I read the wiki article on the holographic principle, but it never answered this question. Can anyone explain the math that leads to this conclusion?


2 Answers 2


Explaining the math behind the holographic principle would be lengthy exercise. Is that really what you want?

A short hand-waving argument would be that you can pack a limited number of qbits (in the form of photons) together in a given space. If you take long wavelength photons, you can pack a lot of them together before a black hole forms. Two fundamental principles limit the number of photons you can put in a sphere of radius $R$:

1) you can't make their wavelength too long as this would prevent you from localizing the photons within the sphere, and

2) if you make their wavelengths too short, the energy content within the sphere becomes too high and a black hole forms that would have a radius larger than $R$.

The bottom line is that you can pack no more than a number of photons proportional to $R^2$ into the sphere of radius $R$, provided these have wavelengths comparable to $R$.

If you would select massive qbits (rather than massless photons) things get worse.

  • 3
    $\begingroup$ You presumed correct, I wanted the handwavy math. The answer satisfies my curiosity sufficiently. $\endgroup$ Dec 14, 2012 at 2:46
  • $\begingroup$ Also, for a black hole, the average density of the volume enclosed by the event horizon goes down as the size increases. This is why R^2 for information density vs R^3 for volume. Can even be applied to universal scales. If you can get an estimate of the mass of the real universe, including outside of our observable universe, convert this to energy, then you know how big the real universe was at the time of the Big Bang. Before that, I guess whatever was there exceeded this limit, but then it didn't exist, or it was under a different set of physical laws, or some other loophole... $\endgroup$
    – user26589
    Jul 3, 2013 at 18:18

Start from the De Donder balance law for the entropy

$$\mathrm{d} S = \mathrm{d}_\mathrm{i} S + \mathrm{d}_\mathrm{e} S$$

where the first term in the sum is due to production within the volume and the second due to flows crossing the surface that encloses that volume. The holographic principle (as string theory and BH thermodynamics) only deals with the special case when the production term is zero [*]

$$\mathrm{d} S = \mathrm{d}_\mathrm{e} S$$

By definition the flow term is

$$\mathrm{d}_\mathrm{e} S \equiv -\int J_S \mathrm{d} A \mathrm{d} t$$

where $J_S$ is the amount of entropy transferred per unit time and unit area [**].

If we further approximate $J_S$ by its average, using the decomposition $J_S = \langle J_S \rangle + \delta J_S$, and integrating, we obtain the traditional proportionality between entropy $S$ and area $A$ reported in the link that you give

$$S = \kappa A$$

with superficial constant $\kappa \equiv - \int \langle J_S \rangle \mathrm{d} t$

Although the "holographic principle" can be considered a true principle in a string theory sense, it can shown to be a theorem derived under the special conditions stated here (and other more technical that I have omitted).

[*] In rigor they can deal with situations where is nonzero, but then only can compute the difference $\mathrm{d}S - \mathrm{d}_\mathrm{i} S$.

[**] This is considered an "areic flow" in the last terminology, but less modern terminology names it "flux density" and older terminology names it "flux".

  • $\begingroup$ The entropy that you calculate is the amount of entropy that can enter the sphere in a given amount of time. You find that this is proportional to the area. However, if you finish the integration and actually integrate from T=0 to T=infinity, the quantity of entropy that will be in the sphere will be infinity. Hence you have not succeeded in proving that the maximum amount of entropy in the sphere is proportional to the Area of the sphere. Thoughts? $\endgroup$
    – Anon21
    Feb 8, 2017 at 15:31
  • $\begingroup$ @AlexandreH.Tremblay. No, because for an infinite time span, <J_s> goes to zero to warrant a finite ĸappa and a finite transfer of entropy. $\endgroup$
    – juanrga
    Feb 19, 2017 at 12:05

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