# Tag Info

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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 ...

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If you insist on thinking of photons as waves (which is fine of course), you can think of all of their amplitudes as being equal. But it is important to remember that they are not some infinitely extending wave (which would imply infinite energy) but rather a wave packet. This is the wave dual of the particle. And the amplitude of this wave does not even ...

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Polarization effects. For EM radiation you have two types of polarization (TEM, TE and TM) whereas in sound you don't have any polarization (it's a scalar field).

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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 ...

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Wonderful question! And it relies on one central concept: virtual particles. Particles that do not necessarily exist, but explain (or would explain) phenomena wonderfully. By assuming that the particles (electrons, protons, etc) are exchanging photons, we can predict their behavior with great accuracy. Imagine that two protons are people, and they are ...

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Though I don't, by any means, have a collegiate degree in Physics or Electromagnetism, I'd like to attempt to give an answer. First, we need to understand what electricity is. Fundamentally, electricity is the flow of particles with any electric charge. There are two common matter particles with a charge: protons and electrons. Since the proton's mass is ...

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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 ...

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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 ...

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You can uniquely define the polarisation of a plane wave from any of the following: The electric field vector as a function of time $\vec{E}(t)$ and the magnetic field (or induction) $\vec{H}(t)$ (or $\vec{B}(t)$; The wavevector $\vec{k}$ and two scalar functions of time, the latter being the transverse components (in the plane at right angles to ...

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The polarization of an electromagnetic wave follows the direction of the electric field. For example, if the electric component is oscillating along the x-axis and the magnetic field is oscillating in the y-axis, the polarization will be along the x-axis.

-1

I haven't done this in a long time, but my understanding from Feynman's QED is that the speed of a photon is unknown (Heisenberg) - photons traveling in a vacuum are around the speed of light +-, but at any instant the speed differs due to uncertainty. The photons going faster and slower than light speed cancel in the same manner that photons bouncing off a ...

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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 ...

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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 ...

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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 ...

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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 ...

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Might find this useful http://upload.wikimedia.org/wikipedia/commons/1/19/Black_body.svg Shows the dominant wavelength that black bodies emit EM waves at. Notice that I said dominant, doesn't mean that it's the only wavelength they emit at, just the one that is emitted the most on average. Might find this a good read too. ...

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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.

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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 $$E(z,t)=E_0\ sin(k z-\omega t)\ .$$ ...

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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 ...

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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 ...

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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)

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The answer to this question is that if you can only see one line or feature in the spectrum then the redshift cannot be measured unless you have some other information that leads you to guess what the line or feature in the spectrum is due to (e.g. the 21cm line of hydrogen at radio wavelengths is so strong and ubiquitous it can usually be identified ...

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This is the electromagnetic spectrum: Note that radio waves are on the other side of the visible spectrum than gamma rays and xrays. The answer though is the same: penetration depends on the frequency. Radio waves are low frequency/large-wavelength, the wavelength much larger than the interatomic distances, and the vibrational energy levels absorb ...

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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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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 ...

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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 ...

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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.

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You're question is one of the canonical questions that inevitably led to Quantum-Mechanics. It is true that in Classical Mechanics the electron is rotating -> radiating, but if that were true the system would be losing energy all the time, until the electron collapses into the proton. by the same token a proton would be radiating since it also rotates ...

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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 ...

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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 ...

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The answer to your question is yes, we can observe beat notes between two different coherent sources of light. This fact underlies almost every precision laser experiment because it allows for lock-in detection. However, there is a subtle difference from audio beatnotes. The difference is that with sound the oscillations are in the air pressure which is ...

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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 ...

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I too am confused on lensing. When light is emitted onto an object outside of space, normally it is absorbed like on earth. I'm guessing since objects creates a curvature in space is what affects it? So lensing really only works outside of a planetary object? Am I just reading too much into it?

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