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Recently I saw some physical problems that can be modeled by equations with fractional derivatives, and I had some doubts: is it possible to write an action that results in an equation with fractional derivatives? For example, consider a hypothetical physical system with the principle of least action. Is there a "wave equation" with the time-derivative $3/2$? Does such a question make sense?

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When I've seen fractional derivatives I've assumed that one place where they would naturally arise is in physical situations where there's a fractional dependency on time.

For example, random walks typically result in movement proportional to $\sqrt{t}$. Googling for "fractional+derivative+random+walk" on arxiv.org finds some papers that explore this:

http://www.google.com/search?q=fractional+derivative+random+walk+site%3Aarxiv.org

So I'm wondering if there's a way of relating some of the diffusion versions of QM with fractional derivatives.

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  • $\begingroup$ Exactly this is indeed done here. $\endgroup$ – Nikolaj-K Apr 9 '15 at 19:21
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Fractional derivatives are nonlocal, but actions are usually assumed to be local.

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  • $\begingroup$ the locality or non-locality of a Lagrangian density depends on the highest order of derivatives in its various terms. Terms with upto second order derivative (eg. kinetic terms) are local. Terms with derivatives of order 3 and higher are non-local, with the degree of non-locality being measured by the order. So a non-local term in an action does not imply that the variational problem is ill-defined. Also why are fractional derivatives non-local? Even better, what is a fractional derivative? Thanks. $\endgroup$ – user346 Jan 27 '11 at 17:01
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    $\begingroup$ Dear @space_cadet, no, what you write is not correct. QGR is right that whenever fractional derivatives appear, the action is nonlocal. It's enough for some terms to appear, whether they're leading or not, and the action has to be nonlocal. At the end of your answer, you admit that you don't really know why they're nonlocal. Well, it's because the you may write $\partial_t^k$ as $E^k$ in the Fourier transformed energy representation, and the transformation of the operator $E^k$ back to the $t$ basis is nonlocal. $\endgroup$ – Luboš Motl Jan 27 '11 at 17:19
  • $\begingroup$ @space_cadet sloppy physicist thinking here, but I think it can be made precise: imagine taking a taylor series of the fractional derivative --- it would need to go to infinite order, and thus it would be highly non-local. $\endgroup$ – genneth Jan 28 '11 at 10:43
  • $\begingroup$ @space_cadet, a fraction derivative is a simple example of "pseudo-differential operator". It's an operator that can be easily defined for example with fourier transform (like Lubos Motl sayed). You can define operator that in Fourier Transform is equivalent to $\sin{k}$...Fraction derivative is simpler, in Fourier Transform is a multiplication for your transform variable $k$ with fractional exponent ;-) $\endgroup$ – Boy Simone Jan 28 '11 at 12:36
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    $\begingroup$ @Boy, @genneth - thanks. @Lubos, all I was saying is that, AFAIK, having non-local terms is fairly common, eg. higher derivative theories of gravity. So being non-local should not be a obstacle for an action with fractional derivatives, as @QGR seems to suggest. $\endgroup$ – user346 Jan 28 '11 at 12:40

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