# Detection Rate

1. Jan 30, 2015

### ChrisVer

1. The problem statement, all variables and given/known data

I am trying to understand how to obtain the estimated detection rate of a WIMP of mass $100~GeV$ into a Germanium detector of $1kg$ and detection efficiency of $P_{eff}=70 \%$.

2. Relevant equations

If you have at hand that the cross section is given $\sigma = \mu_R G_F^2$ (I think something is wrong with my units here, maybe I should have $\mu_R^2$) where $\mu_R = \frac{m_N m_{wimp}}{m_{wimp}+m_N}$ the reduced mass, and $m_N$ the mass of the detection medium nucleus.

3. The attempt at a solution

If I use that the probability of interaction in a width $dx$ in my material with $N$ Germanium atoms is:

$dW = \sigma N dx$

I have that the probability of detection is $D_{etection-rate}= P_{eff} \times W = P_{eff} \times \sigma N L$

Where $L$ is the path taken within the detector for the particle to interact.

In 1kg of Germanium I have $N = \frac{1~kg}{m_N}$ atoms.

So:

$D_{etection-rate}= P_{eff} \times \frac{1~kg}{m_N} \times L \times \mu_R G_F^2$

My problem is that I don't understand how to get rid of this "L"...

2. Jan 30, 2015

### Staff: Mentor

Your detection rate (to get counts/time) still needs the surface area of the detector. Multiply it by L and you get the volume which you can cancel against N.
Also, where is the WIMP flux?

3. Jan 30, 2015

### ChrisVer

So the detection rate would be what I wrote times a $A$ (surface of my germanium detector) and so that $A L = V_{det}$?

Also do you find the expression for the cross section correct? if $[\mu_R] = GeV$ and $[G_F] = GeV^{-2}$ I am getting units of volume and not units of area :(...

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4. Jan 30, 2015

### Staff: Mentor

The rate of detected events depends on the WIMP flux. More particle going through your detector lead to more events.
You can express the rate in dependence of this flux, but you cannot say "I will see 2 particles per year" if you do not know how many WIMP particles are going through your detector.

You might have to estimate the flux based on the dark matter density and the WIMP mass.

I agree with your analysis of the units, but I don't know which part is wrong. Do you get cross-sections in a reasonable range ($10^{-44} m^2$ plus or minus some orders of magnitude)?

5. Jan 30, 2015

### ChrisVer

Another way I thought of using was:
$\Gamma = n_{det} < \sigma u>$
And make some assumption for $u$ ... like $u \sim \frac{p}{E} = \frac{p}{m \sqrt{ (p/m)^2 +1}} \approx x (1 - \frac{x^2}{2} )$ , where $x=p/m \ll 1$

While $n_{det} = \frac{N}{V}$ if my germanium detector is a cube of 1m then I'm having $n_{det} = 8.292 \times 10^{24} m^{-3}$

The cross section is by using the $\mu_R^2$ instead:

$\sigma = 2.215 \times 10^{-7} ~GeV^{-2} \approx 3.5 \times 10^{-37} m^2$

And so:

$\Gamma = n_{det} < \sigma u> \approx 29.2 \times 10^{-13} m^{-1} \times \frac{p}{m_{wimp}}$

Now if I have $n_{wimp}$ WIMP particles flowing in my detector, I will have $n_{wimp} \times \Gamma$ interactions. From these interactions, I will be able to see only the $0.7 \times n_{wimp} \times \Gamma$ because of my efficiency...
If $\Omega_{cdm} = 0.22 = \frac{m_{wimp} n_{wimp}}{\rho_c}$ I can find the $n_{wimp}$.

This approach however, needed-
assumption 1: I used 1m length-cube for detector
unknown quantity: momentum of the wimp

6. Jan 30, 2015

### Staff: Mentor

The size of the detector should cancel out again because it is in your gallium density N as well (I think the volume is missing in your initial definition of N).

7. Jan 30, 2015

### ChrisVer

I think that N are the germanium nuclei. So in 1kg germanium detector, I have N= (1kg)/(m_ger) nuclei...I don't see how the volume enters in here...

8. Jan 31, 2015

### ChrisVer

It's OK. This thread can close. I came in contact with my tutor, and he told me that they didn't expect to give a final result, just to write down the detections rate given by:
N= eff * WIMPs Flux * cross section * number of targets
Which is exactly what I wrote down with:
$N_{detection-rate} = P_{eff} \Gamma N_{wimp} = P_{eff} n_{targ} <\sigma u> N_{wimp} = P_{eff} \sigma N_{targ} J_{wimp}$
with $J_{wimp} = \frac{N_{wimp}}{V} u = n_{wimp} u$ the Wimp flux in #Wimps per area per second....

and indeed there was a typo with the cross section and it should have the reduced mass squared....

Last edited: Jan 31, 2015
9. Jan 31, 2015

### Staff: Mentor

Then your initial formula where the length came in is wrong, because written like this N has to be the density.

10. Jan 31, 2015

### ChrisVer

indeed.... and not only that... the initial definition [that's why I abandoned it by proposing the Gamma] was the probability of interaction, and not the rate ... I had to take the derivative of it w.r.t. time, and end up with the Gamma as the interaction rate which would give me the detection rate afterwards...

Last edited: Jan 31, 2015