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Temperature is defined as \begin{equation} \frac{1}{T} \equiv \left(\frac{\partial S}{\partial U}\right)_{N,V}. \end{equation} However, a system with a fixed temperature does not have a fixed amount energy, rather the ensemble average of the energy of the system is fixed. This means that the definition of temperature is ambiguous unless it's made clear what we mean by the internal energy $U$. In the definition of temperature, does $U$ represent the ensemble average of the energy of the system?

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    $\begingroup$ For a macroscopic system, the energy of the system will be essentially indistinguishable from the ensemble average with probability essentially equal to $1$. $\endgroup$ – Yvan Velenik Apr 17 '18 at 7:37
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The formula you quote for the above is the definition of temperature for a macroscopic system (large number of microscopic components) where, as was mentioned in the comments the ensemble average will be indistinguishable from the energy of the system.

To see where this argument comes from though, consider the statistical mechanics definition of temperature. If we have a system with a degree of freedom, $X$, then we define the multiplicity, or number of states of the system when $X$ takes a value $x$ as $\Omega(x)$.

For large systems, the system tends strongly towards the state of largest multiplicity. So we can solve:

$$ \frac{\mathop{d} \Omega(x)}{\mathop{d}x} = 0 $$

To find the equilibrium value that gives the maxima for the multiplicity.

Then introduce the additional constraint of having a fixed amount of energy, $E$ and let $E(x)$ represent the amount of energy in the system when $X = x$. So we want to maximize $x$ while keeping the constraint $E(x) = E$. To do this, introduce a lagrange multiplier $t(E - E(x))$. For convenience, work with $\ln \Omega(x)$ Then:

$$ L(x) = \ln(\Omega(x)) + t(E - E(x))$$

Solving for $\partial L / \partial x$ to minimize $L$ wrt $x$ and $t$ we can find a expression for for t:

$$ t = \frac{\partial \ln(\Omega(x))}{\partial E}$$

Then looking at the thermodynamic identity as you cited it, and recalling that $S = k_b \ln(\Omega)$ we can see that:

$$ T = \frac{1}{tk_b} $$

is just the reciprocal of the lagrange multiplier we use to maximize the entropy of a system under the condition of fixed energy. Introducing the constant $k_b$ merely provides the historically desired dimensions whereas $t$ itself has dimension of inverse energy.

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