Is the partition function of a 3D vibrating string a sum of discrete energies, an integral of an energy continuum, or both?

$$ Z_{\text{disc}} = \sum_{k=1}^{\infty}g_ke^{-\beta E_k} $$ or $$ Z_{\text{cont}} = \int_{1}^{\infty}g_ke^{-\beta E_k}dk $$

  • $\begingroup$ Why does the integral go from 1 to $\infty$? Also, what are the boundary conditions of the string? $\endgroup$
    – DanielSank
    Commented Jul 6, 2015 at 21:45
  • 1
    $\begingroup$ A good answer to this question would explain the modes of a vibrating string and how they are critically dependent on boundary conditions. The statistical mechanics is trivial once this is understood. $\endgroup$
    – DanielSank
    Commented Jul 6, 2015 at 21:46

2 Answers 2


Ideal string cannot be described in terms of canonical ensemble (with Boltzmann probability distribution).

The fact it is ideal means it has infinity of degrees of freedom and since each would have, according to the Boltzmann probability distribution, average energy

$$\frac{1}{2}k_B T,$$

total energy of the string would be, on average, infinite. This is physically wrong result, so one should not apply the assumption of thermodynamic equilibrium to systems with infinite number of degrees of freedom.

However, if you replace the string by finite number of bodies connected with finite number of strings, this problem vanishes. The partition function of such system in classical statistical physics is

$$ Z = \int_\Gamma e^{-\frac{H(q,p)}{k_B T}}\,dqdp $$

where $H$ is Hamiltonian function for the system and $\Gamma$ is the accessible phase space.

  • $\begingroup$ You stated correctly that an ideal string has infinitely many degrees of freedom, but not the cardinality of phase space (which was the question). I believe the condition that the length is an integer multiple of the wavelength means the phase state must be discrete. $\endgroup$
    – Real
    Commented Jul 6, 2015 at 22:15
  • $\begingroup$ What do you mean by "cardinality of phase space"? The boundary condition does imply every possible state of the string is sum of discretely indexed standing waves, but this does not mean phase space is discrete. Intensity of each standing wave can vary continuously. $\endgroup$ Commented Jul 6, 2015 at 22:30

If you assume linearity and the string is finite, then the partition function is a sum of discrete energies: it is the same case as a string in 2D ( $y=f(x,t)$ ), but with two uncoupled dynamics: $(y,z)=(f(x,t),g(x,t))$.

This can be shown by considering an oscillator in 2 dimensions with dynamics $\ddot{\vec{x}}=-k\vec{x}$: you will see you can separate it into two 1D oscillators (either $x,y$ modes or a radial and an axial mode).

enter image description here

More details on 2D oscillators here: http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node28.html


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