# Why fields are solutions of waves equations?

This could be extremely trivial but I am having problems figuring it out.

I think I understand properly the difference between waves and fields. A field is a function valued on space or spacetime which takes values that can be scalars, vectors, tensors and so on. While waves can be understood as a consequence of fields' perturbations that brings the fields out of their equilibrium state. For example a static mass generates a gravitational field and if the mass oscillates, then it forms gravitational waves.

My problem is why in QFT (and not only) the fields ends up being solutions of wave equations, for example the free scalar field ends up being solution of the Klein gordon equation (a relativistic wave equation)

$$(\Box + m^2)\phi=0$$

Similarly Maxwell equations can be rewritten in the Coulomb Gauge for the potential vector field $$A$$ as $$\Box A=0$$

What to me seems to be a problem is that we name fields objects that are solutions of wave equations, while these are different concepts even thoguh are related. In a not rigorous way at first I thought I could partially solve my doubts thinking to make a perturbation of the field $$A \rightarrow A+\delta A$$ through the presence of a source $$\rho$$ (or through an interaction) and considering this

$$\Box A=0$$ $$\Box(A+\delta A)=\rho \rightarrow \Box \delta A=\rho$$ and so I could see the perturbation as a wave related to the source that generates the perturbation. Even if this is correct (which I am absolutely not sure) the problem stands even from a mathematical point of view as fields and waves are different objects. Fields are sections that goes from a manifold to a bundle while waves I guess are just solutions of certain differential equations.

• Is there an issue with treating the $\phi$ in the KG equation as a field to begin with and say that the equation describes the on-shell condition for the field configuration? The on-shell condition might or might now look like a wave equation, in general. – Dvij D.C. May 22 at 21:29
• I'm not sure I get this. Is the question about semantics? Or is it more of "why do we study fields wtih dynamics that obey wave equations?" – Alfred Centauri May 22 at 21:32
• I don't think it is semantic, and sorry if it isn't clear. what i am tring to find out I think is more related to your second intepretaion, maybe I could state the question more clearer "why a free field is described by waves equations" in terms of the dynamics, I can maybe understand this saying that it happens because waves are the way by which the field propagates after the perturbation (even if am not sure is formally correct) – Ratman May 22 at 22:08
• and the question is even about the mathematical aspect, how the two concepts of fields and waves are related from a mathematical point of view, as thinking of fields as sections doens't allow me to find a relation – Ratman May 22 at 22:08
• Ratman, thanks for clarifying. Could a satifsfactory answer to you be based on the regions of the field exchanging potential and kinetic energy in a way related to such an exchange in the canonical mass-spring (or even a LC) system? – Alfred Centauri May 22 at 22:11

Mathematically, you can define any dynamic law you'd like on a field. A dynamic law is simply a map $$\Phi^t(C)$$ mapping a system configuration $$C$$ - no matter what it is, could be a field, position/velocity pair, state of a game grid, etc. - to an "evolved" one after a lapse interval $$t$$ that satisfies the composition principle $$\Phi^s \circ \Phi^t = \Phi^{s + t}$$ and $$\Phi^0(C) = C$$. There is absolutely no requirement that if $$C$$ is a field (i.e. a function defined at each point of a space), that $$\Phi^t$$ must have anything to do with waves. As a simple example, consider where $$C$$ is any one-dimensional field, i.e. a function $$\phi(x)$$ mapping some field quantity at each point in space $$x$$, and define the action of the map by

$$[\Phi^t(\phi)](x) := \phi(x e^t)$$

. (Simple exercise: Verify this is a valid dynamic law.) This is clearly not a wave behavior - the field simply "dilates" exponentially with time evolution $$t$$.

However, physics is "maths with a constraint of reality": it just so happens that the entities that we like to describe using fields, that exist in our Universe, their dynamic laws are those which can be described using wave equations. There may be a more fundamental reason for this we have not yet discovered, or it could simply itself be just "how it was made".

FWIW, it's also relevant to point out that "solution of wave equation" then is not so much a property of the field, which is just the function that associates a quantity with each point in space, but of the evolution of that field. The initial condition for that evolution, i.e. what you feed into $$\Phi^t$$, so that is returned unaltered at $$t = 0$$, is very free. However the question is phrased as "why fields are solutions of wave equations?" It would be more technically precise to say "why are the histories of fields solutions of wave equations?"

• In particular, the dynamic laws that govern the fields we usually work with in classical and quantum field theory are the solutions of the Euler-Lagrange equations. – probably_someone May 23 at 3:44
• Thanks a lot for the answer, the first part of the answer is perfectly clear. In the last paragraph I believe you are explaining what is my principal doubt, and I think I don't understand it properly. What you mean is that fields per se are not solution of waves equations, while their dynamics (for example the way they propagate after a perturbation) is solution of wave equations? that's what I think is true but not sure if it is what you meant. Thanks a lot – Ratman May 23 at 9:24
• Anyway don't worry anymore, thanks to yours and the others answers I think I solved my problem, (and realized how the question wasn't properly clear). Thanks a lot – Ratman May 24 at 7:31
• @Ratman : Yes, it is how the field changes with time that is the solution of the wave equation. The wave equation does not constrain the configuration at any given time beyond some differentiability requirements. – The_Sympathizer May 24 at 7:59

Fields are not, in general, solutions of a wave equation. They are solutions of the equations of motion derived from a particular Lagrangian density.

The dynamics of a field are described by a Lagrangian density which is specified as part of the foundations of the field theory. For the free scalar field $$\phi$$, this Lagrangian density is:

$$\mathcal{L}=\partial_\mu\phi\partial^\mu\phi-m^2\phi^2$$

For the electromagnetic field, described by the four-potential $$A^\mu=(V/c,\vec{A})$$ and with charges described by a four-current $$J^\mu=(c\rho,\vec{J})$$, the Lagrangian density is:

$$\mathcal{L}=-\frac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu}-A_\mu J^\mu$$

where $$F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$$ is the Faraday tensor.

There are many other possible Lagrangian densities, though. For example, the simplest interacting scalar field theory is described by a Lagrangian density:

$$\mathcal{L}=\partial_\mu\phi\partial^\mu\phi-m^2\phi^2-\lambda\phi^3$$

for some coupling strength $$\lambda$$.

To get the equations of motion from a Lagrangian, you use the Euler-Lagrange equations:

$$\frac{\partial\mathcal{L}}{\partial\phi}=\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)$$

If we evaluate the Euler-Lagrange equation for the free scalar field, we get the Klein-Gordon equation. If we do the same for the electromagnetic field, we get the Maxwell equations.

But if we do this for the interacting field theory I mentioned above, then we get the following:

$$-2m^2\phi-3\lambda\phi^2=2\partial_\mu\partial^\mu\phi$$

Or, if we simplify, using $$\Box=\partial_\mu\partial^\mu$$, we get:

$$(\Box+m^2)\phi=-\frac{3}{2}\lambda\phi^2$$

which definitely isn't a wave equation.

Generally, you will consistently see the $$(\Box+m^2)$$ operator in these equations of motion because it comes from two terms in the Lagrangian density that will almost always be present for any physical field theory: the "kinetic term" $$\partial_\mu\phi\partial^\mu\phi$$ for the field, and the "mass term" $$m^2\phi^2$$.

• This is quite a clarifying answer to a confusingly worded question, +1. I think the OP has an interesting observation that for a perturbation around a solution to satisfy the on-shell condition, the perturbation would have to follow a wave-equation which seems true to first order for weak couplings. – Dvij D.C. May 22 at 21:55
• I think you correctedly pointed out that i should have wrote just free fields and not fields in general (sorry for the mistake), but thinking only about free fields the question becomes clearer? – Ratman May 22 at 22:04
• @Ratman Can you define "free field", in that case? – probably_someone May 22 at 22:08
• as a field with no sources or interactions. For example the fields that satisfy the equations wrote in the answer. Or regarding your answer the field that satisfy the Eulero-Lagrange equations of the first Lagrangian density you wrote, without the interaction term – Ratman May 22 at 22:12
• @Ratman Then the answer is trivial: free fields are defined to come from a specific set of Lagrangian densities that all have wave equations as the equations of motion. – probably_someone May 22 at 22:16

An answer from an experimental physics physicist, who used theories to analyze data of high energy physics experiments:

Fields are similar to a coordinate system on which the behavior of elementary particle interactions can be modeled with mathematical functions. In essence they replace the luminiferus aether with a Lorenz invariant "substance" that has the same function, to allow modelling energy and momentum behavior of elementary particle interactions.

That they are wave equations is because the model developed from the study of waves in matter, to the study of electromagnetism with great success in describing data. It happens that solutions of second order differential equations are adequate for modelling data at the level of measurements for particle physics too.

I think that wave equations are related to pure fields.

If we allow non equilibium situations in theory of elasticity, the general solution is an elastic wave. The static solutions (as a catenary for example) stay there as special cases.

But, by general solution it doesn't means taking in consideration the air vibration and its effect on the string. This is simply considered negligible.

It is the same for other fields. Static electromagnetic fields in the vacuum must be a solution of the EM wave equation, even not being a travelling wave. On the other hand, a travelling wave, (even a pulse) is also a solution, and I don't see why it could not be called a field. The only difference is not being constant in space and time.

But EM fields that result from interaction with electron fields, in conducting wires for example, are also present in the Maxwell equations, and are not solution of the wave equation.