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Physical applications of Riemann zeta function

  1. Feb 17, 2015 #1
    Hi
    I was wondering if there any observations that have only been described using the Riemann Zeta function? What would it mean in physics to assign a divergent series a finite value?
    Thank you

    Edit
    Sorry I overlooked a thread just posted that asked about this so this might need to be deleted.
     
  2. jcsd
  3. Feb 18, 2015 #2

    Svein

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    Well, the Zeta function diverges for z=1, but converges for real values of z with z>1.
     
  4. Feb 18, 2015 #3

    stevendaryl

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    The zeta function is defined for negative and even complex values of [itex]z[/itex], but it isn't equal to the series representation except in the case [itex]Re(z) > 1[/itex].

    The series representation is: [itex]\zeta(z) = 1 + 2^{-z} + 3^{-z} + ...[/itex]. For [itex]z = -1[/itex], this diverges: [itex]1 + 2 + 3 +...[/itex]. But the zeta function is defined for [itex]z=-1[/itex], and has value [itex]\zeta(-1) = - \frac{1}{12}[/itex]. So sometimes people write:

    [itex]1 + 2 + 3 + ... = - \frac{1}{12}[/itex]

    which is not literally true...
     
  5. Feb 18, 2015 #4
    So are the sums

    [itex]1 + 2 + 3 + ... = - \frac{1}{12}[/itex]

    or

    [itex]1 + 1 + 1 + ... = - \frac{1}{2}[/itex]

    ever used? I guess it would be possible in the summation of probabilities?

    Thanks
     
  6. Feb 18, 2015 #5

    Svein

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    Of course they are not used (except in mathematical jokes). They are the values of the Zeta function, not the value of a sum.
     
  7. Feb 18, 2015 #6

    stevendaryl

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    Actually, summing divergent series can come into play in physics. If you have a problem that you only know how to solve perturbatively, you could very well end up with a result such as: [itex]f(\alpha) = \sum_n C_n \alpha^n[/itex] for some physical quantity that depends on some parameter [itex]\alpha[/itex]. In some cases, the coefficients [itex]C_n[/itex] can be computed, and you can actually prove that the sum diverges. However, if we can recognize the terms [itex]C_0 + C_1 \alpha + C_2 \alpha^2 + ...[/itex] as being the Taylor series for a known function, then we can directly evaluate that function, rather than trying to sum the series.

    To give an example, if you have the series: [itex]1 - \alpha + \alpha^2 - ...[/itex], it will be divergent, if [itex]\alpha > 1[/itex], but we can see that it is the Taylor series expansion for the function [itex]f(\alpha) = \frac{1}{1 + \alpha}[/itex] which has a perfectly good value for [itex]\alpha > 1[/itex].

    According to http://motls.blogspot.com/2007/09/zeta-function-regularization.html, the divergent sum [itex]1+2+3+...[/itex] comes up in string theory, and they get sensible answers by letting this equal [itex]\frac{-1}{12}[/itex].

    The claim is not that [itex]1+2+3+...[/itex] converges, in any sense, to [itex]\frac{-1}{12}[/itex], but that a problem whose correct answer is [itex]\frac{-1}{12}[/itex] can lead to such a nonsensical summation when done perturbatively.
     
  8. Feb 18, 2015 #7
    The Riemann zeta function is used in the calculation of the energy of the quantum vacuum and the Casimir force, which is the quantum force between two parallel plates in vacuum. For the ideal case of perfect conductors, the wavelengths of the virtual photons cannot exceed twice the separation of the plates. When summing the energies of all these photons, the zeta function is used. The physically measurable quantity is the change in vacuum energy due to the plates, which can be expressed in terms of a zeta function.
     
  9. Feb 18, 2015 #8

    Svein

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    That's what I like about physicists - they find a use for every mathematical expression, even if it is obvious nonsense.
     
  10. Feb 18, 2015 #9
    In that article it says there are many different ways to arrive at the same result, and that it acts as a consistency check, so does that mean one way to arrive at the result is "more correct" than another (double quotes because I'm not entirely sure what it even means to be more correct in this case)? Or is it purely more evidence that a theory is correct?

    I guess I'm wondering what are the implications of using these results in our understanding of what is really happening. Thanks to everyone replying!
     
  11. Feb 19, 2015 #10
    I think that physicists have a pragmatic side. If we use mathematics, even in nontraditional ways, e.g. Dirac delta functions, or taking the difference of infinite quantities and getting a finite difference, and can get an answer that agrees with experiment, we tend to be happy and believe it. We are a simple folk (I am laughing). As another example, we don't really understand or have a complete mathematical description for measurement in quantum mechanics or in QED, but when you follow all the "rules" of calculation, you always get an answer that, to date at least, agrees with experiment, with superb precision. That is worth a lot. Perhaps this paradox occurs because mathematics is often an idealization, a Platonic ideal, but we are interested in calculations of the real world. Thanks for bringing us this interesting topic that makes us reflect on our process.
     
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