I finally went and got my Ham Radio licence. I've been reading about how radio operators use the ionosphere for long distance communication through ionic reflections of transmitted/received EM waves. I'm trying to visualize how this happens, so that I have more intuition into what frequencies to use under various conditions of the ionosphere. Can someone give an illustration of how EM waves are reflected by ions in the ionosphere, to help convey pictorially why wavelength matters?
123351
says:     ... Unlike high frequencies (HF), the ionosphere does not usually   reflect VHF waves (called skywave propagation) so transmissions are   restricted to the local radio horizon less than 100 miles...   Why does this happen? Is it merely that a VHF signal of say, 100Watts, carries more energy than an HF signal of the same power? Are there any other factors in play here?
53315
I am confused on why would the propagation speed of any EM waves at ANY frequency is constant in the free space (vacuum) but they seem to disperse in any other materials as the propagation speed of EM waves in other materials depends on the frequency? Can someone give me a good proof of this phenomena with appropriate formulas?
49122
I need to understand one seemingly simple thing in wave mechanics, so any help is much appreciated!  When a pulse on a string travels to the right toward an open end(like a massless ring that is free to oscillate only in the vertical direction), then when the wave reaches the end it gets reflected and it becomes a positive pulse traveling to the left. So, my first question is, why do we even have reflection in this case(as far as energy and forces go)? What causes it in physical terms(intuitively)? And why for an open end the reflected wave is a positive wave and why for a closed end the reflected wave is a negative wave? An explanation using the tension of the string and energies will be much appreciated.  Also, for the open end case, the massless ring will have an amplitude of two time the amplitude of the incident wave while the wave gets reflected. Why is this the case? (or to state it differently why does the ratio of the amplitude of the transmitted wave divided by the amplitude of the incident wave equal to +2?).  Thank you!    ΝΟΤΕ: I want an explanation using forces or energy and not "image waves"
16766
How is the following classical optics phenomenon explained in quantum electrodynamics?   Reflection and Refraction   Are they simply due to photons being absorbed and re-emitted? How do we get to Snell's law, for example, in that case?  Split by request: See the other part of this question .
38628
For an object to create different EM waves, it needs to increase the temperature, so what if we or some material could be so hot, that it would emit ultraviolet light, and thanks to that be invisible for the human eye.   I have a lot of questions about this and I would like you to answer me.   At which temperature does an object emit UV light? If an object emits UV light, we wouldn't see anything or we would see some type of violet light? Is there any material that can get to that temperature without melting? Is there any powerful insulator?   Thanks for reading this, and for breaking my dream of making someone invisible, I invite you to day-dream and imagine stupid questions.
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During emission spectrum $$\Delta E=h\nu,$$ where $\nu$ is the frequency. All books write that it is the frequency of photon, but photon is a particle and not a wave.  More than that what this frequency actually is? Is it the frequency at which energy packets are released?  This wave and particle nature is causing conflict!  What is the meaning of frequency of photon? I mean it is a particle and not a wave and frequency is a physical quantity associated with a wave.
5281
The book I'm reading about optics says at some point that "each color (wavelength) contained in the white light interferes only with itself". But why is this so?  Edit: I moved the rest of the question .
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$?
18800
Relative permittivity shows that force of interaction in some media is 3 times less than in vacuum ($\varepsilon = 3$) for example. When looking tables of permittivity I never saw a value less than 1.  Russian wikipedia says that $\varepsilon &gt; 1$ is typical for low frequency. So I guess that under some conditions it is possible that $\varepsilon &lt; 1$.   For a static (electrostatic) electrical field is it ever possible (and for what material) that $\varepsilon &lt; 1$? For the electric component of EMF field - when is it possible that $\varepsilon &lt; 1$?
29191
This asked:   And also this:     The second question is contradictory; maximum frequency -> minimum wavelength.   I am asking the very opposite;    What is the minimum frequency and maximum wavelenght of electromagnetic radiation?   The lowest measured/defined seems to be 3 Hz;  Which means a wavelenght 1/3 of the speed of light; ~100 000 000 m.   But this can't be the physical limit for the wavelenght. Does such a physical limit for the wavelength exist? (Similar limit like the speed of light is for velocity).
47210
A  posted on Youtube.  How does this phenomenon work? I know he is using frequency to propagate water in a sine/cosine wave, but how does it exactly work this way? Why do we see it as if its paused, or in slow motion, or even in reverse?  Can someone please explain this? I am intrigued by the video and want to know it's physics.
47641
I'm just starting to learn physics and I have a question (that is probably stupid.)  I learned that energy levels that the bound electron can have are discrete. I also learned that when an electron transitions from one level to another, a photon with a specific wavelength (energy) is created and released into the wild.  My question is this: are there holes in the electromagnetic spectrum, values of frequency that the created photon can't possibly have?
10270
Both light and radio waves are electromagnetic waves.that means they have almost similar properties. Both are EM waves ,E and H fields are there in both waves.....but My question is "radio waves can pass through walls ,but why can't a light pass through walls or opaque medium???
56339
I don't see why you can derive the black body spectrum from the mode density of the radiation inside a reflecting cavity. I do understand the mathematical side of the derivation and I know that an absorbing cavity with a small hole is the best approximation of a black body. But aren't the standing EM waves inside the cavity only there because the cavity is perfectly reflecting, determining the boundary condition of a vanishing field at the boundaries? How is this releated to a cavity with black, (perfectly) absorbing walls? Again, I do understand that the walls can't reflect all of the incoming waves, because there must be a possibility for the matter and EM waves to interact in order to establish the thermodynamic equilibrium, but then how can you look at standing waves inside the cavity, if it is not perfectly reflecting?
92110
What is the value of $$a/b/c,$$  for $a,b,c&gt;0$?  Is it $ac/b$ or $a/bc$?  I get confused about this when I was studying wireless communications. The unit of the spectral efficiency is defined as $bit/second/Hz$.    if it is $ac/b$ then the unit of spectral efficiency will be $bit Hz/second$ which is $bit/second^2$; and  if it is $a/bc$ then I will have $bit/secondHz$ so it is simply $bit$.    Hence, I guess  $$a/b/c=a/bc.$$  But that's not true, isn't it?
38884
Despite being overall near-netural, I would expect the individual electrons and proton to radiate long EM waves as we accelerate around the Sun or rotate around our poles. Is the acceleration so small that the radiated energy is miniscule?  Or is it that somehow the macroscopic neutrality prevent radiation? If this is so, what mechanism causes single electrons or nuclei to fail to radiate?
16042
The greater the frequency range of a transmission medium, the greater the number of bits per second it can transmit. In other words, the bigger the bandwidth in hertz available, the bigger the bandwidth in bits per second that can be transmitted.  Given that, why we don't use gamma rays, x-rays or ultraviolet to transmit data instead of e.g. visible(light) in optical fiber or microwaves(cell phones)?