# The phase of electric and magnetic forces in a photon.

• merlyn
In summary, the electric and magnetic forces are in phase and perpendicular to each other because of the properties described in the free Maxwell equations and the plane-wave ansatz. This leads to the dispersion relations for electromagnetic waves in a vacuum and shows that the fields are always in phase and perpendicular. Additionally, Feynman's discussion of an infinite plane of charge can provide further insight into this concept.
merlyn
I'm curious if someone help me understand why the electric and magnetic forces are IN PHASE at right angles to one another?
Should they not be 90 degrees out of phase in order to conserve energy? I do understand they are in phase but why?

Thank you all for your time.

Merlyn.

Just solve the free Maxwell equations with the plane-wave ansatz
$$\vec{E}=\vec{E}_0 \exp(-\mathrm{i} \omega t + \mathrm{i} \vec{k} \cdot \vec{x}), \quad \vec{B}=\vec{B}_0 \exp(-\mathrm{i} \omega t + \mathrm{i} \vec{k} \cdot \vec{x}),$$
where it is understood that the physical fields are just the real parts of these complex-valued fields. The reason for using this trick is that it is easier to do the calculation with exponential rather than trigonometric functions.

Now (in Heaviside-Lorentz units) two free Maxwell equations, i.e., with ##\rho=0## and ##\vec{j}=0## read
$$\vec{\nabla} \cdot \vec{E}=\vec{\nabla} \cdot \vec{B}=0.$$
This implies
$$\vec{k} \cdot \vec{E}_0=\vec{k} \cdot \vec{B}_0=0,$$
i.e., both the electric and the magnetic fields are transverse waves.

Then you have
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0$$
and
$$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=0.$$
Taking the curl of the first equation, using again ##\vec{\nabla} \cdot \vec{E}=0## and eliminating ##\vec{B}## with the 2nd equation leads to
$$\left (\frac{1}{c^2} \partial_t^2 -\Delta \right)\vec{E}=\Box \vec{E}=0.$$
In an analogous way you find
$$\Box{\vec{B}}=0.$$
Plugging in our plane-wave ansatz leads to
$$\frac{\omega^2}{c^2}-\vec{k}^2=0 \; \Rightarrow \; \omega = c |\vec{k}|=c k,$$
i.e., the usual dispersion relations for em. waves in a vacuum.

Now you also have
$$\vec{\nabla} \times \vec{E}=-\vec{E}_0 \times \vec{\nabla} \exp(\cdots) = -\mathrm{i} \vec{E}_0 \times \vec{k} \exp(\cdots) \stackrel{!}{=} -\frac{1}{c} \partial_t \vec{B}=\mathrm{i} \frac{\omega}{c} \vec{B}_0 \exp(\cdots).$$
Together with ##\omega=c/k## this leads to
$$\frac{\vec{k}}{k} \times \vec{E}_0=\vec{B}_0.$$
The other Maxwell equation doesn't lead to anything new but
$$\vec{E}_0=\vec{B}_0 \times \frac{\vec{k}}{k}.$$
This is equivalent to the previous equation, given that ##\vec{k} \cdot \vec{E}_0=\vec{k} \cdot \vec{B}_0=0##.

This implies that ##\vec{E}## and ##\vec{B}## are always in phase and perpendicular to each other.

To describe photons rather than classical fields, just read field operators instead of c-number valued fields.

Last edited:
BvU
vanhees71 said:
with the plane-wave ansatz
It looks like your plane wave ansatz assumes the thing to be proven.

Dale said:
It looks like your plane wave ansatz assumes the thing to be proven.
No it doesn't, because ##\vec{E}_0## and ##\vec{B}_0## are arbitrary complex amplitudes. As my calculation shows, however
$$\vec{B}_0=\frac{\vec{k}}{k} \times \vec{E}_0,$$
and thus, since ##\vec{k}/k## is a real vector, there's no phase shift between ##\vec{B}_0## and ##\vec{E}_0##.

merlyn said:
Should they not be 90 degrees out of phase in order to conserve energy?

In a standing wave the fields are 90 degrees out of phase.

You could write down traveling plane wave guesses for the electric and magnetic fields that were 90 out of phase and if you plug your guess into Maxwell's equations your guess won't satisfy Maxwell's equations? Maybe in the process of doing so you might come to an intuitive understanding why they are in phase in a traveling wave?

Feynman discusses in infinite plane of charge given sudden motion. It might help.

http://www.feynmanlectures.caltech.edu/II_18.html

vanhees71

## 1. What is the phase of electric and magnetic forces in a photon?

The phase of electric and magnetic forces in a photon refers to the relationship between the electric and magnetic fields that make up the photon. It describes the oscillation of these fields as the photon travels through space.

## 2. How is the phase of electric and magnetic forces in a photon measured?

The phase of electric and magnetic forces in a photon is typically measured using interferometry techniques, which involve splitting the photon into two beams and then recombining them to observe their interference patterns.

## 3. What is the significance of the phase of electric and magnetic forces in a photon?

The phase of electric and magnetic forces in a photon is important because it determines the polarization of the photon, which in turn affects its interactions with matter. It also plays a role in the propagation and detection of light.

## 4. Can the phase of electric and magnetic forces in a photon be altered?

Yes, the phase of electric and magnetic forces in a photon can be changed through interactions with matter or through manipulation of the photon's path. This is the basis for technologies such as phase shifters and modulators.

## 5. How does the phase of electric and magnetic forces in a photon relate to the wave-particle duality of light?

The phase of electric and magnetic forces in a photon is a manifestation of the wave-like nature of light. It is also closely related to the particle-like behavior of light, as it determines the energy and momentum of the photon. This duality is a fundamental aspect of quantum mechanics and is still an active area of research in physics.

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