In deriving Planck's blackbody formula, the number density of normal modes (per unit frequency$^\dagger$) is found, given by $$ N(\omega)=\frac{V}{\pi^2c^3}\omega^2, $$ where $V$ is the volume of the blackbody. Then the average energy of a mode of frequency $\omega$ is calculated using Planck's quantization that energy of a normal mode with frequency $\omega$ can only be $n\hbar\omega$ for some whole number $n$.

But hold on a minute!

The energy of a linear quantum oscillator with frequency $\omega$ can be $(n+1/2)\hbar\omega$, and since in deriving Planck's formula, it is the analogy of normal modes of the electromagnetic waves inside the blackbody cavity with the normal modes of the linear oscillator is made, I'd expect that we take into account this "zero-point energy", $\epsilon_\omega = \hbar\omega/2$ for each normal mode. Note that this zero-point energy, $\epsilon_\omega$ depends on $\omega$.

But if I do take this into account then I get bizarre results!

The average energy of normal mode of frequency $\omega$ is then given by $$ \langle E_\omega \rangle = \frac{\hbar\omega}{e^{\hbar\omega/k_BT}-1} + \epsilon_\omega, $$ and this leads to the following energy density.

\begin{align} \rho(T, \omega) &:= \frac{1}{V} N(\omega)\langle E_\omega \rangle\\ &\;= \frac{\hbar}{\pi^2c^3}\frac{\omega^3}{e^{\hbar\omega/k_BT} - 1} + \underbrace{\frac{1}{\pi^2c^3}\omega^2\epsilon_\omega}_{\text{additional term}} \end{align}

Now, this result is catastrophic! Unless $\epsilon_\omega = 0$ for all $\omega$'s (in which case this will coincide with the "correct" Planck's formula), $\rho$ diverges as $\omega\to\infty$.

Questions: So what's the way out? Is the often-presented analogy with quantum oscillator plainly wrong? For electromagnetic radiation, is the zero-point energy exactly zero for all $\omega$'s?

$^\dagger$ By frequency, I mean angular frequency.


1 Answer 1


Thermodynamics is all about the study of change. For example, the first law $dU=PdV + TdS$ expresses the change of energy in terms of work and heat flow.

Note that the zero point energy depends only on frequency and fundamental constants. There is no external parameter we can vary to change the zero point energy. So the zero point energy is, thermodynamically speaking, decoupled from the rest of the system, and therefore can be ignored.

Now there are some caveats.

First, it's not quite true that the fluctuations do not couple to an external parameter. If you consider a box with movable, reflecting walls, then there is a difference in how much zero point energy is contained in the box, then there would be in an equivalent amount of empty space without the reflecting walls. The reason is that the reflecting walls imposes boundary conditions which remove some of the modes from the sum over $n$. This difference in energy leads to the Casimir force acting on the walls of the box. However, this is a very small effect.

Second, gravity should couple to these fluctuations. There is an infinite (or at least very large) amount of energy density sourced by the zero-point fluctuations that should be causing the Universe to accelerate at an infinite (or at least very large) rate, but we don't observe this. This is the cosmological constant problem, to which there is no universally agreed-upon solution.

  • $\begingroup$ But here, I'm not looking at changes. I'm finding the energy density at equilibrium. So I don't think that neglect of the zero-point energy is quite justified. $\endgroup$
    – Atom
    Nov 5, 2020 at 5:55
  • $\begingroup$ I could imagine saying there is an energy density of "ghosts" which exists, but does not couple to any experimentally controllable parameter and which does not have any effect on any experiments. On the one hand I could insist that you include the energy density of ghosts in your calculations, on the other hand if you simply leave the ghost contribution out you will get the same answer for any observable quantity. The situation is similar with the zero point energy, except for very special situations where you do need to include it (ie, the Casimir effect, or maybe in gravity). $\endgroup$
    – Andrew
    Nov 5, 2020 at 5:58
  • $\begingroup$ Sorry, but I can't understand this "coupling with experimentally controllable parameter." Can you elaborate (or point me to some resource)? $\endgroup$
    – Atom
    Nov 5, 2020 at 6:04
  • 1
    $\begingroup$ Another way to say this is that you can't convert the zero point energy to any other kind of energy, and vice versa. So from the point of view of thermodynamics there's no point in including this kind of energy. It's like having a box that you can't exchange any energy or entropy with, you can say the box is there but it has no observable consequences. The Casimir effect is the one situation I know of where you actually do need to include it, but for >99% of applications you don't need to worry about the Casimir effect. $\endgroup$
    – Andrew
    Nov 5, 2020 at 6:07
  • $\begingroup$ It's analogous (but not quite the same) as saying the energy is defined only up to an overall constant. $\endgroup$
    – Andrew
    Nov 5, 2020 at 6:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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