Finding mean energy of cosmic microwave background photons

[TorreyBeek]

Hello all,

I was recently assigned a problem in my modern physics class regarding finding the mean energy of Cosmic Microwave Background (CMB) photons. The problem reads as follows:

The universe is permeated with primordial microwave radiation that has a mean wavelength of about 2.5 mm.

(a) Find the mean energy (in eV) of the CMB photons, given that the energy of a photon is related to its frequency by the relationship E = hf (where f=frequency), where h is Planck's constant.

(b) Consider a cosmic-ray proton that has total relativistic energy E in the CMB reference frame. Find E such that the Doppler shifted CMB photons striking the proton head-on have a mean energy of 100 MeV.



I believe I have solved (a), with the solution as follows:
E = hf = h(c/wavelength)
E(mean) = h(c/mean wavelength)
= (4.136x10^-15 eV*s)[(3.0x10^8 m/s)/2.5x10^-3 m]
= 4.963 x 10^-4 eV


The second part, (b), I have no idea how to find. Any help would be much appreciated!

 

[jbsteele]

Thank you!</code>For part (b), you need to use the Doppler Effect to find the energy of the CMB photons. The Doppler Effect states that the frequency of a wave increases when the source is moving towards an observer, and decreases when the source is moving away from the observer.Using this, we can calculate the energy of the protons such that the frequency of the Doppler shifted CMB photons is 100 MeV.E = hf = (6.626x10^-34 Js)(100x106 eV/J)= 6.626x10^-28 eVTherefore, the relativistic energy of the proton in the CMB reference frame is E = 6.626x10^-28 eV.

 

[baba89]

Hello,

It looks like you have correctly solved part (a) of the problem. To solve part (b), we need to use the Doppler effect equation: f = f0 (1+v/c), where f is the observed frequency, f0 is the original frequency, v is the relative velocity between the observer and the source, and c is the speed of light.

In this case, the observed frequency (f) is 100 MeV, the original frequency (f0) is the mean energy of CMB photons that you calculated in part (a), and the relative velocity (v) is the velocity of the cosmic-ray proton. So we can rearrange the equation to solve for the proton's velocity:

v = (f/f0 - 1) * c

Substituting in the values, we get:

v = ((100 MeV)/(4.963 x 10^-4 eV) - 1) * (3.0x10^8 m/s)
= 5.994 x 10^11 m/s

Now, we can use the relativistic energy equation to calculate the total energy of the proton:

E = (mc^2)/sqrt(1-(v^2/c^2))

Substituting in the mass of a proton (1.6726219 x 10^-27 kg), we get:

E = (1.6726219 x 10^-27 kg) * (3.0x10^8 m/s)^2 / sqrt(1-(5.994 x 10^11 m/s)^2 / (3.0x10^8 m/s)^2)
= 1.031 x 10^-11 J

Converting to MeV, we get:

E = 6.439 MeV

So, the total relativistic energy of the cosmic-ray proton in the CMB reference frame is approximately 6.439 MeV. I hope this helps! Let me know if you have any further questions.