Intuition The one-line ideaA thermodynamic potential is a function whose mixed second partial derivatives don't care about the order of differentiation . That single fact of calculus, applied to each potential, spits out four powerful relations connecting quantities you can measure (like ∂ V / ∂ T \partial V/\partial T ∂ V / ∂ T ∂ S / ∂ P \partial S/\partial P ∂ S / ∂ P Maxwell relations are just the equality of mixed partials of the thermodynamic potentials .
Definition Exact differential & the equality of mixed partialsIf z = z ( x , y ) z = z(x,y) z = z ( x , y ) d z = ( ∂ z ∂ x ) y d x + ( ∂ z ∂ y ) x d y ≡ M d x + N d y dz = \left(\frac{\partial z}{\partial x}\right)_y dx + \left(\frac{\partial z}{\partial y}\right)_x dy \equiv M\,dx + N\,dy d z = ( ∂ x ∂ z ) y d x + ( ∂ y ∂ z ) x d y ≡ M d x + N d y ( ∂ M ∂ y ) x = ( ∂ N ∂ x ) y = ∂ 2 z ∂ x ∂ y \left(\frac{\partial M}{\partial y}\right)_x = \left(\frac{\partial N}{\partial x}\right)_y = \frac{\partial^2 z}{\partial x \, \partial y} ( ∂ y ∂ M ) x = ( ∂ x ∂ N ) y = ∂ x ∂ y ∂ 2 z reciprocity / Schwarz / Euler condition. Every Maxwell relation is this one identity in disguise.
WHAT we exploit: Thermodynamic potentials (U , H , F , G U, H, F, G U , H , F , G
HOW we use it: Read off M M M N N N
Potential
Definition
Differential
Natural vars
Internal energy U U U
—
d U = T d S − P d V dU = T\,dS - P\,dV d U = T d S − P d V ( S , V ) (S,V) ( S , V )
Enthalpy H H H
H = U + P V H = U + PV H = U + P V d H = T d S + V d P dH = T\,dS + V\,dP d H = T d S + V d P ( S , P ) (S,P) ( S , P )
Helmholtz F F F
F = U − T S F = U - TS F = U − T S d F = − S d T − P d V dF = -S\,dT - P\,dV d F = − S d T − P d V ( T , V ) (T,V) ( T , V )
Gibbs G G G
G = U − T S + P V G = U - TS + PV G = U − T S + P V d G = − S d T + V d P dG = -S\,dT + V\,dP d G = − S d T + V d P ( T , P ) (T,P) ( T , P )
Why the Legendre transform? To change which variable is "independent." Example: H = U + P V H=U+PV H = U + P V d H = d U + P d V + V d P = ( T d S − P d V ) + P d V + V d P = T d S + V d P . dH = dU + P\,dV + V\,dP = (T\,dS - P\,dV) + P\,dV + V\,dP = T\,dS + V\,dP. d H = d U + P d V + V d P = ( T d S − P d V ) + P d V + V d P = T d S + V d P . − P d V -P\,dV − P d V d P dP d P V V V Why this step? Adding P V PV P V + P d V + V d P +P\,dV+V\,dP + P d V + V d P P d V P\,dV P d V
Each differential has the form d z = M d x + N d y dz = M\,dx + N\,dy d z = M d x + N d y ( ∂ M / ∂ y ) x = ( ∂ N / ∂ x ) y (\partial M/\partial y)_x=(\partial N/\partial x)_y ( ∂ M / ∂ y ) x = ( ∂ N / ∂ x ) y
U U U d U = T d S − P d V \;dU = T\,dS - P\,dV d U = T d S − P d V x = S , y = V , M = T , N = − P x=S,\ y=V,\ M=T,\ N=-P x = S , y = V , M = T , N = − P ( ∂ T ∂ V ) S = − ( ∂ P ∂ S ) V \left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V ( ∂ V ∂ T ) S = − ( ∂ S ∂ P ) V Why this step? M = ( ∂ U / ∂ S ) V = T M=(\partial U/\partial S)_V=T M = ( ∂ U / ∂ S ) V = T N = ( ∂ U / ∂ V ) S = − P N=(\partial U/\partial V)_S=-P N = ( ∂ U / ∂ V ) S = − P ∂ 2 U / ∂ V ∂ S \partial^2 U/\partial V\partial S ∂ 2 U / ∂ V ∂ S
H H H d H = T d S + V d P \;dH = T\,dS + V\,dP d H = T d S + V d P x = S , y = P , M = T , N = V x=S,\ y=P,\ M=T,\ N=V x = S , y = P , M = T , N = V ( ∂ T ∂ P ) S = ( ∂ V ∂ S ) P \left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P ( ∂ P ∂ T ) S = ( ∂ S ∂ V ) P Why this step? No minus sign because d P dP d P + V +V + V d V dV d V − P -P − P
F F F d F = − S d T − P d V \;dF = -S\,dT - P\,dV d F = − S d T − P d V x = T , y = V , M = − S , N = − P x=T,\ y=V,\ M=-S,\ N=-P x = T , y = V , M = − S , N = − P ( ∂ S ∂ V ) T = ( ∂ P ∂ T ) V \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V ( ∂ V ∂ S ) T = ( ∂ T ∂ P ) V Why this step? ∂ ( − S ) / ∂ V = ∂ ( − P ) / ∂ T ⇒ \partial(-S)/\partial V = \partial(-P)/\partial T \Rightarrow ∂ ( − S ) / ∂ V = ∂ ( − P ) / ∂ T ⇒ most useful relation: RHS is easily measured (how pressure rises with T T T
G G G d G = − S d T + V d P \;dG = -S\,dT + V\,dP d G = − S d T + V d P x = T , y = P , M = − S , N = V x=T,\ y=P,\ M=-S,\ N=V x = T , y = P , M = − S , N = V ( ∂ S ∂ P ) T = − ( ∂ V ∂ T ) P \left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P ( ∂ P ∂ S ) T = − ( ∂ T ∂ V ) P Why this step? RHS relates to the thermal expansion coefficient α = 1 V ( ∂ V / ∂ T ) P \alpha=\frac1V(\partial V/\partial T)_P α = V 1 ( ∂ V / ∂ T ) P
Sign pattern (WHY): A minus sign appears exactly when the differential contains a − P d V -P\,dV − P d V V V V U U U F F F which conjugate has the minus ). Practical rule: the relations from U U U H H H independent variable) put T T T F F F G G G S S S measurable ones.
Worked example The "energy equation" — Maxwell relations in actionStart: d U = T d S − P d V dU = T\,dS - P\,dV d U = T d S − P d V d V dV d V T T T ( ∂ U ∂ V ) T = T ( ∂ S ∂ V ) T − P \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial S}{\partial V}\right)_T - P ( ∂ V ∂ U ) T = T ( ∂ V ∂ S ) T − P Why this step? We hold T T T d S → ( ∂ S / ∂ V ) T d V dS\to(\partial S/\partial V)_T\,dV d S → ( ∂ S / ∂ V ) T d V
Now substitute the Helmholtz Maxwell relation ( ∂ S / ∂ V ) T = ( ∂ P / ∂ T ) V (\partial S/\partial V)_T=(\partial P/\partial T)_V ( ∂ S / ∂ V ) T = ( ∂ P / ∂ T ) V ( ∂ U ∂ V ) T = T ( ∂ P ∂ T ) V − P \boxed{\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V - P} ( ∂ V ∂ U ) T = T ( ∂ T ∂ P ) V − P Why this matters: For an ideal gas P = n R T / V P=nRT/V P = n R T / V T ( ∂ P / ∂ T ) V = T ⋅ n R / V = P T(\partial P/\partial T)_V = T\cdot nR/V = P T ( ∂ P / ∂ T ) V = T ⋅ n R / V = P ( ∂ U / ∂ V ) T = 0 (\partial U/\partial V)_T = 0 ( ∂ U / ∂ V ) T = 0 proved from thermodynamics alone that an ideal gas's internal energy is volume-independent — Joule's law derived, not assumed .
Common mistake "The natural variables are whatever I want."Why it feels right: You can write U U U ( T , V ) (T,V) ( T , V ) The fix: Only when U U U natural variables ( S , V ) (S,V) ( S , V ) d U = T d S − P d V dU=T\,dS-P\,dV d U = T d S − P d V ( T , V ) (T,V) ( T , V ) The potential must be paired with its natural variables.
Common mistake Getting the minus signs wrong.Why it feels right: All four look symmetric, so you assume all are "= = = The fix: The sign is inherited directly from the sign in the differential. If the term is − P d V -P\,dV − P d V N = − P N=-P N = − P Always re-derive from the differential, never memorize signs blindly.
Common mistake Confusing the subscript (held-constant variable).Why it feels right: People drop the subscript like in ordinary calculus. The fix: In thermodynamics ( ∂ T / ∂ V ) S ≠ ( ∂ T / ∂ V ) P (\partial T/\partial V)_S \ne (\partial T/\partial V)_P ( ∂ T / ∂ V ) S = ( ∂ T / ∂ V ) P part of the quantity ; the Maxwell relation only holds with the subscripts shown (the other natural variable is held fixed).
Recall Feynman: explain to a 12-year-oldImagine a smooth hill. To get from the bottom to the top, it doesn't matter if you go east-then-north or north-then-east — you climb the same total height. The "height function" doesn't care about the order. Thermodynamic potentials are like that hill: their energy is a smooth function, so when you measure how steeply it changes in two different directions, the answers must match up in a special crossed way. That matching gives us free facts — like figuring out something hard to measure (how messy/disordered a gas gets when squeezed) by measuring something easy (how much it expands when warmed). Same hill, two paths, equal climb.
Mnemonic Remembering the potentials & signs"Good Physicists Have Studied Under Very Fine Teachers" isn't needed — use the square :
Write the square U H F G with S V T P on the corners. The natural-variable square (Born square) "Valid Facts and Theoretical Understanding Generate Solutions "... simplest: remember d U = T d S − P d V dU=TdS-PdV d U = T d S − P d V build the other three by Legendre transform (add/subtract P V PV P V T S TS T S Derive, don't memorize.
Thermodynamic potentials & Legendre transforms — where the four potentials come fromFirst and Second Laws of Thermodynamics — source of d U = T d S − P d V dU=TdS-PdV d U = T d S − P d V Equality of mixed partial derivatives (Schwarz theorem) — the calculus engineJoule expansion and internal energy — application via the energy equationHeat capacities $C_P - C_V$ — derived using Maxwell relationsThermal expansion coefficient and isothermal compressibility — appear on the RHS
What single calculus fact underlies every Maxwell relation? Equality of mixed second partial derivatives of a state function (exact differential):
∂ 2 z / ∂ x ∂ y = ∂ 2 z / ∂ y ∂ x \partial^2 z/\partial x\partial y = \partial^2 z/\partial y\partial x ∂ 2 z / ∂ x ∂ y = ∂ 2 z / ∂ y ∂ x .
Natural variables of U U U ( S , V ) (S,V) ( S , V ) , since
d U = T d S − P d V dU=T\,dS-P\,dV d U = T d S − P d V .
Natural variables of G G G ( T , P ) (T,P) ( T , P ) ;
d G = − S d T + V d P dG=-S\,dT+V\,dP d G = − S d T + V d P .
Maxwell relation from Helmholtz free energy F F F ( ∂ S / ∂ V ) T = ( ∂ P / ∂ T ) V (\partial S/\partial V)_T=(\partial P/\partial T)_V ( ∂ S / ∂ V ) T = ( ∂ P / ∂ T ) V .
Maxwell relation from Gibbs free energy G G G ( ∂ S / ∂ P ) T = − ( ∂ V / ∂ T ) P (\partial S/\partial P)_T=-(\partial V/\partial T)_P ( ∂ S / ∂ P ) T = − ( ∂ V / ∂ T ) P .
Maxwell relation from U U U ( ∂ T / ∂ V ) S = − ( ∂ P / ∂ S ) V (\partial T/\partial V)_S=-(\partial P/\partial S)_V ( ∂ T / ∂ V ) S = − ( ∂ P / ∂ S ) V .
Maxwell relation from H H H ( ∂ T / ∂ P ) S = ( ∂ V / ∂ S ) P (\partial T/\partial P)_S=(\partial V/\partial S)_P ( ∂ T / ∂ P ) S = ( ∂ V / ∂ S ) P .
How do you get H H H U U U Legendre transform
H = U + P V H=U+PV H = U + P V , giving
d H = T d S + V d P dH=T\,dS+V\,dP d H = T d S + V d P .
Why must a potential be in its natural variables for the Maxwell relation? Only then do the differential's coefficients equal clean quantities (
T , − P , . . . T,-P,... T , − P , ... ) whose cross-derivatives give the relation.
State the energy equation and what it proves for an ideal gas. ( ∂ U / ∂ V ) T = T ( ∂ P / ∂ T ) V − P (\partial U/\partial V)_T=T(\partial P/\partial T)_V-P ( ∂ U / ∂ V ) T = T ( ∂ P / ∂ T ) V − P ; for ideal gas it equals
0 0 0 , so
U U U is volume-independent.
Where does the minus sign in a Maxwell relation come from? Directly from the sign of the term in the differential (e.g.
− P d V ⇒ N = − P -P\,dV \Rightarrow N=-P − P d V ⇒ N = − P ).
Exact differential dz eq M dx plus N dy
Equality of mixed partials
Potentials are state functions
Internal energy U vars S V
Intuition Hinglish mein samjhoDekho, Maxwell relations koi naya physics law nahi hai — ye sirf calculus ka ek simple fact hai jo thermodynamics pe apply hota hai. Fact ye hai: agar koi function smooth hai (jaise hamare potentials U , H , F , G U, H, F, G U , H , F , G mixed second partial derivatives order pe depend nahi karte . Matlab pehle x x x y y y
Ab har potential ka ek natural differential hota hai. Jaise d U = T d S − P d V dU=TdS-PdV d U = T d S − P d V T T T − P -P − P ( ∂ T / ∂ V ) S = − ( ∂ P / ∂ S ) V (\partial T/\partial V)_S=-(\partial P/\partial S)_V ( ∂ T / ∂ V ) S = − ( ∂ P / ∂ S ) V H , F , G H, F, G H , F , G chaar relations mil jaate hain. Yaad rakhna: minus sign seedha differential ke sign se aata hai, isliye ratta mat maaro, derive karo .
Ye relations kyun important hain? Kyunki kuch quantities measure karna mushkil hai — jaise entropy ka pressure ke saath kaise change hota hai ( ∂ S / ∂ P ) T (\partial S/\partial P)_T ( ∂ S / ∂ P ) T − ( ∂ V / ∂ T ) P -(\partial V/\partial T)_P − ( ∂ V / ∂ T ) P ( ∂ U / ∂ V ) T = 0 (\partial U/\partial V)_T=0 ( ∂ U / ∂ V ) T = 0 prove kiya, assume nahi kiya. Bas yaad rakho: ek smooth hill, do raste, same height — wahi pura concept hai.
