On Sat, Jul 17, 2010 at 11:26 AM, Henry Baker <hbaker1@pipeline.com> wrote:
Read the paper, then Google the physics blogs for "Verlinde" & download Verlinde's (~1 hour) audio & PDF slides from a lecture that he gave. In the audio file, there are questions from the audience that are interesting. I've listened to it once & will listen to it several more times.
I'd appreciate a link--I haven't been able to find the video you're referring to. The very core of the argument goes like this. Say we have two boxes, one inside the other: +---------------+ | | | +----------+ | | | | | | | | | | | | | | +----------+ | +---------------+ Say the inner box has room for ten bits on its surface and the outer one room for twenty. Each box can use as many "1"s as there are particles inside it: +---------------+ | X | | +----------+ | | | | | | | X | | | | | | | +----------+ | +---------------+ In this case, the inner box has only one particle inside, so there are 10 choose 1 = 10 ways to choose a labeling of the inner box; the outer box has two particles inside, so there are 20 choose 2 = 380 ways. Thus there are 3800 ways to label the system in all. If both particles are in the inner box, though, the number of ways to label the system increases: +---------------+ | | | +----------+ | | | | | | | X X | | | | | | | +----------+ | +---------------+ The inner box now has 10 choose 2 ways = 90, while the outer box still has 380. So using the standard assumption that all labelings are equally accessible, it's 9 times as likely to find both particles in the inner box, and we get an entropic force drawing them together. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com