26 



loses energy through rotation of the dipoles (6). This is related to rotational 

 excitation in the gaseous phase, but is, of course, a quite distinct process. The 

 principle of the method which we used, in physical terms, is to replace the mov- 

 ing electric field of the electron by an equivalent radiation field consisting of 

 virtual photons; once that is done it is comparatively east to calculate the ener- 

 gy loss because one knows the absorption due to the rotating dipoles. This is 

 just the dielectric absorption and is well known, for example, in the case of 

 water, 



The result is: 



\ dt ; di 



2 (£ 



e 



s ir 7 



dip Tdn 4 



Here, e is the electronic charge, (£ s - ^ r ) is the difference between the static 

 dielectric constant, which for water is about 80, and the high frequency or "in- 

 frared" dielectric constant. The latter is the effective dielectric constant in the 

 infrared spectral region, and for water has a value which is comparatively large -- 

 about 5. Further, n is the optical refractive index, d is the molecular diameter, 

 and 1* the relaxation time. 



One striking thing about this result is that it is almost independent of veloc- 

 ity, and of about the same order of magnitude as the oscillational loss, for our 

 electrons. If the electrons have energy below about 0. 2 ev, it is, of course, 

 greater. 



KAMEN: How does it happen that velocity does not come in? I would think 

 that both kinds of excitation would be dependent on velocity. 



PLATZMAN: They are, in a sense. The formula is usually given as (-dW/ 

 dx), and this is l/v times (-dW/dt). 



Once the electron gets below an energy of about |ev, d has to be replaced 

 by the de Broglie wavelength, and this puts a "v" in the numerator which starts 

 to decrease the rate of energy loss below that energy. This effect is not im- 

 portant. 



We have, finally, the elastic scattering, and this I know virtually nothing 

 about. I would only like to take the opportunity to point out that I don't think 

 many other people know much about it either since this is not the elastic scat- 

 tering by single water molecules; once we get down in the low energy region 

 the wavelength is big enough so that the electron always interacts with several 

 water molecules at once. In complete ignorance, one can make the usual as- 

 umption that the scattering is isotropic with a cross section of the order of the 

 molecular cross section. 



KAMEN: Isn't that pretty large, comparing it to the energy loss? 



PLATZMAN: Elastic scattering gives practically no energy loss. It very 

 definitely dominates the picture for the actual path of the electron; however, 

 you can find, in the new book by Massey and Burhop (7), which is a great boon 

 to fundamental work in radiobiology, information on the elastic scattering of 

 electrons by water molecules in the gas; it is relevant but not very relevant. 



ONSAGER: That might affect the distance at which the electron finally comes 

 to a relative rest. 



