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Newton's second law of motion states that $F=\frac{d(mv)}{dt}$. This is a first-order differential equation, in which the order of differentiation of momentum is 1. So we can write it $F=\frac{d^k(mv)}{dt^k}$ where $k=1$.

Newton used the dot notation for differentiation and unlike Leibniz did not conceive of orders of differentiation that were not integers. Perhaps if he had he would have changed his notation, because it's hard to draw half a dot! We, however, have no such difficulty.

Let's make $k$ probabilistic, so that its mean value is 1 but it varies from 1 with decreasing probability as the difference from 1 increases. Let's assume the probability density function of its distribution is normal, so that $f(k)=N(1,\sigma)$, and let's further assume that $\sigma$ is a universal constant.

If $\sigma = 0$, we get $F=\frac{d(mv)}{dt}$, and either a Newtonian universe or, when we take mass to be relativistic, an Einsteinian one.

If we assume a big bang cosmology on the Lambda-CDM model, is there a value $\epsilon > 0$ such that if $\sigma < \epsilon$ we still get the same cosmology, because the tweak we have made to the equation $F=\frac{d(mv)}{dt}$ is too small to make a difference?

And what happens to the currently prevalent version of big bang cosmology when we increase the value of $\sigma$ such that a difference is actually made?

I ask the question in that way because I am not saying let's give $\sigma$ a large value such as 0.5. I am saying let's give it a very tiny value that will just begin to have a cosmological effect. What will that cosmological effect be?

Bearing in mind that non-integer-order derivatives are non-local, do we get matter appearing out of empty space and a possible basis on which to support a steady state cosmology? What other adjustments might be suggested - of big bang theory, quantum theory, or both?

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    $\begingroup$ I'm not sure that Physics SE will take it either since it does not appear to ask a question about mainstream physics @RobJeffries. $\endgroup$
    – Brick
    Commented Oct 7, 2015 at 17:00
  • $\begingroup$ I thought of posting it to Physics but decided on Astronomy because of the focus on cosmology. I realise the two overlap: Physics tags include "astronomy", "astrophysics" and "cosmology". Happy to transfer it if that's what people would prefer. Both Mathematics and MathOverflow have a "fractional calculus" tag, but even though FC may (?) be known to more mathematicians than physicists, I thought people on the maths SEs would find the question too physicsy. Hopefully it will be welcome somewhere. It's not mainstream but it's serious and not crazy. $\endgroup$ Commented Oct 7, 2015 at 18:26
  • $\begingroup$ @RobJeffries etc. - I have edited to clarify the cosmological core of the question. Is it clear enough now? $\endgroup$ Commented Oct 8, 2015 at 7:34
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    $\begingroup$ Cosmology is built upon tensor calculus, not $F=\dot{p}$, so I doubt fractional derivatives of momentum would be useful in the field, though it'd be interesting to see how it could be applied to Newtonian mechanics. $\endgroup$
    – Kyle Kanos
    Commented Oct 8, 2015 at 14:28
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    $\begingroup$ Maybe i am missing the point but the force $F$ is only defined dimensionally correct when $k=1$. For any other $k$, it cannot be considered force but 'something else' which i doubt is physically useful. $\endgroup$
    – nluigi
    Commented Oct 9, 2015 at 9:46

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Let me help you formulate your proposal in QFT terms for a real scalar field $\phi$, $$ Z[j] = \int D\phi dk e^{\frac{i}{\hbar}\int (L + \phi j) dx^4}, $$ where $$ L = \sqrt{-g}\{\phi(\partial_\mu\partial^\mu -m^2)^k\phi + \frac{k^2}{2i\sigma^2}\}. $$ Then you can work out either the classical (focusing on $L$ only) or quantum (tackling $Z[j]$) cosmological consequences as an easy homework as your professor promised ;)

Of course you have to settle on a properer fractional calculus. Actually you can shop around a couple of choices of fractional calculus.

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Your question is equivalent to asking what if gravity (or any other force) did not have the form we think it has. Just keep integrating your definition of force until you are left with second order differentials and first order differentials. If you have the other terms left, it means you have modified gravity for instance. And if you have the first differential, it means you have some sort of viscosity that is acting as friction in your model.

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