Possiblity of break-down of Maxwell's Equations with very low current radiators

• Possiblity of break-down of Maxwell's Equations with very low current radiators by goran d on 28 Sep, 2016 11:15
• The idea is the following. We use very tiny currents on dipole or loop radiators. The currents are so small that there is not enough radiation energy per whole period (as per Maxwell Equations) to emit a single photon. If assuming there is no radiation, the acceleration terms in the field formulas (as derived from Lienard-Wiechert potentials) dissapear. We are left with only the near field components. If this assumption is correct then the Maxwell Stress Tensor conservation of momentum is no longer valid, and we can (potentially) break conservation with only the near field components present. Although it's possible that the non-radiating fields will be something different than simply  using only the near field.
• #1 by Proponent on 28 Sep, 2016 13:06
• Interesting thought.
  
  We need the mean power radiated over a cycle to be much less than the energy of a photon, which, of course, is Planck's constant, h, times the frequency.  In other words, the mean radiated power divided by the square of the frequency must be much less than h.  Plugging into the Larmor formula, dropping coefficients of order unity, and canceling terms, I get that we need e²/c << h, in CGS Gaussian units.  That's not satisfied by many orders of magnitude.
  
  Maybe there's a loophole if you can use a less efficient radiator, but off hand it looks to me like the quantization of charge prevents this from working.
• #2 by goran d on 28 Sep, 2016 17:04
• Why are you assuming acceleration to be equal to the speed of light? it could be less. There is no physical limit on how small the acceleration could be.
• #3 by Proponent on 28 Sep, 2016 17:50
• Quote from: goran d on 28 Sep, 2016 17:04

  Why are you assuming acceleration to be equal to the speed of light? it could be less. There is no physical limit on how small the acceleration could be.
  

  
  I make no such assumption.  The quantity c appears simply because it appears in when the Larmor formula is expressed in CGS Gaussian units.
• #4 by goran d on 29 Sep, 2016 10:06
• In the equation you talk about there are acceleration and frequency terms both to the second power. How did you get rid of them?
  It's like
  constant*e^2*a^2=h*f^2
  But in your answer there is no a and f.
  Please clarify.
• #5 by goran d on 07 Oct, 2016 15:10
• Anyway, there is no limit on how tiny the current can be on a piece of wire. Current in a wire is not quantised.
  
  My take on it is as follows:
  Power by dipole:
  P=R_radI²
  PT<hf
  P<hf²
  R_radI²<hf^2
  I<f(h/R_rad)^0.5
  Since R_rad is independent of frequency if we scale the dipole as a fixed proportion of wavelength, the critical current is proportional to frequency.
  
  Further more, the resultant field tensor does behave as a tensor in special relativity (although the potentials lose their meaning).
• #6 by laszlo on 07 Oct, 2016 15:48
• So if I make a dipole and don't connect it to anything, just count on the internal thermal movements of electrons, and if the dipole feedlines are close enough so that the elements are electrostatically coupled (allowing AC current a la capacitors), that minimal possible current should cause the dipole to accelerate?
  
  I think that really works because I just held a disconnected dipole in my hand and let it go. It immediately accelerated. It's reproducible every time. As far as I've been able to measure, the acceleration is about 344 inches per second squared.
  
  
  
  
  
• #7 by 1 on 07 Oct, 2016 20:32
• Folks, neither Maxwell's equations nor properties derived from them (including the Lienard-Wiechert equations) are expected to hold up at either extreme. In the small extremes, quantum eletrodynamics takes over, and the fact that Maxwell's equations no longer model things correctly is of little surprise to anyone.
  
  Regarding small currents, they most certainly are quantized in small values. See https://en.wikipedia.org/wiki/Probability_current for a standard treatment of such. Note that the current operators are extended from and are more complex than the simpler momentum operators, and thus a different understanding of what 'current' means is needed to understand interactions at the quantum level.
  
  
  
  The derivation is less important than the realization that there can exist solutions where the momentum (p) operator gives non-zero measurement but the current operator does not. Therefore, the fact that a charged particle appears to have motion (has momentum) is not enough to declare that a current exists; nor is the fact that a particle appears to be accelerating enough to declare that it must radiate.
  
  The original, real-world infinitesimal dipole / infinitesimal loop antenna is the case of a simple hydrogen atom; an electron "orbiting" a proton yet not radiating anything. Classically, an accelerating charge must emit something, yet an atom in its ground state does not. Why? Because quantum mechanically, the electron has an equal probability of 'existing' anywhere around that proton; and 'moving' in any direction. Even though it has a non-zero momentum, the fact that all allowed motions are equally likely to occur means that the 'state' of the electron is unchanged. Quantum mechanically, such a system should not radiate because there is, on average, no preferred movement in any one direction, and thus no real current is measured.
  
  Wanna know when you might see a current term appear? When your wavefunction is a superposition between a higher energy state and a lower energy state. Depending on which two states you chose, current can be zero, or non-zero (and oscillating!). If zero, then the transition is forbidden (see https://en.wikipedia.org/wiki/Selection_rule ); but if it's allowed then the system will radiate by the rules of quantum mechanics. This radiation takes the form of an emitted photon. And like a photon, the current operator gives a soliton-y, wave-packet-y behavior over time that eventually settles back to zero once the transition has occurred.
  
  The original question of an oscillator at energy too low to radiate begs the question of how such an oscillator is being driven. Driving the oscillator requires an energy input which, via reciprocity and uniqueness theorems, must then allow the oscillator to have enough energy to emit something in return. But if we allow such a scenario to exist, most likely it would then behave much like the hydrogen atom; existing but never radiating ( or doing anything else for that matter) unless it were perturbed at some point in the future. Most classical theories, or extensions thereof, could be safely disregarded.
• #8 by laszlo on 08 Oct, 2016 11:31
• Beautiful explanation.  Either you're a science writer or should be.