PHY326/426:Lecture 11

Transcription

1 PHY326/426:Lecture 11 Towards WIMP direct detection WIMP Cross Sections and Recoil Rates (1) Introduction to SUSY dark matter WIMP-nucleon collision kinematics Recoil energy in the CM frame Probability density for recoil energy

2 Homework (1) Deadline: Mon 17th Nov Section 1 for PHY326 Section 1 AND 2 for PHY426 For PHY326-15% of overall mark for the module is on homework 1 and 2 together For PHY426-25% of overall mark for the module is on homework 1 and 2 together

3 The New Particle Zoo Here are a few of the candidates on a plot showing cross section vs. mass. An enormous range. We will focus on WIMPs thanks to L. Roszkowski (Sheffield)

4 Why Like Relic CDM Particles? Indicates Weak Scale So cosmology indicates generic WIMPs at W&Z scale that give about the correct value of Ω we need for dark matter: Candidate 1 (SUSY WIMPs, LSP, Neutralino) Candidate 2 (UED WIMPs, LKP)

5 WIMP Candidate 1 Supersymmetric Dark Matter Each particle gets a sparticle counterpart. Bosons get fermions and vice versa. e.g. Photon Photino W Wino Z Zino etc The Lightest Supersymmetric Particle (LSP) is predicted to be stable. This is called the NEUTRALINO.

6 The Neutralino The neutralino is a possibility for the lightest supersymmetric particle, and hence a WIMP candidate. The neutralino is a quantum mechanical superposition of the bino, wino and two higgsinos More in a later lecture on SUSY and dark matter by Dr. Joel Klinger

7 How do WIMPs interact? HALO WIMP WIMP rest energy, GeV. Velocity about 220km/s ELASTIC SCATTERING OF A WIMP OFF A NUCLEUS NUCLEUS rest energy, GeV. At rest in lab. VERY IMPROBABLE FOR NUCLEUS RECOILS ANY SINGLE NUCLEUS DUE TO FEEBLE NATURE OF WEAK INTERACTIONS

8 Worked Example Neutralino Dark Matter Properties Work out the local number density of neutralinos (given that the local mass density is about 0.3 GeV/c 2 ). Make the ʻstandardʼ assumption of an isothermal sphere of dark matter (be aware that this assumption may be an oversimplification, or even wildly incorrect). ANS: between 0.3 and 30 particles per litre.

9 Worked Example Neutralino Dark Matter Properties Work out the de Broglie Wavelength of a neutralino. assume velocity ~ 220km s -1 or about 10-3 c. assume ANS: the de Broglie wavelength is between 100 and 1 fm. working given in lecture

10 The WIMP-induced Recoil Spectrum So in a detector we want to observe: (1) a signal - what form does that take? (2) a background - what form does that take? The recoil energy spectrum draw the form of the expected spectrum given in lecture

11 How Much Energy can be Transferred to the Nucleus? For WIMP mass m W and velocity equal to the virial velocity in our halo, what is energy of the recoil of the target nucleus? BEFORE As for any classical two body elastic collision, maximum energy transfer when the two bodies collide head-on. AFTER WIMP m W m W m T m T NUCLEUS Conservation of momentum: rearrange and square: Conservation of energy:

12 Kinematics of Head-on Collisions from last slide: multiply cons. of energy equ by 2: multiply by m w : Substitute in to momentum conservation equation: Cancel left hand side with 1st term on right hand side: Rearrange: Square: multiply by m T /m W :

13 Energy Transferred vs. Masses of 1 Constituents Notice that the energy transferred to the kinetic energy of the recoiling nucleus is optimal when the nuclear mass equals the WIMP mass, and for a head on collision the WIMP loses all its kinetic energy. This is the best possible (and highly unlikely) case.

14 Worked Example The Best Transferred Energy How much kinetic energy could be imparted to a target nucleus? In an elastic collision, the maximum energy available for conversion into recoil of the nucleus is the kinetic energy of the incident WIMP, IF (when you are lucky), the nuclear mass exactly equals the WIMP mass. ANS working given in lecture It is very hard to detect this amount of recoil energy in a single recoiling nucleus embedded in a huge number of other nuclei.

15 How do you go about detecting WIMPs? (1) Ionisation Charge (2) Scintillation Light (3) Heat Phonons More in a later lecture on WIMP detection

16 The Observed Recoil Spectrum In fact the recoil spectrum actually observed in a detector is quite a complex product of different factors that we will aim to understand. It can be written in simplified form as: dr de OBS = R 0 S(E R )F 2 (E R )I This is an important formula to know R0 the total event rate, determines the over signal rate available S the spectral function that comes from the interaction kinematics of, determined by the relative mass of WIMP and target nucleus F 2 the form factor correction, for high high A nuclei this supresses the spectrum I the interaction type, whether spin-dependent or spin-independent determines the event rate depending on whether the target nucleus has net nuclear spin or not.

17 The Observed Recoil Spectrum dr de OBS = R 0 S(E R )F 2 (E R )I We will aim to derive this formula... Note in the following ER is the same as ET (the energy of the recoiling target nucleus)

18 Breaking Down R(ER) The real target of this analysis is the RATE for collisions for a recoil energy between E R and E R +de R. This is given by is the number of WIMPs incident per unit cross sectional area of the target per second (the WIMP flux) and A is the cross sectional area of the target consider as two parts: (a) The probability, GIVEN that a collision occurs, that it results in a target recoil of kinetic energy E R, and (b) The probability that an elastic collision occurs at all.

19 Bayesʼ theorem Extra Note In words for our case here this says ʻprobability of WIMP colliding with a target nucleus in the detector AND resulting in a recoil in some energy range is equal to the product of : (a) The probability, GIVEN that a collision occurs, that it results in a target recoil of kinetic energy E R, and (b) The probability that an elastic collision occurs at all.

20 WIMP-Nucleon Collision Kinematics WIMP velocity is typically 10-3 c, so we can use non-relativistic dynamics - do the calculation in the CENTRE of MASS FRAME. a fancy term for physics as seen by an observer who is at rest with respect to the centre of mass of the experiment. What is the velocity v c of the centre of mass frame for a WIMP colliding with a nucleus, relative to the nucleus? wimp M W v CM target nucleus mass M T at rest x y

21 Extra Note Particle Momentum in CM Frame Momentum of each particle to an observer in the CM frame TARGET NUCLEUS. Moves at velocity -v c so momentum is -M T v c or WIMP. Moves at velocity v-v c so its momentum is M W (v-v c ). Therefore in the centre of mass frame, the momenta of the WIMP and the target nucleus are equal in magnitude and opposite in sign

22 Two Body Elastic Collisions in CM Use centre of mass scattering centre of mass scattering angle Write Define the reduced mass of the WIMP & TARGET: so that and

23 Conservation of Energy in CM Frame so What we need is the ENERGY TRANSFER - the amount of energy imparted (in the LAB frame) to the nucleus by the WIMP. In CM frame, the TARGET momentum before collision = -m T v C, as the only velocity comes from the motion of the centre of mass. The WIMP momentum before the collision has the same magnitude, m T v C, as the momentum of the target. After collision, the magnitude of the momentum stays the same. Therefore the final velocity of the WIMP is m T v C /m W. This resolves into two CM frame components. In the lab frame the horizontal component of the velocity gets an addition of v C.

24 Final State Velocities in CM Frame We are interested in the velocity of the target nucleus after the collision. Its momentum after the collision in the CM frame has the same magnitude as it did before the collision. So its velocity resolves into two components, still in the CM frame, one parallel to the incident direction of the WIMP and the other perpendicular. In this frame the initial velocity of the target is v C in the -x direction. Therefore its final velocity has these components: horizontal CM frame: vertical CM frame: Returning to the lab frame, we must subtract the centre of mass velocity v c from the x component of the target velocity

25 Final State Target KE in CM Frame The Kinetic Energy in the lab frame is half the mass of the target times the sum of the squares of the lab frame velocity components. Re-write in terms of the reduced mass and using the calculated value of the centre of mass frame velocity v c we get This is an important formula to know

26 Questions Work out the de Broglie Wavelength of a neutralino. Work out the local number density of neutralinos (given that the local mass density is about 0.3 GeV/c 2 ). Describe and draw the expected signal recoil spectrum from WIMP interactions. Why is this form of spectrum going to make WIMPs difficult to identify? Describe the form of the equation for the recoil spectrum (4 parts). Derive an equation that relates the ratio of initial WIMP kinetic energy to that of the (initially stationary) recoiling target nucleus after collision in terms of their masses. Prove by conservation of momentum and energy that the maximum observable energy from a 10 GeV WIMP interacting in matter is 4 kev. To get this energy what nucleus would be needed?

27 Impact Parameter and Theta Derive a relation between the impact parameter and the scattering angle

28 End