# Tag Info

64

In short, the answer is: because gluons behave in a way that makes them useless for this purpose. To understand why, let's back up a little and look at how photons are useful, and then see how gluons behave differently. We (animals pretty broadly) evolved to see photons because they allow us to move around in and respond to our environment more ...

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 ...

16

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 ...

10

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 ...

8

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 ...

5

This is one of the first examples of energy levels for electrons within the atom! If we take the Bohr model, which imagines that electrons circle the nucleus on set orbits Each of these orbits has a corresponding energy. The electrons are more stable at lower energy levels, and thus, prefer to be there. When you provide energy to the electrons (in the ...

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 ...

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 ...

2

The color of the photon is related to its frequency $f$, which can be related to the energy of the photon by the expression $E = hf$, where $h$ is Planck's constant. Thus the different colors of the emitted photons describes their different energies. The next step is to determine why specific elements emit certain colors. This has to do with the different ...

2

The key is in your words "to ... appear". I believe that it's a perceptual issue with how your brain processes the two kinds of images: a smooth rendering or a pixelated rendering. There is another possibility. In order to be sensitive to single photons, the detector is also going to be sensitive to very low-level noise. An image taken with a bright ...

2

As for your last question, a similar experiment has been done though it doesn't involve a double-slit. It's called the Michelson experiment, and using mirrors it tests the interference pattern created by light when the light-waves are combined with time-delayed versions of themselves. By changing the distance of one of the mirrors, the time-delay can be ...

2

The spatial coherence is due to the fact that even for a single emitted photon it's the same wave that reach the 2 slits. I'm noot too sure what you mean by that. Spatial coherence has nothing to do with photons, it comes from the apparent size of the source as seen by the observer. Every source you might want to use in an interference experiment (a ...

2

Short answer: because the Sun emits photons, not gluons. Having a long range sense is vital for finding food and recognizing predators. Seeing light and forming an image of our surroundings is one of the three long range senses we have (the others are hearing and smelling). Gluons are extremely short range; they don't even exist as naked particles. How ...

2

A wave plate is a passive component, and can be modelled as a unitary operator on the quantum state. The state is a superposition of left and right circular polarized photons, and the operator gradually alters the relative phases. The total distance then determines the final polarization state.

2

Total length. The wavelength is already provided as 632.8nm.

2

There is a great misunderstanding here. The photon is an elementary particle (the gamma in the table is the photon) of spin one and mass zero. This has been validated innumerable times in nuclear physics, atomic physics and particle experiments. There is no question about it. Special relativity treats the four-vector of a particle. In this framework the ...

1

Okay, so I am taking this question to mean what is the lowest-energy photon that can be individually detected. This is certainly an interesting technological question. I can't give an authoritative answer, but the lowest energy detectors I am familiar with is at the CMB microwave background energy of ~ $3Kk_B$, which corresponds to a wavelength of about 5 ...

1

Science never answers "why" questions, so in a strict sense there is no such explanation, but one can try to triangulate where we stand, at the moment. In classical physics space, time and the existence of massive bodies are inexplicable pre-physical facts. Inertia then becomes an observed property of massive bodies that allows to differentiate them by ...

1

I think CuriousOne's comment provides most of the answers to your question, but for completeness I'll expand it into an answer. Light is described by quantum field theory and can only be fully understood in this context. We sometimes talk about photons and sometimes talk about light rays, but these are only approximations. As a general principle light ...

1

The point dipole is an approximation from classical physics - note that it also involves an infinite field strength in its center, where the field amplitude is not differentiable. I think such a source is not compatible with the common approach to quantum mechanics. If you take such a very small, subwavelength source, it is true that the evanescent near ...

1

By Wigner's general procedure of representing the little group, the $\theta(\Lambda,p)$ is the angle of rotation associated to the massless little group element $L(\Lambda,p)\in\mathrm{SE}(2) = \mathrm{SO}(2)\ltimes\mathbb{R}^2$ fulfilling $$L(\Lambda,p) = l^{-1}(\Lambda k)\Lambda l(k)$$ where $l(k)$ is the Lorentz transformation carrying the null vector ...

1

Any two points in spacetime are linked by a four-vector that physicists conventionally write as $(x^0, x^1, x^2, x^3)$, where $x^0$ is normally the timelike dimension and the other three components are spatial. If we use the usual Cartesian coordinates in flat spacetime we'd generally write the four-vector as $(t, x, y, z)$. In this case suppose the light ...

1

It's an issue of contrast; in the classical wave experiment there is plenty of data, and the contrast between the peaks and valleys is very clear; but when you are counting one-by-one the pixelation remains obvious. Pixelation can be reduced by (a) more gray levels in each pixel, and (b) more pixels per unit of area. You can simulate this by taking off ...

1

Semiconductor light emitters are made of such materials, which have quite large index of refraction. This makes it hard for light to exit the emitter — due to Fresnel equations and low index of refraction of air. In a laser the light mostly goes back and forth between two mirrors, and reflections only help the lasing. So the light either exits from a tiny ...

1

It does't really make much sense to talk about a tree-level truncation (it helps for calculations, but that's it) or to take the first Feynman diagram as a true representation of reality. By the way, in your $e^- e^- \to e^- e^-$ example, the whole notion of spatial separation is ill-defined since this is a t-channel process. If going from virtual to real ...

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: $$\nabla \cdot \mathbf{E}=0$$ $$\nabla \cdot \mathbf{B}=0$$ ...

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 ...

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Yes, and it is unavoidable. Let's consider an intrinsic semiconductor for simplicity. If the semiconductor is at the absolute zero of temperature, then all electrons will be in the valence band. At any non-zero temperature there is a chance that some electrons will have been promoted by thermal agitation into the conduction band. These electrons will ...

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