# Tag Info

6

You have to realize that when we are speaking of photons, we are speaking of elementary particles and their interactions are dominated by quantum mechanics, not classical mechanics, and in addition special relativity is necessary to calculate anything about them. In general, we know about elementary particles because we observe their traces in detectors for ...

6

The space between atoms depends very much on the medium you are talking about. In solids the typical distance between atoms is about the same as the size of the atoms themselves. In everyday gases at room temperature and pressure the distance between molecules is many times their size, and in deep space you can get densities as low as one proton per cubic ...

5

the wave function of a single photon has several components - much like the components of the Dirac field (or Dirac wave function) - and this wave function is pretty much isomorphic to the electromagnetic field, remembering the complexified values of $E$ and $B$ vectors at each point. The probability density that a photon is found at a particular point is ...

5

Update: I went over this answer and clarified some parts. Most importantly I expanded the Forces section to connect better with the question. I like your reasoning and you actually come to the right conclusions, so congratulations on that! But understanding the relation between forces and particles isn't that simple and in my opinion the best one can do ...

5

Imagine an elastic collision of two protons - something that they also see at the LHC but it's not the most interesting kind of interaction. The two protons will repel because of the electromagnetic interaction. Assume that the distance between them is never too small, relatively to the proton radius. But you may calculate the cross section of this process ...

4

I will address the premise that the electron is in an orbit around the hydrogen atom. This is a classical picture overlayed on the basic quantum mechanical one. The electron around the hydrogen atom is in a "spherical" probability cloud about the proton. The above is the n=1,l=0,m=0 probability distribution, which is the lowest energy state. A single ...

4

The same way you determine if the interval is space-like or time-like. In fact you do it by computing the square of the four-momentum and examining the sign. Which sign is space-like and which time-like is a matter of convention, and varies from source to source. I like to compute the squared-interval as  (\Delta s)^2 = (\Delta t)^2 - (\Delta \vec{x})^2 ...

4

These are just my thoughts as someone who studied the subject for a while: The concept of virtual photons that mediate interaction should not be seen as "what really happens". A virtual photon is not a real object (hence the name "virtual"), but an artifact of perturbation theory. If we knew an effective way (or even "a" way) to do the calculations without ...

3

Your seemingly unrealistic gedanken experiment is in fact a quite realistic. First, one can kick out the proton with help of a fast neutron. Next, to increase your "delay" time, you can consider a Rydberg atom with a high enough $n$, so the electron velocity is rather small with respect to the light (and the maximum proton) velocity. What happens to the ...

3

When thinking about fundamental entities, it's quite easy to ask a question that, upon reflection, is contradictory. The questions of this kind take the form: What is [some fundamental thing] made of? The contradiction here is that there can only be an answer if the fundamental thing isn't fundamental! The electromagnetic field is one such fundamental ...

3

I do not think that this question has an answer. A photon is a quantum mechanical object. a) there is no conservation of photons, virtual or not. b)there is no lower limit to the energy of the photons, so in principle they are infinite ( infrared problem) c)The energy of the virtual photons will depend on the motion of the charge and or the probe that ...

2

The original question seemed highly confused IMO. I wrote a comment to the original question (see above) which I slightly paraphrase as follows: OK, so we've sharpened your question. You want to know 1. What are virtual particles and 2. How, in principal does their exchange give rise to real forces, e.g. a) how does the exchange of the virtual gauge ...

2

Apparently you imagine a charge as a point with a Coulomb field around it. Outside the charge there is no charge but there is a field, you say, and it consists of virtual photons. So how many of them are at the distance $r$ from the charge? I let the others answer this question and here I will give you my vision of that. The charge is not point-like but ...

2

The physical process of Magnetic force in relativistic Quantum Field Theory is much easier to understand non-pertubatively versus the pertubative picture of virtual photons. Just look at the phase change rates of the charge's field induced by the Magnetic field $B$ using the vector potential $A^\mu$. What you see is that the de Broglie wave length becomes ...

2

You are right, real photons always travel at the speed of light and would carry energy away from a magnet. From a field theory point of view, all static fields, whether electric, magnetic, the weak nuclear force or the strong nuclear force can be thought of as being mediated by virtual particles. So for either electric or magnetic fields, that would be ...

2

Virtual particles appear when one wants to calculate cross sections and branching ratios for elementary particle interactions. This is done with the prescription of Feynman diagrams Feynman Diagram of Electron-Positron Annihilation In the above diagram the external "legs" are real particles with the quantum numbers given in the standard model table, ...

2

The field lines in your drawing are not the trajectories of photons. The field lines show the direction of the force on a test magnetic dipole. The force, and therefore its direction, is mediated by virtual photons (or can be described that way) but those photons will travel in straight lines just like ordinary photons.

2

Virtual particles, whether photons or electrons or... are, in the context of QFT, particles that are off-shell, i.e., their associated energy and momentum are not related by the relativistic energy-momentum relation. Please read this to get an idea of how virtual particle exchange can create attractive or repulsive forces. Photons are quanta of the modes ...

2

Suppose a scattering process with a 3- particle vertex : $A \to B + \gamma$. Here we suppose that particles $A$ and $B$ are massive, with the same mass $m$, and $\gamma$ is the "virtual" photon. Let $a,b, q$ be the momenta of the particles $A,B,\gamma$. You have : \$q^2 =(a-b)^2 \\= (a_0 - b_0)^2 - ( \vec a - \vec b)^2 \\=(a_0^2 - \vec a^2) + (b_0^2 - ...

1

A previous question has been signaled by Chris White, and I think that the answer of Arnold Neumaier is great. Now, let us add some hints relatively to your question. In principle, we could describe all physics without EM fields (or photons), as they are mainly a useful tool to describe "action at distance" (which does not mean instantaneous) between ...

1

The short answer is the Heisenberg uncertainty principle allows for the attraction. Suppose you have two opposite charges, and the one on the left emits a virtual photon with momentum directed leftwards. The left charge begins to move towards the charge on the right. Now, where's the virtual photon? It's momentum is some exact value directed left, so ...

1

Virtual photons do not pass from one plate to the other. The leave a plate and then return back to the plate and the electron or positive ion they left from - all in a very short time. The only time the leave a plate and do not return is when they have imparted some momentum to another charged particle (imagine a Feynman diagram for two electrons scattering ...

1

In your case it is a near field that stands for "virtual" photons. The near field does not propagate like EMW in the sense it is "attached to the charges and currents. In fact, it is the Coulomb and quasi-static magnetic interactions of charges and currents. Despite being time-dependent, the near field decays with distance differently (faster) and does not ...

1

There is a difference between the value of a field, and the excitations of that field. If you have a speherically symmetric charge distribution collapse to a black hole (while maintaining the spherical symmetry), there is no electric nor gravitational radiation--Gauss's Law and Birkhoff's Theorem work in tandem to keep the gravitational and electric forces ...

1

In QED, for transverse photons, the electric field doesn't commute with the photon number operator. Neither does the magnetic field. In fact, the electric and magnetic fields don't even commute with each other. To get a state with a fixed electric field — or at least with a small uncertainty in it in the quantum sense — we need a superposition of ...

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