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A sub-thread to Yomammas Infinity issue

  1. Sep 24, 2005 #1
    I`m relativly new to Physics but have heard of a technic used called renomalisation, the cancelling of infinties. If infinity is not a number how does this work?

    (my appologies to the mods if i`ve posted this in the wrong place)
     
  2. jcsd
  3. Sep 24, 2005 #2

    Hurkyl

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    This exercise is something I've actually stumbled through at work -- it is fairly elementary, and I understand is a related sort of thing.


    I wanted to get a Taylor series for the cumulative standard normal distribution. Recall that it's given by:

    [tex]
    f(z) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^z e^{-t^2 / 2} \, dt
    [/tex]

    So, I replaced the exponential with its Taylor series, giving:

    [tex]
    f(z) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^z
    \sum_{n = 0}^{\infty} \frac{1}{n!} \left( -\frac{t^2}{2} \right)^n
    \, dt
    [/tex]

    And then optimistically pulled the sum out of the integral:

    [tex]
    f(z) = \frac{1}{\sqrt{2\pi}} \sum_{n=0}^{\infty}
    \frac{(-1)^n}{n!2^n} \int_{-\infty}^{z} t^{2n} \, dt
    [/tex]

    From which you "get":

    [tex]
    f(z) = \frac{1}{\sqrt{2\pi}} \sum_{n=0}^{\infty}
    \frac{(-1)^n}{n!2^n}
    \left( \frac{z^{2n+1}}{2n+1} + \infty \right)
    [/tex]

    At which point I had the idea that I knew f(0) = 1/2, and since all of the ±∞ "are" constants, I could just collect them together into the correct constant and get the correct formula (assuming I haven't made any silly mistakes here):

    [tex]
    f(z) = \frac{1}{2} + \frac{1}{\sqrt{2\pi}} \sum_{n=0}^{\infty}
    \frac{(-1)^n}{n!2^n} \frac{z^{2n+1}}{2n+1}
    [/tex]

    Of course, once I did all of this, I realized that I could do the thing properly by observing that:

    [tex]
    \int_{-\infty}^z g(t) \, dt = \int_{-\infty}^0 g(t) \, dt + \int_0^z g(t) \, dt
    [/tex]
     
  4. Sep 24, 2005 #3

    selfAdjoint

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    And the renormalization part of Hurkyl's derivation was formally collecting all those infinities into a normalizing constant. This is the way some people think of renormalization, and of course then they think it's just "sweeping the infinity under the rug". Actual renormalization in field theory is a little more sophisticated and is NOT sweeping under the rug.

    First of all, renormalization concerns the infinities that come into a theory at very short scales; the electron interacts with itself and the theory adds up all the ways such interactions can happen and gets infinity. In quantum mechanics very small spaces are associated with very big energies, because of the uncertainty principle (your position is very sharply limited so your momentum can reach very high limits and the energy with it). So the physicists say, my theory is just not complete enough to handle these very high energies; I will put a cutoff in my theory that prevents it from wrongly trying to process them. This use of a cutoff is called regularization. So then they have a theory ( a bunch of integrals) with a cutoff in the form of an unspecified constant in them. And they work out the theory - simplify and transform the integrals to get the numbers they want to calculate - and the constant flows through all this work and will make it impossible to get the actual numbers. So at the last stage they remove the constant by combining it into a normalization constant. This is the process called renormalization. Note that it wasn't infinity that they moved, it was a finite but unknown constant that they got from putting a cutoff into the theory. And the process is philosophically justified by the admission that the theory was incomplete and unable to handle energies higher than some (very high) limit.
     
    Last edited: Sep 24, 2005
  5. Sep 24, 2005 #4

    Hurkyl

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    And, incidentally, note that what I did would have been fully justified if I replaced the lower limit -∞ with some negative constant H.

    Then, instead of infinities, I have finite numbers which add up to a finite value, which I could simply roll up into a constant.

    (And my example is particularly nice, because the limit as H→-∞ worksout nicely)
     
  6. Sep 24, 2005 #5

    JamesU

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    :rolleyes: OMG....
     
  7. Sep 24, 2005 #6
    ditto :surprised

    Thanx for the info and example (Hurkyl) but ...

    I think I follow what your saying selfAdjoint right up until the part
    could you clarify, i don`t understand the term normalization constant.
     
    Last edited: Sep 24, 2005
  8. Sep 24, 2005 #7

    EnumaElish

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    Yeah, is it similar to Hurkyl's having realized that he has a "boundary condition" f(0) = 1/2 and solving for the integration constant using that condition?
     
  9. Sep 24, 2005 #8

    selfAdjoint

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    Right Enuma. In a theory you typically have to set a scale, your integrals or whatever don't come out in any particular scale and you "normalize" them by multiplying them by some number that is chosen to make things come out right. This is the normalizing constant. Once you choose it you can't change it, but what you can do if you want to gent rid of that unknown cutoff constant (call it K) is to ASSUME that your good normalization constant N is composed of a factor M times K: N = MK. So when you normalize your theory, which is after you have done all the computations, you just say "...and N includes the cutoff."
     
  10. Sep 24, 2005 #9

    Mk

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    You were sorry you ever clicked on that link.
     
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