I'm reading Penrose's "The road to reality". I got it for Christmas and after a burst of reading then, haven't got much further. Our just-finished trip up the Old West River (pic: a bit down from the GOBA mooring at sunset yesterday) gave me some computer-free evenings to restart reading it. I discovered I was only 3 pages away from the end of the "maths" sections and the start of the "physics" bits, so finally got to read some of the physics.Anyway, having struggled about 1/2 way through the text (to ch 22, Quantum algebra, geometry, and spin) my review is: very heavy going; good though; probably rather individualistic in places; maths degree or equivalent a bare minimum requirement. On that latter point: I found it very funny that having lead us through various infinities, Ricci tensors, fibre bundles (still not really sure what those are), gauge connections, etc etc with occaisional nods back to "see the defn on page XX if you've forgotten what index-raising is...) in section 21.1 we have the perfectly normal differential equation y + d^2y/dx^2 = x^5 with (see section 6.3 for the meaning of the symbols). Really! If you don't know what a differential equation is, you're totally stuffed, and ref to 6.3 won't help you. Quite possibly the same is true of the back-refs to Ricci tensors, but I just blip over them...
But, to come to the trick, its in solving the ODE above... taking d^2y/dx^2 + y = x^5, we re-write this as D^2 y + y = x^5, ie (D^2 + 1)y = x^5. Now we treat D as a simple number, and write y = (D^2 + 1)^-1 x^5. This being so, we do a formal series expansion, and get y = (1 - D^2 + D^4 - D^6 + ...) x^5. Since D^6 x^5 is zero (as are all higher powers), we get y = x^5 - 20 x^3 + 120 x. Which is the right answer.
This is a neat trick. I can't remember if I was ever taught it, or even if its commonly known. Did you know it? Then tell me!
Incidentally, M took Beethoven Klaviersonaten Band I to read.
I think it can be justified in terms of Laplace (or Fourier) transforms. Needless to say, one needs to be very careful doing tricks like that. There's a neat "derivation" of the Euler-Maclurian summation formula in Concrete Mathematics along the same lines.
ReplyDeleteA fiber bundle should be thought of something that looks locally like a product. Take a line segment and a circle, for example. A global product of the two looks like a cylinder. However, we can also have a Mobius strip. For any small segment of the circle, this looks like a product, but we have some twisting as we go around the circle. So, the Mobius strip is a line segment bundle over the circle.
Hope that helps.
Hmm, for the moment, a fibre bundle is something I blip over when I read about. I may go back and have anoher go at understanding it, possibly next christmas :-). If I stopped to understand *all* the stuff in Penrose, I'd never get anywhere...
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