# Tag Info

6

The only way for a vector field to have strict spherical symmetry is for it to be purely in the radial direction. For, if it had a non-radial component then that component would have to be preserved under rotations, but you cannot construct a vector field which has that property everywhere on the surface of a sphere. I provide a proof below. (This is closely ...

5

There can be, so long as in addition to being constant, it is curl-free. Ampere's law says that the curl of magnetic field is produced by a current density or time-varying electric field. Since neither is present just outside the interface, then the magnetic field is curl-free there. Faraday's law says that a time-varying normal component of the magnetic ...

5

Write Ignatowsky's equations (erroneously referred to as Jefimenko's equations in ) in the form as follows : $$\mathbf{B} = \frac{1}{c}\int d^3\mathbf{x}'\frac{([\mathbf{J}]\times\hat {\mathbf{n}})}{R^2} +\frac{1}{c^2}\int d^3\mathbf{x}'\frac{([\dot{\mathbf{J}}]\times\hat {\mathbf{n}})}{R} \tag{4}\label{4}$$ $$\mathbf{E} = \int d^3\mathbf{x}' \frac{[... 3 From a radio engineer's perspective, a travelling wave sweeps through a fixed point in space and produces an oscilliatory response at that point. The response will repeatedly cross zero in a periodic manner and certainly if you sample only at those time instants, no response will be detected. There is a related argument in the use of synchronous detection. ... 3 Depends. For a general electromagnetic wave, there is no specific relation between the two. You can have a low-intensity, high-frequency wave (which has only a few photons), a high-intensity, high-frequency wave (which has more photons), a low-intensity, low-frequency wave (which has a large number of photons), or a high-intensity, low-frequency wave (which ... 3 On an atomic scale, atoms emit light like a phased array of individual antennas. Then interference happens, and you get all the behavior of Snell’s Law, Fresnel Equations, etc. 3 Why an electron bound to a nucleus does not emit photons when accelerated? Because bound electrons have no individual identity, no orbits, otherwise they would neutralize on the nucleus, but orbitals which are probability loci. The whole atom/molecule is the quantum mechanical entity, i.e. obeys the rules of quantum mechanics. If an atom molecule has an ... 3 The gamma rays released deep in the core of a star are scattered off of ionized atoms there, which adds energy to the atoms and removes it from the photons. The scattering events are so frequent that it takes a time scale of order ~thousands of years for a photon to rattle its way all the way out to where it can escape into space without further scatterings. ... 3 The skin depth in aluminum at 4 GHz is about 1.3 um (calculated here). So it's not surprising that an aluminum block of thickness 0.5 mm shorting together the two conductors of your coaxial transmission line can completely block the signal. The question is, why did you see nonzero transmission in the real world experiment? Without knowing the actual physical ... 3 Here is your equation for large z$$ \begin{split} U_2(x,y) &= \frac{\textrm{exp}(ikz)}{i\lambda z} \iint_\Sigma U_1(\xi,\eta)\; \textrm{exp}\Bigg( \frac{ik}{2z} \bigg[ (x-\xi)^2 + (y-\eta)^2 \bigg] \Bigg)\,d\xi\, d\eta \hspace{2.4cm} (5)\\ &= \frac{\textrm{exp}(ikz)}{i\lambda z} \textrm{exp}\bigg( \frac{ik}{2z}\big[x^2+y^2\big]\bigg)\times\; ... \\...

2

I want to remind fact that $\vec{B} = - \frac{\vec{r}}{r} \times \frac{\vec{E}}{c}$... It is valid for a plane wave. You can have for example a charged capacitor in a table. There is an electric field inside it, and also the magnetic field of the Earth. The angle between $\mathbf B$ and $\mathbf E$ can be changed arbitrarily

2

As the wave passes through both slits the light diffracts into two coherent waves. Try to imagine the sine wave graphic (top picture) but from above. It's the same principle as a ripple; it's just crests and troughs. The lower graphic can be misleading as when you'd do this experiment with a laser there isn't nearly that much spread of the beam before it ...

2

Imagine a large number of atoms or ions, each allowed to oscillate about their equilibrium positions at high frequency and bounce off one another. We bathe those atoms in electromagnetic radiation, so that energy gets shared back and forth between the atoms and the radiation until their vibrations (which absorb and emit radiation in discrete quanta) come ...

2

(I'm going to focus on protein crystallography here) Yes, in fact Small-angle X-ray Scattering, in which (usually) proteins are not in the crystalline state, is used extensively in biochemistry, molecular biology, etc. to determine the size and shape of biomolecules, and to capture large structural changes for which the atomic-scale picture is not necessary. ...

2

Matter is composed of atoms/molecules in the form of lattices, i.e. the atoms/molecules are organized a distance apart maintained by repulsive and attractive forces mainly electromagnetic ones. As photons impinge on the solid they will interact with the electromagnetic interaction with the spill over field in the spaces between atoms, or directly hit atoms. ...

2

To answer your first question, let's figure out the EOM for your simplified system. First, let's consider the system without damping. Electromagnetism is described by the following Lagrangian density \begin{align} \mathcal{L}_\text{EM} = -\frac14 F^2 + J A\,. \end{align} A simple harmonic oscillator is described by the Lagrangian \begin{align} L_\text{SHO} = ...

2

From Maxwell's Equations of Electromagnetism, we know that accelerated charges emit electromagnetic radiation. It can be shown (see here) that the total power radiated by such a charge accelerating with some acceleration $a$ is given by the Larmor formula: $$P = \frac{1}{4\pi\epsilon_0}\frac{2 e^2}{3 c^3}a^2.$$ Classical electrons are considered to be ...

2

The thing you should keep in mind is that nothing can move at the same speed as light. In fact, as weird as it sounds, the light will always move at the same speed relative to you! This was first discovered experimentally by Michelson and Morley and led to the discovery of special relativity, where speeds don't add up in a simple way. Also, there is a ...

2

I think your emphasis on electric only fields is wrong, both semantically, as bremsstrahlun comes from the german root "to brake" and has nothing specific to electric fields, and because Maxwell's equations are symmetric in electric and magnetic fields. As for the objection on "acceleration". when one defines acceleration from the ...

1

It depends on what the wave is scattered from. The simplest case to study (as an exercise to get intuition) is scattering of a monochromatic EM wave in vacuum from an infinite metal plane: in this case the boundary condition is that the component of the electric field along the surface should be zero. This may add a phase of $\pi$ to one or both components ...

1

The nice and short answer from Gilbert could be expanded a bit. Light as part of the electromagnetic radiation consists of photons. These quanta have an oscillating electric field component and an oscillating magnetic field component, both perpendicular to the direction of propagation. In vacuum the electric and the magnetic field components are exact ...

1

A very brief wave would be described by a wave packet, and if you look the animation in that Wikipedia link, you’ll see that as a wave packet propagates past a location, every part of it passes through that location. That is, the detector cannot see only node,* because at the next instance, the detector would see the peak. Another way to think about this is ...

1

Light consists of quanta. These photons have two oszillating fields, an electric and a magnetic, both perpendicular to the direction of propagation. This is the key to the question, how light is redirected at the surface between two media as well as how the double slit experiment works. The prisms surface contains surface electrons. These electrons interact ...

1

The expression f(x – vt) indicates that you have a function moving in the positive x direction with a speed v. But many waves can be represented by sine or cosine functions (or combinations of these) which require angles (usually in radians) as their arguments. The k and ω put the arguments in radians and also introduce properties of the wave (frequency ...

1

Answers to your question: Yes, the tilted plane wave you draw is exactly what is meant by a plane wave component with a non-zero spatial frequency. If you look at the 2D function representing the complex amplitude of that wave on the indicated plane you will see it has uniform amplitude but a spatially varying phase. The phase will vary with a periodicity ...

1

I assume your first drawing is for a specific time t, as in a photograph, and it is showing spatial variation. Spatial frequency is a measure of how often a particular feature of the wave repeats per unit distance. A plane wave in the positive x-direction such as in your first drawing has a kx term, e.g. in a sin function, where $k = 2\pi/L$ where L is the ...

1

I think your confusion stems from referring to "Fresnel" diffraction pattern, rather than just looking for the complete equation for the wavefront at any point in space. Quoting a randomly-selected paper on high-quality approximations, the rigorous solution involves Lommel functions of two variables, which are defined as series expansions in Bessel ...

1

Yes. Even single molecule imaging is possible and will be practical in the near future. More precisely: at modern x-ray free electron laser facilities it is possible to image single particles such as molecules without the need to crystalize them. Currently, there are still many technical limitations, but rapid progress is being made. See for example HN ...

1

It is very important to understand what we mean by thermal equilibrium when we talk about black body radiation. Thermal equilibrium is meant to say that inside the body, all particles energy level distributions, speed distributions, are the same, and can be characterized by a single value. That is, its energy level distributions, particle speed ...

1

The energy we receive from the Sun, in the form of photons, comes from the photosphere. This is the very outer layer of the Sun. If it is in equilibrium, i.e. not getting any hotter or colder, then in terms of what we can see when we look from the outside, it does not matter where the energy comes from that heats the photosphere. The Sun is of course much ...

Only top voted, non community-wiki answers of a minimum length are eligible