Why is entropy expanded only in dV and dT, not all variables?

[Jacob Nie]

Homework Statement

There is an equation in Riley's Mathematical Methods that I am confused about:

Applying (4.43) to $dS$, with variables $V$ and $T$, we find
$$dU = T \ dS - P \ dV = T\left[ \left(\dfrac{\partial S}{\partial V}\right)_T \ dV + \left(\dfrac{\partial S}{\partial T}\right)_V \ dT\right] - P \ dV.$$

Relevant Equations

Eq 4.43:
$$ dU = \left(\dfrac{\partial U}{\partial X}\right)_Y \ dX + \left(\dfrac{\partial U}{\partial Y}\right)_X \ dY$$

What I don't understand is why $dS$ is expanded in only the two differentials $dV$ and $dT.$ Why doesn't it look more like:
$$dS = \left(\dfrac{\partial S}{\partial V}\right)_{T,P,U} \ dV + \left(\dfrac{\partial S}{\partial T}\right)_{V,P,U} \ dT + \left(\dfrac{\partial S}{\partial P}\right)_{V,T,U} \ dP + \left(\dfrac{\partial S}{\partial U}\right)_{V,T,P} \ dU$$
?

 

[vela]

Those other variables, P and U, aren't independent of V and T. For an ideal gas, for example, if you vary V and T, you've determined dP through the ideal gas law.

 

[robphy]

In this problem,
the fuller description of the energy is U(S,V) [thinking of the energy as function on the S-V plane]
and
the fuller description of the entropy is S(V,T) [thinking of the entropy as function on the V-T plane].

 

[Jacob Nie]

Thank you for the responses - that makes sense.

I forgot to read the sentence of the book that said:

These four quantities are not independent, since only two of them are independently variable.

 

[Chestermiller]

According to the phase rule, the thermodynamic equilibrium state of a pure single phase material is determined by only two independent parameters.