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QED- Project

QED- Project. Manfred Hanke, August 2005:. (guided by Prof. A. Schäfer). Compton- scattering of the cosmic background radiation off a ultrarelativsitic cosmic proton. and pair production by a (back-scattered) photon. Contents of this talk:.

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QED- Project

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  1. QED-Project Manfred Hanke, August 2005: (guided by Prof. A. Schäfer) Compton-scatteringof the cosmic background radiation off a ultrarelativsitic cosmic proton andpair productionby a (back-scattered) photon

  2. Contents of this talk: 0. Introduction- Cosmic background radiation- Cosmic rays- Compton-scattering 1. Energy-loss of a cosmic proton due to Compton-scattering- Cross-section- Kinematics- Differential probabilities- Mean energy-loss- Result 2. Mean free path of a back-scattered photon- Cross-section- Differential probabilities and mean free path- Result 3. Summary

  3. Cosmic background radiation • follows Planck‘s formula for black-body-radiation with T = 2,725 K: • predicted by G. Gamow and R. Alpher in the 1940s • discovered by A. Penzias and R. W. Wilson in 1964 (Nobelprize in 1978)

  4. Cosmic background radiation • follows Planck‘s formula for black-body-radiation with T = 2,725 K: • predicted by G. Gamow and R. Alpher in the 1940s • discovered by A. Penzias and R. W. Wilson in 1964 (Nobelprize in 1978)

  5. Cosmic rays - discovered in 1912 by V. Hess (Nobelprize 1936) - high-energy particles (up to 1020 eV) - mostly (97%) nucleons, especially protons, -particles

  6. Fluxes of Cosmic Rays Flux (1 particle per m²·s) Knee(1 particle per m²·year) Ankle (1 particle per km²·year) Energy

  7. Cosmic rays - discovered in 1912 by V. Hess (Nobelprize 1936) - high-energy particles (up to 1020 eV) - mostly (97%) nucleons, especially protons, -particles - origin: solar eruptions, supernovae, cosmic jets (from black holes / pulsars), ..., ? • Nucleons with energies higher than 5·1019 eV loose their energy by the GZK-effect: (Greisen-Zatsepin-Kuzmin) •  + p  +  N +  What is the energy-loss through Compton-scattering?

  8. QED -Compton-scattering e +  e +  Easy calculation of the cross-section in the Dirac-theory: (Klein-Nishina) How to calculate Compton-scattering off a proton?

  9. Contents of this talk: 0. Introduction- Cosmic background radiation- Cosmic rays- Compton-scattering 1. Energy-loss of a cosmic proton due to Compton-scattering- Cross-section- Kinematics- Differential probabilities- Mean energy-loss- Result 2. Mean free path of a back-scattered photon- Cross-section- Differential probabilities and mean free path- Result 3. Summary

  10. The cross-section In Hildebrandt, Griesshammer, Hemmert, Pasquini: “Signatures of Chiral Dynamics in Low Energy Compton Scattering off the Nucleon“ (nucl-th/0307070) one finds: , with To calculate the energy-loss through Compton-scattering,one needs... for Compton-scattering off a proton the Ai‘s defined as page-long integrals over two Feynman parameters (!)

  11. A1 , forexample, is given by:

  12. Where do these expressions come from? • EFT (Chiral Effective Field Theory) „The Heavy Baryon Chiral Perturbation Theory only involves explicit πN degrees of freedom.“

  13. Where do these expressions come from? • EFT (Chiral Effective Field Theory) „The Heavy Baryon Chiral Perturbation Theory only involves explicit πN degrees of freedom, whereas the Small Scale Expansion formalism includes explicit spin 3/2 nucleon resonance degrees of freedom.“

  14. Where do these expressions come from? • EFT (Chiral Effective Field Theory) „The Heavy Baryon Chiral Perturbation Theory only involves explicit πN degrees of freedom, whereas the Small Scale Expansion formalism includes explicit spin 3/2 nucleon resonance degrees of freedom (and within that – in my opinion – very exotic couplings, like  N  or   N N, for which the parameters have been fitted from experimental cross section data).“

  15. Problem: for > 0 - m  130 MeV, the values get imaginarydue to the resonance,and zero-values cause numerical divergencies by the denominators! Here, the following abbreviations and constants are used:

  16. The cross-section 20 nbarn  numericalresults for < 130 MeV

  17. Kinematics To calculate the energy-loss through Compton-scattering,one needs... In the relativistic limit, one gets - for the photon-energy in the center-of-mass-frame: Here is k := energy of the cosmic background photon,  := cos (proton, photon)lab - for the energy-loss of the proton: , z := cos (proton, scattered photon)cm

  18. Differential probabilities  := cos (proton, photon)lab , z := cos (proton, scattered photon)c k := energy of the cosmic background photon, Now, one can calculate...

  19. the differential probability one can look at the spectrum of interacting photons: Now, as one has calculated

  20. Spectrum of interacting photons (Ep = 1019 eV) Do you see any difference to the Planck-spectrum?

  21. spectrum of energy-loss: Now, as one has calculated the differential probability, one can look at the For the numerical simulation, the -function is realized by a histogram.

  22. Spectrum of energy-loss (Ep = 1019 eV)

  23. Spectrum of energy-loss  One can rewrite the -function and perform the integral over z to get an analytic expression for that is only an integral over k and , which can more easily be numerically determined.

  24. Spectrum of energy-loss (Ep = 1019 eV)

  25. The mean energy-loss  5.3 MeV / ly for proton with Ep = 1019 eV ~ Ep2

  26. Result 1. Energy-loss of a cosmic proton • The low energy-loss is due to the small cross-section for Compton-scattering. • A mean energy-loss of 5.3 MeV / ly for 1019 eV- protons corresponds to a mean free path of 1.9 · 1012 ly. (The mean distance between galaxies is of order 106 ly.) • Compton-scattering of the cosmic background radiation off such a ultra-high-energy cosmic proton therefore does not lead to a noticeable decceleration of cosmic rays. The result is, that there is no result. (what concerns the decceleration of cosmic protons)

  27. 2. Mean free path of a back-scattered photon But: The proton‘s energy-loss (up to 1018 eV for Ep = 1019 eV) is added to the photon‘s energy. (This is known as Compton-back-scattering / inverse Compton-scattering,which is one way to produce ultra-high-energy cosmic -rays.) What happens with these high-energetic photons? e+ / e– - pair production from single photons is not allowed, but they can interact with the cosmic background radiation.

  28. Contents of this talk: 0. Introduction- Cosmic background radiation- Cosmic rays- Compton-scattering 1. Energy-loss of a cosmic proton due to Compton-scattering- Cross-section- Kinematics- Differential probabilities- Mean energy-loss- Result 2. Mean free path of a back-scattered photon- Cross-section- Differential probabilities and mean free path- Result 3. Summary

  29. The total cross-section (Breit-Wheeler) It is . As a result from kinematics:  , for e+ / e– - pair production from two photons

  30. The total cross-section

  31. Differential probabilities k0 = 3.21 · 109 MeV kmax(CMB) kmax()  maximum

  32. Differential probabilities k0 = 5 · 107 MeV kmax(CMB) <kmax()  suppression by the exp-factor

  33. Differential probabilities k0 = 1011 MeV kmax() < kmax(CMB) suppression by the k²-factor

  34. Mean free path rapid decrease of probability for k0 < 5 · 108 MeV dW/dL(k0 = 109 MeV) = 2.52 · 10-5/ly dW/dL(k0 = 108 MeV) = 2.28 · 10-9/ly dW/dL(k0 = 107 MeV) = 1.60 · 10-54/ly

  35. Mean free path (slow) decrease of probability for k0 > 1011 MeV dW/dL(k0 = 1010 MeV) = 3.0 · 10-5/lydW/dL(k0 = 1011 MeV) = 9.4 · 10-6/ly dW/dL(k0 = 1012 MeV) = 2.0 · 10-6/ly

  36. Mean free path minimal probability at k0 = 3.21 · 109 MeV • dW/dL = 3.8 · 10-5/ly • maximal mean free path<L> = 26 · 103 ly

  37. Result 2. Mean free path of a back-scattered photon The universe should be almost transparent for very-high-energy -rays with k0 < 1014 eV (at least what concerns e+/e–-pair production) – the mean free paths are billions of lightyears! Photons with ultra-high energies 2·1014 eV < k0 < 1019 eV should interact with the cosmic background radiationand create e+/e–-pairs within less than 3 million ly, what is approximately the mean distance of galaxies. There should be no ultra-high-energy extragalactic -rays! (Back-scattered photons with these energies can‘t be observed.)

  38. 3. Summary Contents of this talk: 0. Introduction- Cosmic background radiation- Cosmic rays- Compton-scattering 1. Energy-loss of a cosmic proton due to Compton-scattering- Cross-section- Kinematics- Differential probabilities- Mean energy-loss- Result 2. Mean free path of a back-scattered photon- Cross-section- Differential probabilities and mean free path- Result

  39. Contents of this talk: 0. Introduction- Cosmic background radiation- Cosmic rays- Compton-scattering 1. Energy-loss of a cosmic proton due to Compton-scattering- Cross-section- Kinematics- Differential probabilities- Mean energy-loss- Result 2. Mean free path of a back-scattered photon- Cross-section- Differential probabilities and mean free path- Result 3. Summary from EFT:  20 nbarn : spectrum of energy-loss

  40. Spectrum of a protons energy-loss due to Compton-scattering

  41. Contents of this talk: 0. Introduction- Cosmic background radiation- Cosmic rays- Compton-scattering 1. Energy-loss of a cosmic proton due to Compton-scattering- Cross-section from EFT:  20 nbarn - Kinematics- Differential probabilities: spectrum of energy-loss- Mean energy-loss - Result 2. Mean free path of a back-scattered photon- Cross-section- Differential probabilities and mean free path- Result 3. Summary : ~ Ep2, but only 5.3 MeV / ly for Ep = 1019 eV

  42. Contents of this talk: 0. Introduction- Cosmic background radiation- Cosmic rays- Compton-scattering 1. Energy-loss of a cosmic proton due to Compton-scattering- Cross-section from EFT:  20 nbarn - Kinematics- Differential probabilities: spectrum of energy-loss- Mean energy-loss: ~ Ep2, but only 5.3 MeV / ly for Ep = 1019 eV - Result 2. Mean free path of a back-scattered photon- Cross-section- Differential probabilities and mean free path- Result 3. Summary :  26 · 103 ly (k0,min = 3.2·1015 eV) : no -rays with 2·1014 eV < k0 < 1019 eV

  43. That‘s it! Thank you very muchfor your attention!

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