a from the quantum Hall effect The everyday Hall effect: A current density jx in the presence of a perpendicular magnetic field Bz induces an electric field Ey which is perpendicular to both. From balance of forces F=qvB and F=qE the field is equal to vB where v is the average velocity of charge carriers. The Hall resistance R is defined to be the ratio of induced field to applied current; R=E/j=vB/Nvq=B/Nq. So the normal Hall resistance is a measure of N, the density of charge carriers in the sample. It is sensitive to material type, impurities, defects, temperature, etc. The quantum Hall effect: If electrons are confined in a thin layer at low temperature in a high magnetic field the Hall resistance vs. B rises in a series of steps which are quantised at levels given by R=h/ie2 where i is an integer.

quantum Hall continued The quantum Hall effect was discovered by von Klitzing in 1980 (Nobel prize 1985). It was totally unexpected and initially unexplained. A partial explanation: A perpendicular magnetic field splits the states in a 2D electron gas into “Landau levels”. The number of current carrying states in each level is eB/h. The position of the Fermi level relative to the Landau levels changes with B. So the number of charge carriers is equal to the number of filled Landau levels, i, times eB/h. Plugging this into the classical expression for R gives R=h/ie2. From our point of view the important features are; The resistance can be precisely measured ( 1 in 108 ). It is simply related to fundamental constants. R=h/ie2, a= e2/2e0hc so R=1/2ie0ca. Both e0 and c are constants without errors: e0= 1/m0c2, m0=4p . 10-7 (NA-2) and c=299,792,458 (ms-1) by definition. The measurement is done at very low energy so higher order corrections are negligible. B.I.Halperin. Scientific American 1986. D.R.Leadley http://www.warwick.ac.uk/~phsbm/qhe.htm

GF from the muon lifetime. Before electro-weak unification, muon decay ( and all other weak interactions ) were described by an effective 4-fermion coupling of strength GF. We now know that weak decays take place via an intermediate W boson like this : However, the Fermi constant GF is still one of the most accurately measurable parts of the standard model. So it is taken as one of the primary constants. It is related to the other “more fundamental” constants by GF=sqrt(2).g2/8.mW2 . Note g is related to other electroweak parameters through sinqw=e/g and cosqw=mW/mZ. The muon lifetime is calculated to be 1/tm= GF2 mm5/192p3 + h.o.c.

The muon lifetime experiment An accurate measurement of the muon lifetime was made back in 1984 at Saclay, Paris. Not improved since. Because GF is accurate enough for present tests of SM? Method: A muon source is built up by stopping 140 MeV/c p+ ( and some m+ ) during a 3 ms burst of the linac. A positron from muon decay is detected as a triple coincidence in one of the 6 scintillator telescopes. Some time after the end of each burst a clock is started and the time distribution of positron signals is recorded.

How do you stop a 140 MeV/c pion or muon ? Some questions How do you stop a 140 MeV/c pion or muon ? Any charged particle passing through matter loses energy by ionisation or excitation of some of the atoms that it passes. If it hits a thick enough piece of matter it will slow down and stop unless something else happens first (e.g.. nuclear interaction, bremsstrahlung or the particle itself decays). The energy loss rate, -dE/dx, of a pion of bg=1 in aluminium is about 2.5 (MeV g-1 cm2) and is rapidly rising as the momentum falls. The energy loss in a thickness t of the sulphur target will be more than; DE = t (cm) . 2.5 (MeV g-1cm2) . rsulphur(g cm-3) Sulphur has a density of 2.1 so we can say that 11 cm will stop the pions, probably half that is enough. ( K.E. = sqrt(p2+m2)-m = 58 MeV ) What happens when they stop? 99.99% decay p  mnn with a lifetime of 26 ns. Where do the muons from pion decay go ? In 2 body decay of M, the energy of decay products in the rest frame of M is E1 = (M2 – m22 + m12)/2M. mp =139.6 MeV, mm = 105.7 MeV so Em = 109.8 MeV. They will stop within a cm or so from where the pion decayed.

Possible sources of error Backgrounds. Cosmic and Beam-induced: The arrival time of cosmic rays will be random so they will simply contribute a flat background which is easy to subtract. The cosmic background level was measured by taking data at long times after the linac burst. The beam particles can create other states with short lifetimes. Their decays may trigger the telescope and mimic a positron from muon decay. They will be correlated with beam bursts in the same way as the muons, so are a potentially a serious bias to the muon lifetime. The paper states that there was only one significant background of this type ( X-rays from thermal neutron capture ) and that it was fitted with a single exponential with a time constant of 160 ms. Random coincidences: If two decays occur within the resolving time of the electronics then only one of them will be seen. This is more likely to happen when decays are happening frequently – it is proportional to instantaneous rate. It is taken into account in the fit: R(t) = R0 [ exp(-lt) + A r exp(-2lt) + B ] l = muon lifetime, r=rate, A=coincidence effect, B=background.

Errors due to muon polarisation The muons which are produced from pion decay in the target are unpolarised, but some muons reach the target after pion decay in flight and these ones are partially polarised (Why?). The direction of the positron from muon decay is correlated with the direction of the muon spin, so if the telescope efficiency is not isotropic a change of spin direction leads to a change of counting rate and therefore a lifetime error. This effect is minimised by Using sulphur for the target which rapidly depolarises the muons (Why?). Some muons will stop in other materials so this does not entirely eliminate the problem. Reducing the magnetic field in the target region with Helmholtz coils to 0.1 G ( ie nearly cancelling Earth’s field ). It is the change of spin direction by precession that is harmful. Measuring the lifetime in each of the 6 telescopes independently and checking that all give the same result.

Lifetime result The result is tm+= 2197.078  0.073 ns. This is converted into GF including corrections of up to (me / mm )8 and (mm / mW )2 and a2. It gives, together with a similar measurements of tm- , the result GF/(c)3 = 1.16637(1)  10-5 GeV-2

mZ from LEP lineshape scan Dr Loebinger has already told how the mass of the Z boson was measured very accurately at LEP-1 by scanning the energy of the accelerator across the resonance and measuring the cross section at each point. The result is mZ=91187.5  2.1 MeV. The uncertainty is mainly 1.7 MeV due to uncertainty of the beam energy and common to all LEP experiments. There is also a common 0.3 MeV due to theoretical uncertainty of initial state radiation.