First of all, I am not an expert. I was reading about the holographic principle and came across the Planck area. It says that Planck area is the square of Planck length and there were some pictures like this: enter image description here source

I know that this Planck area is used for black holes. But doesn't matter if this Planck area is for a triangle or a square?

And there was a clip that Leonard Susskind was saying that Planck area is $10^{-33}$ cm on a side. source 39:44.

But isn't it that the Planck length is $10^{-33}$ cm? If so, what is the "on a side"? Does it mean a face? Please keep it simple so I can understand it.

  • $\begingroup$ The picture is wrong by a factor. 1 planck area squared contains not a 1/4 bit nor 1 bit but 1/4 nat of information which is equal to $\frac{\ln 2}4$ bit. $\endgroup$ – Anixx Aug 29 '19 at 14:12

The Planck length is usually defined as

$$l_P=\sqrt\frac{\hbar G}{c^3}\approx 1.6\times 10^{-35}\,\text{m},$$

where $\hbar$ is the reduced Planck constant, $G$ is Newton’s gravitational constant, and $c$ is the speed of light.

The Planck area is usually defined as the square of the Planck length,

$$A_P=l_P^2=\frac{\hbar G}{c^3}\approx 2.6\times 10^{-70}\,\text{m}^2.$$

There are various quantum gravity theories, none yet accepted. The role of the Planck length and Planck area in physics is therefore still being researched. In particular, bits of information in black holes should not be pictured as being stored in nice little triangles or nice little squares on the event horizon. One bit of information probably “requires” about one Planck area, in some sense, but we don’t yet understand just how the information is stored so we can’t talk about it being in a particular place on the horizon or having a particular shape like a triangle or a square. That diagram is not to be taken literally.

As for the phrase “on a side”, it can he used to talk about the sizes of squares or triangles. A square has four sides and a triangle has three sides, so instead of talking about their area you can talk about the length of their sides.

  • $\begingroup$ G. Smith, thank you for this, it is helpful. If you could, I'd appreciate your perspective on how entropy is "encoded" in the area of the event horizon, this is something that I simply cannot visualize- NN $\endgroup$ – niels nielsen Aug 19 '19 at 2:27
  • $\begingroup$ @nielsnielsen I’m not a quantum gravity expert (I wish I was!) and don’t have a perspective on quantum models of black hole entropy. I am vaguely aware that string theory can explain the entropy by counting stringy microstates, but I don’t understand where those strings are relative to the black hole, or whether that question even makes sense. Maybe a string expert can comment. $\endgroup$ – G. Smith Aug 19 '19 at 4:22
  • $\begingroup$ I'll post this as a question here, if I get my courage up! $\endgroup$ – niels nielsen Aug 19 '19 at 5:18
  • $\begingroup$ The picture is wrong by a factor. 1 planck area squared contains not a 1/4 bit nor 1 bit but 1/4 nat of information which is equal to $\frac{\ln 2}4$ bit. $\endgroup$ – Anixx Aug 29 '19 at 14:13

The Planck length is approximately $1.616255\times 10^{-35}$ m. You get the Planck area by squaring that, just like you get the square metre by squaring a metre. So the Planck area is approximately $2.61228\times 10^{-70}\,\mathrm{m}^2$.

As Wikipedia says, we expect geometry to get weird near that scale, so despite your diagram it's probably not appropriate to think about the Planck area in terms of Euclidean squares and triangles.

The reason I say "expect" and "probably" is that we don't yet know what really happens at that scale, we need a proper theory of Quantum Gravity for that. And even when such a theory is developed we may never be able to verify its predictions at that tiny scale: the energy required to do that is vast. It's essentially equivalent to the energy density in the first few Planck time units after the Big Bang started.

  • $\begingroup$ To test something at the Planck scale you need much less energy. A planck mass, that is, equivalent of 0.02 mg or about an energy of a considerable explosion. The problem is to accelerate an electron to an energy of an aviabomb. $\endgroup$ – Anixx Aug 29 '19 at 14:17
  • $\begingroup$ @Anixx To be fair, there are various indirect ways to probe the Planck scale, as Lubos Motl discusses here. An electron with kinetic energy equal to the Planck energy has $\gamma\approx 2.39\times 10^{22}$, which is rather fast. ;) At that speed, 1 second gets time dilated to around 757 trillion years. $\endgroup$ – PM 2Ring Aug 30 '19 at 1:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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