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13

Let's take a slightly more general case: Consider a wave with wave vector $\vec k=(k_x,k_y,k_z)$, with the electric field given by $$\vec E=\vec E_0\ e^{i(\vec k \cdot \vec r-\omega t)} $$ where $\vec r=(x,y,z)$. Now, we wan't to satisfy Maxwell's equations in the vacuum, including Gauss' law: $$\vec \nabla \cdot \vec E=0$$ The derivative is quite easily ...


11

In quantum mechanics a particle can be treated as a wave and a wave can be treated as a particle. This is the notorious wave particle duality. I won't go into this any further here because it's been discussed to death in lots of previous questions. Search this site for wave particle duality if you're interested in finding out more. Anyhow, assuming I ...


7

In this answer, I'll start with a real expression for $E$, because I think the exposition is clearer. There is no loss of generality in doing that, because the real expression will always be equivalent to the real part of the complex version of $E$, for some appropriate choice of the origin. Thus, my starting point is $$E(z,t)=E_0\ sin(k z-\omega t)\ .$$ ...


7

In fact the light is not split up like in a rainbow or a prism. The colours appear due to thin film interefence - see e.g. http://en.wikipedia.org/wiki/Thin-film_interference The thickness of the petrol or oil is similar to the wavelength of visible light, which is about 380 to 750 nm or 0.38 to 0.75 $\mu$m (- or about 500 atoms think). Different colours ...


7

When did they know that light and other electromagnetic wave doesn't need a medium? Physicists had an inkling early in the 20th century with the development of Planck's law in 1900, Einstein's development of special relativity in 1905, and Einstein's explanation of the photoelectric effect, also in 1905. This strongly suggested that electromagnetism was ...


4

Comment to the question (v4): It seems relevant to mention that there in principle could be a difference between the universal speed limit constant $c$ (which is usually casually referred to as the speed of light in vacuum), and the actual speed of light in vacuum if the photon has a rest mass, at least from an experimental point of view. Of course, no ...


4

No, different EM waves do (classically) not interact, they just pass through each other. There are (tiny) contributions to a photon-photon diagram in quantum electrodynamics which could be seen as photons scattering off each other, but, at the macroscopic level where we usually talk about EM waves, there isn't any interaction at all.


3

Thermal radiation consists of electromagnetic radiation that was produced by the thermal motion of charged particles in matter. In particular, the thermal radiation surrounding an object in thermodynamic equilibrium with its environment is known as black-body radiation, which has a characteristic spectrum that depends only on the object's temperature. Most ...


3

They have various properties that differ, but the differences are quantitative, not qualitative, and there is no sharp boundary. The differences occur because of the difference in frequency. A wave that is a gamma ray in one frame of reference could be an x-ray if observed in a different frame. An example of their different properties is that gamma rays are ...


3

The Michelson-Morley experiment was the first good evidence they had that there is no luminiferous aether, and was conducted in 1887. (Michelson had an experiment in 1881 trying to do the same, but it was flawed)


2

Short answer: yes, they are both electromagnetic waves and differ only in frequency. Slightly longer answer: of course they also differ in all other properties that are a function of frequency: wavelength, energy, momentum.


2

Imagine you have a monochromator that you can tune to give you whatever wavelength you like. Now send light at a wavelength of 568 nm towards the interface. What has been calculated is that destructive interference takes place, so that less of the incident radiation is reflected. Let's take an extreme case where the reflection from the first and second ...


2

It's not so much about "penetrating magnetic fields" as it is about seeing the signal above background noise (assuming your transmission is above the background plasma frequency). The intensity of any given signal drops off as $\propto$ r$^{-2}$. This means that a signal sent from r = 1 will be 16 times as strong as a signal from r = 4. All receiving dish ...


2

EMP is typically a high frequency signal. It is possible to allow low frequency signals to enter your Faraday cage by decoupling the signal - a choke in series and a capacitor in parallel. You need to make sure that the choke does not saturate at the current spikes expected - and that the voltage rating of the capacitor is sufficient to absorb the energy. A ...


2

The wave mechanics dispersion relation you cite is for EM waves propagating in free space. In other media, the dispersion relation is not necessarily linear (it can be quadratic or have some more complex dependence). So in this context, there's nothing special about quantum mechanics. More generally, the dispersion relation tells us about the phase speed ...


2

Typically electrical measurements rely on skin contact electrodes and do not directly provide detailed information on the electrical characteristics of the heart. On the other hand, magnetic fields go through the skin and provide a direct "view" of the heart and its movement. Electrical pickups are easy. Low field magnetic sensors are difficult to implement ...


2

If photons transmit the electromagnetic force, which is observable: the photon or the electron? Do we ever directly measure a photon, or do we only measure it's effect on electrons. For example suppose I shine a laser at a wall Let us clear up that photons ( and also electrons) are quantum mechanical elementary particles, and classical electromagnetic ...


1

That the phase speed can have a dependence on the wavelength/frequency of the wave. For instance, a whistler mode wave can have a cubic dispersion relation at low frequencies. In this limit, the higher(smaller) frequencies(wavelengths) propagate faster than the converse. It results in a sort of "spreading out" of the wave modes. This if often seen ...


1

The Result You Seek The Wikipedia page on angular momentum of light gives the classical angular momentum as: $$\frac{\epsilon_0}{2i\omega}\int_{\mathbb{R}^3} \left(\mathbf{E}^\ast\times\mathbf{E}\right)d^{3}\mathbf{r} +\frac{\epsilon_0}{2i\omega}\sum_{i=x,y,z}\int_{\mathbb{R}^3} ...


1

Generally Tesla coild are fairly safe from the point of view of X-ray generation. You could potentially test for it using a sealed can of camera film with a high ISO rating, placing it in the vicinity of the coil while it is operating, and then developing it to see if it is fogged. However, a bigger problem might be ultraviolet light, ozone and nitrogen ...


1

If you insist on thinking of photons as waves (which is fine as long as you ignore absorption, though you should really think of it as a disturbance in an electromagnetic field), you can more or less think of all of their amplitudes as being equal, and this is why your premise doesn't make sense. More precisely, the amplitude of a single photon isn't ...


1

Some of the other answers have suggested that the amplitude of the photon's wavefunction is well defined, and that it has the same value for any two photons of the same energy. Whatever else we may say, this can't possibly be right. The amplitude of an electromagnetic wave is defined either by its electric field or by its magnetic field. (In a sensible ...


1

The fact that different observers in relative motion can measure the same light ray to move at a speed of c has to do with the fact that each observer defines the "speed" in terms of distance/time on rulers and clocks at rest relative to themselves. It's crucial to understand that different observers use different rulers and clocks to measure speed, because ...


1

Some of my recent results may be highly relevant to your question. For example ( http://link.springer.com/content/pdf/10.1140%2Fepjc%2Fs10052-013-2371-4.pdf - published in the European Physical Journal C, open access; http://akhmeteli.org/akh-prepr-ws-ijqi2.pdf - published in the International Journal of Quantum Information), I showed that the matter field ...


1

The (say, average) frequency depends on the temperature of the body emitting the thermal radiation. It's infrared for room temperature, it can be UV for much higher temperatures.


1

Here is a slightly different take on this using the boundary conditions for electromagnetic fields at an interface. A key boundary condition, that is derived from Faraday's law, is that the component of the E-field tangential to the boundary must be continuous. So take an EM wave travelling at normal incidence with the electric field solely in a direction ...



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