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Q&A

Why are there infinitely many modes of electromagnetic radiation?

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My question.

I have started reading (on my own) B.Hall, "Quantum Theory for Mathematicians (Graduate Texts in Mathematics, 267)", Springer, 2013.

In the introduction section, the author says that the black body cavity has infinitely many modes of electromagnetic radiation.

I am wondering why it is that the black body cavity has infinitely many mode of electromagnetic radiation and not only finite number of them.


My attempt.

The context for the infinity of e/m radiation modes is the following. The author used it to explain the concept of ultraviolet catastrophe which would introduce the reader to the problem which quantization of energy solved.

The author starts with saying that, under the thermal equilibrium condition, the statistical mechanics suggests that the average energy of every mode of e/m radiation is $k_BT$. Then he says that, since there are infinitely many modes, adding up their energy will give an infinite energy of the e/m radiation, which is a contradiction to reality and is called ultraviolet catastrophe.

I could not understand how statistical mechanics, which is concerned with the "particles" constituting a body, and e/m radiation, which is concerned with e/m waves, are connected. Then I learnt that it is customary to think that a body consists of particles, particles oscillate, oscillations result in emissions of e/m waves, the energy of oscillations is converted to the energy of waves without losses, therefore, if statistical mechanics says that the average energy of every mode of the particle's oscillation is $k_BT$, then the average energy of every mode of its e/m wave is $k_BT$ as well.

That explanation satisfied me to some degree and the next logical step for me was to try and understand what a mode of oscillation is. I always loved mechanics, because it is something we can see with our own eyes and understand. So, I decided to learn what a mode is on the example of oscillating bodies - not on the example of e/m waves. I found a nice draft of the book D.Morin, "Waves", Harvard university. He starts with two bodies on a spring moving horizontally, derives a dynamic equation for them and solves it as: $$x_1(t) = A_s cos(\omega_s t + \phi_s) + A_f cos(\omega_f t + \phi_f)$$ $$x_2(t) = A_s cos(\omega_s t + \phi_s) - A_f cos(\omega_f t + \phi_f)$$ Here, $x_1(t)$ and $x_2(t)$ are displacements of the first and the second body respectively (the bodies move along the $x$ axis). Then he proceeds as to say that if $A_f=0$, then $x_1(t) = x_2(t) = A_s cos(\omega_s t + \phi_s)$ which means that the two bodies move in unison and he calls it the first normal mode of the system. If on the other hand, he says, $A_s=0$ , then $x_1(t) = -x_2(t) = A_f cos(\omega_f t + \phi_f)$ which means that the two bodies move in the opposite directions and he calls it the second normal mode of the system.

And I concluded that, mathematically, one just count the number of the $cos$ terms in the solution and get the number of the normal modes of the system.

Morin then says that if the number of the masses on the spring goes to infinity, we, effectively, describe a continuous system. I understand it as the model of any body. In our case, it can be the model of our black body. Morin derives the equation for an infinite number of masses on a spring: $$\xi(x,t) = A_1 e^{i(kx+\omega t)} + A_1^* e^{i(-kx-\omega t)} + A_2 e^{i(kx-\omega t)} + A_2^* e^{i(-kx+\omega t)}$$

If one uses Euler's identity to expand every term, one can count, at most, $8$ $cos$ and $sin$ terms, which means (as I pointed out above) the system has, at most, $8$ normal modes - not infinity.

And since, as I understand it, this equation represents the oscillations of the infinite number of the particles constituting our black body, oscillating at their places and emanating the e/m wave without losses of energy, the same equation describes of our e/m wave, which means that our e/m wave has, at most, 8 normal modes - not infinity of them.

This is where I am stuck and seeking help of Physics Codidact community.

Note that I know that in the above equation, I can substitute $\xi$ with $E$ per eranreches's answer to the Physics SE post titled "Why is electromagnetic wave (light) a sine wave?". Note that I also know that e/m radiation has its own equation based on $E$ and $B$ per freecharly's answer to the Physics SE post titled "What is a mode? Regarding electromagnetic waves in a closed cavity". But I want to build up intuition about normal modes of e/m waves starting from particles emanating e/m waves, because this is how the book that I am reading introduces me to the ultraviolet catastrophe.

Apart from that, the Physics SE post titled "What is a mode? Regarding electromagnetic waves in a closed cavity" embeds my question in itself (i.e., the question about the number of modes of e/m radiation), but none of the answers to that post addresses it.

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1 answer

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In what should come as no surprise, you don't determine the number of normal modes by the number of $\cos$ subexpressions occurring in the solution. (I guess there's a sense where you could say that you're "counting the cosines", but it's not this simple syntactic sense.)

It would help to have the Big Picture behind normal modes and why we care about them. Let $L$ be a constant coefficient, linear differential operator. For example, the early examples in the chapter you link would have $L = m\frac{d^2}{dt^2} + K$ where $K = \begin{pmatrix} k + \kappa & -\kappa \\ -\kappa & k + \kappa\end{pmatrix}$. Solving the differential equation then means finding a function $f$ satisfying the boundary conditions such that $Lf=0$.

Write $\mathcal F[f](\omega) = \int_{-\infty}^\infty f(t)e^{-i\omega t}dt$ for the Fourier transform of $f$. Using the key property $\mathcal F\left[\frac{df}{dt}\right](\omega) = i\omega\mathcal F[f](\omega)$, our differential equation $Lf = 0$ becomes $$Lf = 0 \iff \mathcal F[Lf](\omega) = 0 \iff \tilde L(i\omega)\mathcal F[f](\omega) = 0$$ where $\tilde L(i\omega)$ is a matrix whose components are polynomials in $i\omega$. For example, the above $L$ becomes $\tilde L(i\omega) = \begin{pmatrix}m(i\omega)^2 + k + \kappa & -\kappa \\ -\kappa & m(i\omega)^2 + k + \kappa\end{pmatrix}$ which is equation $(9)$ in the paper. $\tilde L(i\omega)$ for a fixed $\omega$ is a linear operator, and, indeed, $\mathcal F$ is a linear operator. We get that a normal mode corresponds to an orthogonal basis vector of the linear subspace $\tilde L(i\omega)^{-1}(\mathbf 0)$ for some $\omega$. Altogether, they form an orthogonal basis for the linear subspace of solutions to $Lf = 0$. Indeed, they essentially let us build a solution in terms of its Fourier components.

From equation $(10)$, we see if $\kappa=0$, there is only one frequency $\omega$ (modulo sign), and $A_1$ and $A_2$ can be chosen independently. This makes sense since the equations then describe two independent oscillators. This would lead to only one "cosine", but we'd have two distinct normal modes at the single frequency.

For the equation you list at the end, the reason there are infinitely many normal modes is any frequency $\omega$ can be chosen which will lead to $k = \omega\sqrt{\rho/E}$ as written in equation $(84)$. So there's a distinct solution for any frequency. The logic used in the section can again be analyzed in terms of Fourier transforms. In this case, we'd use a two-dimensional Fourier transform satisfying $\mathcal F\left[\frac{\partial f}{\partial x}\right](\omega,k) = ik\mathcal F[f](\omega,k)$ and $\mathcal F\left[\frac{\partial f}{\partial t}\right](\omega,k) = i\omega\mathcal F[f](\omega,k)$ which would transform $E\frac{\partial^2 f}{\partial x^2} - \rho\frac{\partial^2 f}{\partial t^2} = 0$ to the equation $-E k^2 + \rho\omega^2 = 0$ which is just equation $(84)$ again.

If you're not familiar with Fourier transforms, then the above is probably hard to go through. In that case, though, I'll quote from the final paragraph of the chapter you linked:

Fourier analysis plays an absolutely critical role in the study of waves. In fact, it is so important that we’ll spend all of Chapter 3 on it.

Fourier analysis is an extremely valuable tool across all of math and physics, both theoretical and practical. It should be clear from the above that it simplifies and systematizes the computation of normal modes, or, indeed, the solution to this class of differential equations in general.

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