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15

The easiest solution is to use a fiber-optic camera, i.e. one with a fiber-optic connection between the front lens and the actual camera electronics. You can easily bend the fiber (but not too much!). You can now make a small hole in the microwave (smaller than the wavelength, insert the fiber. Bend the fiber and wrap it in aluminium foil. There will be ...


8

The reasoning has to be the other way around: Light acts on the metal and makes the electrons move. This, however, results in an energy loss, as the electrons feel a resistance and thus the radiation loses energy. This can be formulated more precisely with counteracting electric fields. That's why all good conductors are opaque. In insulators this can not ...


7

You should look at the form of the advanced fundamental solution of D'Alembert equation, built up in geodesically convex open sets including the source localized at the event $y$ and the test point localized at the enent $x$ receiving the wave generating by the source. The construction, at least for analytic manifolds with analytic metrics, is obtained by ...


6

Electromagnetic radiation in a medium propagates according to the law $$ \mathbf E,\mathbf B \propto e^{\imath(\pm k_xx-\omega t)} $$ where $$ k_x^2 = \frac{n^2\omega^2}{c^2}\;. $$ The refractive index $n$ can also be complex, in which case its imaginary part describes the absorption of the EM wave in the medium. But the oscillating part is in any case $$ ...


4

Take a look at the conventional form of Maxwell equations. They tell us that Gauss's law actually applies every time. However, to get the field $\vec{E}$ from the charge distribution by the usual methods, we also need to know that $$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} = 0$$ Because otherwise the field could not be generated by the ...


4

Every star or galaxy contains some elements, and each element emits a particular frequency. Here are the lines of the Sun (https://en.wikipedia.org/wiki/Fraunhofer_lines) In particular, Hydrogen is present almost everywhere and Hydrogen lines are visible in most galaxy spectra. The Hydrogen-alpha line is particularly strong in many galaxies. This ...


3

An overview in layman's terms: First, it is important to note that not any electric field will induce current in a conductor, because other than the fact the intensity of the field defines the speed of each charge (bigger difference of potential), the oscillation frequency of the $\mathbf{E}$ also plays a very important role, if the frequency is too high, ...


3

It generally does not work in curved spacetime. There is a quite thick book almost completely devoted to study this issue by P. G√ľnther: Huygens' Principle and Hyperbolic Equations. Some discussions can be found in Friedlander's book about the wave equation in curved spacetime. A necessary condition for the validity of the Huygens principle is that the ...


3

It's because most materials have (many) natural resonances. I assume you are alright with the phase velocity being different in different media, that is I assume you are alright with something of the form $$ \frac{ \omega }{ k } = \frac{c}{n} = \frac{ c}{\sqrt{ \mu \epsilon }} \sim \frac{c}{\sqrt \epsilon} $$ where $k$ is the wavenumber of a wave, ...


3

You correctly surmise that you could make a cage with small holes, so that the visible light could pass through but the microwaves would be blocked. However, you could have some major problems if you decide to put a metal object inside a microwave oven - see the same wikipedia page you got your frequency information from. You would have to construct your ...


3

For the benefit of brevity I'll give a straight answer: the second Weyl scalar is not related to electromagnetic radiation in any way. As per my previous comment, a simple proof is the Schwarzschild (or Kerr) spacetime. It describes a stationary black hole, without electromagnetic (or even gravitational) radiation, yet there is a basis in which all Weyl ...


2

Your picture is ok, so long as you realise that this is a snapshot in time. If you looked "from above" the electric field of the waves (by your definition of the polarisation direction) would be pointing towards or away from you - so this becomes rather difficult to draw in the same way. It might be better to think of the E-field as a field of arrows, ...


2

Yes, it an extremely small effect but it exists in Einsteins general relativity. There is one case of a double star where there rotation around each other seems to lose energy at rate that this phenomena should give according to general relativity


2

Question: Why do accelerating charges radiate, and can moving charges with zero acceleration also produce electromagnetic radiation? Quick Answer: Only accelerating charges (and changing currents) produce electromagnetic radiation. The power of fields generated by charges with zero acceleration dies out at far distances. Full Answer: The full answer to ...


2

Perhaps the best way to understand this is to start simply: Consider a function $f(x)$. Now, let's try to take the Fourier transform of its derivative $f'(x)$. Just use the definition of the Fourier transform: $$\mathscr{F}(f'(x))(k)=\frac{1}{\sqrt{2\pi}}\int dx\ e^{-ikx}f'(x) $$ and now use integration by parts (assuming $f(x) \to 0$ as $|x|\to \infty$, a ...


1

Part of the genius of Maxwell was to realise that it did not require the presence of currents or charges to generate a magnetic field. In analogy to the way that changes in magnetic field generating an electric field (or macroscopically we say that changes in the magnetic flux linked with an electric circuit can produce an EMF and hence a current), it turns ...


1

what does orbits with definite energy mean? It means that each orbit has an amount of energy associated with it. If you move between 2 orbits, it requires a certain amount of energy. It will be the same amount of energy for the same transition in the same atoms. Different transitions have different amounts of energy and different atoms have different ...


1

There is a good explanation of this in Matter and Interactions vol II by Sherwood and Chabay. I no longer have the text; I will try to summarize its explanation as I remember it. The electrons in a substance are analogous to charged masses on springs. The electrons in insulators are relatively tightly bound; those in conductors are loosely bound or unbound. ...


1

No. When matter and antimatter annihilate, they do so particle by particle. Each electron annihilates with an anti-electron, protons with anti-protons, etc. And in each annihilation the total rest mass of the two particles is converted to energy in photons. Even the lightest annihilation, that of an electron with an anti-electron, must put over $1\ ...


1

You could put your camera in an invisibility cloak for microwaves. I don't know how that would distort the images, but it has probably the advantage of not disturbing anything else inside the microwave. If you use a metal cage, there might be interference effects, and the existence of the camera would influence the thing you want to examine.


1

Unless you are using an oversized microwave (like industrial or scientific size) and a large protective shield for the camera, I doubt it would be practical to film the inside of a microwave oven while it is running, without destroying said camera. If you are only looking to simulate the effect, you may disconnect the magnetron circuitry from the microwave ...


1

Electrons and photons definitely are not the same. For instance electrons have rest mass and photons do not. Also, electrons have charge, while photons do not. I could go on about their very different statistical behavior in quantum mechanics (electrons have a quantum mechanical property called "spin" that is half-integer, photons have integer "spin" ) but ...


1

I understand that voltage is the movement of electrons No, the movement of electrons is (one type of) electric current. There can be a voltage without the movement of charge. and that the antenna acts as a light bulb No, an antenna is a resonant system that, ideally, has zero resistance while an incandescent light bulb has a resistive element ...


1

You're asking a classical electromagnetism question. You don't benefit from thinking about photons. If the antenna's electromagnetic field (calculated from Maxwell's equations) is right-circularly-polarized at a certain point, then you can say that every photon is right-circularly-polarized. If the field is linearly polarized, then you can say that every ...


1

The field from a current in a wire is purely magnetic for a static current. When the current varies with time, there will be radiation. What you are missing is that a radio antenna doesn't operate with DC. You can perhaps understand it like this: the magnetic field from a current loop depends on its magnetic moment, which is current times area. When the ...



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