# Electromagnetic wave equation - phase and amplitude

• I
There are some things that confuse me about electromagnetic waves, and I haven't found good answers anywhere.
Consider the following equation: E=E0 e i(wt-kx) (here E and E0 are vectors, I couldn't find the right symbols).
The things that confuse me are the following:

1° We say that the power of the exponential is the phase of the wave. So at every instant it would correspond to the angle on the unit cercle. If the wave is linearly polarised, we could represent it as a cosine, so, on the graph, each phase would have different amplitude. But here is the confusion, the phase is not related to the amplitude - which is also confirmed given that the amplitude of the exponential is always 1, but given the graph of cos, it would seem to me that E0 should be multiplied by a value smaller than 1 of we are not on themaximum point (which is wrong, I know, but please help me visualise this in another way that would help clear my confusion).

2° For a linearly polarised function (the subject of my confusion could be extended to all waves however, but this is simpler to explain) the wave could be expressed as Ecos(wt-kx). How could this function be graphed in dependance of both t and x at the same time, to keep its form? This is what I don't understand. If we take x as a variable, then it's clear that on y (for example) we have its amplitude, but how does t enter the game?

3° (Bonus question) This is a bit unrelated to the previous two, but how could a beam of light be carrying Electric and Magnetic fields? This means that light should interract somehow with itself and other photons due to the field. I suppose it doesn't, because for there to be an interraction there needs to be a force, but since light has no mass there can be no force. Is this why a photon is called the EM force carrier?

## Answers and Replies

jtbell
Mentor
on the graph, each phase would have different amplitude
In English physics-speak, "amplitude" always means the maximum (or "peak") value of the quantity that is "waving" (E in this case).

If we talk about a wave on a string, or a sound wave, where objects (e.g. molecules) are moving back and forth, then we often talk about the displacement of the wave at a certain phase. Then the amplitude is the maximum displacement.

However, in an electromagnetic wave, nothing actually moves back and forth the way molecules do in a vibrating string. What changes is the strength (and direction) of the E field (and the B field) at a given position. Therefore we don't use the word "displacement" here. I can't think of a good single word to use instead. I would just say something like "the value of E at phase 90°"

• nasu
tech99
Gold Member
3° (Bonus question) This is a bit unrelated to the previous two, but how could a beam of light be carrying Electric and Magnetic fields? This means that light should interract somehow with itself and other photons due to the field. I suppose it doesn't, because for there to be an interraction there needs to be a force, but since light has no mass there can be no force. Is this why a photon is called the EM force carrier?
I am not an expert, but my understanding is that the source of radiation is an accelerating charge. This creates a transverse electric wave, and when this passes an observer, for relativistic reasons, a B field is observed. I am not sure why you are saying light should interact with itself, as other sorts of waves do not do this in a linear medium.

jbriggs444
Science Advisor
Homework Helper
2019 Award
how could a beam of light be carrying Electric and Magnetic fields? This means that light should interract somehow with itself and other photons due to the field. I suppose it doesn't, because for there to be an interraction there needs to be a force, but since light has no mass there can be no force. Is this why a photon is called the EM force carrier?
Classically, a beam of light is a disturbance in the electric and magnetic fields. Maxwell's equations call for disturbances to propagate at the speed of light. There are no photons in the classical description.

If you use the quantum mechanical description then you have photons. Photons are massless. But massless entities can have non-zero momentum.

tech99
Gold Member
Classically, a beam of light is a disturbance in the electric and magnetic fields.
But if we start with a charge which is "stationary", where is the magnetic field in which to make a disturbance? By contrast, the stationary charge already possesses a static electric field which can be disturbed when the charge accelerates.

RPinPA
Science Advisor
Homework Helper
1. The real part of ##E_0 e^{i\theta}## is ##E_0 \cos(\theta)##. That is a real-valued oscillating function. And the imaginary part is ##E_0 \sin(\theta)##. That is a purely imaginary oscillating function. Those are the things which can have magnitude less than ##E_0##.

The complex exponential is really a mathematical convenience, and they come in conjugate pairs. If you have a frequency ##\omega## in your solution, you probably also have a frequency ##-\omega##. Since actual fields are real-valued, the coefficients on these two terms will be such that the total field is real-valued. In that case it will be a sine or cosine with some phase that has the oscillating behavior you're expecting.

What I'm saying is really a general statement about differential equations with exponential solutions, which represent real physical quantities. For instance if you solve the equation for a mass on a spring, you can express the solution as complex exponentials (which as I said are mathematically convenient). But when you include the initial conditions to get the specific solution, you better get a real valued function because position is a real value.

2. If you hold x fixed at ##x_0## so treat ##k x_0## as a constant phase term ##\phi## , then you have ##E_0 cos(\omega t - \phi)##. This is oscillating in time with frequency ##\omega##. That expresses that at any fixed point, you'll see a field oscillating in time. If you hold t fixed at ##t_0##, meaning you take a snapshot of the wave at an instant in time, and treat ##\omega t_0## as a constant phase term ##\phi##, you have ##E_0 cos(\phi - k x)##, which is a wave in space with distance ##2\pi/k## between the crests. So the wavelength is ##\lambda = 2\pi/k##.

If you change x by ##\Delta x## and t by ##\Delta t## such that ##\omega \Delta t - k \Delta x = 0##, so ##\omega (t + \Delta t) - k(x + \Delta x) = (\omega t - k x) + (\omega \Delta t - k \Delta x) = \omega t - kx##, then the value of the function will be unchanged. That condition holds if ##\Delta x## and ##\Delta t## are related by ##\Delta x / \Delta t = \omega / k##. That means that the entire function is traveling at the velocity ##\omega / k##.

3. Classically, Maxwell's equations say that a changing electric field induces a magnetic field and a changing magnetic field induces an electric field. Light doesn't interact with itself because electromagnetic fields have no charge, not because there's no mass. Electric fields affect charges. Magnetic fields affect moving charges.

tech99
Gold Member
3. Classically, Maxwell's equations say that a changing electric field induces a magnetic field and a changing magnetic field induces an electric field... .
But it might be worth mentioning again that the two fields are in-phase.

tech99
Gold Member
3. Classically, Maxwell's equations say that a changing electric field induces a magnetic field and a changing magnetic field induces an electric field. Light doesn't interact with itself because electromagnetic fields have no charge, not because there's no mass. Electric fields affect charges. Magnetic fields affect moving charges.
Even if two light rays hit a conductor, containing charges, they do not interact. One ray is not altered by the other. To have interaction we require something with charges and which is non linear, like a diode.