# Why are differential equations for fields in physics of order two?

What is the reason for the observation that across the board fields in physics are generally governed by second order (partial) differential equations?

If someone on the street would flat out ask me that question, then I'd probably mumble something about physicists wanting to be able to use the Lagrangian approach. And to allow a positive rotation and translation invariant energy term, which allows for local propagation, you need something like $-\phi\Delta\phi$.

I assume the answer goes in this direction, but I can't really justify why more complex terms in the Lagrangian are not allowed or why higher orders are a physical problem. Even if these require more initial data, I don't see the a priori problem.

Furthermore you could come up with quantities in the spirit of $F\wedge F$ and $F \wedge *F$ and okay yes... maybe any made up scalar just doesn't describe physics or misses valuable symmetries. On there other hand in the whole renormalization business, they seem to be allowed to use lots and lots of terms in their Lagrangians. And if I understand correctly, supersymmetry theory is basically a method of introducing new Lagrangian densities too.

Do we know the limit for making up these objects? What is the fundamental justification for order two?

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– Qmechanic♦ Dec 21 '11 at 14:24
Please wait until the last moment for putting the 'answered' mark. This is a very interesting question and I want to see what the big ones here have to say. Leave that temping bounty waiting as much as possible. – Eduardo Guerras Valera Nov 13 '12 at 23:26

## 7 Answers

First of all, it's not true that all important differential equations in physics are second-order. The Dirac equation is first-order.

The number of derivatives in the equations is equal to the number of derivatives in the corresponding relevant term of the Lagrangian. These kinetic terms have the form $${\mathcal L}_{\rm Dirac} = \bar \Psi \gamma^\mu \partial_\mu \Psi$$ for Dirac fields. Note that the term has to be Lorentz-invariant – a generalization of rotational invariance for the whole spacetime – and for spinors, one may contract them with $\gamma_\mu$ matrices, so it's possible to include just one derivative $\partial_\mu$.

However, for bosons which have an integer spin, there is nothing like $\gamma_\mu$ acting on them. So the Lorentz-invariance i.e. the disappearance of the Lorentz indices in the terms with derivatives has to be achieved by having an even number of them, like in $${\mathcal L}_{\rm Klein-Gordon} = \frac{1}{2} \partial^\mu \Phi \partial_\mu \Phi$$ which inevitably produce second-order equations as well. Now, what about the terms in the equations with fourth or higher derivatives?

They're actually present in the equations, too. But their coefficients are powers of a microscopic scale or distance scale $L$ – because the origin of these terms are short-distance phenomena. Every time you add a derivative $\partial_\mu$ to a term, you must add $L$ as well, not to change the units of the term. Consequently, the coefficients of higher-derivative terms are positive powers of $L$ which means that these coefficients including the derivatives, when applied to a typical macroscopic situation, are of order $(L/R)^k$ where $1/R^k$ comes from the extra derivatives $\partial_\mu^k$ and $R$ is a distance scale of the macroscopic problem we are solving here (the typical scale where the field changes by 100 percent or so).

Consequently, the coefficients with higher derivatives may be neglected in all classical limits. They are there but they are negligible. Einstein believed that one should construct "beautiful" equations without the higher-derivative terms and he could guess the right low-energy approximate equations as a result. But he was wrong: the higher derivative terms are not really absent.

Now, why don't we encounter equations whose lowest-order derivative terms are absent? It's because their coefficient in the Lagrangian would have to be strictly zero but there's no reason for it to be zero. So it's infinitely unlikely for the coefficient to be zero. It is inevitably nonzero. This principle is known as Gell-Mann's anarchic (or totalitarian) principle: everything that isn't prohibited is mandatory.

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Thanks for the answer. What is the reason that "their coefficients are powers of a microscopic scale or distance scale $L$"? In the last paragraph you use this again, where it's implied that the lower order derivatives are a priori related to a bigger scale, which then outweighs the later ones associated with higher orders. Is there a justification, which goes back to axiomatic assumptions or is it "just" an empirical insight from dealing with effective field theories? – Nick Kidman Dec 21 '11 at 15:08
Dear @Nikolaj, $L$ determining the coefficients is microscopic because microscopic scales are the natural ones for the formulation of the laws of physics. By definition, microscopic scales are the scales associated with the elementary particles. These general discussions talk about many things at the same moment. For example, in GR, the typical scale is the Planck length, $10^{-35}$ meters, which is the shortest one. In other theories, the typical scale is longer. But it's always microscopic because it determines the internal structure/behavior of the fields and particles which are small. – Luboš Motl Dec 21 '11 at 16:06
The comment that the derivatives are not just related, they produce long scale was meant to be as a self-evident tautology. What I mean is that if we consider a field that is changing in space, e.g. as a wave with wavelength $R$, then the derivative will pick a factor of order $1/R$, too. For example, the derivative of $\sin(x/R)$, the wave of length $2\pi R$, is $\cos(x/R)/R$. Cos and sin is almost the same thing, of the same order 1, and we therefore picked an extra factor of $1/R$. All these things are order-of-magnitude estimates. Macroscopic usage of field theory has a macroscopic $R$. – Luboš Motl Dec 21 '11 at 16:09
I'm not sure if I successfully pointed out my problem in the comment. My question is: What is the justification for assuming the coefficient of smaller orders would describe a bigger scale? What speaks against a situation, where the fourth order term has a small coefficient, but the second order term has an even smaller one? Then in the classical limit, just the fourth order expression would survive. – Nick Kidman Dec 21 '11 at 18:36
Dear @Nikolaj, it's likely that I don't understand your continued confusion at all. Whether a term may be neglected depends on the relative magnitude of the two terms, the neglected one and the surviving one. So I am estimating the ratio of higher-derivative terms and two-derivative terms and it scales like $(L/R)^k$, a small number, so the higher-derivative terms may be neglected if the two-derivative terms are there. It doesn't matter how you normalize both of these terms in an "absolute way". What matters for being able to neglect one term is the ratio of the two terms. – Luboš Motl Dec 21 '11 at 18:53
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One can rewrite any pde of any order as a system of first order pde's, hence the assumption behind question is somewhat questionable. Also there exist first order PDE's of relevance to physics (Dirac equation, Burgers equation, to name just two).

However, it is common that quantities in physics appear in conjugate pairs of potential fields and their associated field strength, defined by the potential gradient. Now the gradients of field strength act as generalized forces that try to move the system to an equilibrium state at which these gradients vanish. (They will succeed only if there is sufficient friction and no external force.)

In a formulation where only one half of each conjugate pair is explicit in the equations, a second order differential equation results.

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Here we will for simplicity limit ourselves to systems that have an action principle. (For fundamental and quantum mechanical systems, this are often the case.) Let us reformulate OP's question as follows:

Why does the Euler-Lagrange equations of motion for a relativistic system (non-relativistic system) have at most two derivatives (time-derivatives), respectively?

(Here the precise order depends on whether one considers the Lagrangian or the Hamiltonian formulation, which are related via Legendre transformation. In case of a singular Legendre transformation, one should use the Dirac-Bergmann or the Jackiw-Faddeev method to go back and forth between the two formalisms. See also this Phys.SE post.)

Answer:

The higher-derivative terms are in certain theories suppressed for dimensional reasons by the natural scales of the problem. This may e.g. happen in renormalizable theories.

But the generic answer is that the equations of motion actually doesn't have to be of order $\leq 2$.

However, for a generic higher-order quantum theory, if higher-derivative terms are not naturally suppressed, this typically leads to ghosts of the so-called bad type with wrong sign of the kinetic term, negative norm states and unitarity violation.

Explicit appearances of higher time-derivatives may be removed on the paper by introducing more variables, either via the Ostrogradsky method, or equivalently, via the Lagrange multiplier method. However, the positivity problem is not cured by such rewritings, and the quantum system remains ill-defined. See also e.g. this and this Phys.SE answer.

Hence one can often not make consistent sense of higher-order theories, and this may be why OP seldom faces them.

Finally, let us mention that it is nowadays popular to study effective higher-derivative field theory, with the possibly unfounded hope, that an underlying, supposedly well-defined, unitary description, e.g. string theory, will cure all pathologies.

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Weinberg gives a pretty good answer for this in Volume 1 of his QFT opus: 2nd order differential equations appear in the field theories relevant to particle physics because of the relativistic mass-shell condition $p^2 = m^2$.

If we have a quantum field $\phi$, and we think of its fourier modes $\phi(p)$ as creating particles with 4-momentum $p$, then the mass-shell condition provides a constraint: $(p^2 - m^2)\phi(p) = 0$, because we don't want particle creation off-shell. Fourier-transform this back to position space, and you find that $\phi$ has to obey a 2nd order differential equation.

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This doesn't apply to general relativity, where nevertheless equations are of second order. – Arnold Neumaier Nov 12 '12 at 16:39

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It does tell you that the linearized Einstein equations should be second order. And it explains why the renormalization flow should be defined in such a way that the kinetic term is fixed, which is a important assumption implicit in Lubos' answer. – user1504 Nov 12 '12 at 16:42

First of all, it's not true that all important differential equations in physics are second-order. The Dirac equation is first-order.

This is correct. However, physical evolution equations are second (in time) order hyperbolic equations. In fact, each component of Dirac spinor follows a second order equation, namely, Klein-Gordon equation.

Now, what about the terms in the equations with fourth or higher derivatives?

They're actually present in the equations, too.

Neither the Standard Model (SM) Lagrangian nor the Einstein-Hilbert (EH) action contain higher than second order temporal derivatives. These are the actions which are experimentally tested and these two theories are the most fundamental scientific theories we have. We know that there are physics beyond these two theories and people have good candidates to the underlying theories, but physics is an experimental science and these theories are not experimentally verified. The effective SM Lagrangian (a Lorentz invariant theory with the gauge symmetries of the SM but with irrelevant operators) does contain higher than second order temporal derivatives. Equally for the EH action plus higher order scalars. Two clarifications are however in order:

• These irrelevant terms are not experimentally verified. Almost everyone is sure that neutrino mass terms (which are irrelevant operators but do not contain higher order derivatives) exist in order to explain neutrino oscillations, but so far we do not have direct measurements of neutrino masses thus we are not allowed to claim that these terms exist. Summarizing: the effective SM is not a verified theory.

• The origin of these irrelevant terms is a consequence of integrating out fields with a mass much greater than the energy scale we are interested in. This could be the case of the neutrino mass term and a right-handed neutrino. For instance, in quantum electrodynamics, if one is interested in the physics at much lower energies than the electron mass, one can integrate (or nature integrates-out) out the electron field obtaining an effective Lagrangian (Euler-Heisenberg Lagrangian) with terms with higher order derivatives like $\frac{\alpha ^2}{m_e^4}~F_{\mu\nu}~F^{\mu\nu}~F_{\rho\sigma}~F^{\rho\sigma}$ (which contains four derivatives). These are terms suppressed by coupling constants ($\alpha$) and high-energy scales ($m_e$). There are terms with a number of derivates arbitrarily high, and they come from inverses of differential operators. This makes that the higher order derivatives do not enter in the zeroth-order equation of motion.

However, in a fundamental theory (in contrast to an effective one), finite higher order derivatives are not allowed in interactive theories (there are some exceptions with gauge fields, but for example a generic $f(R)$ theory of gravity is inconsistent). The reason is that those theories are not bounded from bellow (see Why are only derivatives to the first order relevant?) or, in some quantizations, contain negative norm states. These terms are among the forbidden operators in Gell-Mann's totalitarian principle.

In summary, evolution equations are order two because of existence of a normalizable vacuum state and unitarity (including here the fact that physical states must have positive norm). Newton was right when he wrote $$\ddot x=f(x,\dot x)$$

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It was already noted in other answers that fields in physics are not always governed by second order partial differential equations (PDEs). It was said, e.g., that the Dirac equation is a first-order PDE. However, the Dirac equation is a system of PDEs for four complex functions - components of the Dirac spinor. It was also mentioned that any PDE is equivalent to a system of PDEs of the first order.

I mentioned previously that the Dirac equation in electromagnetic field is generally equivalent to a fourth-order partial differential equation for just one complex component, which component can also be made real by a gauge transform (http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (my article published in the Journal of Mathematical Physics) or http://arxiv.org/abs/1008.4828 ). Let me also mention my article http://arxiv.org/pdf/1111.4630.pdf , where it is shown that the equations of spinor electrodynamics (the Dirac-Maxwell electrodynamics) are generally equivalent to a system of PDEs of the third order for complex four-potential of electromagnetic field (producing the same electromagnetic field as the usual real four-potential of electromagnetic field).

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The reason for equations of physics, not being of second order is due to the so-called Ostrogradskian instability. (see paper by Woodard). This is a theorem, which states that equations of motion with higher-order derivatives are in principle unstable or non-local. This is easily shown using the Lagrangian and Hamiltonian formalism.

The key point is that in order to get an equation of motion of third order in the derivatives, we need a Lagrangian that depends on the coordinates and the generalized velocities and accelerations: $L(q,\dot{q},\ddot{q})$. By performing a Legendre transformation to obtain the Hamiltonian, this implies that we need two generalized momenta. The Hamiltonian results to be linear in at least one of the momenta and therefore it is unbounded from below (it can become negative). This corresponds to a phase space in which there are no stable orbits.

I would like to write the proof here, but it was already answered in this post. There the question is why Lagrangians only have one derivative, but it is actually closely related, since one can always find the equations of motion from a Lagrangian and viceversa.

Citing Woodard: "It has long seemed to me that the Ostrogradskian instability is the most powerful, and the least recognized, fundamental restriction upon Lagrangian field theory. It rules out far more candidate Lagrangians than any symmetry principle. Theoretical physicists dislike being told they cannot do something and such a bald no-go theorem provokes them to envisage tortuous evasions. ... The Ostrogradskian instability should not seem surprising. It explains why every single system we have so far observed seems to be described, on the fundamental level, by a local Lagrangian containing no higher than first time derivatives. The bizarre and incredible thing would be if this fact was simply an accident."

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– Qmechanic♦ Mar 24 at 20:24
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