There are solutions to Maxwell's Equations called Standing waves, or so I was told .   If we consider the particle nature of EM waves, what does standing waves mean in terms of photons?  And when electrons excite and then release photons, can these be standing waves? Does that make any sense?  Edit: Most surely, this question is not a duplicate of . The other question is about the general relationship between EM waves and photons. I want to know about this exact same  relationship, but for the case of Standing Waves .
19930
At the end of this nice video (), she says that  is a chain reaction of electric and magnetic fields creating each other so the chain of wave moves forward.  I wonder where the  is in this explanation. What is the relation between electromagnetic wave and photon?
19070
Why do we say that electromagnetic wave is oscillating? Or does light propagate really in a wavy form like this image?    What is making the photons oscillate and how is it oscillating is it oscillating back and forth (and if oscillating back and forth how does it move if its moving back and forth at the same point?)?  Why do we say that the EM field is constantly changing when in the above picture its constant? (we know that a compass is not able to follow an electromagnetic radiation because the EM field is constantly changing)
118042
I'm currently referring to the wave equation derivation given in "Introduction to Electrodynamics" by David J. Griffiths. It follows something like this:  The electromagnetic wave equations are given by the equations: \begin{equation}  v^2_{ph}\nabla^2\textbf{E} = \frac{\partial^2 \textbf{E}}{\partial t^2} \tag{1}\label{eq1} \end{equation} \begin{equation} v^2_{ph}\nabla^2\textbf{B} = \frac{\partial^2 \textbf{B}}{\partial t^2} \tag{2}\label{eq2} \end{equation} Where, $v_{ph} = \frac{1}{\sqrt{\mu\epsilon}} $, is the velocity at which the wave travels in the medium    We have the Heaviside' form of Maxwell's equations in differential form, $$\nabla . \textbf{E} = \frac{\rho}{\epsilon_0}$$ $$\nabla . \textbf{B} = 0$$ $$\nabla \times \textbf{E} = -\frac{\partial\textbf{B}}{\partial t}$$ $$\nabla \times \textbf{B} = \mu_0 \left(\textbf{J} + \epsilon_0\frac{\partial\textbf{E}}{\partial t}\right)$$ In a vacuum and charge-free space ($\rho = 0, \textbf{J} = 0$), these equations are:, \begin{equation} \nabla . \textbf{E} = 0 \tag{3}\label{eq3} \end{equation} \begin{equation} \nabla . \textbf{B} = 0 \tag{4}\label{eq4} \end{equation} \begin{equation} \nabla \times \textbf{E} = -\frac{\partial\textbf{B}}{\partial t} \tag{5}\label{eq5} \end{equation} \begin{equation} \nabla \times \textbf{B} = \mu_0\epsilon_0\frac{\partial\textbf{E}}{\partial t} \tag{6}\label{eq6} \end{equation} We also have the vector identity for any vector function of space, \begin{equation}  \nabla \times \textbf{V} = \nabla.(\nabla\textbf{V}) - \nabla^2\textbf{V} \tag{7}\label{eq7} \end{equation} We thus try to find the curl of $ (5) $ \begin{equation} \nabla \times (\nabla \times \textbf{E}) = \nabla \times -\frac{\partial\textbf{B}}{\partial t} = -\frac{\partial}{\partial t}(\nabla \times \textbf{B}) \tag{8}\label{eq8} \end{equation} Plugging $(6)$ into $(8)$, we get \begin{equation} \nabla \times (\nabla \times \textbf{E}) = -\frac{\partial}{\partial t}\left(\mu_0\epsilon_0\frac{\partial\textbf{E}}{\partial t}\right) = -\mu_0\epsilon_0\frac{\partial^2\textbf{E}}{\partial t^2} \tag{9}\label{eq9} \end{equation} Using the identity (7), \begin{equation} \nabla.(\nabla\textbf{E}) - \nabla^2\textbf{E} = -\mu_0\epsilon_0\frac{\partial^2\textbf{E}}{\partial t^2} \tag{10}\label{eq10} \end{equation} Plugging (3) into (10), we get \begin{equation} 0 - \nabla^2\textbf{E} = -\mu_0\epsilon_0\frac{\partial^2\textbf{E}}{\partial t^2} \tag{11}\label{eq11} \end{equation} With a slight bit of rearrangement we can see that, $$- \nabla^2\textbf{E} = -\mu_0\epsilon_0\frac{\partial^2\textbf{E}}{\partial t^2} $$ $$\nabla^2\textbf{E} = \mu_0\epsilon_0\frac{\partial^2\textbf{E}}{\partial t^2}$$ \begin{equation}  \frac{1}{\mu_0\epsilon_0}\nabla^2\textbf{E} = \frac{\partial^2\textbf{E}}{\partial t^2} \tag{12}\label{eq12} \end{equation} (12) and (1) correspond to the same equation under the conditions imposed. Following a similar line of reasoning for equation (6), we will get the equation; \begin{equation}   \frac{1}{\mu_0\epsilon_0}\nabla^2\textbf{B} = \frac{\partial^2\textbf{B}}{\partial t^2} \tag{13}\label{eq13} \end{equation} which corresponds to equation $(2)$. Thus, we have obtained the electromagnetic wave equation from Heaviside' form of Maxwell's equations in differential form.   Is there a way to derive the same wave equations without assuming to be in a vacuum and charge-free space i.e $ \ \textbf{J} = 0, \rho = 0$?
23955
Do we have any experimental evidence to confirm that light is an electromagnetic wave? Or is it confirmed simply by Maxwell's equations showing a similarity in speed?
43688
We know a sound wave is a disturbance that moves through a medium when particles of the medium set neighboring particles in motion. And using the pressure variations we can plot a  pressure/time graph which looks like a 'wave'in the drawing sense.  What do we mean when we say light is a wave and particles? using the above system, can we  say that light is the disturbance that moves through a medium of photons( forgive me that makes no sense). When we draw a light 'wave', what does it mean? a photon goes up and down?  Please keep the answer as 'not complicated math' as possible.
114051
As I understand, light is what is more generally called "electromagnetic radiation", right? The energy radiated by a star, by an antenna, by a light bulb, by your cell phone, etc.. are all the same kind of energy: electromagnetic energy, i.e. photons traveling through space.  So far, so good? (if not please clarify)  On the other hand, there are these things called "electromagnetic fields", for example earth's magnetic field, or the magnetic field around magnets, the electric field around a charge or those fields that coils and capacitors produce.  Now here is my question:     Are these two things (electromagnetic fields and electromagnetic radiation) two forms of the same thing? or they are two completely different phenomena?   If they are different things, What does light (radiation) have to do with electromagtetism?   I'm not asking for a complex theoretical answer, just an intuitive explanation.
34885
I have a life science degree and even worked in research for a few years. So I feel I should be able to answer this question for myself, but yesterday my daughter blindsided me by asking why, if sound and light are both waves, light can travel through the vacuum of space but sounds can't?  So, I told her that one was a mechanical wave and other was an electromagnetic wave and ... then realised that was the extent of my knowledge. I was parroting stuff I'd learn years ago without any real understanding.  Can someone explain to me, ideally with science but without maths:   what the difference is? why one requires a medium for propagation and the other doesn't? how come electromagnetic waves can still propagate via a medium (i.e. heat waves still carry heat energy through air, or the body)? whether all electromagnetic waves are quantized (i.e. light waves are made of photons but ... heat waves are too, right?) whether it's possible for a wave of each type to exist at the same wavelength?   EDIT: stepping back, it appears I have caused confusion born of ignorance. Essentially, while I can understand the properties of a mechanical wave simply by observing it, I am struggling to understand how intangible things like heat and light can have the same properties.
110732
Now, I don't really mean to say that Maxwell's equations are wrong. I know Maxwell's equations are very accurate when it comes to predicting physical phenomena, but going through high school and now in college, Maxwell's equations are seen as the equations of electricity and magnetism. Now, it's common knowledge among students that, while Newton's laws are generally accurate when applied to everyday experiences, they are also replaced at high velocities by special relativity (and general relativity for very large gravitational fields).  But this is less so the case with Maxwell's equations. I have read that Maxwell's equations are replaced with quantum electrodynamics (which to me has all the effect of mere buzzwords, since I don't know what quantum electrodynamics is) as a more accurate way of describing electromagnetic waves, but what are the limitations of Maxwell's equations?  Or let me phrase this differently. I'm currently an electrical engineering major. I know NASA scientists and engineers can still get away with using Newtonian physics for their calculations, because it's that accurate. I also know, however, that relativity does have to come into play with GPS. So, in what situation as an electrical engineer would Maxwell's equations ever fail me? When (assuming I'm working on such a sufficiently advanced project) would I have to resort to more accurate ways to describe electromagnetic waves?
65939
If photons are spin-1 bosons, then doesn't quantum mechanics imply that the allowed values for the z-component of spin (in units of $\hbar$) are -1, 0, and 1?  Why then in practice do we only use the $\pm 1$ states?  I have been told that this is directly related to the two polarizations of the photon.  This seems to be more of a classical argument however, arising from the fact that Maxwell's equations do not permit longitudinal EM waves in a vacuum.  I have also heard that it is related to the fact that photons have no rest mass, although I understand far less of this reasoning.  What I'm looking for are elaborations on these two arguments (if I have them correct), and perhaps an argument as to how these two are equivalent (if one exists).
28403
Photons are massless, but if $m = 0$ and , then $E = 0c^2 = 0$. This would say that photons have no energy, which is not true.  However, given the formula $E = ℎf$, a photon does have energy as $ℎ$ is a non-zero constant and photons with a zero frequency do not exist.  What am I missing here, or does the famous formula not apply to photons?
66731
Are photons , quantum waves, or both?  If I subdivide an electromagnetic field into smaller , should I eventually find an electromagnetic wave of a ?   How can individual quantum waves combine to form the macroscopic observable of an electromagnetic field?
75540
Pardon me for my stubborn classical/semiclassical brain. But I bet I am not the only one finding such description confusing.  If EM force is caused by the exchange of photons, does that mean only when there are photons exchanged shall there be a force? To my knowledge, once charged particles are placed, the electromagnetic force is always there, uninterruptedly. According to such logic, there has to be a stream of infinite photons to build EM force, and there has to be no interval between one "exchange event" to another. A free light source from an EM field? The scenario is really hard to imagine.  For nuclei the scenario becomes even odder. The strong interaction between protons is caused by the exchange of massive pions. It sounds like the protons toss a stream of balls to one another to build an attractive force - and the balls should come from nothing.  Please correct me if I am wrong: the excitations of photons and pions all come from nothing. So there should be EM force and strong force everywhere, no matter what type of particles out there. Say, even electrical neutral, dipole-free particles can build EM force in-between. And I find no reason such exchanges of particles cannot happen in vacuum.   Hope there will be some decent firmware to refresh my classical brain with newer field language codes.
78766
For example, has anyone has directly observed charges oscillating due to standing EM waves? I am particularly interested because it'd demonstrate that radiation has a transverse electric component to it. Anything else (historical or modern) that shows that light has a transverse electric component would also be gladly invited.
105315
What exactly is causing the electric discharge coming from the clouds to emit light while traveling through the air. I've read and thought about it a little but with my current knowledge I cant really figure what is happening at a subatomic level. In other words, what causes the emission of photons during lightning. What I want to know is, is there some sort of collisions taking place that produces the photons we see, or is it some other interaction that is causing it?. I don't know too much about particle physics, I've only read a couple of pages in the Introduction to Elementary Particles by Griffihts. So if you could keep the answer simple then that would be great.
73549
I understand how one associates the spin of a quantum particle, e.g. of a photon, with intrinsic angular momentum. And in electromagnetism I have always understood the polarization of an EM wave as the oscillations of the E and M field, not necessarily being  aligned with the direction of propagation of the wave.     Questions:   But when one talks about the polarization of a photon in Quantum Mechanics, how does it really differ from its spin?  Are they somehow related and what's the physical idea behind photon polarization in contrast to photon-spin? Feel free to use mathematical reasonings as well if you see it fit!
1974
This is perhaps a total newbie question, and I will try to formulate it the best I can, so here it goes. How does an electromagnetic wave travel through for example, the vacuum of space?   I usually see that waves are explained using analogies with water, pieces of rope, the strings of a guitar, etc, but it seems to me that all those waves need a medium to propagate. In fact, from my point of view, in those examples the wave as a "thing" does not exist, it's just the medium that moves (involuntary reference to The Matrix, sorry).   But in space there is no medium, so how does a wave travel? Are there free particles of some sort in this "vacuum" or something? I believe , so supposedly there isn't a medium for the wave to travel through.   Moreover, I've seen other answers that describe light as a perturbation of the electromagnetic field, but isn't the existence of the field, potential until disturbed? How can it travel through something it does not exist until it's disturbed by the traveling light in the first place? (this last sentence is probably a big misconception by me).
60483
Suppose an electron is created somehow in the universe -- through particle-pair production or what have you. Suppose it is stationary so that electrodynamics (and specifically propagating electromagnetic waves) is unavailable for explanation. Certainly distance charges will feel Coulomb's force, but how quickly?   I'm wondering if, since electromagnetism is manifestly relativistic (Lorentz rather than Galilean invariant) we can use that to always move to a frame where the electron is travelling and then claim that there are in fact EM waves? I don't know.