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16

No. There is no void left by the lack of an aether. The very notion of aether should serve as a warning as to how catastrophically analogical reasoning can fail. "Water waves are in water, sound waves are in air, therefore there must be something in which light propagates." This is flawed logic, and decades of physics were arguably hindered by adhering to ...


13

It's a rather late answer, but the paper Visible optical beats at the hertz level has just appeared on the Arxiv and this describes exactly the phenomenon you ask about. This image from the paper shows the experimental setup: The light frequency is modified using acousto-optic modulators (labelled AOM in the diagram), and to make it look pretty a lens is ...


7

The entire premise of this question is false, neither do electrons orbit atoms with a well-defined speed, nor does this, in any way, correspond to the temperature, since that is a property of systems in thermal equilibrium, not of single particles. Also, whether the Planck length signifies really a shortest achievable wavelength is...debatable.


2

Normally the "data" that is modulated onto the radio frequency of the carrier has a lower frequency than the carrier (this is not an absolute requirement, but few systems exceed it). In case of cell phones, the carrier wave is around 800-1900MHz, while the data rate is on the order of a few Mbit/s, i.e. two orders of magnitude slower than the carrier ...


2

In the case of frequency modulation, the information ("data") is contained in slight modulations of the carrier frequency over time. A "1" could mean that the frequency gets a bit higher and a "0" could mean it gets a bit lower. The modulation in frequency must be tiny enough as to not overlap with the next carrier frequency. And as long as the next higher ...


2

Photons have spin 1, hence they can be treated inside the cavity as a Bose gas. Their distribution obeys the Bose-Einstein statistics: $$\bar{n}_r = \frac {1}{e^{\beta\epsilon_r}-1}$$ where $\bar{n}_r$is the mean occupation number of the energy level $r$, $\epsilon_r$ is the energy of the energy level $r$, and $\beta = \frac{1}{k_bT}$, $k_b$ being Boltzmann ...


2

The change in electronic excitation represents both a potential and a kinetic energy term in classical physics, but there is no simple correspondence to classical physics terms, when you are looking at quantum systems. All we really care about is the total energy difference between electronic states. Those energy differences correspond to the energies at ...


2

What constituent of internal energy does an electron excitation represent? You can think of electrons as just like planets orbiting the sun and get the correct answer to this question. An electron in a higher energy level has less kinetic energy, but more potential energy as it is (generally) farther from the nucleus. The net result is more energy. ...


2

Your math does check out: \begin{align} r&=vt \\ &=0.05\cdot2.9979\times10^{10}\frac{cm}s\cdot4\cdot604800\,s\\ &=3.63\times10^{15}\,cm\\ &=36.3\times10^9\,km\\ &=0.012\,pc \end{align} When a supernova explodes, it enters the free expansion phase, it's position is linear in time ($r=vt$, as used above). It stays in this phase for a few ...


2

Vacuum magnetic birefringence basically involves the same loop diagram as light-light elastic scattering except that two of the four photons come from a magnet. Detecting this effect is the aim of the PVLAS experiment in Ferrara, Italy. See arXiv:1406.6518 and references within. The experiment is running at the moment but the sensitivity is not good enough ...


2

Assuming that we have a uniform electric field between the plates, we can find its intensity to be (using Gauss' Law) $E(t)=\frac{\sigma(t)}{\epsilon_0}=\frac{q(t)}{\epsilon_0 A}$, where $q(t)$ is the absolute value of the charge in each plate of the capacitor which is not constant in time, since the capacitor is being charged. Now, considering a circular ...


2

In term of Maxwell's equations what you need to do is take the two curl equations, and then isolate and substitute: $$ \nabla\times E +\frac{\partial B}{\partial t}=0\\ \nabla\times B =\frac{1}{c^2}\frac{\partial E}{\partial t} $$ So you take a time derivative of, say, the first, exchange nabla and differentiation by time (linear operators that are ...


2

Observable simply means that with the right detector, we could see it. As you have realised, opacity refers to what percentage of the radiation is blocked. The thing is, we can 'see' radio waves - just not with our eyes. Our eyes are only sensitive to a very small range (on the diagram it's the rainbow). If you have something that can detect radio waves, ...


1

On this diagram, why is the atomspheric opacity shaped as it is? Different parts of the atmosphere are responsible for the shape of that curve. Electromagnetic radiation impinging on some object can be reflected by the object, absorbed by the object, or transmitted through the object. An ideal black body absorbs all incoming radiation, regardless of ...


1

An important factor in determining how much energy electromagnetic radiation carries is its intensity, which is just the power per area. In fact, it can be found in the wikipedia article that intensity is sometimes taken to be synonymous with 'level' (even though it is not really correct), so surely this is the right variable to look at.


1

It is believed that high-level microwave radiation is harmful to human ... It is not just believed but well known that extremely intense microwave radiation will cook people. We use microwaves to cook meat, after all, and a good portion of our bodies is in the form of meat. Lesser intensities can cause survivable burns, even lesser intensities might ...


1

I agree with @DavidHammen here, the 'high' and 'low' most likely refer to intensity. A microwave oven works by causing water molecules to oscillate (i.e., shake back-and-forth), which results in the generation of energy (i.e., heat) due to intermolecular "friction." Every molecule in the universe has a resonant frequency. Think of "rocking" on a swing ...


1

If we have a wave of a well-defined direction and frequency, the dependence of the field $F$ (something that is waving) on position and time is $$ F = F_0 \cos (\omega t - k_x x - k_y y - k_z z ) $$ Adult physicists would use complex exponentials instead of the cosine but I decided to remove this potentially difficult piece of maths. The argument of the ...


1

The statement "group velocity represents energy or information transmission" is not entirely accurate. And that is a big can of worms, in fact. But let us start from the beginning. What is the dispersion relation? Suppose that you have some quantity $u$ that depends on a coordinate $x$ and on time $t$. Dispersion relation $\omega(k)$ is basically a ...


1

This is primarily a biological question. We (humans in particular, mammals specifically) can't see radio waves because our bodies do not have the sensors to detect them. We can detect light in the visible spectrum because the rods and cones of our retina that constitute our light sensors absorb photons in that spectrum, ultimately activating neurons in our ...


1

Electromagnetic waves travelling in chiral media will have some parallel component of their electric and magnetic fields, although they are usually not exactly parallel. The quantity $\mathbf{E}\cdot\mathbf{B}$ is proportional to the helicity of light, which basically tells you how close the beam is to having left or right-handed circular polarisation. Tang ...


1

We cannot multiply light by mere reflections, because the very definition of "reflection" means that the same light comes out. We can however multiply light by letting it pass through special materials which we "pumped" into a certain state, that's called Laser. And yes, a Laser design includes a sequence of reflections, but it is not the series of ...


1

Unfortunately, in quantum mechanics "ordinary" reasoning does not get you anywhere. The photon, like any other particle, is neither a particle nor a wave; it is an entity that we can only describe mathematically. It's only when we observe it that it shows up as either particle or wave. Or senses, and hence our logic, evolved to make sense of the real ...


1

You can think of light as the carrier of the electromagnetic interaction. The particles interact with light, not directly with each other. It is an experimental fact that light does not interact with itself. Note that this is not the case with quantum chromodynamics (the theory of nuclear matter). This theory is built along the same lines as quantum ...


1

I am going to address only your second point: It seems that the maximum temperature an object could have is when: B) When the speed of the electrons nears the speed of light. In fact this does not impose any limit to temperature. When you add more and more heat to a body, its atoms and molecules move faster. At this stage, the electrons are not ...



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