Laboratory 1



Yüklə 14,42 Kb.
tarix18.04.2018
ölçüsü14,42 Kb.
#39178

Student’s name____________________________

AstroParticle Physics 2004/05 (De Angelis)

Laboratory 2

The GZK cutoff


1. Purpose and Introduction


The purpose of this laboratory exercise is to determine what is the energy beyond which protons, colliding with photons from the Cosmic Microwave Background (CMB), enter in the region of the resonances (unstable baryons). When the threshold for the process

p + CMB ->  -> N  (1)

(where N is a generic nucleon) is reached, the proton energy is degraded. This energy is thus a cutoff for the propagation of protons through astronomical distances; it is called the GZK cutoff, from the initials of the authors who first foresaw the effect [1].

Despite the fact that we can demonstrate the existence of such a cutoff, the AGASA experiment [2] has observed a dozen of events beyond the limit. Several explanations have been proposed; you will see for example that the cutoff calculation probes the theory of relativity at extreme limits.



2. Procedure

First one has to determine the peak energy for photons in the CMB. Taking as a central value for the temperature of the Universe T = 2.72 K [3], by applying Wien’s law


peak = 2.9 106/T
( in nm, T in K) one can obtain the peak value for the wavelength , and then, from E = hc/(where hc=1240 eVnm is Planck’s constant times the speed of light) the central value for the energy
peak = _________________ nm

Epeak = ___________________eV


Note: compare the expression obtained with the relation obtained in thermodynamics for a free gas
E = 3/2 kB T
where kB = 8.6 10-5 eV/K is Boltzmann’s constant.

Then one can write the 4-vectors of a proton and of a  from CMB in the most favourable case for the production of a ; (head-on collision), and compute the minimum energy Ecutoff for a proton, at which a  can be produced.

Ecutoff ~ ___________________ GeV
(note: take for the  mass a value of1.23 GeV/c2, and for the proton mass a value of 0.94 GeV/c2).
This cutoff energy is quite large: as an example, a single proton of this energy it might lift a mass of 1 kg by ________ cm.

References

[1] K. Greisen, Phys. Rev. Lett. 16 (1966) 748;

5. G. T. Zatsepin and V. A. Kuz'min, JETP Lett. 4 (1966) 78.

[2] http://www-akeno.icrr.u-tokyo.ac.jp/AGASA/

[3] http://lambda.gsfc.nasa.gov/product/map/wmap_parameters.cfm
AstroParticle Physics 2003/04 (De Angelis)

Laboratory 2

The GZK cutoff

SOLUTION

Taking as a central value for the temperature of the Universe T = 2.72 K [3], by applying Wien’s law


peak = 2.9 106/T
( in nm, T in K) one can obtain the peak value for the wavelength , and then, from E = hc/(where hc=1240 eVnm is Planck’s constant times the speed of light) the central value for the energy
peak = 1.07 106 nm

Epeak = 0.0012 eV (2)


The relation obtained in thermodynamics for a free gas, E = 3/2 kB T,

where kB = 8.6 10-5 eV/K is Boltzmann’s constant would give E peak = 0.0004 eV (one third of the above value).


The 4-vector of the proton can be written as

(Ep, p, 0, 0) ~ (Ep, Ep, 0, 0)

while the 4-vector of the CMB photon with the most favorable direction is

(Epeak, - Epeak, 0, 0)


The square of the invariant mass M is thus

M2 = (Ep + Epeak)2 – (Ep -Epeak)2 ~ 4Ep Epeak


One must have

(1.23 109)2 ~ 4Ep Epeak => Ecutoff~ 1.5 1018/0.005 eV ~ 3 1020 eV



(such a particle has an energy of 20J: it could lift a mass of 1kg by 2 meters).


Yüklə 14,42 Kb.

Dostları ilə paylaş:




Verilənlər bazası müəlliflik hüququ ilə müdafiə olunur ©genderi.org 2024
rəhbərliyinə müraciət

    Ana səhifə