This week's physics puzzler.

This question is prompted by a passage from Brian Greene's Fabric of the Cosmos. Greene is discussing the fact that in our usual experience entropy is proportional to volume (other things being equal) whereas for black holes the entropy is proportional to the area of the horizon. On page 479 of chapter 16 he says
Were you to double the radius of a black hole, its volume would increase by a factor of 8 while its surface area would only increase by a factor of 4.
This is misleading. His radius, area, volume relations are only valid for flat space, in curved space the ratios are different. The radius circumference relations for a circle on the surface of a two sphere illustrate the point.

So here's the question, in two parts:

a) What are the radius, area, volume relations (for a surface r=constant) in a Schwarzschild space-time?

b) Do the notions of radius and volume even make sense for a black hole?

Comments

  1. Anonymous5:41 PM

    I don't remember enough GR to calculate the 'volume' relations for Schwarzschild spacetime, but for (b) I think it's fair to say that familiar notions of radius and volume are pretty much kaput for a black hole.

    We can obviously see the event horizon, and thus a familiar surface area, but when we go inside there our radial coordinate becomes timelike and conversely.

    So for a BH radius, I think people just talk about the pseudo-radius that would correspond to the observed surface area were it embedded in a flat spacetime.

    BH "volume" is more confusing. To give a familiar measurement of volume, you need to convert the within-horizon radial measurement into a timelike measurement for outside guys. Googling this a neat answer comes up:
    BH_volume = surface area * lifetime
    http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970808.html

    ReplyDelete
  2. Anonymous6:56 PM

    Hartle calculates length, area, volume and four-volume on ppgs. 146-148 of his book Gravity for general diagonal metrics (including Schwarzschild spacetime), so that solves (a). The calculation should break down at the horizon for a BH, on account of the coordinate singularity. Inside the BH, all calculations collide with the singularity at the "center."

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