Bob_for_short

An example of atmosphere-pressure heat-to-motion transformation

Yes, I can. One particle acts on another. We may say that one particle is a source of a field that exists everywhere. But the true sense of this field is the force term appearing in the second particle equation as an external force. Similarly we may say about the second particle in respect to the first one. In both equations we have the same force depending on the distance between particles. Working with such forces does not cause any physical and mathematical problems. However working with proper fields, for example, calculating their energy, gives infinities and some conceptual problems. I want to say that when we assign an independent meaning to the field, it becomes overly complicated mathematically and physically. Another example is a self-action - when we insert the proper field into the first particle equations of motion.

A field is nothing else but a force standing in the equation of motion of a probe particle. It cannot be explained differently. As soon as we can put our particle anywhere in the space, we may assign to the space points certain fields (forces). When our particle goes from x1 to x2, the force changes from F(x1) to F(x2). The concept of field is quite similar to the concept of space - of all possible particle positions. But is is always a certain "number" in the particle equation of motion.

Yes, of course. A still body with respect to you may be moving with respect of a moving guy and may hit his head.

Nohow. It is not the transverse photons that act along the line. It is the electrostatic Coulomb force = the property of charges, not photons. The pointing vector for electrostatic field equals zero (B=0). It makes sense only for propagating waves, not for static fileds.

In order to calculate the Compton effect or e-p scattering in the first Born approximation we use m = m_e. We constructed QED with the observable m_e, m_f, and e. It is our interaction term that gives divergences, not the original parameters. So the renormalizations are modification of the interaction term jA, namely, removing the self-action effects (=corrections to masses and charges). In the end we have results of another, renormalized theory with another interaction term. You know, it is a too speculative reasoning about physics at short distances because we judge about it from our classical notions, namely from a point-like electron. We speak of "vacuum polarization" that "screens" too singular potential, etc. At the same time the electron is very tightly bound with the quantized electromagnetic field so it is not point-like but smeared quantum mechanically. So there is no singularity at r = 0. A point-like nucleus bound in an atom creates a positive charge cloud with no singularity at the atomic center. You may find the corresponding effective potential in my paper in Fig. 1 and Fig. 4, see http://www.springerlink.com/content/h3414375681x8635/?p=78336560cf3d4e3f98e6fb8eac587340π=0 (available also in arXiv). Yes, the physical predictions of QED will be different at short distances but the theory as a model does need any scale distance to be renormalizable.

Again, it is a consequence of the original equations of theory which is gauge invariant by construction. WT identity is an identity, something like a = a. It's a banality, not a fundamental feature, I would say. Yes but for photon mass they say "it is protected" by this or that as if it was not the sequence of the theory basics. The electron mass is also "protected" then. So what is the mass values before and after renormalizations? We do not have any other value but m_e. I do not think so. After renormalizations there is no trace of the cut-off or any other regularization parameter. QED is not an effective theory. Renormalizations are necessary with an interaction term containing a self-action. Renormalizations remove the self-action "effects". In particular, the self-action terms contain a "perturbation" of a kinetic nature so such terms modify the original masses. If one uses the right experimental values as the original masses, the perturbatibe corrections are not necessary so they are discarded (= renormalizations). With a potential interaction term no corrections to the masses arise and no renormalizations are necessary, in my opinion.

The photon mass is zero because of Maxwell equations for the field strengths E and B. Of course it is so in terms of four-vector potential too. Do "quantum effects" (you mean interaction) push the electron mass? It is numerically m_e in any order of perturbation theory, isn't it? It is a coefficient, not a photon feature.

Yes, please, think it over. I need your insight. By the way, does your answer mean that the electron mass is "not protected"?

I was told recently that there is no photon mass renormalization. It brings up an interesting question: what is then renormalized in photon? Which of photon features? (What is wrong in the bare photon that needs redefining?)

No, it cannot since currently m=11 and our universe is 3-dimensional. We need a better m-theory - with m=3.

Magnetic field is a field determining the magnetic force in the equation of motion of a charge ((q/c)[vxB]) or in equation of motion of a neutral magnet (dipole interaction force). It can be measured and given experimentally or calculated from a given current/magnet data. It is an inter-charge interaction force, if you like. No propagating photons are necessary to explain it.

No. It is virtual photons, not real. Real propagate, virtual do not. That is why many avoid employing the term "virtual photons". The good and unambiguous term is a "variable magnetic near-field", not a "virtual photon".

No. Virtual photons are "attached" to charges. They are not propagating to infinity unlike real photons (wave packets). They are rather different because they are electric and magnetic quasi-static charge interaction terms in the Hamiltonian. They are known also as a "near" field.

Your question is not clear. If you mean static magnetic fields, they are classical and do not need any particles to be explained. A classical static magnetic field B is a solution of a static equation with a known current j: B = B(j). It determines the field distribution in space. If you speak of variable magnetic fields that is calculated from QED equations, they are automatically taken into account when one considers charge interaction. It may be "explained" as due to virtual particle exchange. This "explanation" is similar to the Coulomb time-dependent interaction "interpretation" in terms of virtual photons. Factually, however, the magnetic and Coulomb time-dependent interaction terms can be separated from radiation and be still considered as properties of charges rather than virtual photons. This is achieved in the so called Coulomb gauge. It is the charge wave-functions that overlap, not virtual particles.

I mean the same thing as an atomic size - the electron elastic form-factor determining the charge distribution in space for ealstic scattering. When we write 1/r, we mean a point-like electron with Coulomb law. In QED the charge is in permanent interaction with the quantized electromagnetic field that smears the charge over space. I wonder - what size of charge smearing does QED predict for a real electron? In order to calculate the corresponding cross section (expressed via elastic form-factor), one has to add the quantized EMF in the electron equation.

Yes, I also speak of a physical theory with physical predictions. Currently, however, the theory operates with "bare" stuff, "counterterms", etc., and the final results are not really clear to most of researchers. For example, the size of a real (or dressed) electron, what is it according to QED? I mean theoretical estimation in terms of e, m, h-bar (not an experimental one).

Thus it is clearly a "difficulty" of the present theory. It is not the only difficulty, unfortunately. My second question is whether it is desirable to have a theory without any conceptual and mathematical difficulties, just in the spirit of QM?

Can you give me references, if any, to the justification of adiabatic hypothesis in QFT? Why should we switch off the interaction in asymptotic states? Because we cannot solve coupled equations?

Yes, I can. I am not an experimentalist but I think any radio-wave attenuation in a conducting medium is a simple example of that. A photon has a finite wave-train to be more or less of certain frequency (10^4 oscillations, for example) so its absorption time is finite.

Apart form this language (creation-annihilation operators) there is another, more appropriate one: the state amplitude depending smoothly on time. While absorbing the photon wave the old state amplitude fade and the excited atomic state amplitude grows up. In the end there is not the initial states but the final only. Similarly for scattering a photon. The state populations depending on time is more physical since there is no instant creation or destroying states in reality.

I am not sure if the electron in higher orbits moves faster.

A photon is a long wave-train propagating is space. An atom is a compound system also existing in space as a de Broglie wave. When two waves meet, the photon wave starts to push and pull the atomic electron. If the resonance conditions are approximately satisfied, the atom may get excited in the end and the photon energy is thus spent on increasing the atomic internal motion energy. The quantum mechanics is a wave mechanics so this works as I have just described.

No, for example, 3/2 + 1 = 5/2.

I have no idea about it.