Week 3 Homework 2

1. What percentage of the energy budget does 4He occupy? Recall that from a previous week’slecture, for every one helium nucleus there are twelve hydrogen nuclei.

(a) 0.5%

(b) 4.4%

(c) 1.1%

(d) 8.3%

2. Here we explore how much money it would cost to create anti-matter. Let’s say electricitycosts 0.2$

kW hr . Naively imagining you can power a machine that creates anti-matter at 100%efficiency from the electric power, about how much would it cost to create a gram of anti-protons? Comments:

• “kW hr” is kilowatt hour, and is a measure of energy—finally a unit that older peoplewho pay bills will understand and younger students won’t :)

• Also, the reason this is “naive” and the real cost is explored in the next problem

(a) $5 million (5× 106 $)

(b) $200 million (2× 108 $)

(c) $50 billion (5× 1010 $)

(d) $20 trillion (2× 1013 $)

3. Continued from problem (2). The actual cost of creating a gram of antiprotons is around200-400 million billion dollars, $2-4×1017 (number taken from a CERN cost, found here:http://en.wikipedia.org/wiki/Antimatter#Cost). How much larger is this than the amountyour calculated about from problem (2)?

Not necessary text, but you might be interested: You might wonder why the costis so much larger than our naive estimate. The reason for this is that anti-protons are pro-duced by colliding protons at high energies into nuclei, and then trapping the anti-protons.First the colliding protons are accelerated to about 25-100 GeV (note that this is a largeenergy cost itself—rememeber, protons have a mass of about 1 GeV). Next, for every 105

or so collisions, only one to two anti-protons are produced and successfully trapped. Bettertechniques for trapping anti-protons would lead to the biggest improvement on cost. Finally,only a fraction of a percent of the electricity out of the wall “ends up” with the acceleratedproton.

(a) 104

(b) 107

(c) 109

(d) 1011

4. The CDMS experiment is a dark matter direct detection experiment of the sort describedin lecture. It is composed of two types of detectors, silicon (Si) and germanium (Ge). Thispast April they observed three nuclear recoil events at near the low energy threshold in

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their Si detectors. They observed no events in their Ge detectors. (See their press releasehere: http://cdms.berkeley.edu/press.html) These three events could be possible dark matterinteractions; if so, they would correspond to a dark matter mass of about 10 GeV. Based onthe discussion of direct detection in “Nature of Dark Matter II”, why is it not surprising thatthey only observed events in their Si detectors and not Ge detectors? By the way, in problem(6) below we examine a theoretical model which motivates this size of dark matter mass.

(a) The Si detectors in CDMS have been running longer than the Ge detectors. Therefore,they are more likely to have seen recoil events.

(b) The Si detectors use both phonons and scintillation light to search for events, while Gedetectors only use phonons.

(c) There is a minimum threshold of recoil energy needed for an event to be detected. SinceSi is lighter than Ge, this minimum recoil energy can be acheived for lower dark mattermasses with Si than with Ge.

(d) Ge detectors are better for measuring spin-dependent interactions. Since the Si detectorsregistered the events, this indicates that if these are dark matter events, they are spin-independent interactions.

5. How is a positron to be used in PET?

(a) The positive charge repels strongly off of protons in nuclei, allowing one to map thetissue.

(b) The positron is captured by a neutron in a nucleus, turning it into a proton (so-calledinverse beta decay). The number of these changed nuclei are then counted.

(c) A positron PET, unlike dogs and cats which are friendly, will try and annihilate a smallpiece of you.

(d) The positron annihilates with an electron producing two back-to-back gamma rays withenergies E = mec

2 = 511 keV which are then detected and used to map the tissue.

6. Disclaimer: I know this problem looks long. While the text is long, the question at the endis not too challenging. The idea of the problem is to explain another possible model of darkmatter besides the one discussed in lecture. If you have confusion over the language or ideas,please use the discussion boards as I am more than happy to help clarify!

The lectures focused on the possibility that dark matter is a so-called WIMP. This restson the assumption that dark matter was once in thermal equilibrium with the rest of matterin the early universe but is not any longer, and hence is also called a thermal relic. One nicefeature of this paradigm was that it naturally predicts a scale for the dark matter mass tobe around 100-1000 GeV, an energy range currently being studied at the LHC. This is a nicefeature since it might be related to other new physics which could show up at this energyscale. Physicists like these sorts of things because it ties multiple questions together; we haveother reasons to believe there is new physics at the energy scales probed by the LHC andwe also know there is dark matter in the Universe—wouldn’t it be great if the two were related!

Besides the WIMP model, another possibile model of dark matter that has gained popu-larity in the past several years is so-called “asymmetric dark matter (ADM)”. Some peoplewonder why the energy budgets of matter and dark matter in the Universe—4.4% and 25%,respectively—are so close to each other; there isn’t a reason a priori for them to be of the

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same order of magnitude. As we explain below, one appealing feature of ADM models isthat they give a mechanism for generating dark matter in the Universe while also providinga reason for why the matter and dark matter energy budgets are of a similar size.

ADM models purport that the same mechanism responsible for the surplus of matter overanti-matter in our Universe also generates more dark matter particles than anti-dark matterparticles. Then, just as in the case for regular matter, the energy budget of dark matter isdo to the extra amount of dark matter particles. Since the mechanism for producing thisasymmetry is the same for regular matter and dark matter, ADM models predict that thereare about the same number of dark matter particles as regular matter particles. Given thisand the energy budgets in the previous paragraph, about what mass of dark matter particlesdo ADM models predict? Recall that protons and neutrons have a mass of about 1 GeV.

(a) 200 MeV

(b) 1 GeV

(c) 5 GeV

(d) 50 GeV

7. Here we elaborate on the lower bound for the dark matter mass that Professor Murayamatalked about in “Nature of Dark Matter I”. I will post some notes or a video to explain a bitmore of the physics and thoughts behind this problem shortly. In the meantime, we can stilldo the exercise.

The bound came from saying that dark matter has to “fit” in galaxies. Basically the “Bohrradius” for the dark matter is

rB =~2

GNMm2

where m is the dark matter mass, M is the total amount of dark matter mass contained insidethe “gravitational atom”, and ~ and GN are constants as usual. We have observed featuresin the dark matter profile in galaxies down to about a kiloparsec (kpc= 3.1× 1019 m). If theBohr radius were larger than this distance, these features would not exist. Since lowering thedark matter mass increases the Bohr radius, this puts a lower bound on the dark matter mass.

Taking the dark matter to be of constant density ρ0 = 1.3 × 10−22 kg/m3, calculate thelower bound on the dark matter mass m. To do this, you will need to calculate how muchmass M is inside this atom using the given density. Give your answer to two significantfigures.

8. In the lecture, Professor Murayama talked about how we can use cosmic ray muons to mapotherwise invisible things. A particularly novel example he discussed was Luis Alvarez lookingfor a hidden chamber in an Egyptian pyramid. In this problem we explore the basic idea abit further.

Muons, since they are charged, lose energy as they pass through material. Energetic muons(1000 GeV & E & 1 GeV) are in the so-called “minimum ionizing case” where they lose aconstant amount of energy for most of their passage. Just before they stop, they rapidly losetheir enery (just like the protons of last week’s proton beam therapy question). If you wantto see a plot of energy loss per distance of muons as a function of energy, see figure 30.1 (page

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4) here: http://pdg.lbl.gov/2013/reviews/rpp2012-rev-passage-particles-matter.pdf

In the minimum ionizing case, the “stopping power” of a material on muons is

dE

dx≈ 2

MeV cm2

g

(don’t worry about the derivative sign—there is no calculus in this problem!). What stoppingpower means is that for a given density of material, ρ× dE

dxis the energy lost per distance of

the muon passing through the material (check the units to make sure this is the case!).

(a) How far will a 100 GeV muon travel through rock? Take the density of rock to be2.7 g/cm3 and give your answer to two significant figures.

(b) Now imagine you have three different detectors as shown in figure 1. Each detector hasan area of 1 m2. We will be asking how many muons reach each detector. Imagine theflux of muons (here flux = number per area per time per energy, #

area time energy) to beconstant up to 100 GeV, and then zero for energies above that.

φ =100 m−2s−1GeV−1 for E < 100 GeV0 for E > 100 GeV

(note that this constant flux in energy is not true in the real world—see the optionalpart of this problem). How many muons reach detectors (B) and (C) after one hour?Give your answer to two significant figures.

To help you do this, let me show you how many reach detector (A). Since there isnothing stopping them, all the muons headed for detector (A) reach it. Dimensionalanalysis essentially tells you the answer:

NA = φ︸︷︷︸

flux of muons

×Adetector × ∆E︸︷︷︸

energy range reaching detector

× ∆t︸︷︷︸

length of time detector is running

= 100muons

m2 s GeV× 1 m2

× 100 GeV× 3600 s

= 3.6× 107 muons

(c) Optional Note: below are two variations of the same problem. One requires calculus,the other doesn’t. Either answer will be graded as correct.

The actual flux is energy dependent, decreasing with energy as φ ∝ E−2.7. For con-creteness, the flux is

φ =100

m2 s GeV

(GeV

E

)2.7

for E > 1 GeV

where E is the energy measured in GeV (so that at E = 1 GeV the flux is the valueused in part (b) of the problem).

Variation of problem requiring calculus: Using the flux equation above, how manymuons with initial energies between 1 and 100 GeV reach detectors (B) and (C) afterone hour? Give your answer to two significant figures.

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Figure 1: Muons impingent on three different detectors. Each detector has an area of 1 m2. In(A) there is no material above the detector. Detector (B) is beneath 50 m of rock with a densityof ρ1 = 3 g/cm3. Detector (C) is beneath a tower of two different types of rock, ρ1 = 3 g/cm3 andρ2 = 2 g/cm3

Hints/tips/possible problems: If we were really doing this properly, we should spec-ify a detector resolution, i.e. what range of muon energies can our detector measure.We won’t worry about this (I’ve avoided this problem by not asking you about the fluxat detector A where you might notice a problem if you tried to send the energy to zero,as well as by asking you only about the flux for given range of initial energies—not theirenergies at the detector). As a check so you don’t need to worry about if you are enteringanswers into the website correctly, if we let the maximum initial energy extend from 100GeV to infinity, you would find 653 muons at detector B after one hour.

Variation of problem without calculus: An approximate form of the muon fluxis shown in figure 2. Using this graph, how many muons reach detectors (B) and (C) af-ter one hour? Give your answer to two signficant figures. Hint: part (b) of this problemis like having the histogram at the height of 100

m2 s GeVfor 1 GeV < E < 100 GeV.

To those that have done both ways of the problem, the histogram is an overestimation onthe actual answer using calculus. For a better approximation, you could do the height ofthe histogram at the midway points, i.e. take the height as φ(E) at E = 5, 15, 25, 35, . . .GeV.

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Figure 2: Approximate flux of muons. The height of each block in the histogram is determined byφ(E) at the points E = 1, 10, 20, 30, 40, 50, 60, 70, 80, and 90 GeV. The first few values at E = 1,10, and 20 are shown in the figure. Use the equation for φ to get the values at the other points.

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