Download PHY323:Lecture 12

PHY323:Lecture 12
WIMP Cross Sections and Recoil Rates
(1)
•WIMP-nucleon collision kinematics
•Recoil energy in the CM frame
•Probability density for recoil energy
Candidates for Dark Matter
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/c2).
Make the ‘standard’ 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.
Worked Example
Neutralino Dark Matter Properties
Work out the de Broglie Wavelength of a neutralino.
assume rms velocity = virial 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
Notes
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
Notes
e.g. describe and draw the expected signal recoil spectrum from
WIMP interactions. Why is this form of spectrum going to make
WIMPs difficult to identify?
How Much Energy can be
Transferred to the Nucleus?
For a WIMP nucleus of mass mW and velocity equal to the virial
velocity in our halo, what energy can we expect from the recoil of
the target nucleus?
m
m
W
As for any classical
two body elastic
collision, maximum
energy transfer when
the two bodies collide
head-on.
T
WIMP
NUCLEUS
mW
Conservation of momentum:
rearrange:
square:
Conservation of energy:
mT
Notes
i.e. 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.
8
Kinematics of Head-on Collisions
multiply conservation of
energy equation by 2:
multiply by mw :
Substitute in to momentum conservation equation:
Cancel left hand side (LHS) with 1st term on RHS:
Rearrange:
Square:
multiply by mT/mW :
Energy Transferred vs. Masses of
Constituents
1
0
0
1
5
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.
Worked Example
The Best Transferred Energy
Work out 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.
Notes
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?
How do you go about detecting
WIMPs?
(1) Ionisation Charge
(2) Scintillation Light
(3) Heat Phonons
see later lectures
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
2
= R0 S(E R )F (E R )I
dE OBS
!
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
F2 – 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.
The Observed Recoil Spectrum
dR
= R0 S(E R )F 2 (E R )I
dE OBS
We will aim to derive this formula....
!
Note in the following ER is the same as ET
Breaking Down R(ER)
The real target of this analysis is the RATE for collisions for a
recoil energy between ER and ER+dER. 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:
The probability, GIVEN that a collision occurs, that it
results in a target recoil of kinetic energy ER, and
(b) The probability that an elastic collision occurs at all.
(a)
Extra Note
Bayes’ theorem
In words for our case here this says ‘probability 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 :
The probability, GIVEN that a collision occurs, that it
results in a target recoil of kinetic energy ER, and
(b) The probability that an elastic collision occurs at all.
(a)
WIMP-Nucleon Collision Kinematics
Extra Note
WIMP velocity is typically 10-3c, 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 vc of the centre of mass frame for a WIMP
colliding with a nucleus, relative to the nucleus ?
wimp
MW
CM
v
target
nucleus mass
MT at rest
x
y
Particle Momentum in CM Frame
Extra Note
Momentum of each particle to an observer in the CM frame
TARGET NUCLEUS. Moves at velocity -vc so momentum is -MTvc
or
WIMP. Moves at velocity v-vc so its momentum is MW(v-vc).
Therefore in the centre of mass frame, the momenta of the WIMP
and the target nucleus are equal in magnitude and opposite in sign
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
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 = -mTvC,
as the only velocity comes from the motion of the centre of mass.
The WIMP momentum before the collision has the same
magnitude, mTvC , as the momentum of the target.
After collision, the magnitude of the momentum stays the same.
Therefore the final velocity of the WIMP is mTvC/mW.
This resolves into two CM frame components. In the lab frame the
horizontal component of the velocity gets an addition of vC.
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 vC 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 vc from the x component of the target velocity
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 vc
we get
This is an
important
formula to know
Notes
Does this make sense ?
Check (DIY) that for a head-on collision, where the incident
particle bounces back in its original direction, in the case where
the WIMP and target masses are the same, the TARGET takes
all the kinetic energy from the WIMP
What is it good for ?
We have found that we can express the final kinetic energy
of the target in terms of known quantities, AND the center of mass
frame scattering angle theta. BUT how can we make use of this,
since we don’t know what this angle is going to be for any given
collision ?
The angle is related to the IMPACT parameter.
Probabilities
Start by assuming that the WIMP is destined to hit the target.
(Later we will talk about the probability of this happening)
If it does, then the collision is somewhere between the slightest
possible glancing blow, for which b = R and a head-on collision
with maximum energy transfer, for which b = 0
Assume that the WIMP is equally likely to strike the target nucleus
anywhere in the circular disk radius R seen from the perspective of
the incoming WIMP
Therefore the probability
of an impact parameter b
is proportional to the
circumference of a ring of
radius b:
where A is a constant
Interaction Probabilities and Cross
Sections
Extra Note
Of course, the scattering of a WIMP off a nucleus is not really
the scattering of two hard spheres. If it was, then the
interaction of a WIMP with a target would be guaranteed once
the impact parameter was less than some effective radius.
Quantum mechanical interactions don’t work like this. The
best we can do is to give a probability for a WIMP nuclear
interaction. A classical picture is again useful. Imagine the
target really is a sphere with a cross sectional area, viewed
from off to one side, of
Classically, whenever an incoming particle has an impact
parameter on the target less than R, the interaction will occur.
Quantum mechanical cross sections
Extra Note
Quantum mechanically, the analog might be pictured as this:
The target has a cross sectional area of pi*R^2. Whenever
the incident particle has an impact parameter less than R,
the interaction which we are interested in has a probability
of ‘p’ of occuring. 0<p<1.
area A
area B
But the quantum mechanical probability of the interaction you
are interested in being less than 1 is equivalent to reducing the
effective cross sectional area of the target.
Probability Density for the Impact
Parameter b
where A is a constant
Now require that the WIMP hits the target nucleus, so p(b)
should integrate to 1 over the domain 0 < b < R
therefore
and
What we really want, though, is the probability density for
the cosine of the scattering angle.
Impact Parameter and Theta
Derive a relation between the impact parameter and the scattering
angle
Probability Density for Recoil Energy
Back to the recoil energy
We know that we can relate the CM frame scattering angle
theta to the impact parameter b
here I have defined x as the ratio of b to R. We can use some
trig identities to express 1-cos(theta) in terms of theta/2:
Recoil Energy in Terms of x=b/R
We can write the recoil energy as a function of x = b/R
To get a probability distribution for the energy of the recoiling
nucleus, use the formula for converting probability densities again:
therefore:
we have
and we can evaluate
Extra Note
Relating probability densitites
Suppose some physical outcome is defined by a range of impact
parameters, and this range corresponds to a range of values of
some other parameter, say x. Then (see the diagram below), for
the probability of physical outcome to be given by both probability
distributions we must have
db
b
dx
x
The modulus signs ensure
that the relative signs of
db and dx in the
transformation
do not result in negative
probabilities.
Extra Note
Relating probability densitites example
Again if two parameters can be used to describe the same set of
possible outcomes, the probability densities for these parameters
(e.g. here b and x) are related by:
For example, suppose we want the probability distribution for b/R
and we have the probability distribution for b. Then
so that
therefore
Extra Note
How to convert p(b) to p(cos(theta))
The quantity that appears in the formula for the kinetic energy is
the cosine of half the centre of mass scattering angle, theta. We
would like a probability distribution for this.
Suppose some physical outcome is defined by a range of impact
parameters, and this range corresponds to a range of values of
some other parameter, say x. Then (see the diagram below), for
the probability of physical outcome to be given by both probability
distributions we must have
db
b
dx
x
The modulus signs ensure
that the relative signs of
db and dx in the transformation
do not result in negative
probabilities.
Probability Density for Recoil Energy
Therefore
Rather surprisingly, the probability distribution for the recoil
energy of the target nucleus is UNIFORM. This means that all
recoil energies between zero (for the lightest possible glancing
blow) to the maximum for a head on collision are equally likely.
Check that the normalization works out. The maximum possible
recoil energy is for b=0, or x=0, for which the recoil energy is
Now draw p(ET) over the domain of allowed ET values.
Worked Example
What is the maximum recoil energy possible?
ANS:
given in lecture
The p(ER) vs. ER Spectrum
Check normalization for WIMP recoil energies.
p(ER)
ER
otherwise
where
The p(ER) vs. ER Spectrum
Check normalization for WIMP recoil energies.
p(ER)
Notice:
ER
(1) the area of the rectangle is 1, so the integral of the probability
density over the domain of allowed values of the random variable
(ET) is unity, as required by statistics.
(2) the maximum possible target kinetic energy and the probability
density are both independent of R, the scattering angle theta, and
all numbers except those which are in principle measurable - the
velocity of the incoming WIMP (from your model of WIMP energy
distributions) and the masses of the WIMP and the target nucleus.
Probability of Any Collision
So far we have calculated the probability distribution for the recoil
energy GIVEN that there is a collision. We have assumed that a
collision took place in the first place. So we need to look at the
probability of any collision taking place as well, consider two parts:
The probability, GIVEN that a collision occurs, that it
results in a target recoil of kinetic energy ER, and
(b) The probability that an elastic collision occurs at all.
(a)
The real target of this analysis is the RATE for collisions for a
recoil energy between ER and ER+dER. This is given by
Here
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
Conclusion
We are part of the way along figuring out the probability
distribution for the energy transfer to the target nucleus in an
arbitrary WIMP nuclear collision.
Note that our hard sphere model does not account for any
complicated particle physics at the collision. In particular,
we have assumed that there is no spin-spin coupling of the
WIMP to the target. In the case that there IS such a coupling,
things get more complicated. The simple case we are
considering is s-wave scattering, and the coupling between
the WIMP and the nucleus is known as spin independent.