# Why are higher order Lagrangians called 'non-local'?

And in what sense are they 'non-local'?

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Higher order in what? Both higher powers of fields and higher powers of derivatives are not non-local. Non-locality always means interaction between fields at points separated by some distance (instead of at a single point). This can be introduced e.g. by non-polynomial functions in the Lagrangian but I am not sure what your higher-order means. In any case, you should spend some effort on making the question precise. – Marek Aug 16 '11 at 8:30
Marek: But that's exactly my question. I've repeatedly encountered referral to theories with higher order Lagrangians (I mean higher order in derivatives) as 'non-local' despite these not actually having a 'non-local' interaction of the type you are referring to. So what do they mean? – WIMP Aug 21 '11 at 5:42
@WIMP: Please provide reference to support the claim in the title (v3). – Qmechanic Jun 25 '14 at 20:16

The higher the number of derivatives the more initial data you have to provide. If you have some Lagrangian that contains an infinite number of derivatives (or derivatives appearing non-polynomially, such as one over derivative) then you have to provide an infinite amount of initial data which amounts to non-local info, in the sense explained below.

If you think in terms of Taylor expansions around your initial value, then you have to provide the full function (and thus non-local information) if you have an infinite number of derivatives. This is to be contrasted with cases where you provide only the field and its first derivative as initial values (and thus rather local information).

Personally, I would not call any higher-derivative Lagrangian "non-local", but only those theories where the number of derivatives is formally infinite in the Lagrangian.

In any discretization scheme you literally see the non-locality induced by higher derivatives: to define the first derivative you need to know the function on two adjacent lattice points, to define the second derivative on three and to define the n-th derivative on n+1 lattice points. Thus, the more derivatives the more non-locality. If you have an infinite number of derivatives you need to know the function on an infinite set of lattice points.

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I do not like a discretization ("lattice") explanation. According to it, the first derivative is non local. – Vladimir Kalitvianski Aug 16 '11 at 16:01
No, the first derivative is not non-local, it is just "less local" than the zeroth derivative. As I said, I would no call any higher derivative theory "non-local", only those where you need infinitely many derivatives - and thus infinitely many lattice points. – Daniel Grumiller Aug 16 '11 at 17:25
Yes, okay, thanks. I still find it confusing to call that 'non-local' but if that's what they mean, then, well, that's what they mean. (I believe what you say is true only for a finite number of derivatives, but then that wasn't my question.) – WIMP Aug 21 '11 at 5:47
I also didn't get the reason why or where "non-local" terminology came from. Regarding the number of points needed to define derivatives, if the points are closer and closer to each other as we go to higher derivatives, they will be "local", don't they? – Revo Aug 23 '11 at 2:12

Disillusionment with systems described by higher order Lagrangians harks back to a 1950 paper by Pais and Uhlenbeck, in which they showed that such systems were prone to pathologies, including states with negative energy and states with negative norm. There's a more recent discussion of this in arXiv:hep-th/0408104.

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I know the Pais paper. But my question wasn't why people don't like these theories. I really just wanted to know why they're called 'non-local' as different authors seem to mean very different things with the word. – WIMP Aug 21 '11 at 5:48

Clearly, an interaction involving $\phi(x+h)$ deserved to be called nonlocal. But since $\phi(x+h)=\sum_{k=0}^\infty \phi^{(k)}(x) h^k/k!$, any nonlocal interaction can be expressed as a power series involving arbitrarily many derivatives. Therefore an action (or Lagrangian) is called nonlocal if it involves infinitely many derivatives.

If there are only finitely many derivatives, they are not nonlocal. Indeed, by introcing additional fields for the derivatives togetyher with squares of the differences between the new fields and their definition, one can rewrite these in terms of a new action/Lagrangian which leads to identical equations of motions. (Unfortunately, this doesn't help with quantization, as the corresponding Legendre transformations to the Hamiltonian form is singular.)

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This is the only correct answer. Higher order theories are not nonlocal, I don't know anyone who calls them nonlocal--- they give rise to local equations of motion (of high order), and local simulation methods. – Ron Maimon May 14 '12 at 1:49
@Arnold: I am fine with the fact that any non-local interacn can be expressed as a power series involving arbitrarily many derivatives, but how can you say that any infinite series of higher order derivatives, you give you such an interaction? How can we generally assume, any infinite power series of higher order derivatives would give a non-local interaction? – user7757 Dec 21 '12 at 13:33
@ramanujan_dirac: Non-local is just defined this way. There is no other way to tell local from nonlocal. – Arnold Neumaier Sep 12 '13 at 10:18