Calculating WIMP Detection Rate in a Germanium Detector

In summary, the author is trying to calculate the detection rate of a WIMP of mass 100~GeV into a Germanium detector of 1kg and detection efficiency of P_{eff}=70%. 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, the author attempted to solve for the probability of detection D_{et
  • #1

ChrisVer

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Homework Statement



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

Homework Equations



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

The Attempt at a Solution



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

[itex]dW = \sigma N dx [/itex]

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

Where [itex]L[/itex] is the path taken within the detector for the particle to interact.

In 1kg of Germanium I have [itex]N = \frac{1~kg}{m_N} [/itex] atoms.

So:

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

My problem is that I don't understand how to get rid of this "L"...
 
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  • #2
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
I am not given a WIMP flux...neither any more information about the scales of my detector...
So the detection rate would be what I wrote times a [itex]A[/itex] (surface of my germanium detector) and so that [itex] A L = V_{det}[/itex]?

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

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  • #4
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
Another way I thought of using was:
[itex]\Gamma = n_{det} < \sigma u> [/itex]
And make some assumption for [itex]u[/itex] ... like [itex]u \sim \frac{p}{E} = \frac{p}{m \sqrt{ (p/m)^2 +1}} \approx x (1 - \frac{x^2}{2} ) [/itex] , where [itex]x=p/m \ll 1[/itex]

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

The cross section is by using the [itex]\mu_R^2 [/itex] instead:

[itex] \sigma = 2.215 \times 10^{-7} ~GeV^{-2} \approx 3.5 \times 10^{-37} m^2 [/itex]

And so:

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

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

This approach however, needed-
assumption 1: I used 1m length-cube for detector
unknown quantity: momentum of the wimp
 
  • #6
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
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
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:
[itex]N_{detection-rate} = P_{eff} \Gamma N_{wimp} = P_{eff} n_{targ} <\sigma u> N_{wimp} = P_{eff} \sigma N_{targ} J_{wimp}[/itex]
with [itex]J_{wimp} = \frac{N_{wimp}}{V} u = n_{wimp} u [/itex] 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...
 
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  • #9
ChrisVer said:
I think that N are the germanium nuclei.
Then your initial formula where the length came in is wrong, because written like this N has to be the density.
 
  • #10
mfb said:
Then your initial formula where the length came in is wrong, because written like this N has to be the density.

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...
 
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1. What is a WIMP?

A WIMP (Weakly Interacting Massive Particle) is a hypothetical particle that is believed to make up dark matter, which is a type of matter that does not interact with light and makes up about 85% of the total matter in the universe.

2. How does a Germanium detector detect WIMPs?

A Germanium detector is a type of particle detector that uses the properties of germanium crystals to detect particles, including WIMPs. When a WIMP interacts with a germanium nucleus, it produces a small amount of heat and ionization, which can be detected by the detector.

3. What factors affect the WIMP detection rate in a Germanium detector?

The WIMP detection rate in a Germanium detector can be affected by several factors, including the sensitivity and efficiency of the detector, the mass of the detector, the energy threshold for detection, the exposure time, and the flux of WIMPs in the surrounding environment.

4. How is the WIMP detection rate calculated in a Germanium detector?

The WIMP detection rate in a Germanium detector is calculated by using the following formula: R = N x σ x ε x ρ x v, where R is the detection rate, N is the number of target nuclei in the detector, σ is the WIMP-nucleus scattering cross section, ε is the detector efficiency, ρ is the local WIMP density, and v is the WIMP velocity. This formula takes into account the various factors that affect the detection rate.

5. What is a typical WIMP detection rate in a Germanium detector?

The WIMP detection rate in a Germanium detector can vary depending on the specific parameters of the detector and the surrounding environment. However, a typical detection rate is on the order of a few events per kilogram of detector material per year, due to the extremely low interaction probability of WIMPs with matter.

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