Why does a photon with wavelength equal to the radius of the blackhole it's falling into have 1 bit of information?

In this lecture: https://www.youtube.com/watch?v=yMRYZMv0jRE&t=1260s Susskind derives the entropy of a black hole by assuming a bit of information could be added by a photon with wavelength equal to that of the black hole's radius. I'm asking for the reasoning behind that.

• I'm not an expert by any means, but the situation described sounds a lot like entropy, $dS = dQ / T$ where the radius of the blackhole relates to its temperature, $T$, and the heat added by radiation absorption of the referenced photon corresponds to $dQ$. Commented Oct 4, 2022 at 0:39
• Susskind clarifies briefly by claiming that as stated, the thermodynamic process of adding a single photon with radius $R$ matching that of the blackhole only has 1 bit of information because it only has two states in the model described, $\text{true} \implies \text{inside the blackhole}$, $\text{false} \implies \text{not inside the blackhole}$ Commented Oct 4, 2022 at 0:47
• Could you pls explain why? Commented Oct 4, 2022 at 0:52

The original argument that particles falling into a black hole led to an increase in entropy was due to Bekenstein, who argued that once a particle has passed into a black hole, we (as external observers) have lost the information about whether the particle still exists. That is the answer to one yes-or-no question and thus one bit of information. Loss of information is then equivalent to increase in entropy.

That one bit was actually, Bekenstein argued, the minimum information loss caused by a particle falling beyond the event horizon. If particle is in motion when it passes the event horizon, then all information about the specific trajectory of the particle is also lost—subsumed into the macroscopic energy-momentum of the black hole. However, if the particle is captured at a turning point of its classical orbit,* no velocity information is lost, since the particle is stationary at its capture. That also corresponds to the kind of classical process by which the area of the event horizon grows the least.

This is only a hand-waving argument. However, Bekenstein was able to derive the correct functional forms for the entropy and temperature of a black hole, although there were unknown constants involved. To get more precise expressions for the information lost when a particle enters the black hole requires a much more careful treatment. To get the precise values of expressions in black hole thermodynamics, with no unknown $$\mathcal{O}(1)$$ factors, requires Hawking's more careful semiclassical reasoning about propagating fields in the black hole background.

*This obviously requires there to be other gravitating bodies potentially pulling the test particle away from the black hole.

• So, the Susskind's "derivation" was only for illustrative purposes, and the assumptions made were made to correspond to the correct entropy equation, which Bekenstein derived through an alternative method. Commented Oct 4, 2022 at 1:15
• @Matrix001 Yes, I would say that's correct.
– Buzz
Commented Oct 4, 2022 at 19:08

Well, my argument will be totally different.

Black hole dimensionless Bekenstein-Hawking entropy (information amount) can be expressed as : $$S_{BH} = \frac {4 \pi G M^2}{\hbar c} ~~~~(1)$$

Now, lets assume information change of $$1~\text{bit}$$ in $$(1)$$ equation for getting what mass portion must fall into black hole for increasing it's information content just by 1 bit :

$$\Delta M _{+1bit} = \sqrt{\frac {\hbar c}{4 \pi G}} ~~~(2)$$

Substituting all constants we get that $$\Delta M _{+1bit} = 10^{−9}~kg \approx 0.3~m_p ~~~(3)$$

We can get what photon wavelength corresponds to a given rest mass by : $$\lambda = \frac {h}{m_0c} ~~~(4)$$

In $$(4)$$ equation assuming rest mass which we've got from $$(3)$$, gives that:

$$\lambda_{+1bit} \approx 10^{−33}~m \approx 136~\ell_P,$$

In other words,- only very energetic photons can add 1 bit of information to black hole. Otherwise if very weak photons could add 1 bit to a black hole content (these of the wavelength order of black hole radius) - then black hole would be growing too fast.

EDIT

My fault. I've derived $$(2)$$ equation badly, due to not managing quadratic relationship between mass and entropy correctly. Correct expression should be : $$\Delta M _{+1bit} \approx \frac {\hbar c}{8 \pi G~M_0} ,~~~(5)$$

where $$M_0$$ is initial black hole mass (with -1 bit info). Now it is seen that when black hole mass increases,- $$M_0 \to \infty$$, then $$\Delta M _{+1bit} \to 0$$. That is with bigger mass less and less additional mass we need to add to supply black hole total information with additional 1 bit new info.

So it seems that when black hole grows, it looses it's ability to compress information, i.e. it compresses information less and less effectively, probably due to increasing entropy of black hole.

When I've checked the numbers, I've got that for middle size black holes, indeed for 1 bit addition we need to send photon approximately with the wavelength of black hole radius to be absorbed by black hole. (Within wavelength mismatch error of 1 order of magnitude).

Conclusion is that my argument was faulty.

• Your second equation isn't correct because entropy doesn't increase linearly with mass. Your result only holds for a blackhole with 0 initial mass, but for larger blackholes the LHS of (2) should be $\sqrt{M_{1}^{2}-M_{0}^{2}}$. Commented Oct 4, 2022 at 23:20
• Also, I think in the lecture Susskind was referring to the angular wavelength being equal to the radius. Commented Oct 4, 2022 at 23:31
• (2) eq is derived not for initial mass, but for mass change which results in additional 1 bit of information. Commented Oct 5, 2022 at 8:08
• Yeah, the mass change for 1 bit of entropy changes with the "initial mass" $M_{0}$ (i.e.: Mass before absorption). In my previous equation $M_{1}=M_{0}+ΔM_{+1bit}$. Commented Oct 5, 2022 at 8:10
• Your equation approaches mine as the initial mass approaches positive infinity. Commented Oct 5, 2022 at 22:15