I was reading about gamma ray bursts and read something along the lines of 1 MeV gamma rays corresponding to a fireball above 2 billion degrees Celsius. How do scientists get temperature from that? Does it have something to do with using eV as a measure of temperature? If so, does that mean gamma ray bursts have a temperature?

EDIT: Quick follow-up. Could a burst of gamma rays heat the air in the atmosphere significantly?


1 Answer 1


For any photon you can calculate it's energy with the formula $E = h \nu$, where $h \simeq 6.626 \times 10^{-34}\:\mbox{J s}$ is Planck constant and $\nu$ the frequency of the photon.

At the same time you can crudely convert energy to temperature with $E = k_B T$, where $k_B \simeq 1.381 \times 10^{-23}\:\mbox{J/K} = 8.617 \times 10^{-5}\:\mbox{eV/K}$ is Boltzmann constant.

This means that $T = \frac{E}{k_B} = \frac{1 \times 10^6\:\mbox{eV}}{8.617 \times 10^{-5}\:\mbox{eV/K}} \simeq 1.16 \times 10^{10}\:\mbox{K}$, i.e. 10 billions Kelvin (or degrees Celsius, at this point the difference is irrelevant) in one single point. If you distribute that energy over a larger radius, you'll have your "fireball", although at these temperatures speaking of fire may be misleading.

A gamma ray burst could completely wipe out the atmosphere if it happens near enough. Fortunately the sources of such powerful events are also distant enough to not have any direct consequence on us.

  • $\begingroup$ So, a sufficiently powerful blast of gamma rays could heat the air to the point it burns away? Also, could you clarify what you mean by 10 billion Kelvin in "one single point?" $\endgroup$ Feb 17, 2016 at 16:57
  • $\begingroup$ Gamma ray burst are always powerful enough, but are fortunately far away, they come from activity of other galactic nuclei. What matters for the consequences is the distance, not the power. About the single point, I mean that a single 1 MeV photon can hit the atmosphere in one single point, transferring all the heat in that specific place. Then the heat will be transferred around radially and the temperature lowers. $\endgroup$
    – GRB
    Feb 17, 2016 at 17:17
  • $\begingroup$ How quickly would the temperature lower with distance from the source, if I may ask? $\endgroup$ Feb 17, 2016 at 17:46
  • $\begingroup$ Inverse square law for intensity ... just like ordinary light and gravity. $\endgroup$ Feb 17, 2016 at 18:05
  • $\begingroup$ The inverse square law works for temperature, too? I thought that was for radiation intensity. $\endgroup$ Feb 17, 2016 at 18:59

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