is also adiabatic but leaves its temperature unchanged.

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Past exam of the physics B course of the University of Cambridge / 2021 / Paper 2 / Section A / A2 / Solution Created 2026-10-18 Updated 2026-10-20

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The third law of thermodynamics, in its perfect-crystal form, states that the entropy of a pure, perfectly ordered crystal with a unique ground state tends to zero as

T \to 0

. Its more general entropy-difference formulation says that the entropy difference between equilibrium states related by a finite change of a regular external control parameter tends to zero at absolute zero:

lim_{T \to 0} [S (T, X_{2}) - S (T, X_{1})] = 0.

(1)

The common limiting entropy is independent of

X

. Ground-state degeneracy or frozen disorder can produce residual entropy, so zero entropy is not an unconditional assertion about every sample.

In the usual regular thermodynamic setting, absolute zero cannot be attained from positive temperature by a finite sequence of cooling operations with finite resources. An entropy-temperature plot shows why a finite reversible adiabatic step cannot reach the common zero-temperature endpoint from a state with positive entropy above that endpoint: an isentropic process follows a horizontal line, while all the regular entropy curves meet only at

T = 0

Figure 1.
Entropy curves and a cooling staircase approaching absolute zero
. **Details**.

The figure illustrates this with two possible settings having

S - S_{0} = a_{1} T

and

S - S_{0} = a_{2} T

, where

0 < a_{1} < a_{2}

. First change the setting isothermally from the upper to the lower curve, rejecting heat; then change it reversibly and adiabatically back to the upper curve. Entropy conservation on the latter step gives

a_{1} T_{n} = a_{2} T_{n + 1}, T_{n + 1} = \frac{a _{1}}{a _{2}} T_{n}, T_{n} = T_{0} (\frac{a _{1}}{a _{2}})^{n} > 0

(2)

for every finite

n

. The staircase approaches the origin but reaches it only after infinitely many such stages. Irreversibility generates entropy and cannot improve on the reversible endpoint.

This is a thermodynamic illustration of unattainability of absolute zero, not a claim that every cooling apparatus has this particular linear entropy law. Its conclusion presumes regular equilibrium states and bounded control parameters; dynamical statements about finite cooling time additionally require physical assumptions about the available interactions and cooling rates.

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Past exam of the physics B course of the University of Cambridge / 2022 / Paper 2 / Section C / C8 / e / Solution 2026-10-18

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This is the Brayton cycle. Assume fixed gas amount, constant heat capacities, and internally reversible adiabatic processes, as intended by the ideal engine model. Put

r_{p} = p_{h} / p_{l} = p_{2} / p_{1} > 1

and

a = \frac{γ - 1}{γ}, k = r_{p}^{a} .

(1)

Combining

p V^{γ} = constant

with the ideal gas law gives

T \propto p^{(γ - 1) / γ}

along either reversible ideal-gas adiabat. Therefore

\frac{T _{2}}{T _{1}} = k, \frac{T _{3}}{T _{4}} = k, T_{3} - T_{2} = k (T_{4} - T_{1}) .

(2)

During a constant-pressure process the heat supplied equals the enthalpy increase. If

C_{p}

is the molar heat capacity at constant pressure, the positive heat input and output are

Q_{in} = n C_{p} (T_{3} - T_{2}), Q_{out} = n C_{p} (T_{4} - T_{1}) .

(3)

The insulated branches exchange no heat, so the first law of thermodynamics gives

W = Q_{in} - Q_{out}

. Hence

η = 1 - \frac{Q _{out}}{Q _{in}} = 1 - \frac{1}{k} = 1 - (\frac{p _{2}}{p _{1}})^{(1 - γ) / γ} .

(4)

An adiabatic process alone need not be an isentropic process; the relations used here also require internal reversibility. Dissipative compression or expansion would require extra information and would not, in general, obey this efficiency formula.

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Stagnation pressure 2026-10-14

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The pressure obtained by bringing a flow to rest without losses; in compressible flow the usual definition specifies an isentropic process for stopping. For the constant-mass density, same-height Bernoulli equation discussed under dynamic pressure,