0
$\begingroup$

wiki says:

Photon–photon scattering limits the spectrum of observed gammas to a photon energy below 80 TeV, that is, a wavelength of more than ~ 1.5×10−20 m. The other photon is one of the many photons of the cosmic microwave background. In the frame of reference where the invariant mass of the two photons is at rest, both photons are gammas with just enough energy to pair-produce an electron–positron pair.

80TeV means frequency in the area of 10^27 Hz, can you explain the meaning of "observed" how, where? what I do not understand is if they are sayng that scattering can take place below that threshold, can you explain if that is the case and why?

One would expect that such scattering is only possible at very high frequencies, because ordinarily photons do not interfere with one another. What am I missing?

$\endgroup$
  • $\begingroup$ There should be references on the wiki page explaining the mechanism. "only... at very high frequencies". $10^{25}$ Hz is very high. $\endgroup$ – my2cts Aug 4 '18 at 8:38
2
$\begingroup$

What they are talking about here is very high energy gamma ray photons from cosmologically distant sources (distant Active Galactic Nuceli for example) undergoing photon-photon pair production interactions on the CMBR. This is referred to as scattering (somewhat inaccurately) in astrophysics because it modifies the AGN gamma ray spectrum in a way analogous to e.g. Raleigh scattering changing visible spectra in the Earth's atmosphere.

The interaction happens when in the center of momentum frame of the two interacting photons, the total energy is greater than $2mc^2$. As it happens, this translates to other reference frames as the rule that for photons of energy $\epsilon_1$ and $\epsilon_2$,

$\epsilon_1\epsilon_2>(mc^2)^2$

(Very approximately). $m$ is the electron mass. $(mc^2)^2$ therefore has a value of $~0.25 \mathrm{TeV}^2$

CMBR photons have energies of the order of meV, and so pose a signature scattering (absorption, really) source for $~100 \mathrm TeV$ photons.

$\endgroup$
  • $\begingroup$ Thanks, so, there can't be a real scattering, where the weaker photon bounces off with increased energy? $\endgroup$ – user157860 Aug 4 '18 at 9:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.