In astrophysics, what is referred to as "entropy" is actually the adiabatic constant derived as follows. Using the first law of thermodynamics for a quasi-static, infinitesimal process for a hydrostatic system
d Q = d U − d W . {\displaystyle dQ=dU-dW.} For an ideal gas in this special case, the internal energy, U, is a function of only the temperature T; therefore the partial derivative of heat capacity with respect to T is identically the same as the full derivative, yielding through some manipulation
d Q = C v d T + P d V . {\displaystyle dQ=C_{\text{v}}dT+P\,dV.}
Further manipulation using the differential version of the ideal gas law, the previous equation, and assuming constant pressure, one finds
d Q = C p d T − V d P . {\displaystyle dQ=C_{\text{p}}dT-V\,dP.}
For an adiabatic process d Q = 0 {\displaystyle dQ=0\,} and recalling γ = C p / C v {\displaystyle \gamma ={C_{\text{p}}}/{C_{\text{v}}}\,} , one finds
One can solve this simple differential equation to find
P V γ = constant = K {\displaystyle PV^{\gamma }={\text{constant}}=K}
This equation is known as an expression for the adiabatic constant, K, also called the adiabat. From the ideal gas equation one also knows
P = ρ k B T μ m H , {\displaystyle P={\frac {\rho k_{\text{B}}T}{\mu m_{\text{H}}}},}
where k B {\displaystyle k_{\text{B}}} is the Boltzmann constant. Substituting this into the above equation along with V = [ g ] / ρ {\displaystyle V=[\mathrm {g} ]/\rho \,} and γ = 5 / 3 {\displaystyle \gamma =5/3\,} for an ideal monatomic gas one finds
K = k B T ( ρ / μ m H ) 2 / 3 , {\displaystyle K={\frac {k_{\text{B}}T}{(\rho /\mu m_{\text{H}})^{2/3}}},}
where μ {\displaystyle \mu \,} is the mean molecular weight of the gas or plasma; and m H {\displaystyle m_{\text{H}}} is the mass of the hydrogen atom, which is extremely close to the mass of the proton, m p {\displaystyle m_{p}} , the quantity more often used in astrophysical theory of galaxy clusters. This is what astrophysicists refer to as "entropy" and has units of [keV⋅cm2]. This quantity relates to the thermodynamic entropy as
Δ S = 3 / 2 ln K . {\displaystyle \Delta S=3/2\ln K.}
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