3
$\begingroup$

I was watching "Leonard Susskind on The World As Hologram" ( youtube ).

At one point he describes the way Bekenstein calculates the entropy of a black hole. Paraphrasing: Take a minimally sized black hole. Add some "information" to it. Measure how the horizon changes. Rinse repeat several times, It will become clear that entropy is proportional to the area of the black hole.

OK. Makes sense, and for a Schwarzschild black hole seems straightforward. However things I began to think about: what units is the information measured in, and how you convert the amount of energy into the entropy?

Also what books are there that replication this method in great detail?

$\endgroup$
1
$\begingroup$

The following argument is from Susskind's presentation in The Black Hole War, 2008, p. 152. There is a simple algebraic proof that adding 1 bit to a black hole increases its area by on the order of one Planck unit. (This is omitting factors of order unity; there is actually a factor of 1/4 involved.) Drop in a photon of energy $E$ that encodes one bit. If the photon was well localized, then there might be a lot more than 1 bit carried, since it would enter the black hole at a certain location. So to get just one bit, make $\lambda \sim R$, where $R$ is the Schwazschild radius. (Making $\lambda > R$ doesn't work, because the photon just scatters without being absorbed.) The photon's energy $E=hc/\lambda$ is equivalent to a mass $dm=E/c^2=h/c\lambda$. The Schwarzschild radius grows according to $R=2(G/c^2)m$, causing the area to grow as $4\pi R^2$. Neglecting factors of order unity, the growth in area is $R dR\sim m dm \sim Gh/c^3$, which is the Planck area.

what units is the information measured in

Entropy can be defined as the log of the number of accessible states, $\ln M$, or, in SI units, as $k\ln M$, where $k$ is Boltzmann's constant. If you let $k=1$, then essentially you're measuring entropy in something pretty close to units of bits. (The actual factor is that one bit is $(1/2)\ln 2$ units of entropy if $k=1$.)

$\endgroup$

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.