# In a canonical ensemble why does change in energy not cause the temperature to change?

I understand that energy of a system in a canonical ensemble is given by the Boltzmann distribution and the temperature is fixed by the heat bath. If the temperature is a measure of kinetic energy, shouldn't fixing temperature cause the energy to remain constant?

## 3 Answers

The temperature is the measure of the average kinetic energy.

In the canonical ensemble, different from the microcanonical ensemble, the total energy $E= E_\text{kin} + V = p^2/2m+V(x)$ fluctuates. The average kinetic energy is given by $$\left\langle E_\text{kin} \right\rangle = \frac{\int d^3x\,d^{3}p E_\text{kin} e^{-E/k_B T}} {\int d^3x\,d^{3}p e^{-E/k_B T}} = \frac{\int d^{3}p E_\text{kin} e^{-E_\text{kin}}/ k_B T} {\int d^{3}p e^{-E_\text{kin}/k_B T}} = \frac{3}{2}k_B T .$$ For $N$ particles, we have the result $$\left\langle E_\text{kin} \right\rangle = \frac{3}{2} N k_B T.$$

The fluctuations in the kinetic energy are described by the standard deviation $\delta E_\text{kin}$. They describe the effect that the kinetic energy is not fixed but fluctuates due to the presence of the reservoir. Statistical mechanics determines the fluctuations to be given by $$(\delta E_\text{kin})^2 = \left\langle (E_\text{kin} -\tfrac32 k_B T)^2 \right\rangle =\frac{\int d^{3}p (E_\text{kin} -\tfrac32 k_B T)^2 e^{-E_\text{kin}}/ k_B T} {\int d^{3}p e^{-E_\text{kin}/k_B T}} = \frac{3}{2} (k_B T)^2 .$$ For $N$ particles, we have the result $$\left\langle E_\text{kin} \right\rangle = \frac{3}{2} \sqrt{N} k_B T.$$

Thus we have that the kinetic energy is a stochastic variable. It is fluctuating and given by $$E_\text{kin} = \tfrac32 Nk_B T \left(1 \pm \sqrt{\tfrac2{3N}} \right)\,.$$ For large systems ($N\to\infty$) the fluctuations are small and thus the kinetic energy of the system is well described by the average.

The bath is assumed to be so large that the small amount of energy it exchanges with the system is a negligible fraction of its total energy, and so results in negligible fluctuations in its average energy, which determines the temperature.

• I think the OP is asking about the system, not the bath. – Aaron Stevens Mar 8 '18 at 2:42

The relation $$\langle E \rangle = \langle E_{kin} \rangle = \frac 3 2 N k_B T$$ is valid only for an ideal gas. Notice the $\langle \cdot \rangle$s, which means that we are considering average quantities.

However, the answer to

shouldn't fixing temperature cause the energy to remain constant?

is yes, if the number of particles $N$ is large enough.

That is because it can be shown that in general the relative energy fluctuation satisfies (see for example this answer of mine):

$$\frac{\sqrt{\langle E^2 \rangle - \langle E \rangle ^2}}{\langle E \rangle} \equiv \frac{\delta E}{\langle E \rangle} \propto \frac 1 {\sqrt N}$$

which means that

$$\lim_{N \to \infty} \frac{\delta E}{\langle E \rangle}= 0$$

i.e., in the limit of large system size fixing the temperature is equivalent to fixing the energy.