Virial temperatures - why does this equality hold?

[gomboc]

I've done a problem which makes some basic assumptions about an gas of protons and electrons to find that given each electron or proton has kinetic energy KE = \frac{3}{4}m_p\sigma^2 and also that each particle has average energy \frac{3}{2}kT, then we can show that the temperature as

T \approx \frac{m_p \sigma^2}{2k}

This is fine - it was easy enough to show. But the question asks to show that

T \approx \frac{m_p \sigma^2}{2k} = 5\times10^6 \left(\frac{\sigma}{3\times 10^5\ m/s}\right)^2

I can't figure out why the right half of the equality is true. It's easy to test, but I can't seem to show why it might be analytically true. Note that \sigma is the average particle velocity and m_p is the proton mass.

Basically, I need to show why \frac{m_p}{2k} = 5\times 10^6 \cdot \left(\frac{\sigma}{3\times 10^5\ m/s}\right)^2 and I have no idea how to do that.

 

[gannordean]

Hi Mr. gomboc,

This is how to procede:

1. First convert the masse mp to MeV, divide it by the 2 and you'll get roughly 500 MeV.
   
   m_p/2 \approx500 MeV
   
   for K:
   
   K_B\approx8.6*10^{-5}eV*Kelvin^{-1}
   
   and for c:
   
   c\approx3*10^8 m/s
   
2. Separate the results into 5 MeV and 100 /c^2. Then put 100 /c^2 into a fraction with ρ^2 and K. Rise up the square to the entire fraction.
   
   \frac{m_p*ρ}{K_B}\approx5 *10^6 eV*\left(\frac{10*ρ}{(3*10^8 m/s)*\sqrt{8.6*10^{-5}eV}}\right)^2[Kelvin]
   
3. Simplify eV with eV and you will end up with the needed expression.
   
   T=5 *10^6*\left(\frac{ρ}{3*10^5}\right)^2[Kelvin]^*

Notice that if you make the assumption that the gaz is degenerate, it allows you to find out the energy as:

E\approx K_B*T=\frac{1}{2}\left(\frac{ρ}{3*10^5}\right)^2 KeV

If you deal with ultra-relativistic fermions then it gives:

E\approx\frac{1}{2} KeV

which is clearly neglected in comparaison to the energy at rest of one electron:

E_{0e^+}= 512 KeV

even more compared to that of a nucleon:

E_{0n}= 0.9 GeV

It's interesting to put our results in contrast with observations. What happen in the core of a cooler white dwarf star is, with the decreasing of temperature, the core become thermically "static" because it's difficult to "squeech" the matter inside, without any internal heat provided by the fusion reactions of hydrogen. The star dies.


Sorry to be too explanatory. Good luck.


*Note: assuming that ρ is in m/s!