Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

124 submissions , 105 unreviewed
3,647 questions , 1,242 unanswered
4,638 answers , 19,680 comments
1,470 users with positive rep
437 active unimported users
More ...

Bekenstein bound and “dimensionality” of entropy

+ 2 like - 0 dislike
43 views

I submitted this question on physics.stackexchange a couple of weeks ago; it received some upvotes but no answer, so I thought I would resubmit it here, by analogy with the standard practice for maths questions. I apologize for any breaches of etiquette.

***

The Bekenstein bound says that the maximum entropy which can be contained in a (spherical) volume of space with a given amount of energy is proportional to the amount of energy multiplied by the "length scale factor" (radius) of the volume - up to some dimensionless constant, it is just the aforementioned quantity expressed in natural units.

  1. Where does this "dimensionality" come from? Is it of any particular significance? (I understand that the entropy itself is dimensionless, but the dimensionless terms making up the bound do nevertheless correspond to the energy multiplied by the length scale as expressed in natural units)

  2. Wikipedia says "Note that while gravity plays a significant role in its enforcement, the expression for the bound does not contain the gravitational constant G." If the bound is independent of the value of G, in what sense is general relativity "necessary" for deriving it (as opposed to merely having been used in the standard derivation)? Is it possible that the bound could in fact be derived using only special relativity and statistical mechanics?

  3. On a much more speculative note, I note that the "coupling constant" associated with an interaction of inverse-square type has the same "dimensionality" of force multiplied by distance squared. It's tempting to wonder whether this might somehow measure information shared between two entities via the interaction. Might this line of speculation have any merit?

asked Feb 10 in Theoretical Physics by Robin Saunders (10 points) [ revision history ]

1. Energy and length have inverse dimensions, in natural units where $\hbar=1$ and $c=1$. Thus something dimensionless depending on energy must also depend on a length or another dimensionful quantity.

3. Speculations of any kind have merit only if they are accompanied by more solid conclusions.

Regarding your response to 1.: through appropriate choices of which dimensions to set to $1$, any type of quantity can be made inverse to any other. For example (sorry for the wall of text):

Setting $\hbar = 1$ alone, we see that something dimensionless depending on energy must also depend on something with units of time (or inverse temperature, or the inverse square root of power...), and indeed we could choose this time to be the characteristic length scale divided by $c$ (and similarly for temperature or power, by appropriate choices of dimensionful constants); but we could also ignore the length scale and refer instead to $\hbar$ divided by the energy, or to another characteristic timescale (temperature or power scale...) of the system, or even to some other characteristic quantity altogether multiplied by suitable constants to give the right dimensionality.

Setting $c = 1$ alone, we see that something dimensionless depending on energy must also depend on something with units of inverse mass (or inverse momentum, or inverse temperature again...), and again we could choose this mass to be the characteristic length scale multiplied by $c^2/G$ (and similarly for momentum or temperature); but again we could also ignore the length scale and refer instead to the energy divided by $c^2$, or to another characteristic mass scale (momentum or temperature scale), or to some other characteristic quantity multiplied by suitable constants.

Setting both $\hbar = 1$ and $c = 1$ as you suggested, we could ignore the length scale and just use the Planck length itself (or the Planck time, or $\sqrt G$, or any number of other alternatives). And we could instead have chosen to set $c = 1$ and $G = 1$, or $\hbar = 1$ and $G = 1$, or something else altogether, and in each case we would have had many other options involving combinations of characteristic scales and dimensionful constants, any of which would have given an appropriate "inverse" to energy with which to putatively bound the maximum entropy of the system.

So I could reword 1. as: why does the Bekenstein bound depend on energy and length specifically? (Edit: and to the particular power to which they appear, as opposed to some other?)

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\varnothing$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...