How do we explain the discrepancy? What hides the fine 10^-15 m detail in the process?
Classically, a relativistic charge which makes a sharp turn within 10^-15 m will produce at least some electromagnetic waves whose wavelength is ~ 10^-15 m. What prevents us from seeing such 1 GeV photons?
sensor for the
scattered electron
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\ digital camera
\ for photons
\
Z+ ●
^
|
|
e-
| b | impact parameter
Let us again look at Coulomb scattering of an electron from a nucleus. We may initially model the incoming electron as a wave packet whose size is > 2 * 10^-12 m. The encounter we have to calculate using a particle model. The electron may pass the nucleus at a distance ~ 10^-15 m. An electromagnetic wave is produced by the close encounter.
The measured result is an excitation in a cell of a digital camera, and an excitation in a cell of an electron detector.
The Wikipedia page gives a heuristic way to calculate the amount of energy radiated: we put a minimum value for the possible impact parameter b. The minimum value is the de Broglie wavelength of the electron.
Suppose that the impact parameter b is very small, ~ 10^-15 m.
When the electron turns abruptly at the nucleus, it will send an electromagnetic wave which looks like the Edward M. Purcell diagram at the top of the linked page above.
______ electric field line
_______|
sharp
turn at time t
Let W(t) be the produced classical electromagnetic wave if the electron passes the nucleus at a time t.
If we assume that the digital camera will see a sum of waves W(t) for a range of different t, then the sum will have less sharp turns in the electric field lines. That is, there is destructive interference which reduces the energy of the wave.
Classically, the camera would see the wave for one value of t.
When should we use a wave model and when a particle model?
Let us recapitulate what we have found out in the past few blog postings.
A. Coulomb scattering of an electron. We can use either the wave model or the classical particle model to calculate the deviation of an electron which passes a nucleus.
B. Photon produced in Coulomb scattering. A classical particle model predicts photons of a very short (10^-15 m) wavelength, which is wrong. We have to use a wave model where the electron is a wave packet of the size of de Broglie wavelength.
We may simply cut off the short wavelengths, or calculate a sum of of waves where destructive interference removes the short wavelengths. A path integral might be one way to calculate the sum.
Demanding energy conservation of quanta is one way to cut off short wavelengths.
C. Pair production. We do not know how to model the sharp 1 / r Coulomb potential of an electron if we represent the electron as a wave. The energy 1.022 MeV which is needed in pair production requires very steep potential wells. We have to use a particle model.
We need to check how Bethe and Heitler (1934) handled this.
A specific problem is how wide a wave packet we should use in various cases?
Equivalently, what should be the cutoff wavelength if we use a cutoff?
Or, how many paths we should sum if we use a path integral?
In case B, we may model the electron as a continuous charge distribution which flows past the nucleus. Some of the infinitesimal charges are scattered by the nucleus. If the wave packet is very large, say 1 meter across, it is like an electric current flowing past the nucleus. The electromagnetic field would be almost static. There would be essentially no photon radiation. This is clearly a wrong model.
Feynman diagrams in case B are a wave model where we use energy and momentum conservation to implement a cutoff. The Feynman recipe sounds the most rational.
But if there is a loop in the diagram, then we need some other method.
Why the electron in the 1s state in the hydrogen atom does not radiate photons?
The most common explanation is the uncertainty principle. If we try to restrict the electron to a volume which is smaller than the Bohr radius, we need to use large momenta p to build the wave packet, and the total energy of the electron becomes larger than on the 1s orbital.
Let us try to apply the observations of the previous section to this problem. Let us send an electron slowly toward a proton. We may model the electron as a wave packet whose size is the de Broglie wavelength of the electron.
The wave packet passes by the proton and loses some 13.6 eV of energy as radiation. After that, the wave packet is wrapped around the proton.
Let us assume that the wave packet is completely symmetric around the proton at this stage.
Let us use a path integral to calculate outgoing radiation from a point x close to the proton. If in a path, there is a point particle electron with a velocity vector v, there is another path with the exactly opposite velocity vector -v there. The sum of the radiation from these two paths is zero.
Another way to look at this is that the charge density at x stays constant, and there is no reason why the current at x would change either, as a function of time. This implies that the electromagnetic field of the various paths stays constant. There is no radiation.
A basic assumption in our argument is that we should sum all paths and their effects to the electromagnetic field.
Why does the hydrogen atom appear electrically neutral to the outside world? If we must calculate the electric field using the above method of path integral sums, and the wave packet is symmetric around the proton, we arrive to the conclusion that the electric field is zero far away from the atom.
Note that in this case we can treat the electron like its charge would be spread over a relatively large volume, and the charge density would be continuous. We may claim that the electron does not have a particle nature in this situation.
But if we shoot the hydrogen atom with a photon or another electron, the particle nature of the orbiting electron becomes evident again.
Can we somehow explain stationary states of the electron with these observations? If the orbital would be time-dependent, then there obviously would be electromagnetic radiation, and the electron would decay into a lower energy state.
But why does a stationary state 2s or 2p eventually decay into 1s? What causes the electromagnetic radiation?
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