1
$\begingroup$

I encountered several times with the term of natural variables but I'm not sure I totally understand what does that mean. For example, in Hamiltonian mechanics we calculate the differential of the hamiltonian and see that $$d\mathcal{H}=\sum_i (\frac{\partial{\mathcal{H}}}{\partial{q_i}}\cdot dq_i+\frac{\partial{\mathcal{H}}}{\partial{p_i}}\cdot dp_i)+\frac{\partial{\mathcal{H}}}{\partial{t}}\cdot dt$$

so we say that the natural variables of the hamiltonian are $\{q_i,p_i\}$. Another example taken from thermodynamics is the helmholtz free energy. Again, the same calculation of it's differential gives: $$dF=-SdT-PdV$$

and so we say that the natural variables of the helmholtz free energy are $T$ and $V$ (disregarding the number of paricles). What exactly is the mathematical meaning of these natural variables and does it have a mathematical formal definition? does it relate to independence of these variables?

$\endgroup$

marked as duplicate by By Symmetry, ZeroTheHero, user259412, Jon Custer, SchrodingersCat Jun 9 '17 at 17:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

I shall answer the thermodynamics part.

In thermodynamics if you have the fundamental relation $U=f(S,V)$, where $U$ is internal energy, you can calculate every other thermodynamic property of a thermodynamic system in equilibrium. This is an axiom and cannot be derived. Now by performing a Legendre transformation, you may replace $S$ by $T\equiv\frac{\partial f}{\partial S}$, and so get a new function $F=g(T,V)$, and this new function contains exactly the same information as the previous one for internal energy. In this sense $T,V$ are natural variables for $F$, just as $S,V$ are natural variables for $U$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.