Here's my guess, ($157\, \textrm{bits}$) and how I got there. Please feel free to disregard completely and give your own answer. I am required to give an attempted answer or the question gets deleted, so this one assumes a blackhole is a sphere in euclidean space.

My understanding (please correct any wrong assumptions as there may be many): The smallest black hole is a "Planck particle", which is a Planck mass in size, and has a theoretical radius of $5.72947 \times 10^{−35}\, \mathrm{m}$ (or so Wikipedia says).

A "Planck area" is the smallest area to encode a (digital) bit of data. In the Holographic principle, it's believed the information content of a black hole is equal to its surface area (or more specifically, it encodes one bit for every "Planck area" which can cover its surface/horizon).

Now, assuming a Planck particle is spherical and uses normal euclidean geometry and there aren't relativistic or quantum effects that I've completely ignored, this means you can calculate a Planck particle's surface area (from its radius), and thus its information content. Which I calculate as being $157\, \textrm{bits}$.

Is this correct? Does this make any sense at all? Is it possible to calculate the information content of the smallest hypothetical black hole? How would you do it? Is there any meaning to the data capacity of a Planck mass of black hole?

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    $\begingroup$ I think the assumption of Euclidean geometry will be violated in quite an extreme way, because a tiny black hole will be a very highly curved region of space-time. My own wild guess is that the smallest black hole will have an entropy of one bit, and therefore a surface area of four Planck areas -- but I have no idea if that is correct. It's an interesting question: +1. $\endgroup$
    – N. Virgo
    Commented Nov 13, 2013 at 8:51
  • 1
    $\begingroup$ you rounded to the nearest whole number. Think about the actual result of your calculation. There's so such thing as 2.5 bits. This leads me to believe that something is wrong. $\endgroup$
    – psitae
    Commented Oct 22, 2018 at 7:09
  • 1
    $\begingroup$ @psitae there is such a thing as 2.5 bits, two certain bits and a less certain one. $\endgroup$
    – codeshot
    Commented Aug 25, 2019 at 8:15
  • $\begingroup$ Computers use partial bits all the time, like 3 1/3 bits for decimal character encoding, similar to Base64 using 3/4th's of a byte for character information. Partial bits means the bit represents more than one thing. $\endgroup$
    – Zamicol
    Commented Nov 2, 2019 at 4:57

2 Answers 2


I decided to do the calculation and see what happens. According to Wikipedia, the radius of a Planck particle is $$ r = \sqrt{\frac{2Gh}{c^3}}. $$ Apparently, the surface area of a Schwarzchild black hole is given by $4\pi r^2$, the same as a Euclidean sphere. I found this surprising until user10001 confirmed that it's because $r$ is defined as $\sqrt{A}/4\pi$ and need not correspond to the distance from the centre, the meaning of which would be unclear in the case of a black hole.

This means that the surface area of a Planck particle is $$ A_\text{PP} = \frac{8\pi Gh}{c^3}. $$ The Bekenstein-Hawking entropy is given by $$ S_\text{BH} = \frac{\pi c^3 A}{2Gh}, $$ which for $A=A_\text{PP}$ is equal to 4. That's 4 natural units (or nats), which is is equal to $4/\ln 2$ bits.

So this argument would imply that the smallest black hole can encode about $5.77\;\mathrm{bits}$ on its surface. However, intuitively this feels pretty wrong, for a couple of different reasons. Firstly, integer numbers of nats basically never appear in physics or information theory. We use nats because they're a convenient unit to work with (natural logarithms are simpler to differentiate), not because you ever get a round number of them. Secondly, I don't much like the number 4. If you express the Bekenstein-Hawking formula in terms of Planck length you get $\frac{A}{4L_\text{P}^2}$, which might be related -- but I tend to think that 4 is just the result of the Planck length being defined as half what it really should have been. (In other words, with the benefit of hindsight, it would have been better to define $L_\text{P}$ so that a black hole's entropy is equal to its surface area in Planck lengths.)

But perhaps something can be done to bring the result more into line with intuition. The radius of a Planck particle (the first equation above) is defined by setting the Schwarzchild radius equal to the Compton wavelength, which to me (as a complete non-expert) sounds quite heuristical, so maybe there's some wiggle room. For example, it might make more sense to set the diameter equal to the Compton wavelength rather than the radius, which would give a figure of one nat rather than 4. If we can also somehow justify a factor of $\ln 2$ somewhere along the way we will end up with what (to me) seems the intuitively right answer, of $S=1\;\mathrm{bit}$.

Alternatively we can do the whole thing with a Planck-mass black hole (which has a slightly different radius, equal to $2\sqrt{G\hbar/c^2}$), in which case we get $2\pi\:\mathrm{nats}$, or about $9.06\:\mathrm{bits}$, which doesn't match intuition any better than the Planck particle case.

You also asked about what the meaning of the figure would be. If it's one bit then the meaning is reasonably clear: it means there are only two distinct states that the black hole's surface can be in. Presumably these are quantum states, meaning that the black hole can be in a superposition and there are various different measurements that can be performed. In this sense the smallest black hole would be something a little bit like a particle with a spin.

If it's one of the other answers then it's much less clear. There's no fundamental problem with non-integer numbers of bits, but (unless they're something of the form $\log_2 n$, which these aren't) they usually only appear in situations where there's noise or partial knowledge involved. For an irregular number of bits to turn up as a fundamental constant would be pretty weird, and I'm not at all sure what it would mean.

Finally, we can work backwards to deduce that a black hole with one bit of entropy must have a mass of $\sqrt{\ln 2}/2\pi$ Planck masses. I conjecture that this is the mass of the smallest possible black hole.

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    $\begingroup$ The Planck length 'should' have been nothing but what it is. It is the natural unit obtained by setting the fundamental constants to one and combining them in the dimensionally correct fashion. $\endgroup$
    – Danu
    Commented Nov 13, 2013 at 9:54
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    $\begingroup$ @Danu there are still some choices you have to make in defining them, since it depends which constants you consider the most fundamental. If I consider the ratio of black hole entropy to surface area to be an important fundamental constant, I will want to set it to one, at the expense of setting some other constant to something other than 1. It is, of course, fairly pointless to argue about which constants are the most fundamental, since this is purely a matter of taste. $\endgroup$
    – N. Virgo
    Commented Nov 13, 2013 at 10:00
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    $\begingroup$ "Perhaps $r$ is defined so as to make this so" Yeah, $r$ is defined to be square root of surface area/$4\pi$. In a curved geometry it need not be equal to the distance from the center. $\endgroup$
    – user10001
    Commented Nov 13, 2013 at 10:06
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    $\begingroup$ @user10001 many thanks for the clarification! $\endgroup$
    – N. Virgo
    Commented Nov 13, 2013 at 10:17
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    $\begingroup$ @user23660 'bit' can mean two subtly different things. It can either mean a component of computer memory with two distinct states, in which case the quantum eqiuivalent is a qubit, or it can mean a unit for measuring an amount of information. In QM this latter is still called a bit, and we say that the information capacity of a qubit is one bit. It's the amount-of-information sense that I'm using here. $\endgroup$
    – N. Virgo
    Commented Nov 13, 2013 at 15:09

A black hole with placnk length radius will have a 0.00546 bits. Two ways for calculating radius from nº of bits:

  1. double rs= sqrt(8.0 * nbits * pi * plankarea);

  2. double rs2 = 2.0 * planklen * sqrt(pi*(nbits + sqrt(nbits *(1+nbits))));

Second is more preccise. To get nº of bits from radius use:

aux1=32.0 * pi* pi * plankarea;

aux2 = (64.0 * pow(pi,3) * pow(plankarea,2))/pow(radiometros,2);



Here is a link to a google book that have a visible free page with that information: https://books.google.es/books?id=9oRVEAAAQBAJ&pg=PA432&lpg=PA432&dq=blackhole+information+bits+64+32&source=bl&ots=WdZ--pwzo7&sig=ACfU3U3Xl-EttEe5HEIFiCjG06EtWnzJ8Q&hl=es&sa=X&ved=2ahUKEwiC4ay6g-z1AhUM76QKHbHmD_0Q6AF6BAgDEAE#v=onepage&q=blackhole%20information%20bits%2064%2032&f=false

I searched in google: blackhole information bits 64 32

  • $\begingroup$ So what would that even mean? That you'd need 184 blackholes (1/0.00546) to store one bit of information? $\endgroup$
    – Qubei
    Commented Feb 8, 2022 at 2:32

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