# Why light is a sinusoidal wave?

1. Jul 22, 2011

### lovetruth

Light is always considered to be a sinusoidal or harmonic wave without giving any reason. Maxwell's equation says that light is an EM wave but does not say its a harmonic wave or even a periodic wave.
People say that every periodic or non-periodic function can be broken down as a sum of series of sine waves by using fourier analysis. But any function can be broken down as sum of series of other function. For example, any function can be broken down into a sum of series of polynomial functions using taylor series. Sine wave can be broken down as a sum of polynomial functions using taylor series. A simple sine wave can also be broken down into a sum of series of other periodic waves with various frequencies and amplitudes. So why other periodic waves such as triangle, saw tooth, square waves or other periodic waves of arbitary shapes are not considered while talking about light.
While finding intensity in Young double slit experiment, the light is considered as a sine wave. If light is considered as pure triangle wave or other periodic wave, the formula of intensity as a function of distance on the screen will be very different.

2. Jul 22, 2011

### Staff: Mentor

That is the only reason. There is nothing fundamental going on.

You could certainly use other basis functions, but they are usually less convenient.

This is factually incorrect. If you do the math right you will get the same answer in any basis. Of course, it is far easier to make a mistake working in a bad basis.

3. Jul 22, 2011

### lovetruth

A simple sine wave of a certain frequency and amplitude when broken down in triangle wave results in sum of infinite number of triangle waves with various amplitudes and frequencies different than that of a single sine wave. So a single frequency sine wave should give a spectrum of triangle waves with various frequencies. This mean that the wavelength of light really depends on the type of wave being discussed. If all periodic waves were equivalent then a red color sine wave can be split into a infinite spectrum of triangle wave with varying wavelength.

4. Jul 22, 2011

### Staff: Mentor

Sure. When people are talking about wavelengths and frequencies they are talking about sin waves, so doing that would cause all sorts of communication problems and confusion. But there is nothing physically wrong with what you are saying.

What do you mean by "equivalent"?

5. Jul 22, 2011

### lovetruth

By equivalent, I mean no bias for using sine wave when talking about light.

A triangle and sine wave of same wavelength are not similar mathematically. Yet we say color only depends upon the wavelength of the periodic wave. A triangle wave can be broken down into an infinite series of sine waves with varying wavelength.

6. Jul 22, 2011

### mathfeel

Mathematically, there is really no distinction between a sinusoidal and a triangular basis. Any field of the form $E(x,t) = f(k x - c k t)$ is a good solution to the wave equation in vacuum. Any set of basis is just as good. In practice, we have more tools to deal with sinusoidal basis with the whole subject of Fourier transform.

But when you are talking about color, you are talking about interaction with atoms (or color receptors in your eye). That corresponds to discrete atomic energy levels. So transition by light occurs for light is an eigenstate of the energy operator $i\partial/\partial t$, and that means a wave of the form: $E(x, t) \propto e^{i (kx - ckt)}$.

Plus, it is easier to generate a sinusoidal wave than a triangular wave.

7. Jul 22, 2011

### Staff: Mentor

Not sure what you mean by "no bias".

No, "color" refers specifically to the wavelength in the sinusoidal basis, not an arbitrary basis. A pure red filter passes only sinusoidal wavelengths of about 650 nm. A square or triangle wave of 650 nm will have some of the "edges" softened.

You cannot assume that filters or molecular transitions with a given spectrum in one basis will have the same spectrum in another basis.

8. Jul 22, 2011

### lovetruth

We can talk about color without using eye or color receptor. We can instead use spectrophotometer or prism.
My question is that does a triangle wave of wavelength 700 nm will look red? A triangle wave of wavelength 700 nm can be broken down mathematically into sum of sine waves with wavelength varying from zero to infinity.

9. Jul 22, 2011

### lovetruth

By bias I mean that light is always considered sinusoidal without giving an explanation

How can you prove that red color is sinusoidal?

10. Jul 22, 2011

### HallsofIvy

Elementary texts don't because they don't want to go into Fourier Analysis. Advanced texts explain what you have been told here.

?? No one has said it is. There is no such thing as "red color". There is a range of wavelengths for colors that we call "red". If we were to use some other basis rather than sinusoids, those same "colors" would have a different range of wave lengths.

11. Jul 22, 2011

### Staff: Mentor

In that case, there certainly is a bias. Most introductory physics textbooks will not bother to give an explanation about different basis functions.

I think that most textbook authors consider it a fairly minor point that would detract from the more important points, such as the fact that EM waves carry energy and momentum. They cannot cover everything, after all.

Last edited: Jul 22, 2011
12. Jul 22, 2011

### lovetruth

By red color I mean 700 nm and not a range.

13. Jul 22, 2011

### lovetruth

What happens when triangle wave of wavelength 700 nm strikes a prism?

14. Jul 22, 2011

### Staff: Mentor

It gets separated into a rainbow of non-uniform intensity.

15. Jul 22, 2011

### nasu

The frequency or period of the wave is independent of the basis.
The fact that you can decompose a triangular or square wave in sines does not mean you get a different frequency. The frequency of the wave is given by the fundamental.

If you have a triangular wave with frequency F0 and look at the Fourier decomposition the first component in the series will have frequency F0. You will have harmonics with frequencies larger than F0, true. What will be the effect of these? In the case of sound waves, these harmonics will determine the tone of the sound. A "triangular" wave will sound different than a sine wave of the same frequency. But they do have the same frequency and the same pitch. And this effect is independent of the basis.

For visible light, the harmonics may not have any influence as the range of visible frequencies is very narrow (less than an octave). The red receptors are sensitive to light of a specific frequency (actually a quite wide range). A wave with the frequency in this range will excite the red receptors. It does not matter if it is pure sine or it has many components. Suppose you take a triangular wave and decompose it in sines. If the fundamental is in the range of red receptors, the first harmonic will be outside the visible range. So it should look red and there will be no effect similar to the tone of the sound.

16. Jul 22, 2011

### nasu

Shouldn't rather be a discrete series of lines? And all of them but one will be invisible. The first harmonic will have a wavelength of 350 nm which is in the near UV.

17. Jul 22, 2011

### lovetruth

Why is the prism prejudiced for sine wave. Why cant it allows single frequency triangle wave to pass as single frequency triangle wave. But the prism allows single frequency sine wave to pass as single frequency sine wave.

18. Jul 22, 2011

### sophiecentaur

If you pass what is defined as monochromatic light through a prism or diffraction grating then all you get is a single line spectrum. The photons of light will have a particular energy, given by hf. If the 'shape' of the waves were anything other than a sine wave, there would also be harmonics of this fundamental f. These would also have to be photons and would have energies of 2hf, 3hf, 4hf etc.. You will notice that these energies get bigger and bigger. The photons cannot exist with less energy than this - at least according to QM. Photons are released by transitions between certain energy levels in a charge system. We usually talk in terms of a gas, because its atoms have discrete energy levels. All this structure would go wonky if we were to have extra photons emitted, with even higher energies than the fundamental.
Edit: And I just realised that the amplitudes of these harmonics would need to be impossible values, too, because there would have to be photons of the harmonics for EVERY photon of the fundamental frequency. It just couldn't work.

So we have to assume that the emitted light does in fact have a sinusoidal wave shape. It's the only shape that QM will allow.
For em waves with frequencies, much lower than light (i.e. the more manageable RF frequencies), it is quite possible to produce a wave with any shape you want because the energy levels involved in an electronic circuit are spread into a continuous band and any frequency can be produced.

Another reason for favouring sine waves when studying waves in general is that a sine wave can be produced by the very simplest form of oscillator - the Simple harmonic oscillator - which oscillates in a 'sinusoidal' fashion. I think you can say that the sine wave is a pretty fundamental waveform, even though it is possible, as a mathematical exercise (and even in practice sometimes), to synthesise any repeating wave shape with many different families of harmonically related wave shapes.

19. Jul 22, 2011

### Staff: Mentor

Only if it runs with perfect stability forever.

20. Jul 22, 2011

### mathfeel

The reason that prism is "prejudice" toward sine wave is that what scatters light into different direction in the prism are motions of charged particles such as atom or electron. The electric field of light drives them slightly away from their equilibrium position. And they responds by oscillating sinusoidally just like any small oscillator. These oscillating charge then omits sinusoidal EM waves.

When you send in a triangular wave, the charge particles that it interacts with do the same thing. But the superposition of the incoming wave and the induced wave will no longer be triangular.