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Chandrasekhar's Transitivity Logic

  1. Jul 24, 2013 #1
    I've just started reading Chandrasekhar's Introduction to the Study of Stellar Structure, and I'm having trouble following one of his mathematical assertions. Rather than quote the relevant parts in their entirety here, I've typeset them and linked them here. (For those interested, the entire book is available from the Internet Archive). I hope using an outside link isn't bad manners in forums.

    What I don't understand, and would like someone to explain, is why the transitivity of thermal equilibrium is both sufficient and necessary (cf "this is then, and only then, possible...") for the condition of thermal equilibrium to have the form

    t1(p1,V1) - t2(p2,V2) = 0

    (same as Eq (4) in Chandrasekhar, but where I've used subscripts instead of bars). Clearly anything of that form implies transitivity, but I don't understand why transitivity implies that form. Any help?
  2. jcsd
  3. Jul 24, 2013 #2

    Stephen Tashi

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    You'll probably get more help if you define "transitivity of thermal equilibrium" - or post in a physics section where the audience might be familiar with that phrase.
  4. Jul 24, 2013 #3
    Transitivity is a purely mathematical concept. I think physicists would be less likely to understand it than mathematicians. If ~ is some relation, then transitivity is the statement that if a~b and b~c, then a~c. "Transitivity of thermal equilibrium" just means that if some physical system A is in thermal equilibrium with system B, and B is in thermal equilibrium with C, then A is in thermal equilibrium with C. The fact that in this case the relation happens to have a physical interpretation doesn't make my question anything other than a purely mathematical one. Chandrasekhar himself gives the mathematical definition of thermal equilibrium in Eq (1) which I included in the link above.
  5. Jul 25, 2013 #4
    Let me recast the statement of my question in purely mathematical terms.

    Let [itex]S = \{(a,b)|a,b \in \mathbb{R}; a>0; b>0\}[/itex].

    Define a function [itex]F:(S \times S) \rightarrow \mathbb{R}[/itex], and define a relation [itex]\sim[/itex] on [itex]S[/itex] such that two elements [itex]s_1,s_2 \in S[/itex] are related [itex]s_1 \sim s_2[/itex] iff [itex]F(s_1,s_2) = 0[/itex].
    If [itex]\sim[/itex] is transitive (assume reflexive and symmetric only if necessary), show that [itex]F(s_1,s_2) = 0[/itex] can then and only then be expressed in the form
    [tex]T_1(s_1) - T_2(s_2) = 0,[/tex]
    where [itex]T_1,T_2:S \rightarrow \mathbb{R}[/itex] can be any arbitrary functions.
  6. Jul 25, 2013 #5

    Stephen Tashi

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    I doubt that's true. We can try [itex] a = 1/2, b = 7/2, f(x,y) = y((x-1)(x-2)(x-3))^2 + x((y-1)(y-2)(y-3))^2 [/itex], which has zeroes at (1,1)(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3).

    The physics problem must make stronger assumptions - perhaps something about physical laws being invariant under a linear transformation of the quantity that measures [itex] s_1,s_2 [/itex].
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