# Classical point particles to classical fields

I often hear that in the continuum limit we can study large numbers of particles as fields. I always imagined that by removing all bounds on the number of particles (while keeping total energy, momentum, charge etc. conserved), we can generate a field. Of course I have no idea how this is formally accomplished or whether it is even meaningful to do such a thing.

Without appealing to any sort of quantum theory, how can one show that in this 'continuum limit' a large collection of indistinguishable classical point particles gives rise to a classical field? If not, why?

I think this question might have been repeated, but all related questions I found involved taking limits of quantum systems.

• A point charge classically has a well defined electric field, for example. No need of limits of any kind. en.wikipedia.org/wiki/Electric_field – anna v Jan 9 '14 at 19:56
• When you use the terminology "photon" classical fields are out the window. Photons are quantum mechanical entities and have to be treated as such. In this link the build up of classical electromagnetic fields from photons is demonstrated: motls.blogspot.com/2011/11/… . It is not simple. When dealing with classical problems stick to classical electromagnetic theory. It gives the correct answer one would get going through the contortions of the quantum level. – anna v Jan 10 '14 at 4:52

Although I am not 100% sure what you are asking, I believe you are talking about considering a lattice of discrete points and taking the continuous limit such that the distance between the lattice points tend to zero.

The simplest example is to consider an 1-dimensional elastic rod with mass density $\mu$. The force applied on it due to Young's modulus is given by: $$F = -Y \xi$$ where $\xi$ denotes the deflection of the elastic rod from its equilibrium position. We can think of the rod as an infinite amount of equally spaced spaced particles at rest where we will let $m$ denote the mass of each particle (which is of course the same for each particle) and we let $a$ denote the distance between the particles. Now, we can write the mass density as follows: $$\mu = \frac{dm}{dx} = \displaystyle\lim_{a\to 0} \frac{m}{a}$$

Now we will assume that each particle only interacts with its nearest neighbours and so the force between the particles can be approximated by using Hooke's law: $$F = - \kappa \left(y_{i+1}-y_i\right) = - \left( \kappa a \right) \frac{y_{i+1}-y_i}{a}$$ Furthermore, by writing the force expressed in terms of Young's modulus in terms of the relative distance $a$: $$F = - Y \frac{y_{i+1}-y_i}{a}$$ we can write: $$Y= \displaystyle\lim_{a\to 0} \left(\kappa a \right)$$ To sum up, we have related Hooke's constant to Young's modulus.

Furthermore, by ordinary classical mechanics we can write the potential energies in all the springs as: $$V= \sum\limits_{i} \frac{1}{2} \kappa \Delta y_i^2 = \sum\limits_{i} \frac{1}{2} \kappa \left(y_{i+1}-y_i\right)^2$$ and the kinetic energy of all particles is given by: $$T= \sum\limits_{i} \frac{1}{2} m \dot{y}_i^2$$ Therefore, the Lagrangian of the system is given by: $$L=T-V = \sum\limits_{i}\left[ \frac{1}{2} m \dot{y}_i^2 - \frac{1}{2} \kappa \left(y_{i+1}-y_i\right)^2 \right]$$ Using the Euler-Lagrange equation, we can easily find the equations of motion for each particle $j$ in the discretized rod: $$\frac{\partial L}{\partial y_k} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{y}_k} \right) = 0$$ $$\Rightarrow \frac{\partial }{\partial y_k} \left( \sum\limits_{i} \frac{1}{2} \kappa \left(y_{i+1}-y_i\right)^2 \right) + \frac{d}{dt} \left[ \frac{\partial }{\partial \dot{y}_k} \left( \sum\limits_{i} \frac{1}{2} m \dot{y}_i^2 \right) \right] = 0$$ $$\Rightarrow \frac{\partial}{\partial y_k} \left( \frac{1}{2} \kappa \left(y_{k+1}-y_k\right)^2 + \frac{1}{2} \kappa \left(y_{k}-y_{k-1}\right)^2 \right) + \frac{d}{dt} \left[ \frac{\partial }{\partial \dot{y}_k} \left( \frac{1}{2} m \dot{y}_k^2 \right) \right] = 0$$ $$\Rightarrow - \kappa \left(y_{k+1}-y_k\right) + \kappa \left(y_{k}-y_{k-1}\right) + m \ddot{y}_k = 0$$ Let us now consider the limit that the spacing in the discretized rod tends to zero: \begin{align} y_k(t) & \rightarrow y(x,t) \\ y_{k+1}(t) & \rightarrow y(x+a,t) \\ y_{k-1}(t) & \rightarrow y(x-a,t) \\ y_{k+2}(t) & \rightarrow y(x+2a,t) \\ y_{k-2}(t) & \rightarrow y(x-2a,t) \\ & \dots \end{align} The equations of motion can now be written as: $$- \kappa \left[ y(x+a,t) - y(x,t) - y(x,t) + y(x-a,t) \right] + m \ddot{y}(x,t) = 0$$ $$\Rightarrow - \kappa \left[ \frac{y(x+a,t) - y(x,t)}{a} - \frac{y(x,t) - y(x-a,t)}{a} \right] + \frac{m}{a} \ddot{y}(x,t) = 0$$ $$\Rightarrow - \left( \kappa a \right) \left[ \frac{\left(y(x+a,t) - y(x,t)\right)/a}{a} - \frac{\left(y(x,t) - y(x-a,t)\right)/a}{a} \right] + \frac{m}{a} \ddot{y}(x,t) = 0$$ Taking the limit $a \rightarrow 0$, we can write: $$\displaystyle\lim_{a\to 0} \frac{y(x+a,t) - y(x,t)}{a} = \frac{\partial y(x,t)}{\partial x}$$ and: $$\displaystyle\lim_{a\to 0} \frac{y(x+a,t) - 2y(x,t) + y(x-a,t)}{a^2} = \frac{\partial^2 y(x,t)}{\partial x^2}$$ Therefore, we obtain the equations of motion for the vibration of the field $y(x,t)$: $$-Y \frac{ \partial^2 y(x,t)}{\partial x^2} + \mu \frac{ \partial^2 y(x,t)}{\partial t^2} = 0$$

Edit:

• Good example of a continuous limit, works even without probabilistic assumptions. – Ján Lalinský Jan 9 '14 at 22:28
• @JánLalinský thanks for your comment. Happy to hear that. – Hunter Jan 9 '14 at 22:42
• @dj_mummy The photon is par excellence a quantum mechanical entity/particle . One has to go through quantum mechanical calculations when using the terminology "photon" . One can postulate point charges in classical electromagnetism but photons do not exist in that framework. – anna v Jan 10 '14 at 4:57
• @dj_mummy That is the way I understand it, including that there does exists a mathematically smooth transition from the underlying quantum level to the classical set ups, but the classical treatment is adequate as long as elementary particles are out of the framework under study. – anna v Jan 10 '14 at 5:20
• @dj_mummy I don't think you can derive the Klein-Gordon equation this way explicitly. The way I know the Klein-Gordon equation is derived is by considering the relativistic energy-momentum relation: $E^2 = \mathbf{p}^2+m^2$. And then replacing variables by their appropriate operators acting on the wave function $\phi(\mathbf{x},t)$. However, the above example will (conceptually) help us through the table I have added in the original post. – Hunter Jan 10 '14 at 11:50

There are attempts to derive hydrodynamics from kinetic theory of molecules, for example. I do not know much about details, but it has to do with averaging, smoothing and probabilistic assumptions. For example, one can begin with $N-$ particle distribution function that satisfies Liouville equation (particle description) and under certain assumptions and after many complicated steps derive new equations for field quantities such as density or velocity field that describe the particles in probabilistic way.

You can take a look what is it about in R. Balescu: Statistical Dynamics, Matter out of equilibrium, Imperial College Presss, 1997.

• In chap.10, page 151 the author discusses hydrodynamics, but I do not know to what extent this is derivation of hydrodynamics (I did not read it, but generally I expect some hard problems). – Ján Lalinský Jan 10 '14 at 13:49