Xtended tangent bundles turn out to become isomorphic that’s T Q TQ R2n1 . Within this certain instance, by assuming c in (159) can be regarded as a certain case of in (210) if the m = 2n 1, the mapping canonical coordinates (qi , pi , z) are decomposed as ( x a) = (qi , z) exactly where a = 1, . . . , n 1, and ( x) = ( pi) exactly where = 1, . . . , n. 5.two. Equilibrium Thermodynamics Obeying the geometry exhibited in the previous section, we take m = 3 with coordinates (S, V, N) R3 . Here, S stands for the entropy, V could be the volume, and N is the mole quantity of classical perfect gas. The conjugate variables ( T, – P, (R3) would be the temperature, the stress, plus the chemical potential, respectively. By employing the internal power U as the fiber coordinate, we full the following realization of the extended cotangent bundle (S, V, N, T, – P, U) T R3 . Take into consideration the make contact with one-form R3 = dU – TdS PdV – N. As a specific instance, we select the internal power U (S, V, N) = U0 V -1/c N (c1)/c exp( S), cNR (213) (212)as a function according to the base coordinates (S, V, N) R3 . Right here, U0 can be a positive continuous, c may be the heat capacity and R may be the universal gas continuous. The initial prolongation T U is a Legendrian submanifold N from the speak to manifold (T R3 , R3). By considering that the temperature T = U/S and the pressure P = -U/V, we’ve the following set of equations cV 1/c RT = U0 N 1/c exp( S), cNR PV = NRT, = (c 1) RT – TS/N (214)those realizing N . The Legendre Transformations. In the light of Theorem six, and the transformation (211), we now present the Legendre transformation involving the internal power, the enthalpy, the Helmholtz function, and also the Gibbs function. For a equivalent discussion, but in the framework of symplectic geometry, see [12]. We commence with all the Legendrian submanifold determined by the internal power U in (213). (1) We Lactacystin custom synthesis decompose the base variables as (S, N) and V and apply Theorem 6 towards the volume variable. This benefits using a quantomorphism computed to be 1 : T R3 – T R3 ,(S, V, N, T, – P, U) (S, – P, N, T, -V, U PV). (215)Mathematics 2021, 9,35 ofNote that, around the image space, the fiber element may be the Oltipraz supplier enthalpy function B = U PV. If we resolve the pressure in the equation P = -U/V, the enthalpy function may be written as a function with the new base variables (S, P, N), that’s B(S, P, N) = c U(c1)/c 1/(1c)PN exp(S), (c 1) NR(216)where c is really a continual defined to be c1/(1c) c-c/(1c) . In order that, the enthalpy function is yet another generator with the similar Legendrian submanifold. Certainly, the initial prolongation of T B is given the program of equations in (214) in order that 1 T B = N . (two) We get started as soon as much more together with the internal power, but this time we carry out the transformation towards the entropy variable S. For this case, we’ve got the quantomorphism 2 : T R3 – T R3 ,(S, V, N, T, – P, U) ( T, V, N, -S, – P, U – ST). (217)Within this case, the fiber term F = U – ST would be the Helmholtz function. Making use of the identity T = U/S, we write the Helmholtz function as a function in the base elements ( T, V, N) from the image space, which is F ( T, V, N) = cNRT 1 1 1 U0 log N – log V log . c c cRT (218)Hence, F is a further generator in the exact same Legendrian submanifold determined by the equations (214), that is certainly 2 T F = N . (three) This time, we contemplate the Helmholtz function F in (218) and apply the transformation provided in Theorem 6 for the volume variable, that may be( T, V, N, -S, – P, F) ( T, – P, N, -S, -V, F PV). (219) The fiber term G = F PV would be the Gibbs function.