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Most people seem pretty much content with classical electromagnetic theory. And most of the applications use classical EM theory. Hence, I would like to know what was the necessity of Quantum Electrodynamics. Is it an advanced version of classical electrodynamics? Does it remove any flaws in classical electrodynamics? Are there any practical applications of QED?

Mikayla Eckel Cifrese
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Primeczar
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Quantum electrodynamics is needed to describe Nature instead of classical electrodynamics because quantum phenomena are observed – and have been observed at least since 1900 – which prove that classical physics in general and classical electrodynamics in particular is incorrect as a description of Nature and a better theory is needed.

The quantum phenomena include the fact that the energy carried by frequency $f$ electromagnetic wave isn't continuous. Instead, it is an integer multiple of $E=hf$, the energy of a "photon". This may be seen (and was seen by Max Planck) e.g. by analyzing the black-body radiation whose total output is finite and not infinite as classical electrodynamics would predict. So the right theory (which turns out to be QED) must agree with classical electrodynamics whenever classical electrodynamics was tested but it must also be compatible with the existence of photons.

The interactions of the electromagnetic field with other charged particles which follow quantum mechanics – e.g. electrons orbiting the nuclei in the quantum way – make the transition to QED necessary, too. QED has lots of new implications relatively to classical electrodynamics – the existence of antiparticles (and the possilibity to create or annihilate pairs), corrections to Coulomb's law at short distances, and many many others. In all of them, it may be seen by experiments that classical electrodynamics is wrong and QED is right.

QED also helps to solve some puzzles of classical physics. Classical electrodynamics predicted that atoms would collapse in less than a picosecond as they emit electromagnetic waves. The quantum theory makes atoms in the ground state stable. As I mentioned, the ultraviolet catastrophe – the wrong classical prediction that heated bodies emit an infinite amount of energy per second – was fixed and made finite by QED, too. Also, the intrinsic self-interaction energy carried by a charged particle may be made finite, thanks to renormalization (and related effects and their interpretation), in QED.

So yes, QED is an "advanced" version of classical electrodynamics, it removes lots of flaws of classical electrodynamics, and it's needed pretty much for every single physical phenomenon in Nature that was unknown in the 19th century but known since the 20th century.

Luboš Motl
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  • Thanks a lot ,sir for your comment . I am also eager to know whether there are any flaws in the classical EMT description of interaction between uncharged particles where an assumption of communication via fields is made without anything propagating in between them . – Primeczar Feb 13 '15 at 15:02
  • Did you mean interaction between charged particles? Even in classical electrodynamics, CED, the interactions are due to the field in between. This force in may be described in QED using virtual particles, and different gauges, etc., this makes no sense to explain this aspect of QED in isolation from a comprehensive introduction to QED as a whole, sorry. – Luboš Motl Feb 13 '15 at 15:57
  • Motl : Yes, I meant charged particles. – Primeczar Feb 13 '15 at 17:25
  • Sir, I realize that interactions like electrostatic interactions are due to fields. But I wanted to know if anything actually travels from one particle to another so as to convey the force. – Primeczar Feb 13 '15 at 17:37
  • Motl : I have read that these field pictures , in case of gravitation gave the impression of the field propagating faster than the speed of light and it has been an inspiration for general theory of relativity ,for example when a particle moves from one point to another its gravitational force on another particle seems to change instantaneously but actually some time delay occurs .An analogous situation an be realized in case of charged particle interaction . How is this resolved ?How to explain the time delay ? – Primeczar Feb 13 '15 at 17:43
  • Electrodynamics (just like general relativity) is compatible with special relativity so no information or impulse or true matter may propagate faster than light or instantaneously, and quantum electrodynamics doesn't change anything about this fact. There are various descriptions in classical and quantum electrodynamics where it may "look" that something is instantaneous but this is always a result of using nonlocal or unphysical degrees of freedom. Elmg and gravitational waves/impulses are propagating exactly at the speed of light in the vacuum, as the equations show (both in class+quantum). – Luboš Motl Feb 14 '15 at 07:47
  • Motl : Sir, could you tell me what exactly did you mean by nonlocal or unphysical degrees of freedom ? – Primeczar Feb 14 '15 at 07:51
  • An unphysical degree of freedom, like the electrostatic potential, is one that is used in the equations we use to describe physics but that can't be measured. For example, the potential and the whole vector $A_\mu$ is unphysical because of gauge invariance, and only some derivatives like $F_{\mu\nu}$ are physical. A nonlocal variable is a functional of fields that depends on values of fields at many points in the space or spacetime, whose distance is strictly positive. – Luboš Motl Feb 14 '15 at 07:54
  • E.g. you may define the field $H(x,y,z)$ which is the average of $|\vec E|^2$ in a ball of radius 1 meter around $x,y,z$. The equations describing the evolution of $H$ may look nonlocal and allow superluminal propagation of the signal but it's because $H$ is "complicated" in terms of the more elementary fields - and in terms of the elementary ones, the interactions may be seen to be local (the action is never faster than the speed of light). – Luboš Motl Feb 14 '15 at 07:55
  • Let us continue this discussion in chat. – Primeczar Feb 14 '15 at 07:59
  • Motl :I am extremely inspired by this enlightening reply of yours ,sir. Although I do admit that I have more things to learn so as to understand gauge invariance .

    Could we provide a more physical picture of this time delay ? And I am very eager to know as to why there is this time delay when actually there is nothing that is travelling in between the charges,say for the case of the electric field due to a charged particle having uniform velocity?

     – Primeczar Feb 14 '15 at 08:11
  • Dear @Primeczar, good if something has been englightening. The time delay - already existing in classical electrodynamics - is the time needed for the wave to get from one place to another. If you say something, the sound also spreads by a finite speed so there is a time delay. If the charged source of the field is static, you can't say at which moment the charge created the fields we feel now. But when things are moving, the answer is just one - it was after a delay $\Delta t = \Delta x/c$. – Luboš Motl Feb 14 '15 at 18:00
  • Something is traveling in between the source and the other charge that reacts. It's the perturbations of the fields. If the charges are static (time-independent), then the fields inside are also static, so they appear not moving. But if the charged source starts to wiggle, there are literally electromagnetic waves propagating in between, by the speed of light. In QED, electromagnetic waves are reinterpreted as a collection of an integer number of photons, they also move by the speed of light. – Luboš Motl Feb 14 '15 at 18:02
  • Motl :Thanks a lot again sir. Let us consider the time delay due to a charge moving with uniform velocity . In this case ,as per the postulates of classical EM theory, there cannot be any waves as the charges are not accelerated. – Primeczar Feb 14 '15 at 18:10
  • Motl : Is this propagating perturbation an entity like EM wave? – Primeczar Feb 14 '15 at 18:16
  • @motl: " The equations describing the evolution of H may look nonlocal and allow superluminal propagation of the signal but it's because H is "complicated" in terms of the more elementary fields - and in terms of the elementary ones, the interactions may be seen to be local " Sir, may I know what is being meant by elementary fields ? – Primeczar Feb 14 '15 at 18:33