# Tag Info

17

Brief answer: Read only the bold part (and ignore grammar then). The answer you already mentioned lies in Quantum Field Theory (QFT). But to fully understand it, you must give up a particle as a point-like thing that is well-localized. There is one Quantum Field per sort of particle, e.g. the electron field for all electrons, and the photon field for all ...

14

Let me give a second, more technical answer. Observable particles. In QFT, observable (hence real) particles of mass $m$ are conventionally defined as being associated with poles of the S-matrix at energy $E=mc^2$ in the rest frame of the system (Peskin/Schroeder, An introduction to QFT, p.236). If the pole is at a real energy, the mass is real and the ...

14

The short answer for why gravity is unique is that it is the theory of a massless, spin-2 field. To contrast with the other forces, the strong, weak and electromagnetic forces are all theories of spin-1 particles. Although it's not immediately obvious, this property alone basically fixes all of the essential features of gravity. To begin with, the fact ...

8

For example, how do two charged particles know that they are to move apart from each other? Do they communicate with each other somehow through some means? Yes, specifically the electromagnetic field. To give a simplistic view, a charged particle produces an electric field to indicate its presence and a magnetic field to indicate its motion. Any ...

6

When you ask "Why is gravity such a unique force?" then you should know that in the framework of General Relativity gravity is not a force at all. In General Relativity energy (for example the mass of an object) cause curvature. The movement of other objects is influenced by this curvature - they travel along the path of shortest distance between two points ...

6

Photons are force carrying bosons and come in both virtual and real varieties. There is nothing wrong with that. Virtual means off-shell, and real means on-shell. Even on-shell weak bosons decay very quickly, however, because there are plenty of modes with the right quantum numbers and much lower total mass (and thus lots of phase space). I want to ...

5

The short answer in quantum field theories is "by the exchange of virtual particles". Look up "virtual particles" on Wikipedia to get a sense of it. The ancillary questions you asked about the energy budget, and the derivation of Coulomb's law are very much the right ones. They should all be answered there. For example, energy is conserved except for ...

4

You say: Gravity depends on mass but this is not so. The source of the gravitational field is an object called the stress-energy tensor. One element of this object is the energy density, and mass contributes to this through Einstein's well known equation $E = mc^2$, but mass is not required to generate a gravitational field. Even massless particles ...

4

All observed particles are real particles in the sense that, unlike virtual particles, their properties are verifiable by experiment. In particular, W and Z bosons are real but unstable particles at energies above the energy equivalent of their rest mass. They also arise as unobservable virtual particles in scattering processing exchanging a W or Z boson, ...

2

The notion of a point-like but charged particle is somewhat contradictory: on one hand we imagine it as localized in space (like a neutral particle); on the other hand, such a "particle" is "long-handed", i.e., it is felt far away from it "position". One may think that the charge is not point-like but is a "part" of a complicated, extended system. When you ...

2

Hello and welcome to the cite. Yes -- the carrier for any electromagnetic interaction is a photon. Including ordinary magnets. You might ask: "Why then can't I see those photons?" Well, even from classical point view, the photons can have different wavelenghts. And with the magnet you deal with really large wavelenghts, while one cannot "see" radiowaves. ...

2

One could write a novel about those questions... I'll try to nail down the most important facts. Regarding what you figured out so far: Basically correct. I would say: Every system of atoms has a quantum mechanical ground state. You can approximately assign an energy to each of the electrons (depending on the approximation you are using, e.g. Hartree-Fock ...

2

They can be real, no problem with that. However, all Z and W observed are virtual. And yes, they are off-shell, whatever that means to you. We actually measure their width, for instance, http://arxiv.org/pdf/0909.4814 (it's more or less 2 GeV around a mass of more or less 80-90 GeV). What is observed is a pole in the S matrix for some final states in ...

1

I think you can get a estimate like this. For a semiconductor with no split in the quasi-Fermi levels, the electrons and holes take their intrinsic values (carrier density) $n_0$ and $p_0$ ($cm^{-3}$). The charge carriers are in equilibrium with the thermal photons being absorbed and emitted inside the material. So if we calculate the emission rate of ...

1

the injection level is defined as $\delta n/p_0$ where $\delta n$ is the minority carrier (e.g. electrons') density excess at non-equilibrium while $p_0$ is the equilibrium density of the majority carriers (e.g. holes). From the formula which is a ratio of two very similar densities, it's clear that the injection level is dimensionless; "high" and "low" ...

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