# Why do the high frequency waves have the most number of modes?

While reading the Wikipedia page of Ultraviolet Catastrophe, I came across how Rayleigh and Jeans applied the equipartition theorem. They told that each mode must have same energy. Now as the number of modes are greatest in small wavelengths or large frequency, energy radiated will be infinite.

What is mode actually? And why do the large frequencies have the most modes? Please help me explaining these. I need a math-free explanation.

Suppose we have a drum. When you bang on the drum it will vibrate. When you look at any point on the surface of the drum head, you will see it go up and down, similar to if to take a mass hanging from a spring attached to the ceiling and watch it bob up and down.

However, there is also an important difference between these two situations. In the situation with the mass hanging from a spring, the motion is periodic, which means the mass just repeats the same motion over and over. On the other hand, in the case of the drum you will not usually see periodic motion; the motion will be irregular. But there are very special ways the drum can vibrate so that the motion at a single point is periodic and in fact every point on the drum oscillates with the same period. This type of mode is called a normal mode. What makes normal modes very useful is that any motion can be written as a sum of these normal modes.

Now why are there more higher frequency normal modes than lower frequency ones? Lets look at some normal modes of a drum, and it should be more obvious.

This is the motion of just the drum head. The rest of the drum is not showing. We are looking at six different normal modes. Notice that each point in a given drumhead oscillates up and down with the same frequency (but a different amplitude).

This is a nice illustration, but there is one inaccuracy: in the illustration, each normal mode appears to have the same frequency; however, in real life, the ones towards the bottom would have a higher frequency (that is, a shorter period of oscillation). The reason for this is that they have more wrinkles. The more wrinkles they have, the more the drumhead has to stretch to make the wrinkles. This means that once the amplitude is at a maximum, the high stretching will cause a high restoring force. Since there is a high restoring force, there will be a high oscillation frequency. I have now explained why modes with many wrinkles have a large oscillation frequency.

Also, there are more modes with lots of wrinkles than modes with few wrinkles. This is just because, for example, there is only one way to have no wrinkles, but there are many ways you can put in 10 wrinkles.

So we have found that there are a lot more modes with many wrinkles than with few wrinkles, and the ones with many wrinkles oscillate at a higher frequency. This implies that there are more modes at higher frequency than at lower frequency.

• +1 . Sir do modes mean standing waves?? Then high frequency waves have larger standing waves??? Sir, what do the wrinkles mean??? Thanks. – user36790 Nov 25 '14 at 16:44
• yes, the modes are just standing waves. It is a standing wave because the features (the peaks and troughs) stay in the same place over time. This is different from a wave in the ocean, where the crest of the wave travels along the surface. The high frequency waves do not have a larger amplitude (how how the peaks go), but they move up and down faster. Also the high frequency waves have more wrinkles. – Brian Moths Nov 25 '14 at 18:41
• To see what wrinkles mean, look at this picture of wrinkled cloth. You can see the cloth doesn't lie on a flat plane; there are places where the cloth goes above the plane and places where it is below. In fact, in the picture there are many wrinkles, meaning that there are many places where the cloth switches between being up and down. This is similar to what I mean when I say the drum head is wrinkled. If you look at mode 2,1 you can see there are places where the drum head goes above the plane and places where ... – Brian Moths Nov 25 '14 at 18:47
• the drumhead goes below the plane. Now we can also say that mode 0,1 is less wrinkled that mode 2,1, because mode 0,1 has just one region that switches between being up and down, while mode 2,1 has four regions. – Brian Moths Nov 25 '14 at 18:48
• What are these (0,1) (2,1) numbers? – user101134 Dec 28 '16 at 12:02

The "modes" in this case refer to the standing waves that can exist in a cavity. A very nice diagram / explanation is given at http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html

To summarize: if you consider a cavity of dimension $L$, the modes that fit inside the cavity have wave numbers $n_1$, $n_2$, $n_3$ such that

$$n_1^2 + n_2^2 + n_3^3 = \frac{4L^2}{\lambda^2}$$

It then goes on to show that the number of these modes (represented as dots on the surface of a sphere) increases as the radius of the sphere increases (which it does with the inverse of the wavelength).

See the link for all the details.

Edited and simplified on behalf of the crowd (useless at this point of other very good answers):

Consider a cube of edge length L in which radiation is being reflected and re-reflected off its walls. Standing waves occur for radiation of a wavelength λ only if an integral number of half-wave cycles fit into an interval in the cube. In other words, electromagnetic standing waves in a cavity at equilibrium with its surroundings cannot take just any shape.The solution to the wave equation must give zero amplitude at the walls, since a non-zero value would dissipate energy and violate our supposition of equilibrium. The basic solutions are sinusoidal standing waves, that is, it has peaks and valleys. The peacks become valleys and valleys peaks as time gows ones, but the "edges" between peaks and valleys remains at a constant spatial position. Also, at the walls there is always an "edge". More energetic waves have larger frequencies, that is, more peaks and valleys along a given length. The energy actually depends on the total number of peaks when yoy consider the three prependicular directions in space, because the wave is tridimensional. You can have 100 peaks in the vertical directions 10 in the horizontal and 25 in depth. The larger the numbers you choose, the larger the combinations of the three numbers that can result, when squared and added, into a given specific number. A single mode consist of one specific wave with three specific number (and thus one total larger combined number). And there are many ways to choose the individual numbers to get a given combined obtai different waves of the same energy by combining the trhree numbers into the same large number.

Thus, a larger number means more energy, and more potential combinations of smaller numbers that can combine into that single number. The larger the energy the larger the number of modes (remember, one mode is a specific combination of three numbers).

• Sir. Thanks for the answer. But if you please kindly just use words to describe what mode is and why it is more at high frequencies, I will be very grateful. Remember your other answers? They were great. I just want the intuition. Please, please! – user36790 Nov 25 '14 at 11:39
• Sir, please, i really need a math-free explanation. I earnestly urge you , please. – user36790 Nov 25 '14 at 12:23
• @Wolphram jonny: do not think that statistical mechanics has much to do with the question as it stands. Could you trim irrelevant parts and focus attention on modes? – Incnis Mrsi Nov 25 '14 at 13:39
• @IncnisMrsi yes, thanks, I'll do it later today. I will also modify it it for OP's request. – Wolphram jonny Nov 25 '14 at 13:40

You need to look at the idea of Separation of Variables for Partial Differential Equations.

You consider a toy universe comprising oscillators in a box: let's think of a cuboid microwave cavity with electromagnetic fields losslessly confined within perfectly reflecting walls.

The Cartesian components of the electric field all fulfill Helmholtz's equation:

$$\left(\nabla^2 + \frac{\omega^2}{c^2}\right)\psi = 0$$

and they must vanish at all the boundaries. If you work though the standard separation of variables technique, you get a set of allowable modes with standing plane wave variation for the vibration and any general solution is superposition of these allowedd modes. So, the $x$, $y$, and $z$ variations are all of the form $\sin(k_j\,x_j)$.

The electric fields must vanish at the boundaries, this means that the $k_j$ can only be discrete values: $\frac{pi}{L_x},\,\frac{3\,pi}{L_x},\,\frac{5\,pi}{L_x},\,\cdots$ for the $k_x$ values, where we suppose the reflecting walls are at $x=0$ and $x=L_x$. Likewise, the $k_y$ and $k_z$ are restricted to $\frac{(2 n_j + 1)pi}{L_j}$ for $n_j = 0,\,1\,2,\,\cdots$. So we have modes defined by mode numbers: triplets of integers $(n_x,\,n_y,\,n_z)$, and the total variation of each mode is of the form:

$$\psi_{(n_x,\,n_y,\,n_z)}(x,\,y,\,z,\,t) = \sin\left(\frac{(2\,n_x+1)\,x}{L_x}\right)\,\sin\left(\frac{(2\,n_y+1)\,x}{L_y}\right)\,\sin\left(\frac{(2\,n_z+1)\,z}{L_z}\right)\\\,\cos(\omega(n_x,\,n_y,\,n_z)\,t + \delta)$$

But now, substitute this back into the Helmholtz equation and we get:

$$\omega(n_x,\,n_y,\,n_z)^2 = c^2 \left(\frac{(2\,n_x+1)^2}{L_x^2}+\frac{(2\,n_y+1)^2}{L_y^2} + \frac{(2\,n_z+1)^2}{L_z^2}\right)$$

and the above equation can only be fulfilled for discrete $\omega(n_x,\,n_y,\,n_z)$, given that the $n_j$ are integers. Now think of the integer triplets as points in 3D space: the number of modes of frequency $\omega$ or less is the number of these discrete points in the positive $x,\,y\,z$ eighth of the sphere with radius less than $\omega/c$ - in other words, roughly the volume of this eigth, as $\omega$ gets large. Thus the rate of increase of allowed modes with $\omega$ is roughly proportional to the surface area of the spherical sector. In other words, proportional to $\omega^2$. There are roughly $100$ times more modes between frequency $10\,\omega_0$ and $10\,\omega_0 + \Delta$ as there are between frequency $\omega_0$ and $\omega_0 + \Delta$.

• Separation of variables has nothing to do with the question. Domains of dim > 1 are not necessarily rectangles, cuboids, parallelepipeds, or anything where spatial variables can be separated in a meaningful way. – Incnis Mrsi Nov 25 '14 at 13:43
• @IncnisMrsi not in itself, but it is one means of getting to and grasping a model of state density. – WetSavannaAnimal Nov 25 '14 at 13:50

Spectrum of waves in 1D is described with one numerical parameter. It may be either frequency, or wavelength (propagation speed divided by frequency *), or wave vector (inverse wavelength, multiplied by 2π ; hence proportional to frequency *). In three spatial dimensions the wave vector becomes 3-dimesional and frequency is proportional to its magnitude.

Which choice of parameter is more sensible? Frequency is preserved during refraction, whereas wavelength isn’t. Wave vector “behaves reasonably” under changes of reference frames (at least for electromagnetic waves), whereas wavelength does not. Mathematically, frequency and wave vector are also more sensible because naturally arise from Fourier transform. Hence, we should assume that, in 1D, there are 10 times “more waves” between, say, 100 MHz and 1 GHz that between 10 MHz and 100 MHz. In three spatial dimensions “measure of waves” depends on the maximal frequency cubically, not linearly.

* Of course, Ī’m aware of media with dispersion, but it doesn’t change the reasoning much.