prev

next

out of 1

Published on

27-Mar-2017View

214Download

0

Embed Size (px)

Transcript

<ul><li><p>986 L E T T E R S T O T H E E D I T O R </p><p>Cosmic Rays from Supernovae </p><p>Among a number of effects which would seem to be capable of generating cosmic rays in supernovae a particu-larly simple one may here be sketched. </p><p>From direct observations we know1,2 that the total energy Et liberated in a supernova outburst is at least lO*8 </p><p>ergs to 1049 ergs. We may assume with a high degree of probability that in the course of such an outburst the mass M of the original star is split into a number of component parts. It is not essential for our considerations to know the exact details of this fission. For the sake of definiteness we picture M to be divided into a stellar remnant M\ and an expanding shell of gases of mass M2, where Mi-\-M2=M and M2 yM with, presumably, 7 ^ 10~</p><p>4. Since M is made up of electrically charged particles, the </p><p>two parts Mi and M2 cannot be expected to emerge un-charged from the fission. All distributions of atoms, ions and electrons must be admitted as equally probable, as long as the electric potential energy characteristic for these dis-tributions is small compared with Et. We set M=Nmp and Mi = Nimp where mp is the mass of a proton. For simplicity we assume that all of the atoms involved are singly ionized. The numbers of the positively and negatively charged particles in the stellar remnant will be, respectively, m+=Npi+5i+ and nr = Npi+8r where Npi = NMi/M is the most probable value of ni+ or ni~t and 5i</p><p>+, 8r are the actual deviations in a given distribution. The stellar remnant then carries a total charge (5i~ di+)e, where e is the charge of an electron. Setting A = r~ 5i+ we are consequently interested in the value of A2 averaged over all possible distributions of positively and negatively charged particles e over Mi and M2. A simple calculation shows that </p><p> Emin/R = e(M/tnp)*/50R. (5) </p><p>We write M=fiM(Q) and R = rR(Q), where Af(0) = 2Xl033g and i?(O) = 7Xl01 0 cm are the mass and the radius of the sun, and we obtain </p><p> = 4 . 7 X l 0 y / > (6) </p><p>or in volts, for /* = 1 and r=\ </p><p>^300X4.7X106 = 1.4X109 volts. (7) </p><p>The breakdown of these potentials will take place at an advanced stage of the supernova outburst when the ex-panding gaseous shells are of a density sufficiently low to permit the free escape of the initially trapped excess </p><p>charges. The total electric potential energy E&/2 turns out to be very small compared with Et and our assumption regarding equally probable distributions is therefore justified. </p><p>Although considerations of the above type appear suffi-cient to demonstrate that cosmic-ray particles are gen-erated in supernovae, the fact that e$/2</p></li></ul>