The equipartition theorem states that if $x_i$ is a canonical variable (either position or momentum), then

$$\left\langle x_i \frac{\partial \mathcal{H}}{\partial x_j}\right\rangle = \delta_{ij}\ k T.$$

Now, say one has a continuous medium (a field) $\phi(t,x)$ with ergodic properties.

What would the equipartition theorem be for this field?

My (naive?) guess:

since it is such that $x_i(t)\mapsto\phi(t,x)$, then

$$\left\langle\phi(x,t)\frac{\delta\mathcal H}{\delta \phi(y,t)}\right\rangle=\delta(y-x)kT,$$

$$\left\langle\dot\phi(x,t)\frac{\delta\mathcal H}{\delta \dot\phi(y,t)}\right\rangle=\delta(y-x)kT,$$

$$\left\langle\dot\phi(x,t)\frac{\delta\mathcal H}{\delta\phi(y,t)}\right\rangle=0=\left\langle\phi(x,t)\frac{\delta\mathcal H}{\delta \dot\phi(y,t)}\right\rangle,$$

where $\frac{\delta}{\delta \phi(y,t)}$ refers to functional derivative and $\delta(y-x)$ is the Dirac delta distribution.

The Hamiltonian in question is for Lorentz invariant 1+1 dim real scalar field with potential not differentable at minimum. Simplest case:

$$\mathcal H=\int dx\left[ \frac12\left(\frac{\partial\phi}{\partial t}\right)^2 + \frac12\left(\frac{\partial\phi}{\partial x}\right)^2 + |\phi| \right]$$

  • 1
    $\begingroup$ Applying equi-partition to continuous systems with infinite number of degrees of freedom leads to infinite energy, which means wild states that have no or problematic representation, and thus this also likely leads to (regarding physics) misleading results. $\endgroup$ Commented Jul 13, 2022 at 0:26
  • $\begingroup$ What kind of Hamiltonian in terms of $\phi$ are you considering? $\endgroup$ Commented Jul 13, 2022 at 0:26
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    $\begingroup$ Discrete version is fine, because it has finite number of degrees of freedom. The problem is with the continuous field, because it has infinite number of degrees of freedom. In the hypothetical ensemble obeying equi-partitioning, most field functions at arbitrarily small scales contain infinite energy, which is problematic. $\endgroup$ Commented Jul 13, 2022 at 11:56
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    $\begingroup$ @JánLalinský isn't that what the Dirac delta exists for? $\endgroup$ Commented Jul 13, 2022 at 18:07
  • 1
    $\begingroup$ I don't see how Dirac delta has anything to do with this problem. This problem is also called ultraviolet catastrophe. $\endgroup$ Commented Jul 13, 2022 at 22:16


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