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I am currently reading Demirel's Nonequilibrium Thermodynamics in order to learn more about entropy and nonequlibrium thermodynamics

However I and stumped on how the author's do the following step:

The pressure $P$ and temperature $T$ define the values at each point of the system and are therefore called intensive properties, some of which can be expressed as derivatives of extensive properties, such as temperature $$T=\left(\frac{\partial U}{\partial S}\right)_{V,N}$$ where $U$ is the energy and $S$ is the entropy. If $X$ denotes any extensive property (not necessarily a thermodynamic propert) of a phase, we may derive intensive properties denoted by $X_i$ and called as partial properties $X_i=\left(\partial X/\partial n_i\right)_{T,p,n_i}$ ($i\neq j$). For any partial property we have $dX=\sum_i\left(\partial X/\partial n_i\right)dn_i=\sum_iX_idn_i$ at constant $T$ and $P$. The Euler theorem shows that $X=\sum_iX_in_i$

Is the result from the text obtained from direct integration, or it just looks like it is integrated because of the consequence of Euler Homogeneous Function Theorem?

That is, is this relation $$ X=\sum_i X_in_i \tag{1} $$ the integration of this relation $$ dX=\sum_i X_idn_i? \tag{2} $$ But just because of Euler Homogeneous Function Theorem, if (2) is true, then (1) must also be true by the theorem?

Kyle Kanos
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Secret
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  • The print in the picture is tough to read. Also, why (specifically) are you confused by this derivation? – Sean Nov 18 '14 at 14:17
  • There is no derivation. That is what's confusing. There's a derivation of The Euler Theorem, but not of why the Euler theorem implies the result given on the left. – Brian Moths Nov 18 '14 at 14:26
  • At least I think so, otherwise it would be a math question. – Brian Moths Nov 18 '14 at 14:59
  • Yes, that's precisely what I am confused of, the author is saying that by Euler's Theorem, dX=sum (blah d(blah)) goes to X=sum (blah d(blah)) but Euler Theorem only relates f=sum(partials * x), not df=sum (partials * dx)

    Thus I don't understand how it is carried out exactly

    .If I am just integrating both sides, then I have to rationalize why the partials can be pulled out from the integral and the integral only integrate the differential dX terms

     – Secret Nov 18 '14 at 15:03
  • I just found something that is sort of related http://physics.stackexchange.com/questions/60084/why-does-u-t-s-p-v-sum-i-mu-i-n-i?rq=1

    .So using this question as a reference, are we actually not integrating, but simply using the property of X being 1st order homogenous so that by Euler Theorem X= sum paritals . n?

     – Secret Nov 18 '14 at 15:12
  • so in the end it LOOKS like as if the expression is being integrated both side, but is not? – Secret Nov 18 '14 at 15:59

1 Answers1

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Yes, it looks like it's integrated (and sometimes it is even called integrating), but really it is due to the Euler theorem.

Suppose $X$ is a first order homogeneous function of $n_1, n_2, \dots$ (this is equivalent to saying it is extensive) and that $\mathrm{d}X(\mathbf{n}) = \sum_i \frac{\partial X(\mathbf{n})}{\partial n_i}\textrm{d}n_i$. This is to say $\mathrm{d}X(\lambda\mathbf{n}) = \sum_i \frac{\partial X(\lambda\mathbf{n})}{\partial (\lambda n_i)}\textrm{d}(\lambda n_i)$.

Now it is straightforward to write $$X(\mathbf{n}) = \frac{\partial(\lambda X(\mathbf{n}))}{\partial\lambda} = \frac{\partial X(\lambda\mathbf{n})}{\partial\lambda} = \sum_i \frac{\partial X(\lambda\mathbf{n})}{\partial (\lambda n_i)}\frac{\partial (\lambda n_i)}{\partial\lambda} = \sum_i \frac{\partial X(\lambda\mathbf{n})}{\partial (\lambda n_i)}n_i$$

With $\lambda = 1$ we have $$X(\mathbf{n}) = \sum_i \frac{\partial X(\mathbf{n})}{\partial n_i}n_i = \sum X_i n_i$$

alarge
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