As explained in more detail on the main page (under Continuum Radiation), the usual interpretation of synchrotron radiation due to the acceleration of charges is logically inconsistent because it would make the emission observer dependent. An alternative mechanism is therefore required. In the following, the possibility is examined that this radiation could actually be due to recombination of air molecules which have been ionized by the synchrotron beam.

In order to make a reasonable estimate, the following values for a typical synchrotron have been taken from the book Synchrotron Radiation (Ed.: C.Kunz, Springer Verlag): air pressure = 10^{-9} - 10^{-10} Torr (i.e. one can assume a density of 10^{7} cm^{-3} )

electron current j = 100 mA (which corresponds to an electron flux F= 6.2^{.}10^{17} /cm^{2}/sec).

beam cross section = 1 cm^{2}

electron energy E = 1GeV

synchrotron radius R = 10 m. The total energy loss of a synchrotron is quoted as:^{10} erg/sec (rounded).

As the complete circle of the synchrotron contains 2^{.}π^{.}1000 =6.3^{.}10^{3} cm^{3}, the volume energy loss rate is therefore
_{ion}(ΔE) has strictly speaking to be treated as a differential cross section (as given by the Rutherford formula), but if one assumes schematically that the ionization energy ΔE has a fixed value, one can use the well known relationship for Coulomb scattering
^{-24}cm^{2}.

However, the above electron energy rests on the assumption of a relativistic mass increase as the electron approaches the speed of light, which is a flawed concept (see Relativity). If one assumes instead that the electromagnetic forces are merely reduced if the relative velocity approaches c (see A Newtonian Relativistic Electrodynamics), the maximum possible energy E of the electrons in the synchrotron is in fact only 250 keV. Additionally, one has to take into account that each molecule contains about 30 electrons. With this the cross section becomes

From Eq.(3), one finds then that N≈10^{18}cm^{-3}. This corresponds to a much higher pressure than actually measured (in fact it is almost 10% of the atmospheric pressure), but one has to realize that the vacuum pump would only be able to reduce the pressure of the neutral air, but not the ionized air which is trapped by the strong magnetic field and hence can build up to very high densities. Collisions with this plasma could then cause the observed energy loss for the electron beam (which leads to a corresponding amount of further (higher degree) ionization). As in a state of equilibrium ionization and recombination rates have to be equal, this will be associated with a corresponding amount of recombination radiation (i.e. synchrotron radiation) The observed pronounced angle dependence of this radiation is likely to be caused by the strong magnetic field of the synchrotron and/or the strongly directional velocities of the recombining charges.

It would require further theoretical and experimental study to explain how the strong increase of the energy loss for an increasing (assumed) electron energy (Eq.(1)) can be explained, but it is likely that this is actually caused by the increase of the magnetic field strength which should very much reduce diffusive ionization losses and hence cause an increase in the plasma density (which in turn would increase the collisional energy loss).

In order to make a reasonable estimate, the following values for a typical synchrotron have been taken from the book Synchrotron Radiation (Ed.: C.Kunz, Springer Verlag): air pressure = 10

electron current j = 100 mA (which corresponds to an electron flux F= 6.2

beam cross section = 1 cm

electron energy E = 1GeV

synchrotron radius R = 10 m. The total energy loss of a synchrotron is quoted as:

(1) I[Watt] =88.5 ^{.}E^{4}[GeV]^{.}j[mA]/R[m] .

As the complete circle of the synchrotron contains 2

(2) L = 1.6^{.}10^{6} erg/cm^{3}/sec .

(3) N^{.}F^{.}σ_{ion}(ΔE)^{.}ΔE = 1.6^{.}10^{6} erg/cm^{3}/sec .

(4) b^{.}tan(θ/2) = e^{2}/(2E) ,

(5) ΔE/E = sin^{2}(θ/2) ,

(6) b^{2} = e^{4}/(4E^{.}ΔE) .

(7) σ_{ion}(ΔE) = πb^{2} =π^{.}e^{4}/(4E^{.}ΔE) .

However, the above electron energy rests on the assumption of a relativistic mass increase as the electron approaches the speed of light, which is a flawed concept (see Relativity). If one assumes instead that the electromagnetic forces are merely reduced if the relative velocity approaches c (see A Newtonian Relativistic Electrodynamics), the maximum possible energy E of the electrons in the synchrotron is in fact only 250 keV. Additionally, one has to take into account that each molecule contains about 30 electrons. With this the cross section becomes

(8) σ_{ion}(ΔE=20eV) = 5^{.}10^{-20}cm^{2} .

From Eq.(3), one finds then that N≈10

It would require further theoretical and experimental study to explain how the strong increase of the energy loss for an increasing (assumed) electron energy (Eq.(1)) can be explained, but it is likely that this is actually caused by the increase of the magnetic field strength which should very much reduce diffusive ionization losses and hence cause an increase in the plasma density (which in turn would increase the collisional energy loss).

Thomas Smid (M.Sc. Physics, Ph.D. Astronomy)

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See also my sister site http://www.plasmaphysics.org.uk