# Why while deriving Rayleigh Jeans law in this book, an EM wave is divided into its components along the three axes & how is the math formed?

1. An EM wave making arbitrary alpha, beta and gamma angle with x,y and z axes respectively is divided into its components along the three axes? What does it even mean? If the components along the three axes are the components of the oscillating electric fields at different time, then these components cannot be considered separate EM waves because the direction of the changing electric fields along the axes would be like an oscillating spring along respective axis, not longitudinal as needed.

2. It was assumed that the arbitrarily directed EM wave makes a standing wave. I don’t understand how an EM wave falling not perpendicular to a surface might reverse back along the same direction it hit the surface? (The wave is in a cube with each side length ‘a’.)

3. Let’s say the wave makes a standing wave somehow. In the proof, different planes were imagined through the nodes and perpendicular to the direction of the wave so the oscillating electric field is parallel with the planes. Why was it important? And how equations 1-13 were written?

1. In the next page, why the three electric field components had to have their values zero at ‘a’ distance away? What good will it do? How can it guarantee that in this way the original (initial) wave would have its node upon hitting the wall of the cube if the electric field components had their values zero at ‘a’ distance away?
• Ins't this from Resnick & Eisberg Quantum Mechanics? Mar 5 at 11:06
• @Petrini well yes, i feel terrible for saying otherwise, thanks Mar 5 at 11:23

You seem to have overlooked an important piece of context, which is that this derivation is only deriving a special case of the Rayleigh-Jeans law. It is deriving the spectrum of allowed frequencies within a cube with side $$a$$. And allowed frequencies must be the frequencies of standing waves because any EM wave that is not a standing wave will quickly interfere with itself and die away (and we are studying the steady state here, so we are not interested in transients).

So we choose $$x,y,z$$ axes perpendicular to the faces of the cube and we break each wave down into components along the $$x,y,z$$ axes by projecting its wave function $$f(x,y,z,t)$$ onto each of these three axes in turn. Equation 1-13 simply expresses the wavelengths $$\lambda_x, \lambda_y, \lambda_z$$ of the three components in terms of the wavelength $$\lambda$$ of the whole wave. The factor of $$\tfrac 1 2$$ is introduced here because we are actually interested in the distance between nodes, which is half of the wavelength.

The reason that we can consider each component separately is that we know EM waves interact in a linear fashion. So we can find the allowed wavelengths of the $$x,y,z$$ components and then combine them to find the allowed wavelengths of the whole wave. If we were studying non-linear waves than the analysis would be much more complicated - and may not even be possible.

The condition for a wave component, say $$f_x(x, t)$$, being a standing wave within the cube is that $$f_x(0,t)=0$$ for all values of $$t$$ and $$f_x(a,t)=0$$ for all values of $$t$$. And if each of the three components is a standing wave then when we add the components back together we will find that the wave as a whole is also a standing wave i.e. $$f(x,y,z,t)$$ satisfies $$f(0,y,z,t)=0$$ for all values of $$y,z,t$$ and $$f(a,y,z,t)=0$$ for all values of $$y,z,t$$ (and the same for the other axes). Once again, this is a consequence of the linear nature of EM waves.

In the second section there is a switch from counting wavelengths to counting frequencies (Equation 1-14a) and number of allowed frequencies in a small interval between $$\nu$$ and $$\nu + d\nu$$ is approximated by calculating the volume between two concentric spheres (Equation 1-14b). This approximation assumes that $$\lambda$$ is small compared to $$a$$.

So we have calculated the spectrum of allowed frequencies $$N(\nu)$$ for a cubical cavity - but what about other shapes of cavity ? The real "hand-waving" is in the final sentence where it says "It can be shown that $$N(\nu)$$ is independent of the assumed shape of the cavity and depends only on its volume".

• And if each of the three components is a standing wave then when we add the components back together we will find that the wave as a whole is also a standing wave i.e. 𝑓(𝑥,𝑦,𝑧,𝑡) f ( x , y , z , t ) satisfies 𝑓(0,𝑦,𝑧,𝑡)=0 f ( 0 , y , z , t ) = 0 for all values of 𝑦,𝑧,𝑡 y , z , t and 𝑓(𝑎,𝑦,𝑧,𝑡)=0 f ( a , y , z , t ) = 0 for all values of 𝑦,𝑧,𝑡 y , z , t (and the same for the other axes). How would I be sure that the wave hits the wall exactly when the component waves travel distance ‘a’? Mar 5 at 15:13
• @AYMShahriarRahman What you mean by "hits the wall exactly" ? Remember this is a standing wave so in general it resonates throughout the cavity with frequency $\nu$. We only require that its $x$ component on the faces $x=0$ and $x=a$ is always zero - and the same for the other pairs of faces. This condition restricts the values of $\lambda_x, \lambda_y, \lambda_z$ to a set of discrete values. Mar 5 at 17:03
• I also think it is nonsense to decompose a wave into orthogonal "components" because there is no such $\mathbf{k, k_1,k_2,..}$ for which one can write in general that for some amplitudes $Ae^{i k \cdot r} = A_x e^{i k_x \cdot r} +A_ye^{i k_y \cdot r} +A_ze^{i k_z\cdot r}$. To me the argument only make sense for a "brick " shaped cavity and along sides standing waves. Mar 6 at 16:48