Introduction to relativistic quantum mechanics and maybe QFT

In summary: I disagree. For example, in relativistic QM you can see how to solve the Dirac equation for the hydrogen atom, which gives you atom energies which are in better agreement with experiments than those from the non-relativistic Schrodinger equation. It's good to know that something useful and important can be obtained from relativistic QM without QFT.Relativistic QM without QFT is also interesting from the point of view of string theory (for those who are not strictly...physics majors).
  • #36
Demystifier said:
What about its generalizations, such as gauge/gravity or bulk/boundary correspondence?

As far as I understand the generalizations also have trouble describing our universe, because they are restricted to asymptotically AdS space.

But there are attempts like http://arxiv.org/abs/1108.5732
"FRW solutions and holography from uplifted AdS/CFT". I'm not sure of this proposal's status, nor if it can be extended to positive cosmological constant.
 
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  • #37
ddd123 said:
I don't see a problem in teaching relativistic QM right before QFT. You can do the Dirac equation, bilinear covariants, charge conjugation, chirality and helicity with the neutrino parity violation example, the hydrogen atom, Klein's paradox, all stuff that is used later or is instructive anyway and they're basically all exercises. The student won't think that the theory is consistent because of Klein's paradox and won't really waste time in the process.
Yes, but as well you can teach the Dirac equation in a modern way from the beginning, namely as part of QFT. Also the fermionic nature of the Dirac field is very important. This you don't get with the single-particle picture, and Klein's paradox shows precisely that this single-particle picture is not consistent but you need to invoke Dirac's sea hypothesis, which immediately introduces a many-body picture (even with infinitely man particles occupying the negative-frequency states). It's much more natural to describe this state of affairs by QFT, and then no inconsistencies occur.

Also students appeciate this approach. I got a good evaluation for a QM 2 lecture, where I did not take the traditional path to introduce relativistic wave mechanics, but started the relativistic part immediately with QFT. The lecture ended with tree-level evaluations of QED Feyman diagrams (electron-electron, electron-positron scattering, pair annihilation, Compton scattering). The professor who did the didicated QFT lecture in the next semester was also very happy about what these students already knew :-).
 
  • #38
vanhees71 said:
Yes, but as well you can teach the Dirac equation in a modern way from the beginning, namely as part of QFT.
The standard way of teaching Dirac equation, including that in QFT, cannot be directly generalized to curved spacetime. So another direction of modernization is to teach Dirac equation in a way which can more naturally be generalized to curved spacetime, even if the curved spacetime itself is not explicitly mentioned. I have proposed such a modernization in
http://arxiv.org/abs/1309.7070 [Eur. J. Phys. 35, 035003 (2014)]
 
  • #39
I know that paper. If I remember right, it's just another way to introduce the representation ##(1/2,0) \oplus (0,1/2)##, which is described in a local way by the Dirac field, but that's not the point here. I still think that one should teach relativistic QT as many-body theory with a non-fixed particle number, i.e., in terms of QFT, because that's the modern understanding of it.

Another question is the issue about QFT in a GRT background space-time. That's much more complicated, even for free fields/particles. I'd not consider this a standard topic for the introductory QFT lecture.
 
  • #40
Demystifier said:
What about its generalizations, such as gauge/gravity or bulk/boundary correspondence?

Just to add to my thoughts in post #35. If we believe that without hidden variables and assuming one world, the only valid interpretation is a Copenhagen-style interpretation, then one has an informal argument that non-perturbative quantum gravity does not exist except in AdS space, because in quantum gravity there is no classical spacetime, but Copenhagen requires a classical observer, who persumably lives in classical spacetime. AdS space is the exception in that the boundary provides a classical place for the observer to stand. In this case the boundary QM is primary,and the bulk QM is emergent or approximate.

Of course, this is just an informal argument and people have tried dS/CFT. The informal argument is also why I find LQG difficult to understand. Rovellian LQG seems to make much more sense if one can accept proposals like his relational interpretation or MWI.
 
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  • #41
atyy said:
Copenhagen requires a classical observer, who persumably lives in classical spacetime. AdS space is the exception in that the boundary provides a classical place for the observer to stand.
Already QED has enough nontrivial classical boundary behavior (superselection rules), and the superselection structure of the standard model is even richer.
 
  • #42
vanhees71 said:
I know that paper. If I remember right, it's just another way to introduce the representation ##(1/2,0) \oplus (0,1/2)##, which is described in a local way by the Dirac field.
Do you think so? :smile: I remember it as well and think the following about it:
1) The author confuses Pauli’s theorem (about the representation of Dirac algebra) with the spinor representation of the Lorentz group.
2) The paper is based on mathematically wrong statement: Given any non-singular [itex]4 \times 4[/itex] matrix [itex]S[/itex] and Dirac spinor [itex]\psi[/itex], one can easily show that [itex]S \psi[/itex] and [itex]S^{-1} \psi[/itex] are also Dirac spinors. The author claims to “conclude” that [itex]S^{-1}\psi[/itex] is Lorentz scalar.
3) The author should have a look at the following theorem:
The generator matrices of an algebra [itex][T^{a},T^{b}] = i C^{abc}T^{c}[/itex] are group invariant tensors, i.e. for any representation [itex]R[/itex] and its conjugate [itex]\bar{R}[/itex], we have [itex][R] \otimes [\bar{R}] \otimes [A] = [1][/itex], where [itex][A][/itex] is the adjoint representation. The proof is easy and short. The group action leaves the generators invariant: [itex](T^{a})^{i}{}_{j} \to (T^{a})^{i}{}_{j}[/itex].
Similarly, one can show (without reference to Dirac equation) that [tex](\sigma^{\mu})_{A \dot{B}} \to S_{A}{}^{C} \ \bar{S}_{\dot{B}}{}^{\dot{D}} \ \Lambda^{\mu}{}_{\nu} \ (\sigma^{\nu})_{C \dot{D}} = (\sigma^{\mu})_{A \dot{B}} ,[/tex] and [itex](\bar{\sigma}^{\mu})_{A \dot{B}} \to (\bar{\sigma}^{\mu})_{A \dot{B}}[/itex]. This implies that the four matrices, [itex]\gamma^{\mu}[/itex], are Lorentz invariant matrices not Lorentz 4-vector (as the author claims).

Sam
 
  • #43
samalkhaiat said:
This implies that the four matrices, γμ\gamma^{\mu}, are Lorentz invariant matrices not Lorentz 4-vector (as the author claims).
If you were right, then all the textbooks treatments of spinors in curved spacetime would be wrong.
 
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  • #44
@A. Neumaier , you have a strong opinion on most foundational aspects of quantum theory, and I very much respect your opinion, even though sometimes I disagree with you. So I would like to ask you what do you think about the Hegefeldt paradox?

 
  • #45
Indeed, in the standard treatment, you have
$$S^{-1}(\Lambda)\gamma^{\mu} S(\Lambda)={\Lambda^{\mu}}_{\nu} \gamma^{\nu},$$
where ##S(\Lambda)## is the bi-spinor representation of the Lorentz group and ##\Lambda## a Lorentz-transformation matrix. Written the 6 parameters of the Lorentz group in the usual way as ##\omega_{\mu \nu}=-\omega_{\nu \mu}## it turns out that
$$S(\Lambda)=\exp \left (-\frac{1}{4} \omega_{\mu \nu} \sigma^{\mu \nu} \right) \quad \text{with} \quad$$
$$\sigma^{\mu \nu}=\frac{\mathrm{i}}{2} [\gamma^{\mu},\gamma^{\nu}].$$
All this should be consistent with what samalkhaiat wroten in #41, maybe in a somewhat different convention, but I haven't checked this carefully.
 
  • #46
Demystifier said:
what do you think about the Hegefeldt paradox?
What is paradox about Hegerfeldt's paradox? Already classically, a multi-particle relativistic picture is inconsistent. So one wouldn't expect the quantum situation to be better. The particle concept itself is highly problematic, creating many difficulties that are absent if one interprets everything in terms of fields.
 
  • #47
A. Neumaier said:
Already classically, a multi-particle relativistic picture is inconsistent.
Why? What is inconsistent with having many classical relativistic particles?
 
  • #48
vanhees71 said:
Indeed, in the standard treatment, you have
$$S^{-1}(\Lambda)\gamma^{\mu} S(\Lambda)={\Lambda^{\mu}}_{\nu} \gamma^{\nu},$$
where ##S(\Lambda)## is the bi-spinor representation of the Lorentz group and ##\Lambda## a Lorentz-transformation matrix. Written the 6 parameters of the Lorentz group in the usual way as ##\omega_{\mu \nu}=-\omega_{\nu \mu}## it turns out that
$$S(\Lambda)=\exp \left (-\frac{1}{4} \omega_{\mu \nu} \sigma^{\mu \nu} \right) \quad \text{with} \quad$$
$$\sigma^{\mu \nu}=\frac{\mathrm{i}}{2} [\gamma^{\mu},\gamma^{\nu}].$$
All this should be consistent with what samalkhaiat wroten in #41, maybe in a somewhat different convention, but I haven't checked this carefully.
Is this supposed to be an argument that gamma matrix does not change with Lorentz transformation? This would be like using the Schrodinger picture to argue that position operator does not change with time. And yet it changes with time, but in another picture - the Heisenberg one. Likewise, as shown in the paper, there are two different pictures regarding the behavior under Lorentz transformations. In one picture the gamma matrix does not change with a Lorentz transformation, but in another it does.
 
  • #49
It was just the argument, why the ##\gamma## matrices behave like vectors under proper orthochronous Lorentz transformation in the above given sense. This is how it should be, because you like to build the usual bilinear forms behaving like scalar, pseusoscalar, vector, axialvector, and tensor fields,
$$\bar{\psi} \psi, \quad \bar{\psi} \gamma^5 \psi, \quad \bar{\psi} \gamma^{\mu} \psi, \quad \bar{\psi} \gamma^5 \gamma^\mu \psi, \quad \bar{\psi} \sigma^{\mu \nu} \psi.$$
 
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  • #50
Demystifier said:
What is inconsistent with having many classical relativistic particles?
For classical multiparticle Hamiltonian mechanics there is a no-go theorem that excludes the ''natural'' situation; see Jordan-Currie-Sudarshan, Rev. Mod. Phys. 35 (1963), 350-375. Nobody has so far come up with a nice substitute.
 
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  • #51
A. Neumaier said:
For classical multiparticle Hamiltonian mechanics there is a no-go theorem that excludes the ''natural'' situation; see Jordan-Currie-Sudarshan, Rev. Mod. Phys. 35 (1963), 350-375. Nobody has so far come up with a nice substitute.
At the moment I do not have access to this journal, so I will respond another day. But thanks!
 
  • #52
A. Neumaier said:
For classical multiparticle Hamiltonian mechanics there is a no-go theorem that excludes the ''natural'' situation; see Jordan-Currie-Sudarshan, Rev. Mod. Phys. 35 (1963), 350-375. Nobody has so far come up with a nice substitute.

But are there cases where there is a Lagrangian formulation without a Hamiltonian formulation? I have often seen the Feynman theory mentioned as using this loophole.

Eg. http://www.nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html where he briefly says, "We also found that we could reformulate this thing in another way, and that is by a principle of least action. Since my original plan was to describe everything directly in terms of particle motions, it was my desire to represent this new theory without saying anything about fields. It turned out that we found a form for an action directly involving the motions of the charges only, which upon variation would give the equations of motion of these charges."
 
  • #53
atyy said:
But are there cases where there is a Lagrangian formulation without a Hamiltonian formulation? I have often seen the Feynman theory mentioned as using this loophole.

Eg. http://www.nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html where he briefly says, "We also found that we could reformulate this thing in another way, and that is by a principle of least action. Since my original plan was to describe everything directly in terms of particle motions, it was my desire to represent this new theory without saying anything about fields. It turned out that we found a form for an action directly involving the motions of the charges only, which upon variation would give the equations of motion of these charges."

This is the general form of the action of such theories:
##\displaystyle S=-\sum_{a=1}^N m_a c \int_{\lambda_a^-}^{\lambda_a^+} \sqrt{-g_{\alpha \beta} \dot{x}^\alpha_a \dot{x}^\beta_a} d\lambda_a+\sum_{a=1}^N \sum_{b=a+1}^N q_a q_b \int_{\lambda_a^-}^{\lambda_a^+} \int_{\lambda_b^-}^{\lambda_b^+} K(x_a^\alpha,\dot{x}_a^\alpha,x_b^\alpha,\dot{x}_b^\beta) d\lambda_a d\lambda_b ##.

I don't think you can isolate a Lagrangian. It still counts as a Lagrangian formulation though!
 
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  • #54
Shyan said:
This is the general form of the action of such theories:
##\displaystyle S=-\sum_{a=1}^N m_a c \int_{\lambda_a^-}^{\lambda_a^+} \sqrt{-g_{\alpha \beta} \dot{x}^\alpha_a \dot{x}^\beta_a} d\lambda_a+\sum_{a=1}^N \sum_{b=a+1}^N q_a q_b \int_{\lambda_a^-}^{\lambda_a^+} \int_{\lambda_b^-}^{\lambda_b^+} K(x_a^\alpha,\dot{x}_a^\alpha,x_b^\alpha,\dot{x}_b^\beta) d\lambda_a d\lambda_b ##.

I don't think you can isolate a Lagrangian. It still counts as a Lagrangian formulation though!

Thanks! Do you have a reference on this? It's something I've seen in bits and pieces only - I guess I could read Feynman's paper - but is there something like an easy textbook presentation?
 
  • #55
atyy said:
Thanks! Do you have a reference on this? It's something I've seen in bits and pieces only - I guess I could read Feynman's paper - but is there something like an easy textbook presentation?
Special Relativity in General Frames by Eric Gourgoulhon, IMHO the best SR textbook out there(not introductory), containing everything you need to know about SR, maybe even more!
 
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  • #57
atyy said:
But are there cases where there is a Lagrangian formulation without a Hamiltonian formulation? I have often seen the Feynman theory mentioned as using this loophole.
Shyan gave the formulation. But (as Eric Gourgoulhon mentions in his book on p. 375) this theory has the disadvantage that it leads to integro-differential equations that have no well-defined Cauchy problem. This leads to interpretational difficulties.
 
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  • #58
A. Neumaier said:
Shyan gave the formulation. But (as Eric Gourgoulhon mentions in his book on p. 375) this theory has the disadvantage that it leads to integro-differential equations that have no well-defined Cauchy problem. This leads to interpretational difficulties.

Can it describe anything "physical"?
 
  • #59
atyy said:
Can it describe anything "physical"?
Yes, formally it is still one of the best models and with the right (''absorbing'') boundary conditions you get more or less classical electrodynamics. But (like any multiparticle picture I have seen) its interpretation defies any rational sense of physics. In the present case, the dynamics of any particle depends on the past and future paths of all other particles!
 
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  • #60
Demystifier said:
If you were right, then all the textbooks treatments of spinors in curved spacetime would be wrong.
I can mathematically prove all the statements I made including the one you quoted. The textbooks treatments are correct and consistent with every thing I said. I just believe that you need to work on your understanding of the representation theory of group in general and Lorentz group in particular.

Regards
Sam
 
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  • #61
vanhees71 said:
Indeed, in the standard treatment, you have
$$S^{-1}(\Lambda)\gamma^{\mu} S(\Lambda)={\Lambda^{\mu}}_{\nu} \gamma^{\nu},$$
where ##S(\Lambda)## is the bi-spinor representation of the Lorentz group and ##\Lambda## a Lorentz-transformation matrix. Written the 6 parameters of the Lorentz group in the usual way as ##\omega_{\mu \nu}=-\omega_{\nu \mu}## it turns out that
$$S(\Lambda)=\exp \left (-\frac{1}{4} \omega_{\mu \nu} \sigma^{\mu \nu} \right) \quad \text{with} \quad$$
$$\sigma^{\mu \nu}=\frac{\mathrm{i}}{2} [\gamma^{\mu},\gamma^{\nu}].$$
All this should be consistent with what samalkhaiat wroten in #41, maybe in a somewhat different convention, but I haven't checked this carefully.

Yes, I used 2-spinor notation. The group-theoretic treatment is generally clearer in the 2-spinor notation. This is because [itex]\chi_{A} \in (1/2 , 0)[/itex] and [itex](\varphi_{A})^{*} \equiv \bar{\varphi}_{\dot{A}} \in (0 , 1/2)[/itex] are the irreducible representations of the Lorentz group, while Dirac bispinors are reducible [tex]\Psi = \chi_{A} \oplus \ \epsilon^{\dot{A}\dot{B}}\bar{\varphi}_{\dot{B}} .[/tex] The transformation laws for the fundamental spinor and it’s conjugate are given by [tex]\chi_{A} \to S_{A}{}^{B} \ \chi_{B} = \left( e^{- \frac{i}{2}\omega_{\mu\nu}s^{\mu\nu}} \right)_{A}{}^{B} \ \chi_{B} ,[/tex]
[tex]\bar{\varphi}^{\dot{A}} \to \bar{S}^{\dot{A}}{}_{\dot{B}} \ \bar{\varphi}^{\dot{B}} = \left( e^{- \frac{i}{2}\omega_{\mu\nu}\bar{s}^{\mu\nu}} \right)^{\dot{A}}{}_{\dot{B}}\ \bar{\varphi}^{\dot{B}} ,[/tex] where [tex](s^{\mu\nu})_{A}{}^{B} = \frac{i}{4} ( \sigma^{\mu}\bar{\sigma}^{\nu} - \sigma^{\nu} \bar{\sigma}^{\mu})_{A}{}^{B} ,[/tex] and [tex](\bar{s}^{\mu\nu})^{\dot{A}}{}_{\dot{B}} = \frac{i}{4}(\bar{\sigma}^{\mu}\sigma^{\nu} - \bar{\sigma}^{\nu}\sigma^{\mu})^{\dot{A}}{}_{\dot{B}} ,[/tex] are the corresponding Lorentz generators. From these transformations, and with no reference to Weyl or Dirac equations, we obtain the transformation law for the Dirac bispinor [tex]\Psi \to ( S \otimes \bar{S} ) \Psi = (e^{-\frac{i}{2}\omega_{\mu\nu}(s^{\mu\nu} \oplus \bar{s}^{\mu\nu})}) \Psi ,[/tex] with the generators in block diagonal form
[tex]\sigma^{\mu\nu} = \begin{pmatrix} s^{\mu\nu} & 0 \\ 0 & \bar{s}^{\mu\nu} \end{pmatrix} = \frac{i}{4}[\gamma^{\mu},\gamma^{\nu}] ,[/tex] where [tex]\gamma^{\mu} \equiv \begin{pmatrix} 0 & \sigma^{\mu} \\ \bar{\sigma}^{\mu} & 0 \end{pmatrix} .[/tex]

Sam
 
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  • #62
samalkhaiat said:
I can mathematically prove all the statements I made including the one you quoted.
I am sure you can. But to prove them, you must use some mathematical axioms, that's how mathematical proofs work. But axioms, as you certainly know, cannot be proved. So one is free to choose different axioms (as long as they are not logically inconsistent), and then prove different statements. And that's what I do; I take different axioms (which I call a different "picture") and obtain different theorems. With different axioms I obtain that new gamma matrices do depend on the Lorentz frame.

From the physical point of view I show that the new axioms lead to the same physics because the new axioms do not affect the physical currents such as the Dirac current.
 
  • #63
Demystifier said:
But axioms, as you certainly know, cannot be proved.
Yes we can. A an axiom of the Theory T, then A |-- A ; prove A (A is a theorem). I agree this tautology isn't useful to physics.

Patrick
 
  • #64
microsansfil said:
Yes we can. A an axiom of the Theory T, then A |-- A ; prove A (A is a theorem). I agree this tautology isn't useful to physics.
It's not useful even in mathematics (except in pure logic, which some mathematicians (such as Poincare) do not even consider to be a part of "real" mathematics).
 
  • #65
A. Neumaier said:
Already classically, a multi-particle relativistic picture is inconsistent.
I took a look at the paper by Currie, Jordan and Sudarshan that you suggested, and also at the book
https://www.amazon.com/dp/0521143624/?tag=pfamazon01-20
(Sec. 16.4 The no-interaction theorem in classical mechanics).

What appears to be inconsistent is not to have relativistic interacting particles, but to have only relativistic interacting particles. It is still possible to have a classical relativistic interaction that involves particles and fields. This, indeed, is what we have in classical electrodynamics, but also, in a certain sense, in Bohmian mechanics.
 
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  • #66
samalkhaiat said:
Yes, I used 2-spinor notation. The group-theoretic treatment is generally clearer in the 2-spinor notation. This is because [itex]\chi_{A} \in (1/2 , 0)[/itex] and [itex](\varphi_{A})^{*} \equiv \bar{\varphi}_{\dot{A}} \in (0 , 1/2)[/itex] are the irreducible representations of the Lorentz group, while Dirac bispinors are reducible [tex]\Psi = \chi_{A} \oplus \ \epsilon^{\dot{A}\dot{B}}\bar{\varphi}_{\dot{B}} .[/tex]
Sam
Sure, this makes the Lie algebra/group structure clearer. If I remember right, the Dirac representation is a reducible representation for ##\mathrm{SO}(1,3)^{\uparrow}##, but irreducible, if you take space reflections (parity) in addition, i.e., as a representation for ##\mathrm{O}(1,3)^{\uparrow}##. Thus you have Dirac spinors in QED and QCD, because the electromagnetic and the strong interaction respect parity symmetry. The approximate chiral symmetry in QCD in the light-quark sector is an "accidental gift" for hadron-model builders :-), and parity is not broken.

Of course for the weak interaction, parity is broken, and the left- and right-handed quark transform differently under the gauge group (as a WISO doublet and singulet, respectively).
 
  • #67
samalkhaiat said:
The textbooks treatments are correct and consistent with every thing I said.
Please answer each of the following questions by either "yes", "no", or "I don't know"!
1. Do textbooks (which consider spinors in curved spacetime) say that there is gamma matrix which transforms as a vector under general coordinate transformations? (yes/know/I don't know)
2. Is Lorentz transformation a special case of a general coordinate transformation? (yes/know/I don't know)
3. If both answers above are "yes", does it logically imply that (according to those textbooks) there is gamma matrix which transforms as a vector under Lorentz transformations? (yes/know/I don't know)
 
  • #68
vanhees71 said:
Another question is the issue about QFT in a GRT background space-time. That's much more complicated, even for free fields/particles. I'd not consider this a standard topic for the introductory QFT lecture.
By the same kind of argument, one might say that QFT is much more complicated than relativistic QM, even for free fields/particles.
Just as you point out that there are still good reasons to teach QFT before relativistic QM, similarly one may point out that there are also good reasons to teach QFT in curved spacetime before QFT in flat spacetime.

In particular, to learn QFT in curved spacetime you must unlearn some overemphasized concepts that looked so fundamental in flat spacetime. For instance, Poincare invariance is no longer fundamental for definition of particles, the number of particles is no longer observer independent, the gamma matrix is no longer invariant under coordinate transformations (including the Lorentz ones), and so on...

My point is - there is no one correct way to teach physics. There are many complementary ways to do that, each with its advantages and disadvantages.
 
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  • #69
Demystifier said:
It is still possible to have a classical relativistic interaction that involves particles and fields.
Yes, but once you admit classical fields as fundamental, it is more parsimonious to reard only these as fundamental. You get classical particles for free as the idealization where a field is supported on worldlines only. These are usually not exact solutions of the field equations, whence the particle concept is approxiamte only. Indeed, nothing hat we know about real particles suggests that they should be moving points.
 
  • #70
A. Neumaier said:
Yes, but once you admit classical fields as fundamental, it is more parsimonious to reard only these as fundamental. You get classical particles for free as the idealization where a field is supported on worldlines only. These are usually not exact solutions of the field equations, whence the particle concept is approxiamte only. Indeed, nothing hat we know about real particles suggests that they should be moving points.

The description of interaction in the particle theory via fields does not necessarily mean fields are "fundamental", although I am not sure what you meant by this. One possible view is that charged particles are material points and their *interaction* has advantageous description in terms of fields, or less advantageously and less generally, in terms of particular solution to Maxwell equations.

The view that all particles are just continuous fields concentrated to small region of space is common and deserves exploration, but it is mathematically problematic.

For example, in classical physics, having one or few fields only without reference to discrete particles more often than not leads to UV catastrophe. Even if we assume there is some acceptable configuration of the field with finite energy, there does not seem to be a way to convincingly choose a model of relativistic continuum out of endless possibilities available that would faithfully describe what we know as structureless electron.

On the other hand, theories of particles with interaction are free of this problem (Newtonian gravity, EM theory of charged particles considered by Frenkel, Tetrode, Fokker where there is no self-interaction). Also Wheeler-Feynman, although their absorber condition is unrealistic and has questionable benefits.
 

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