By R. Courant, D. Hilbert
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4-4), one may then conclude that the distribution function for the gas in this state may be found from the relation, f cf1c ff1 . This equation is equivalent to: ln f c ln f1c ln f ln f1 . (4-6) If the distribution function satisfies Eq. (4-1), one can obtain that wf wt 0 also, so that such a state of the gas is steady as well as uniform. 41 Chapter 4. The Uniform Steady-State of a Gas Now, consider the form of this distribution function. Eq. (4-6) shows that ln f is a summational invariant of encounters and, therefore, must be a linear combination of the three summational invariants.
Solution: For the time interval, dt , the position of the phase element, d * dvdr , is changed from v, r to vc v Fdt , r c r vdt . Then: d* c w vc, r c d* w v, r w vc, r c w v, r c d* w v, r c w v, r w v , v , v w x c, y c, zc w x , y , z w vcx , vcy , vzc x y z r c const d* d* . v const The external force, F , is assumed to be independent of v . 2. Prove that the relative motion of two interacting molecules with mutual potential energy depending only on the distance between the molecules may be considered as the motion of a single particle in a central force field.
DERIVATION OF THE BOLTZMANN EQUATION. It has been established previously that knowledge of the distribution function gives all the necessary information for a gas. To obtain the basic equation for the distribution function, consider a balance of the number of molecules that are located in the element, dvdr , of the six-dimensional phase space for the time interval, dt . Consider a gas in which each molecule is subject to an external force, mF , that is a function of r and t , but not a function of v .