# Tag Info

23

Yes! In fact, this kind of phenomenon is very common. For example, the mass of a proton is much greater than the sum of the masses of the constituent quarks. Much of the extra mass comes from the gluons that bind the quarks together; like photons, gluons are massless, but they contribute to the inertia.

19

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 ...

13

Nothing is happening. At least, nothing except that a new generalized quantity suggestively called "angular momentum" was defined and subsequently measured. But nothing we know about the usual angular momentum of photons is changed by this in any way. Standard total angular momentum is $J = L + S$, where $L$ is the orbital and $S$ the spin angular momentum. ...

12

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 ...

11

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

Yes, mass and energy are equivalent. A more competent relativist might be able to give you the complete description, but to first order you can say that the mass of an object is simply the total energy in its volume divided by c^2. That mass is equivalent to the inertial mass by the weak equivalence principle, which is a cornerstone of GR. That is to say, ...

6

It probably does not mean anything. That paper concerns the quantization of electromagnetic waves in less than three spatial dimensions. In fact, there are a number of decades-old results showing that it is often possible to evade the spin-statistics relationship in lower-dimensional systems. While these kinds of results (including this new one) may be ...

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

You misunderstand special relativity. For objects that are moving at large speeds, the time runs more slowly for the object compared to the observer who measured the speed. To observe the motion of the object, you don't have to go to its coordinate frame and observe from there. You observe from the outside, that's how you measured the speed in the first ...

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 ...

3

Yes, both the internal potential energy and the internal kinetic energy of a bound system (in the rest frame of its center of mass) contribute to the bound system's inertial mass according to $E=mc^2$. For a paper discussing the evidence that this is true for internal kinetic energy in particular, see Kinetic Energy and the Equivalence Principle.

3

Just as a supplement to ACuriousMind's answer, it is worth noting that buried in the bottom of their paper they actually show what the "spin 1/2" eigenstates are in terms of the regular basis: $|j=1/2\rangle=\frac{1}{\sqrt{2}}(|1, -1 \rangle + |0,1\rangle$) $|j=-1/2\rangle=\frac{1}{\sqrt{2}}(|-1, 1 \rangle + |0,-1\rangle$) where $|l, \sigma\rangle$ is the ...

3

This sounds like the "retrocausation" in the Wheeler–Feynman absorber theory. Since the only invariant quantity in relativity is the relativistic interval, which is zero along light like curves, all "place-instants" of photon's existence are technically not separated from each other in the (pseudo) metric, and hence causal, sense. This means that photon ...

2

"Photons experience no time" is a conclusion drawn from taking the limit of time dilation as $v$ approaches $c$. This transformation never gets you to light speed, i.e. never represents a massless particle, only gets asymptotically closer. Taking that same limit, length contraction causes distances to reduce to nil. A particle moving at the speed of light ...

2

Let me explain light. Classically, light is electromagnetic radiation. There exists a field permeating all of spacetime called the electromagnetic field. Charges create curvature in this field. When charges accelerate, waves are created in this field. These waves are what we perceive as light. A little more specifically, let us examine Maxwell's equations ...

2

Theoretically, the answer is yes. However, looking at the practicality of the situation, the answer is no. The photons can not be contained inside the box unless they are either 1. Created inside the box itself, or 2. They are trapped beforehand, and then brought inside the box. In 1., they do not add to inertia because, they are created using energy ...

2

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 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

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

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 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

You should remember one thing : electromagnetic field is just a spatial representation of how electric charges interact with each other, and by "interact" I actually mean "exchange some energy". Electrostatic and magnetostatic energies Lets imagine that we want to build "from scratch" a given charge distribution $\rho(\textbf{x})$. That means that we ...

2

So the entire electromagnetic force can be described as having these objects which interact by exchanging virtual photons. Photons -- light -- in some sense are the electromagnetic force. It is therefore unsurprising that light can be absorbed by a particle -- an electron, say -- as a sudden "push" which launches the electron in some new direction with ...

2

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 ...

2

Photon experience infinite time dilation and hence, time is stationary for it. Does photon experience time  photon travels at c through the three spatial dimensions. All of its velocity is directed through the three spatial dimensions. Thus Brian and Einstein are stating that a photon must be stationary in the fourth dimension. For if the photon had any ...

1

Borrowing the concept of a "black box" from engineering, we have a photon "going in" into the box and a photon coming out of the box. We take a "picture" of the photon going in (its amplitude and wavelength), and we take a "picture" of the photon coming out (its amplitude and wavelength), and we compare the "pictures." If the "pictures" are equal, then we ...

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

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'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 ...

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