Potential Temperature: A Step-by-Step Derivation from the First Law of Thermodynamics

Potential temperature is a fundamental variable in atmospheric thermodynamics. It helps meteorologists compare air parcels at different pressures, diagnose static stability, and interpret approximately adiabatic motion. Its familiar equation is:

θ=T(p0p)Rd/cp\boxed{\theta=T\left(\frac{p_0}{p}\right)^{R_d/c_p}}

Why does pressure appear in this formula? How can an air parcel cool without losing heat? And why does potential temperature stay constant during reversible dry adiabatic motion? This tutorial answers those questions by deriving the equation directly from the first law of thermodynamics, without skipping the algebra.

You will need basic differentiation, logarithms, and the ideal gas law. Every physical assumption is identified along the way.

1. What Is Potential Temperature?

1.1 Why ordinary temperature is not enough

Imagine a small parcel of air rising through the atmosphere. Ambient pressure generally falls with height. As the parcel rises, it expands and performs pressure–volume work on its surroundings. If no heat enters the parcel, its internal energy decreases and its temperature falls.

The reverse occurs during descent: higher ambient pressure compresses the parcel, work is done on it, and its temperature rises. These temperature changes do not necessarily indicate external heating or cooling.

Thus, comparing the actual temperatures of parcels at different pressures can be misleading. We need a way to compare them at the same pressure.

1.2 The reference-pressure idea

Potential temperature is the temperature a dry-air parcel would attain if brought reversibly and adiabatically to a specified reference pressure. Atmospheric scientists conventionally use 1000 hPa (100 kPa).

The operation is hypothetical. The air parcel need not actually move to that pressure. We calculate the temperature it would have under a well-defined thermodynamic transformation.

Important distinction: “adiabatic” means no heat transfer; “reversible adiabatic” additionally excludes entropy production. The standard potential-temperature equation describes the reversible reference transformation.

2. The First Law of Thermodynamics

For a closed system, the first law states that a change in internal energy equals heat added minus work done by the system:

dU=δQ−δWdU=\delta Q-\delta W

Here, U is internal energy, δQ is heat supplied, and δW is work performed by the system. We adopt this sign convention throughout. A positive work term therefore describes expansion.

2.1 Expressing the equation per unit mass

For a parcel of fixed mass, define specific internal energy and specific volume:

u=Um,α=Vm=1ρu=\frac{U}{m},\qquad \alpha=\frac{V}{m}=\frac{1}{\rho}

Specific volume α is the reciprocal of density ρ. For quasi-static pressure–volume work, the work per unit mass is:

δw=pdα\delta w=p\,d\alpha

Consequently, the first law takes the form:

δq=du+pdα\boxed{\delta q=du+p\,d\alpha}
  • δq: heat received per unit mass (J kg⁻¹).
  • du: change in specific internal energy (J kg⁻¹).
  • p dα: pressure–volume work performed per unit mass (J kg⁻¹).

Why use du but δq? Internal energy is a state function, so its differential is exact. Heat is energy transferred during a process, not a state variable, so δ emphasizes that its integral depends on the path.

3. Connecting Internal Energy and Temperature

Treat dry air as an ideal gas, and approximate its specific heat capacities as constant over the temperature range of interest. For an ideal gas, specific internal energy depends only on temperature:

du=cvdTdu=c_v\,dT

Here, cᵥ is specific heat capacity at constant volume. At constant volume, dα = 0, so the first law gives δq = du. The definition of cᵥ then leads to du = cᵥ dT. Because an ideal gas has internal energy determined by temperature alone, this last differential relationship also applies when its volume changes.

Substitute it into the first law:

δq=cvdT+pdα\boxed{\delta q=c_v\,dT+p\,d\alpha}

We now have temperature in the equation, but we still need to remove specific volume.

4. Using the Ideal Gas Law: Every Algebraic Step

The equation of state for dry ideal air is:

pα=RdTp\alpha=R_dT

Rᵈ denotes the specific gas constant of dry air; its conventional approximate value is 287 J kg⁻¹ K⁻¹.

4.1 Differentiate both sides

Both p and α can change. Apply the product rule to the left-hand side:

d(pα)=pdα+αdpd(p\alpha)=p\,d\alpha+\alpha\,dp

Since Rᵈ is constant:

pdα+αdp=RddT\boxed{p\,d\alpha+\alpha\,dp=R_d\,dT}

4.2 Isolate the expansion-work term

Subtract α dp from both sides:

pdα=RddT−αdpp\,d\alpha=R_d\,dT-\alpha\,dp

Insert this expression into the temperature form of the first law:

δq=cvdT+RddT−αdp\delta q=c_v\,dT+R_d\,dT-\alpha\,dp

Collect the terms multiplying dT:

δq=(cv+Rd)dT−αdp\delta q=(c_v+R_d)\,dT-\alpha\,dp

4.3 Why the constant-pressure heat capacity appears

For an ideal gas, Mayer’s relation is:

cp−cv=Rd⟹cp=cv+Rdc_p-c_v=R_d\qquad\Longrightarrow\qquad c_p=c_v+R_d

Therefore, the first law can be written as:

δq=cpdT−αdp\boxed{\delta q=c_p\,dT-\alpha\,dp}

This is the most useful form for deriving potential temperature because it relates heating directly to changes in temperature and pressure.

4.4 Optional check: the enthalpy interpretation

Specific enthalpy is defined as h = u + pα. Differentiating and applying the first law gives:

dh=du+pdα+αdpdh=du+p\,d\alpha+\alpha\,dp
δq=dh−αdp\delta q=dh-\alpha\,dp

For an ideal gas, dh = cₚ dT. We recover exactly the same pressure–temperature equation. This alternative shows why cₚ naturally appears after the expansion-work term is rearranged.

5. The Adiabatic Pressure–Temperature Relation

In an adiabatic process there is no heat exchange, so δq = 0. For the reversible reference process, our first-law equation becomes:

0=cpdT−αdp0=c_p\,dT-\alpha\,dp

Move the pressure term to the other side:

cpdT=αdpc_p\,dT=\alpha\,dp

Use the ideal gas law to express specific volume in terms of temperature and pressure:

α=RdTp\alpha=\frac{R_dT}{p}

Substitution yields:

cpdT=RdTpdpc_p\,dT=\frac{R_dT}{p}\,dp

Divide by cₚT on both sides and cancel common factors:

dTT=Rdcpdpp\boxed{\frac{dT}{T}=\frac{R_d}{c_p}\frac{dp}{p}}

Notice the signs. T, p, Rᵈ, and cₚ are positive. Thus a negative pressure change implies a negative temperature change: a parcel expanding during ascent cools. Compression during descent produces the opposite result.

6. Integrating from the Actual Pressure to 1000 hPa

Define the dimensionless Poisson exponent:

κ=Rdcp\kappa=\frac{R_d}{c_p}

The differential relationship becomes:

dTT=κdpp\frac{dT}{T}=\kappa\frac{dp}{p}

Let the actual parcel state be (T, p). Let the final reference state after reversible adiabatic transformation be (θ, p₀), where p₀ is fixed. Assume κ is constant. Integrate between these two states:

∫TθdT′T′=κ∫pp0dp′p′\int_T^\theta\frac{dT’}{T’}=\kappa\int_p^{p_0}\frac{dp’}{p’}

The prime marks merely distinguish a dummy integration variable from its limits. Recall that the integral of 1/x is ln|x|. Absolute temperature and absolute pressure are positive, so the arguments below are positive.

ln⁡θ−ln⁡T=κ(ln⁡p0−ln⁡p)\ln\theta-\ln T=\kappa\left(\ln p_0-\ln p\right)

Use the logarithm identity ln(a) − ln(b) = ln(a/b):

ln⁡(θT)=κln⁡(p0p)\ln\left(\frac{\theta}{T}\right)=\kappa\ln\left(\frac{p_0}{p}\right)

Apply the power rule of logarithms:

ln⁡(θT)=ln⁡[(p0p)κ]\ln\left(\frac{\theta}{T}\right)=\ln\left[\left(\frac{p_0}{p}\right)^\kappa\right]

Exponentiate both sides:

θT=(p0p)κ\frac{\theta}{T}=\left(\frac{p_0}{p}\right)^\kappa

Finally, multiply by T and substitute the definition of κ:

θ=T(p0p)Rd/cp\boxed{\theta=T\left(\frac{p_0}{p}\right)^{R_d/c_p}}

We have derived potential temperature from the first law. The corresponding pressure–temperature power law is known as Poisson’s relation for an ideal gas under a reversible adiabatic transformation.

7. What Does the Exponent 0.286 Mean?

For dry air, two commonly used approximations are:

Rd≈287 Jkg−1K−1,cp≈1004 Jkg−1K−1R_d\approx287\ \mathrm{J\,kg^{-1}\,K^{-1}},\qquad c_p\approx1004\ \mathrm{J\,kg^{-1}\,K^{-1}}

Therefore:

κ=2871004≈0.286\kappa=\frac{287}{1004}\approx0.286

With the conventional reference pressure p₀ = 1000 hPa, the operational form is:

θ≈T(1000p)0.286\boxed{\theta\approx T\left(\frac{1000}{p}\right)^{0.286}}

Unit check: T and θ must be in kelvin. When the numerator is 1000, p must be in hPa. The pressure ratio must be dimensionless; using pressure in pascals requires p₀ = 100000 Pa.

8. Worked Example: A Parcel at 850 hPa and 10 °C

Suppose the dry-air parcel is at p = 850 hPa with an actual temperature of 10 °C.

Step 1: Convert to absolute temperature

T=10+273.15=283.15 KT=10+273.15=283.15\ \mathrm{K}

Step 2: Substitute into the potential-temperature equation

θ=283.15(1000850)287/1004\theta=283.15\left(\frac{1000}{850}\right)^{287/1004}

Step 3: Evaluate

θ≈296.61 K\boxed{\theta\approx296.61\ \mathrm{K}}

The corresponding Celsius value at the reference pressure is about 23.46 °C.

296.61−273.15≈23.46 ∘C296.61-273.15\approx23.46\ ^\circ\mathrm{C}

Interpretation: if the parcel were compressed reversibly and adiabatically from 850 hPa to 1000 hPa, its temperature would increase from 10 °C to about 23.46 °C. No external heat supply is required.

9. Why Is Potential Temperature Conserved?

Take the natural logarithm of the potential-temperature equation, holding p₀ and κ fixed:

ln⁡θ=ln⁡T+κln⁡p0−κln⁡p\ln\theta=\ln T+\kappa\ln p_0-\kappa\ln p

Differentiate:

dθθ=dTT−κdpp\frac{d\theta}{\theta}=\frac{dT}{T}-\kappa\frac{dp}{p}

Along the reversible dry adiabat derived in Section 5, the two terms on the right are identical. Their difference is zero:

dθθ=0⟹dθ=0\boxed{\frac{d\theta}{\theta}=0\quad\Longrightarrow\quad d\theta=0}

Following a moving parcel, the same result is written with the material derivative:

DθDt=0\boxed{\frac{D\theta}{Dt}=0}

This means that pressure and actual temperature may change substantially while potential temperature remains constant, provided the parcel follows the assumed reversible, dry, adiabatic process.

Example: a single 300 K dry adiabat

PressureActual temperaturePotential temperature
1000 hPa300.00 K (26.85 °C)300.00 K
850 hPa286.38 K (13.23 °C)300.00 K
700 hPa270.92 K (−2.23 °C)300.00 K

As the parcel moves from 1000 to 700 hPa, its temperature falls by nearly 29 K, yet potential temperature remains 300 K. The calculation is purely thermodynamic; it does not require specifying the parcel’s height.

10. The Deeper Connection: Potential Temperature and Entropy

The reversible thermodynamic relation between heat and specific entropy s is:

δqrev=Tds\delta q_{\mathrm{rev}}=T\,ds

Combine this with the first-law expression δq = cₚ dT − α dp and the ideal gas law:

Tds=cpdT−αdpT\,ds=c_p\,dT-\alpha\,dp
ds=cpdTT−Rddppds=c_p\frac{dT}{T}-R_d\frac{dp}{p}

With constant cₚ and Rᵈ, take the logarithmic differential of potential temperature:

dln⁡θ=dln⁡T−Rdcpdln⁡pd\ln\theta=d\ln T-\frac{R_d}{c_p}\,d\ln p

Multiply by cₚ and compare the two expressions:

ds=cpdln⁡θ\boxed{ds=c_p\,d\ln\theta}

Therefore potential temperature is a monotonic measure of specific entropy for the idealized fixed-composition dry gas. Along an isentropic process, ds = 0 and θ remains constant.

Does adiabatic always mean isentropic?

No. An adiabatic process involves zero heat exchange with the surroundings. An isentropic process has constant entropy. An adiabatic but irreversible process can generate entropy internally (for example through dissipative processes). Consequently, conservation of θ requires the appropriate idealized conditions and should not be extended uncritically to every adiabatic motion.

11. Connection to the Dry Adiabatic Lapse Rate

Consider a parcel remaining in pressure equilibrium with a hydrostatic environment. The surrounding vertical pressure gradient is:

dpdz=−ρg\frac{dp}{dz}=-\rho g

Since α = 1/ρ, we have:

αdpdz=−g\alpha\frac{dp}{dz}=-g

Combine this relation with cₚ dT = α dp for a reversible dry adiabatic parcel displacement:

cpdTdz=−gc_p\frac{dT}{dz}=-g

Define the dry adiabatic lapse rate as the positive magnitude of the parcel’s temperature decrease with height:

Γd=−dTdz=gcp\boxed{\Gamma_d=-\frac{dT}{dz}=\frac{g}{c_p}}

With g ≈ 9.81 m s⁻² and cₚ ≈ 1004 J kg⁻¹ K⁻¹:

Γd≈9.8 Kkm−1\boxed{\Gamma_d\approx9.8\ \mathrm{K\,km^{-1}}}

This is the temperature gradient followed by an approximately reversible dry adiabatic parcel in hydrostatic pressure balance. It is not necessarily the measured lapse rate of the surrounding atmosphere.

12. Understanding Atmospheric Static Stability

Potential temperature is especially useful for analyzing the response of an unsaturated parcel to small vertical displacements. Consider the environmental profile θ(z). A displaced parcel approximately conserves its θ as long as its motion is reversible and dry adiabatic.

  • Stable: environmental potential temperature increases upward. An upward-displaced parcel is colder and denser than the environment at its new pressure; buoyancy tends to restore it.
  • Neutral: environmental potential temperature is constant with height. An ideal small dry adiabatic displacement creates no restoring buoyancy.
  • Unstable: environmental potential temperature decreases upward. An upward-displaced parcel is warmer and lighter than its new environment; buoyancy reinforces the displacement.
dθdz>0 (stable),dθdz=0 (neutral),dθdz<0 (unstable)\frac{d\theta}{dz}>0\ \text{(stable)},\qquad \frac{d\theta}{dz}=0\ \text{(neutral)},\qquad \frac{d\theta}{dz}<0\ \text{(unstable)}

For a dry hydrostatic background, the squared Brunt–Väisälä frequency is:

N2=gθdθdz\boxed{N^2=\frac{g}{\theta}\frac{d\theta}{dz}}

Positive N² indicates dry static stability. Negative N² indicates static instability rather than an oscillatory buoyancy frequency. Atmospheric moisture and condensate can alter buoyancy, so this dry result must be applied with appropriate care.

13. Common Misconceptions

Is potential temperature the same as actual temperature?

Only when the actual pressure equals the reference pressure, p = p₀. Otherwise, the two temperatures generally differ.

Is potential temperature always warmer than the actual temperature?

No. If p < p₀, then θ > T. If p = p₀, then θ = T. If p > p₀, then θ < T, assuming T > 0 and κ > 0.

Does adiabatic mean constant temperature?

No. Adiabatic means no heat exchange; isothermal means constant temperature. An air parcel can cool or warm through expansion or compression without exchanging heat.

Why not put Celsius temperatures directly into the formula?

The ideal gas law and integration use absolute temperature. Substituting degrees Celsius into a ratio or power law gives incorrect results. Always convert to kelvin first.

Does potential temperature remain conserved during condensation?

Ordinary dry-air potential temperature is not generally conserved during moist processes involving latent heating. Moist thermodynamics uses related variables, such as equivalent potential temperature and liquid-water potential temperature, with assumptions appropriate to the process.

14. Test Your Understanding

Question 1: At the reference pressure

A parcel has T = 293.15 K at p = 1000 hPa. What is its potential temperature?

Question 2: A parcel at 900 hPa

A dry parcel has T = 280 K at p = 900 hPa. Calculate θ using κ = 287/1004.

Question 3: Ascent along a dry adiabat

A parcel has θ = 300 K and moves reversibly and adiabatically to p = 700 hPa. What is its actual temperature?

Answers

1. At p = p₀, the pressure ratio is one, so θ = 293.15 K.

2. Substitute T = 280 K and p = 900 hPa:

θ=280(1000900)287/1004≈288.56 K\boxed{\theta=280\left(\frac{1000}{900}\right)^{287/1004}\approx288.56\ \mathrm{K}}

3. Rearrange the potential-temperature equation to solve for the actual temperature:

T=θ(pp0)κ=300(7001000)287/1004≈270.92 K\boxed{T=\theta\left(\frac{p}{p_0}\right)^\kappa=300\left(\frac{700}{1000}\right)^{287/1004}\approx270.92\ \mathrm{K}}

15. Frequently Asked Questions

What is the simplest definition of potential temperature?

The temperature a dry-air parcel would attain after being brought reversibly and adiabatically to a chosen reference pressure, conventionally 1000 hPa.

What is Poisson’s equation in meteorology?

It is the power-law relation between temperature and pressure along a reversible adiabat for an ideal gas with constant specific heat. The potential-temperature formula is one way of writing that relation.

What is the difference between potential temperature and virtual potential temperature?

Potential temperature is based on the dry ideal-gas transformation. Virtual potential temperature accounts for the effect of moisture on density and is especially useful for analyzing buoyancy.

How is potential temperature used in forecasting?

It is useful for diagnosing dry static stability, identifying stratification and air masses, comparing thermodynamic states across pressure levels, and analyzing approximately isentropic motion.

16. Summary: The Complete Derivation in Five Equations

Begin with the first law per unit mass:

δq=du+pdα\delta q=du+p\,d\alpha

For an ideal gas, apply du = cᵥ dT and differentiate pα = RᵈT:

δq=cpdT−αdp\delta q=c_p\,dT-\alpha\,dp

For a reversible dry adiabatic change, set δq = 0 and eliminate α:

dTT=Rdcpdpp\frac{dT}{T}=\frac{R_d}{c_p}\frac{dp}{p}

Integrate from (T, p) to (θ, p₀):

ln⁡(θT)=Rdcpln⁡(p0p)\ln\left(\frac{\theta}{T}\right)=\frac{R_d}{c_p}\ln\left(\frac{p_0}{p}\right)

The result is the potential-temperature equation:

θ=T(p0p)Rd/cp\boxed{\theta=T\left(\frac{p_0}{p}\right)^{R_d/c_p}}

Core insight: the temperature of an air parcel can change because its pressure changes. Potential temperature removes the reversible adiabatic pressure effect by expressing the parcel’s temperature at a common reference pressure.

References and Further Reading

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