Recent content by Thiwanka Jayasiri

solving part by part,\ Equation (1)\\ Need to show that the below Equation (2) \small $$\Omega_dm =0.3 h^{-2} (\frac{x_f}{10})(\frac{g_*(m)}{100})^\frac{1}{2} \frac{10^{-37}cm^2}{\langle\sigma\nu\rangle}$$ Equation(2 )\ follows from the first equation when baryons are neglected. \small...

Reuslt:

$$\sqrt[]{\frac{4\pi^3 G}{45}*(100)} = \sqrt[]{\frac{4\pi^3 *6.707 11(86)×10^{−39} (GeV/c^{2})^{−2 }}{45}*(100)} $$

I'm new to physics hence please excuse me, I guess you are referring to, ##6.707 11(86)×10^{−39} ~c (GeV/c2)−2##

part 1, $$\sqrt[]{\frac{4\pi^3 10^{19} GeV }{45}*(100)} = 6.64513 \frac{(kg^{\frac{1}{2}})m}{s}$$ $$\sqrt[]{\frac{4\pi^3}{45}*(100)} * (G^\frac{1}{2})*\frac{G^\frac{1}{2}}{G^\frac{1}{2}} $$ $$\sqrt[]{\frac{4\pi^3}{45}*(100)} * (\frac{G}{G^\frac{1}{2}}) $$ else I can write it as...

What if I ignore the units of part 2 and part 3 since $$\rho_{cr} \propto T_0^3 $$

solving part by part,\ part 1, $$\sqrt[]{\frac{4\pi^3 10^{19} GeV }{45}*(100)} = 6.64513 \frac{(kg^{\frac{1}{2}})m}{s}$$ part 2, $$10*(10^{-13}GeV)^{3} = 1*10^{-38}GeV^{3} = 1 * 10^{-11} eV^{3}= 4.112739 *10^{-68}J^{3}$$\ $$= 4.122739 *10^{-68}\frac{kg^{3}m^{6}}{s^{6}}$$ part 3...

Hi add the values the equation but I'm still not getting the numerator part, because I'm missing the ##g_*##, and ##x_f##. For me to add that is it fine to include those to the equation ? or what's the best move?

$$g_{x0} $$ value at T = 0.1 MeV $$\approx 100$$

$$\Omega_m =\sqrt[]{\frac{4\pi^3 10^{19} GeV }{45}*(100)}\frac{1}{\langle \sigma\nu\rangle}\frac{(10)*(10^{-13}GeV)^3}{(30*1.054*h^{2}10^{-5} \frac{GeV}{cm^3})}$$

ok, if I plug in the nominal values into the equation $$\Omega_dm =\sqrt[]{\frac{4\pi^3G}{45}*g_*(m)}\frac{1}{\langle \sigma\nu\rangle}\frac{x_fT_0^3}{30\rho_cr}$$ $$\Omega_m =\sqrt[]{\frac{4\pi^3 10^{19} GeV }{45}*g_*(m)}\frac{1}{\langle...

Noted, but 0.3 comes as the magnitude of the when applying to the equation. As per the notes given I took magnitude of that to apply into the equation. This is when the freeze-out begins. The fraction of critical density due to dark matter today, $$\Omega_{X0}$$ , equation given as...

Hi, would you be able to help me to check whether Units being used is in order?

then Equation(1.1) becomes,\ $$\Omega_dm = 0.3 h^{-2} (\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}(\frac{x_f}{10}) \sqrt[]{\frac{4\pi^3}{45}}10^{19}Gev[\frac{(10^{-13} GeV)^3}{30*1.054*10^{-5}\frac{GeV}{cm^3}}] $$ Equation( 1.2)Converting the GeV to eV and Joules got the...

Please refer to the below, $$\Omega_x = \sqrt[]{\frac{4\pi^3G}{45}*g_*(m)}\frac{1}{\langle \sigma\nu\rangle}\frac{x_fT_0^3}{30\rho_cr}$$\\ Considering the facts at WIMP dark matter , $$\Omega_X \approx\Omega_dm\approx0.3$$ considering $$\rho_cr = 1.054h^2 10^{-5} \frac{GeV}{cm^3}$$...