WIMP , dark matter

Main Question or Discussion Point

I'm trying to figure out how to arrive at the final expression, as given in the lecture notes. I tried to work this out by myself but getting different figure and units. I've considered the followings,

normalize the equation from its values e.g. spins and x parameter, as g~100, x~10(m/t~10) also I considered the (G)^1/2 (from m_planck)= 10^19 GeV when solving. Also converted, Critical density = 1.054 h^2 10^-5 GeV/cm^3 (cubed).

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Orodruin
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First of all, you cannot link to a google drive image like that. You need the image URL, not the URL where google drive will display it for you along with a google drive frame.

It is also unclear to me exactly what you have a problem with because it is unclear exactly what you have done.

Thiwanka Jayasiri
First of all, you cannot link to a google drive image like that. You need the image URL, not the URL where google drive will display it for you along with a google drive frame.

It is also unclear to me exactly what you have a problem with because it is unclear exactly what you have done.
Thanks for the tip on the google link. Frist time posting a question here. Let me clarify the question a bit. I want to arrive at the point where 10^-37 cm^2 , for that I used the mentioned values "normalize the equation from its values e.g. spins and x parameter, as g~100, x~10(m/t~10) also I considered the (G)^1/2 (from m_planck)= 10^19 GeV when solving. Also converted, Critical density = 1.054 h^2 10^-5 GeV/cm^3 (cubed)." Hence I need some support to arrive at to that point of 10^-37 cm^2.

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Please use Latex and not a PDF of scribbles.

Thiwanka Jayasiri
Please use Latex and not a PDF of scribbles.

$$\Omega_x = \sqrt[]{\frac{4\pi^3G}{45}*g_*(m)}\frac{1}{\langle \sigma\nu\rangle}\frac{x_fT_0^3}{30\rho_cr}$$\\

Considering the facts at WIMP dark matter ,

$$\Omega_X \approx\Omega_dm\approx0.3$$

considering
$$\rho_cr = 1.054h^2 10^{-5} \frac{GeV}{cm^3}$$

$$G^\frac{1}{2} = 10^{19} GeV$$

$$x = \frac{m}{T}$$
$$x_f \approx 10$$
$$g_m \approx 100$$
$$T_0 = 10^{-13} GeV$$

Hence the equation for $$\Omega_dm =\sqrt[]{\frac{4\pi^3G}{45}*g_*(m)}\frac{1}{\langle \sigma\nu\rangle}\frac{x_fT_0^3}{30\rho_cr}$$

\clearpage

Need to show that the

$$\Omega_dm =0.3 h^{-2} (\frac{x_f}{10})(\frac{g_*(m)}{100})^\frac{1}{2} \frac{10^{-37}cm^2}{\langle\sigma\nu\rangle}$$

follows from the first equation when baryons are neglected.

considering the normalizing factors, the equation can be rewritten as, taken 0.3 as the magnitude.

$$\Omega_dm =0.3\sqrt[]{\frac{4\pi^3G}{45}}*\sqrt[]{\frac{g_*(m)}{100}}\frac{1}{\langle \sigma\nu\rangle}\frac{x_f}{10}[\frac{T_0^3}{30*1.054h^2*10^{-5}\frac{GeV}{cm^3}}]$$

$$\Omega_dm =0.3 \sqrt[]{\frac{4\pi^3G}{45}} (\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}\frac{x_f}{10}[\frac{(10^{-13} GeV)^3}{30*1.054h^2*10^{-5}\frac{GeV}{cm^3}}]$$

Organizing the above equation to tally with the formula needs to be proven,

$$\Omega_dm = 0.3 h^{-2} (\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}(\frac{x_f}{10}) \sqrt[]{\frac{4\pi^3G}{45}}[\frac{(10^{-13} GeV)^3}{30*1.054*10^{-5}\frac{GeV}{cm^3}}]$$

applying the value of the $$(G^\frac{1}{2})= 10^{19} GeV$$

$$\Omega_dm = 0.3 h^{-2 }(\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}(\frac{x_f}{10}) \sqrt[]{\frac{4\pi^3}{45}}(G)^\frac{1}{2}\frac{(10^{-13} GeV)^3}{30*1.054*10^{-5}\frac{GeV}{cm^3}}]$$

\small $$\Omega_dm = 0.3 h^{-2} (\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}(\frac{x_f}{10}) \sqrt[]{\frac{4\pi^3}{45}}(10^{19})Gev\frac{(10^{-13} GeV)^3}{30*1.054*10^{-5}\frac{GeV}{cm^3}}]$$

Last edited:

$$\Omega_x = \sqrt[]{\frac{4\pi^3G}{45}*g_*(m)}\frac{1}{\langle \sigma\nu\rangle}\frac{x_fT_0^3}{30\rho_cr}$$\\

Considering the facts at WIMP dark matter ,

$$\Omega_X \approx\Omega_dm\approx0.3$$

considering
$$\rho_cr = 1.054h^2 10^{-5} \frac{GeV}{cm^3}$$

$$G^\frac{1}{2} = 10^{19} GeV$$

$$x = \frac{m}{T}$$
$$x_f \approx 10$$
$$g_m \approx 100$$
$$T_0 = 10^{-13} GeV$$

Hence the equation for $$\Omega_dm =\sqrt[]{\frac{4\pi^3G}{45}*g_*(m)}\frac{1}{\langle \sigma\nu\rangle}\frac{x_fT_0^3}{30\rho_cr}$$

\clearpage

Need to show that the

$$\Omega_dm =0.3 h^{-2} (\frac{x_f}{10})(\frac{g_*(m)}{100})^\frac{1}{2} \frac{10^{-37}cm^2}{\langle\sigma\nu\rangle}$$

follows from the first equation when baryons are neglected.

considering the normalizing factors, the equation can be rewritten as, taken 0.3 as the magnitude.

$$\Omega_dm =0.3\sqrt[]{\frac{4\pi^3G}{45}}*\sqrt[]{\frac{g_*(m)}{100}}\frac{1}{\langle \sigma\nu\rangle}\frac{x_f}{10}[\frac{T_0^3}{30*1.054h^2*10^{-5}\frac{GeV}{cm^3}}]$$

$$\Omega_dm =0.3 \sqrt[]{\frac{4\pi^3G}{45}} (\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}\frac{x_f}{10}[\frac{(10^{-13} GeV)^3}{30*1.054h^2*10^{-5}\frac{GeV}{cm^3}}]$$

Organizing the above equation to tally with the formula needs to be proven,

$$\Omega_dm = 0.3 h^{-2} (\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}(\frac{x_f}{10}) \sqrt[]{\frac{4\pi^3G}{45}}[\frac{(10^{-13} GeV)^3}{30*1.054*10^{-5}\frac{GeV}{cm^3}}]$$

applying the value of the $$(G^\frac{1}{2})= 10^{19} GeV$$

$$\Omega_dm = 0.3 h^{-2 }(\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}(\frac{x_f}{10}) \sqrt[]{\frac{4\pi^3}{45}}(G)^\frac{1}{2}\frac{(10^{-13} GeV)^3}{30*1.054*10^{-5}\frac{GeV}{cm^3}}]$$

\small $$\Omega_dm = 0.3 h^{-2} (\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}(\frac{x_f}{10}) \sqrt[]{\frac{4\pi^3}{45}}(10^{19})Gev\frac{(10^{-13} GeV)^3}{30*1.054*10^{-5}\frac{GeV}{cm^3}}]$$
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 units of $$[kg][meters^{2}][seconds^{-2}]$$

Hence, we can write the GeV numbers in above equation as follows,

$$10^{19} GeV = 1 *10^{28} eV = 1.602177 *10^{9} J$$
$$10^{-13} GeV = 1 *10^{-4} eV = 1.602177 *10^{-23} J$$
$$316200 eV = 5.066 *10^{-14} J cm^{-3}$$

applying and solving the values in the equation(1.1) reads as follows,

$$\Omega_dm = 0.3 h^{-2} (\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}(\frac{x_f}{10}) (2.1593228 * 10^{-46} J^{3} cm{^3})$$

$$\Omega_dm = 0.3 h^{-2} (\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}(\frac{x_f}{10}) (2.1593228 * 10^{-46} kg^{3} m^{6} s^{-6} cm{^3})$$

$$\Omega_dm = 0.3 h^{-2} (\frac{g_*(m)}{100})^\frac{1}{2}\frac{1}{\langle \sigma\nu\rangle}(\frac{x_f}{10}) (2.1593228 * 10^{-34} kg^{3} s^{-6} cm{^9})$$

First of all, you cannot link to a google drive image like that. You need the image URL, not the URL where google drive will display it for you along with a google drive frame.

It is also unclear to me exactly what you have a problem with because it is unclear exactly what you have done.
Hi, would you be able to help me to check whether Units being used is in order?

Orodruin
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Homework Helper
Gold Member
considering the normalizing factors, the equation can be rewritten as, taken 0.3 as the magnitude
It is unclear to me what you are doing here. Where are you getting the 0.3 from? Note that this is not the 0.3 from the observed $\Omega_{DM}$. You need to insert the nominal values for your parameters into the equation. As a simpler example, consider the typical distance-speed-time relationship $s = vt$ and imagine that you have a typical speed $v = 1$ km/h and a typical time of 1 minute. Your distance $s$ could then be written as
$$s = vt = \left(\frac{v \cdot \mbox{1 km/h}}{\mbox{1 km/h}}\right) \left(\frac{t \cdot \mbox{1 min}}{\mbox{1 min}}\right) = \left(\frac{v}{\mbox{1 km/h}}\right) \left(\frac{t}{\mbox{1 min}}\right) \cdot 1~\mbox{min km/h} \simeq 0.167~\mbox{km} \cdot \left(\frac{v}{\mbox{1 km/h}}\right) \left(\frac{t}{\mbox{1 min}}\right),$$
since 1 min ≈ 0.167 h. You need to do the corresponding thing for your equation.

It is unclear to me what you are doing here. Where are you getting the 0.3 from? Note that this is not the 0.3 from the observed $\Omega_{DM}$. You need to insert the nominal values for your parameters into the equation. As a simpler example, consider the typical distance-speed-time relationship $s = vt$ and imagine that you have a typical speed $v = 1$ km/h and a typical time of 1 minute. Your distance $s$ could then be written as
$$s = vt = \left(\frac{v \cdot \mbox{1 km/h}}{\mbox{1 km/h}}\right) \left(\frac{t \cdot \mbox{1 min}}{\mbox{1 min}}\right) = \left(\frac{v}{\mbox{1 km/h}}\right) \left(\frac{t}{\mbox{1 min}}\right) \cdot 1~\mbox{min km/h} \simeq 0.167~\mbox{km} \cdot \left(\frac{v}{\mbox{1 km/h}}\right) \left(\frac{t}{\mbox{1 min}}\right),$$
since 1 min ≈ 0.167 h. You need to do the corresponding thing for your equation.
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 in to the equation. This is when the freeze-out begins.
The fraction of critical density due to dark matter today,

$$\Omega_{X0}$$ , equation given as
$$\Omega_{x0} = \sqrt[]{\frac{4\pi^3G}{45}*g_*(m)}\frac{1}{\langle \sigma\nu\rangle}\frac{x_fT_0^3}{30\rho_cr}$$\\ then we need to apply the result obtained from the observations and the predictions of the BBN. so I used the 0.3 as magnitued , normalized the $$x_f \approx10$$ and etc. Could you confirm whether I've used it correctly ? about the $$h^{-2}$$ it comes from $$\rho_{cr}$$

Ref to what you quote; are u saying that if I use the normalizing factors, I've to add $$\frac{10}{1} (\frac{1}{100})^\frac{1}{2} \frac{1}{0.3}$$ respectively to the equation?

Last edited:
Orodruin
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You should not be "adding" anything to the equation. It is a matter of multiplying by ones (e.g., 1 = 1 km/ 1 km) and then extracting the numerator.

Thiwanka Jayasiri
You should not be "adding" anything to the equation. It is a matter of multiplying by ones (e.g., 1 = 1 km/ 1 km) and then extracting the numerator.
ok, if I plug in the nominal values into the equation
$$\Omega_dm =\sqrt[]{\frac{4\pi^3G}{45}*g_*(m)}\frac{1}{\langle \sigma\nu\rangle}\frac{x_fT_0^3}{30\rho_cr}$$

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

Still didnt apply the followings,

$$g_*(a_1) = 2 + 16 + 7+8+(30 + 30 + 12 + 12) = 91.5$$
$$g_*\approx100$$

if I'm getting this properly at CDM, WIMP $$X\approx 100GeV$$

Orodruin
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Homework Helper
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You also need to plug in the nominal values for the cross section, $g_*$, and $x_f$. In addition, your result is dimensionally inconsistent because you inserted $\sqrt{G}$ instead of $G$.

You also need to plug in the nominal values for the cross section, $g_*$, and $x_f$. In addition, your result is dimensionally inconsistent because you inserted $\sqrt{G}$ instead of $G$.
$$\Omega_m =\sqrt[]{\frac{4\pi^3 10^{19} GeV }{45}*(100)}\frac{1}{\langle \sigma\nu\rangle}\frac{(10)*(10^{-13}GeV)^3}{(30*1.054*h^{2}10^{-5} \frac{GeV}{cm^3})}$$

$$\Omega_m =\sqrt[]{\frac{4\pi^3 10^{19} GeV }{45}*(100)}\frac{1}{\langle \sigma\nu\rangle}\frac{(10)*(10^{-13}GeV)^3}{(30*1.054*h^{2}10^{-5} \frac{GeV}{cm^3})}$$
$$g_{x0}$$ value at T = 0.1 MeV $$\approx 100$$

You also need to plug in the nominal values for the cross-section, $g_*$, and $x_f$. In addition, your result is dimensionally inconsistent because you inserted $\sqrt{G}$ instead of $G$.
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?

You should not be "adding" anything to the equation. It is a matter of multiplying by ones (e.g., 1 = 1 km/ 1 km) and then extracting the numerator.
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, $$30*1.054*10^{-5} \frac{\small GeV}{\small cm^3} =5.066*10^{-8}\frac{kg}{ms^{2}}$$

summing (part 1,2 and 3 up we get,

$$\frac{(6.64513*4.112739)*10^{-68}}{(5.066)*10^{-8}} \frac{kg^{\frac{5}{2}}m^{8}}{s^5}$$
adding the same to the equation ()

$$\frac{1}{h^2}\frac{1}{\langle \sigma\nu\rangle}\frac{(6.64513*4.112739)*10^{-68}}{(5.066)*10^{-8}} \frac{kg^{\frac{5}{2}}m^{8}}{s^5}$$

could further simplified,

$$\frac{1}{h^2}\frac{1}{\langle \sigma\nu\rangle} (1.000...) \frac{kg^{\frac{5}{2}}m^{8}}{s^5}$$

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

Orodruin
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Again, you are inserting $G^{1/2}$ instead of $G$. I am out of here until you fix that.

Again, you are inserting $G^{1/2}$ instead of $G$. I am out of here until you fix that.
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,

$$\sqrt[]{\frac{4\pi^3}{45}*(100)} * (\frac{G^{\frac{3}{2}}}{G})$$

Orodruin
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No. I am not sure I can say it in another way, you are using $\sqrt G$ instead of $G$. G is not 1e19 GeV.

See http://pdg.lbl.gov/1998/consrpp.pdf

I cannot help you if you are unable to insert values into a given equation properly.

No. I am not sure I can say it in another way, you are using $\sqrt G$ instead of $G$. G is not 1e19 GeV.

See http://pdg.lbl.gov/1998/consrpp.pdf

I cannot help you if you are unable to insert values into a given equation properly.
I'm new to physics hence please excuse me, I guess you are referring to, $6.707 11(86)×10^{−39} ~c (GeV/c2)−2$

No. I am not sure I can say it in another way, you are using $\sqrt G$ instead of $G$. G is not 1e19 GeV.

See http://pdg.lbl.gov/1998/consrpp.pdf

I cannot help you if you are unable to insert values into a given equation properly.

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

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

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It is unclear to me what you are doing here. Where are you getting the 0.3 from? Note that this is not the 0.3 from the observed $\Omega_{DM}$. You need to insert the nominal values for your parameters into the equation. As a simpler example, consider the typical distance-speed-time relationship $s = vt$ and imagine that you have a typical speed $v = 1$ km/h and a typical time of 1 minute. Your distance $s$ could then be written as
$$s = vt = \left(\frac{v \cdot \mbox{1 km/h}}{\mbox{1 km/h}}\right) \left(\frac{t \cdot \mbox{1 min}}{\mbox{1 min}}\right) = \left(\frac{v}{\mbox{1 km/h}}\right) \left(\frac{t}{\mbox{1 min}}\right) \cdot 1~\mbox{min km/h} \simeq 0.167~\mbox{km} \cdot \left(\frac{v}{\mbox{1 km/h}}\right) \left(\frac{t}{\mbox{1 min}}\right),$$
since 1 min ≈ 0.167 h. You need to do the corresponding thing for your equation.
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
$$\Omega_m =\sqrt[]{\frac{4\pi^3 10^{19} GeV }{45}*(100)}\frac{1}{\langle \sigma\nu\rangle}\frac{(10)*(10^{-13}GeV)^3}{(30*1.054*h^{2}10^{-5} \frac{GeV}{cm^3})}$$
\clearpage
\small
solving part by part,\

part 1,

$$\sqrt{\frac{4\pi^3G}{45}*(100)} = \sqrt[]{\frac{4\pi^3*6.70711(86)*10^{-39}(\frac{GeV}{c^{2}})^{-2}*(100)}{45}}$$

solving the part 1 equation,

$$\sqrt[]{\frac{4\pi^3*6.70711(86)*10^{-39}(\frac{GeV}{c^{2}})^{-2}*(100)}{45}} = 3.137*10^{-21} \frac{kg^{2}m^{6}}{s^{6}}$$

part 2, $$10*(10^{-13}GeV)^{3} = 10*10^{-39}GeV^{3} = 1 * 10^{-29} eV^{3}= 1.602177*10^{-49}J^{3}$$\
$$= 1.602177*10^{-48}\frac{kg^{3}m^{6}}{s^{6}}$$

part 3, $$30*1.054*10^{-5} \frac{\small GeV}{\small cm^3} =5.066*10^{-8}\frac{kg}{ms^{2}}$$
$$\frac{P1 * P2}{P3}$$

summing part 1,2 and 3 up we get,

\large
$$\frac{(5.078909*10^{-69}\frac{kg^3 m^6}{s^6})}{(5.066*10^{-8} \frac{kg}{s^{2}}\frac{m^{6}kg^{2}}{s^4})}(\frac{s^4}{m^7 kg^2})(\frac{1}{kg})$$

$$1.00254 * 10^{-61} \frac{1}{ms^{2}}$$

adding the same to the equation (2)

$$\frac{1}{h^2}\frac{1}{\langle \sigma\nu\rangle}1.00254 * 10^{-61} \frac{1}{ms^{2}}$$
Hi, could you check and adv.

Last edited:
PeterDonis
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I'm new to physics
You have marked this thread as "A" level (and its subject matter seems to be consistent with that). That means you are expected to have graduate level background in the subject matter. If you are really new to physics, then you are trying to tackle a problem that is well beyond your background knowledge, and you should probably take the time to get the necessary background first.