Assume we have a classical harmonic oscillator $$ \ddot{x} = -k^2x.$$ Then the general solutions are of the form $x(t) = x_0cos(kt) + \frac{v_0}{k}sin(kt)$ where $x_0$ and $v_0$ are initial conditions. Lets assume that $x_0=0$. The time average of any quantity $f$ is given by $$ \frac{1}{T}\int_0^Tf(\frac{v_0}{k}sin(kt))dt. $$ For instance average oscillation amplitude as $T\rightarrow \infty$ is $$ <x^2>=\lim_{T\rightarrow \infty}\frac{1}{T}\int_0^Tv^2_0sin^2(kt)dt = \frac{v_0}{2k^2}. $$ However if you consider the ensemble average using Boltzmann density $e^{-\beta(\frac{1}{2}mv^2 + k^2x)}$ you get that the average value is $$ <x^2>=\frac{1}{\beta k^2}. $$ Now this is a system near equilibrium (i.e the measure above is ergodic for this dynamics is what I am assuming) so shouldnt one get that the space average is equal to time average for all most all initial conditions.I could say that I am just choosing bad initial conditions but I can not see how one would generally produce a term like $\beta$ by choosing initial conditions correctly.


First, your Hamiltonian is wrong; you should be getting $(\beta~k)^{-1}$ for the force $k x$.

Second, your definitions of $k$ are going to be slightly off; the Hamiltonian $\frac12 m v^2 + \frac12 k x^2$ corresponds to the dynamics $\ddot x = -\omega^2 x$ only for $\omega^2 = k/m.$ Best then to write $\frac12 m (v^2 + \omega^2 x^2)$ with $\langle x^2\rangle = (\beta m \omega^2)^{-1}.$

Third, your interpretation is wrong; the Boltzmann factor comes from the canonical ensemble which is derived from the microcanonical ensemble in the limit where it is connected to a large thermodynamic system maintained at a constant temperature; your spring in the first case is not connected to any such system but rather is undergoing simple harmonic motion as if undisturbed by anything.

  • $\begingroup$ Thanks for the corrections but the reason why I get $(\beta k^2)^{-1}$ is because my potential is $\frac{1}{2}k^2x^2$ instead of $\frac{1}{2}kx^2$. Can you explain a bit more on the physical meaning of appearance of $\beta=k_bT$? What do you mean by a thermodynamic system? For instance would a hot both suffice whose equations are given by $\ddot{x} = kx - \mu\dot{x} + R(t)$ where $R(t)$ is random force.? As far as I remember even if $\beta$ appears in the variance of $R(t)$, it does not appear for the average quantities. How would one produce the average (\beta k)^{-1} from some time average? $\endgroup$
    – Sina
    Feb 20 '16 at 0:00
  • $\begingroup$ How would one physically explain the fact that the two averages differ by a constant $k_bT$ Does thermodynamic enviroment mean replace $v_0$ by the average speed of all oscillators in the thermodynamic enviroment? $\endgroup$
    – Sina
    Feb 20 '16 at 0:16
  • $\begingroup$ @Sina: I have to confess, I'm not 100% sure! However, you're right that the introduction of $R(t)$ such that $\langle R(t)~R(t')\rangle = 2(\mu/\beta)\delta(t-t')$ should thermalize a harmonic oscillator. I don't think you're right about "it does not appear for the average quantities" in general; it seems like it probably disappears for first moments $\langle x \rangle$ but probably persists for second moments $\langle x^2 \rangle.$ Maybe I'm wrong about that, but it would mean that, e.g., the Brownian motion variance wouldn't increase with $T$ which seems crazy. $\endgroup$
    – CR Drost
    Feb 21 '16 at 0:54
  • $\begingroup$ Your answer has been quite helpful for me to understand the subtle difference. And you are right that it appears in the second moments. I found some nice text that does alot of examples including oscillator. Here it is: web.phys.ntnu.no/~ingves/Teaching/TFY4275/Downloads/kap6.pdf $\endgroup$
    – Sina
    Feb 21 '16 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.