New answers tagged

6

It is a matter of definition of "same". Classically one can define "same" condition of particles by labels stuck on them. Light classically is a wave, and same needs a new definition. We apply the everyday definition by identifying the light beam with the source. The light leaving the sun is the same light arriving on earth. The light reflected from the ...


3

Photons are boson, so it follows the Bose-Einstein statistics which is only true if the particles are truly indistinguishable. If you can distinguish between two photons, then it will follow the classical Boltzmann statistics which is not what happen in experiments. That means photons with same properties are the same. Even in your situation with photon ...


6

Your question is based on the assumption that a photon is a fundamental object i.e. that photons are something we can point to and say here is photon 1, here is photon 2, and so on. The trouble is that quantum field theory particles are somewhat elusive objects. This is particularly so for particles like photons that are their own antiparticles because such ...


0

White light is made of visible photons of many colors (frequencies). Our eyes mix the different frequency and interpret it as white. Photons come in all frequencies and through evolution our eyes have evolved to register visible photons from red to violet. Blue photons are higher frequency than red photons. White light is not a photon but a mix of photons


0

The microscopic mechanism of emitting photon in a solid is the transformation of kinetic energy of atoms into EM energy. If an atom is in an excited state due to collisions among other atoms, then it will emit photon when it jumps into the ground state, and the energy of the photon is $$ E=\varepsilon(\text{excited state})-\varepsilon(\text{ground ...


1

No, each color in the spectrum has a characteristic frequency. Every light source has a so called spectrum of frequencies. The relative intensity of these frequencies determines what color you see (or not). For example, the sun looks yellow because it's peak intensity is in the yellow wavelength. White light comes from a source consisting of a very broad ...


1

What are photons? Photons get emitted every time when a body has a temperature higher 0 Kelvin (the absolute zero temperature). All bodies, surrounding us (except black holes) at any time radiate. They emit radiation into the surrounding as well as the receive radiation from the surrounding. Max Planck was the physicist who found out that this radiation has ...


-2

The electromagnetic wave contain oscillating electric and magnetic fields in it. These electric and magnetic fields can store energy. Otherwise, the electromagnetic energy of a wave is stored in the electric and magnetic fields. These energies will be proportional to the amplitude of electric and magnetic fields. Now. if you increase the frequency of ...


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EM Waves are basically spinnings of photons. To create a EM Wave, you basically move some electrons in a directional way(think of an antenna). Electron is a charged particle as well as protons. Charged particles emit photons, and if you emit photons in an ordered way such as this: You are seeing a dipol antenna. When you apply negative voltage(intense ...


-2

The energy is carried in individual photons. A photon with twice the frequency has twice the energy. x-rays are made of photons with higher frequencies. The energy is transferred as kinetic energy as with the photoelectric effect. The energy of a photon is calculated as E=hf (Energy=Plank's constant x the frequency).


0

The key thing is that the surface have facets. That is, it has to have smooth flat parts that can reflect light like a mirror. If the surface is just amorphous then the scattering will tend to be too disorganized to see the polarization. I have seen polarized light coming off quite surprising surfaces. A manhole cover for example. It had been polished fairly ...


0

In the classical theory of reflection (and refraction) of electromagnetic waves, there are equations which describe the reflection of light in two specific orientations. They are known as the Fresnel equations. However, the polarizations of light lie in a 2D vector space, so as long as you decompose any incoming wave of light into the two linearly ...


1

To be clear, Maxwell's equations are known as "Lorentz-invariant" equations, which means that they take the same form in every Lorentz-transformed frame of reference. Special relativity actually came about from studying Maxwell's (classical) equations without charges or currents. Then we get: $$\nabla \cdot \mathbf{E}=0$$ $$\nabla \cdot \mathbf{B}=0$$ ...


0

With a classical wave model for light and a classical mechanics model for matter, the energy which is absorbed by one particle will be a function of the wave amplitude $E$. If we reduce the wave amplitude the absorbed energy $W$ will smoothly go to zero. In mathematical terms: $$ W(E\rightarrow0,\omega) \rightarrow 0 $$ If an electron requires a minimum ...


4

Classical electromagnetism is perfectly compatible with special relativity. In classical E&M, light is an electromagnetic wave and there is generally no useful formulation in terms of particles. The most widely used technique to combine quantum mechanics with special relativity is relativistic quantum field theory. The relativistic QFT that ...


2

Cort and Ilmari have given good answers about the practical issue: the inverse square law is for point sources, and so a non-point source (like an emergency light) will only appear to have the same properties at some minimum distance that depends on the geometry of the real source. However, it seems nobody has mentioned a different "minimum distance" that ...


2

The quote from the reference says it all: (I added caps) "The minimum test distance IN PHOTOMETRY of these sources is called the 'minimum inverse-square distance.'" The minimum distance is therefore a photometry issue, in other words, a measurement problem. The essence of the measurement problem is how far away you have to be before you can approximate the ...


3

The inverse square law says that the intensity of incident light falls off in proportion to the inverse of the square of the distance from the light source. The important word here is "the distance" — the inverse square law implicitly assumes that all parts of the light source are at the same distance from the measurement point, or at least ...


13

As many have said, the inverse square law applies to point-sources. These are idealized light sources which are sufficiently small compared to the rest of the geometry that their size is of no importance. If a light source is larger, it is typically modeled as a collection of idealized light sources, potentially using integration. The exact definition of ...


12

The inverse square law applies to point sources. For extended sources becomes accurate at distances that are large compared to the size of the source. At large distances the source looks like a point. What "large" means depend on the application. In the case of light fixtures, the Illuminating Engineering Society and other organizations have made ...


6

The inverse square law applies to point sources. A real emergency light is not a point source, and therefore the law appears to not apply at close distances, because any real point is at a varying distance from different parts of the emergency light.


1

It comes about by assuming that the wavelength ($\sim k^{-1}$) is much larger than the typical atomic length scales ($r$).


0

The very word, photon, belongs to the quantum mechanical regime. It is one of the elementary particles in the standard model of particle physics. Elementary particles are described with quantum mechanical wave functions, which are complex function. The complex conjugate square gives the probability of finding the particle at (x,y,z,t). In the case of the ...


0

It has to do with the total energy or power of the EM wave you're interested in, as well as the frequency of the wave. As a simple example, a 3mW laser at 500nm wavelength will produce roughly 7.55*10^15 photons per second. From how large this number is, it's not difficult to see how light will usually be made up of an extremely large number of photons. For ...


3

In quantum mechanical domain these type of question does not have meaning. Every single photon is associated with a wave and vice versa. But to talk whether an electromagnetic wave contains a single photon or not is an ambiguous statement. When people say an electromagnetic wave necessarily contains many photons it only means that a incident beam of ...


-1

The photoelectric effect is explained by a cool experiment which exactly displays why the wave model doesn't apply to the idea of photons. Here's a diagram of the experiment: The wave model insinuates that the energy of an electromagnetic varies by the intensity (square of the amplitude of the wave), i.e. $ E \propto I \propto A^2 $ Physically, this ...


1

Let us first describe a relevant experiment: You have a photomultiplier tube, hooked to a loudspeaker for convenience. If you shine on the detector with light you hear noise, which is louder if the light source is brighter. But if you only take a very feeble light, you'll notice a peculiar thing: The loudspeaker does not make noise anymore but produces ...


-1

Classical electromagnetism has no way of linking energy of a wave to its frequency. In classical E-Mag energy density is usually given by $$ u = \frac{1}{2}( \epsilon E\cdot E + \frac {B\cdot B}{\mu} )$$ As you can see this has no frequency dependence in it at all. And in photoelectric effect experiment suggests that frequency affects the energy carried by ...


-2

My friend once had the same question. We came up with this: Glass is transparent. It is transparent due to the fact that we can see light from the other side of the glass. In other words, there is little absorption by the glass particles of any wavelength of light (visible) . So all (most) of the light gets transmitted to our eyes. Hence transparent.


1

Contrary to what the other answer assumes, fog is not made of vater vapour, and light attenuation is not the reason why you can't see through fog. Fog is a suspension of microscopic droplets of liquid water in air. This is the same material we know as "cloud" when it doesn't reach all the way down to the ground. The fog is opaque not because light is ...


3

This is a common misconception about what boundary conditions do and how they do it (for example here). You discussed two types of boundary conditions, Neumann and Dirichlet. In Neumann boundary conditions, we impose that the derivative of the variable normal to the boundary is specified, generally to be zero. With Dirichlet, we impose the value that the ...


0

Yes, you are taking the reversal thing too seriously. If you used Kruskal-Szekeres coordinates then those coordinates don't flip from timelike to spacelike. The flip is just because you chose bad coordinates. If you had flat boring Minkowski spacetime you could pick a coordinate system where a coordinate flips from spacelike to timelike across some surface. ...


1

People are not understanding your question. I think you want someone to verify explicitly that the fields produced in your special case do (or don't) obey the generally valid wave equation. After all, the field produced does not look like a wave. A general solution of the wave equation for a disturbance traveling in the $x$ direction is ${\bf{E}}({\bf ...


0

The mug is made up of two materials, ceramic (an insulator) and metal (a conductor). These two types of materials certainly have different heat coupling coefficients, and metals in general both emit and absorb heat faster than insulators. So it definitely makes sense that metals, by virtue of more efficient radiative heat transfer with the environment, are ...


1

A hot material will radiate heat to a colder, that is to say it will radiate more heat outward than it absorbs from the colder object. The problem is only that the radiation RATE, as well as the absorption rate, is not determined by temperature alone, but by the coupling of the material to light of any given wavelength. Metals are electrically conductive, ...


0

There is a relationship between emissivity and absorptivity, wherein materials that don't absorb well are also poor emitters. Therefore, shiny metals typically don't emit thermal radiation at as high a rate as other objects, and that's what you see in the image. This can all be reasoned out by considering the need for conservation of energy in thermal ...


2

One can use an easy to understand picture. Suppose, you have an antenna composed from 6 rods along the axes of a cartesian coordinate system. At one moment electrons together get accelerated outwards in all 6 rods. Using a hand rule one could see that the magnetic fields around the rods cancel each other out exactly. So you are free to make an experiment ...


0

Well if we neglect the hidden momentum the conservation law of momentum in electromagnetism is simple: The momentum can be stored in static fields ($D\times B$); the mechanical momentum ($mv$) + electromagnetic momentum ($D\times B$) $= constant$. The similar formula is valid for angular momentum (where it is not hidden momentum) See Feynman's Lectures ...


24

A result known as Birkhoff's theorem forbids spherical electromagnetic radiation. The statement of the theorem is that any spherically symmetric vacuum solution to Maxwell's equations must be static. It is rather simple to prove. In a spherically symmetric solution $\mathbf E$ and $\mathbf B$ must be radial. Make an Ansatz, $$\mathbf E = E_0 \exp(i(\mathbf ...


1

Here's a simple explanation based on the dipolar nature of the medium: For most of the materials, we can assume that the source of the reflected and refracted waves are the induced tiny dipoles in the dielectric medium. In an isotropic medium, polarization vector is proportional to the (total) electric field vector with a constant (as opposed to a tensor ...


2

I think this is an interesting question. Unfortunately, many hasty sketches of the history of physics, as they are taught, tend to draw somewhat biased conclusions for the sole purpose of avoiding delving into these types of questions (some people consider it to be a waste of time apparently). As far as I can tell, the classical scattering theory at the ...


1

The definition of s-polarised light is that the electric field is polarised so that it is perpendicular to the plane of incidence. Where there is a specular reflection, the plane of incidence contains the k-vector of the incoming wave and the reflected wave. Since the electric field of an EM wave must be perpendicular to the k-vector. This then leaves the ...


1

There is a simple answer: Symmetry. Suppose the material is isotropic, and consider the initial condition of the p-polarized case, with a p-polarized light wave about to hit the surface. In this case, reflecting in the plane that contains the incident and scattered wave vectors leaves both the (vector) electric field and (pseudovector) magnetic field ...


0

In your layout, imagine the "antenna/you"-capacitor being parallel to the existing one, the antenna being the upper plate. Parallel capacitors add up their capacity. So how do you become the plate although you are not connected to the wires? The first step is to understand is that this setup (inductor + capacitor) will generate frequencies, as you could ...


0

The player's hand acts as a grounded plate remembering that the player is a reasonable electrical conductor. The capacitor is part of an inductor-capacitor circuit, as you have shown above, which control the frequency of an oscillator. So what is missing is a clear indication that the bottom part of the circuit is connected to the earth/ground.


1

Wave velocity, when not otherwise specified, usually means the phase velocity: the rate at which the phase travels in space. This site animates the differences between phase and group velocities. The group velocity is the speed at which energy is transported. So if the context implies that the wave is carrying energy, the wave velocity is also the group ...


17

Very reasonable question. I will try to answer it in an intuitive way. If you have a scattering medium, photons are reflected in random directions; but when you have a refractive medium, something else happens. The photon is not absorbed and re-emitted: instead, the photon interacts with the electrons in the medium, and since these electrons are somewhat ...


3

Saying "the light wave splits" is not an accurate description of what we understand about the theory of light. That's why you haven't seen it discussed. The shape of the electromagnetic field (mode) fills all of space, subject to boundaries (to include containers, obsticles, etc. That is, the mode has a shape determined by the boundaries.). Some of the ...


1

The Fresnel equations are derived by matching the electric and magnetic fields of the incident, reflected and transmitted waves at the interface. In this process only the instantaneous value of the fields is used not their rate of change with time. This means the frequency of the wave simply doesn't enter the calculation. However, as a comment notes, the ...


1

International Agency for Research on Cancer groups cell phone in group 2B, which is for possibly carcinogenic to humans. Not even probably. So, it is not that harmful . But, you can protect your by Electromagnetic shielding. You can figure out a way to connect conductive or magnetic materials either to your or the phone, but if you connect it to your ...



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