Natural Units APPENDIX A [.I [MI = [LJ- = [TI-. (A4. [El = [MI = [PI = [kl, (A.1) 1 e2 1 47rhc 137

Transcription

1 APPENDICES

2 GAUGE FIELD THEORIES MIKE GUIDRY WILEY-VCH Verlag GmbH & Co. APPENDIX A Natural Units The use of h=c= 1 units (natural units) can simplify particle physics notation considerably. Since one typically deals with particles that are both relativistic and quantum mechanical, a multitude of h s and CIS will encumber the equations if natural units are not adopted. Let us consider a few examples of how this works. Set h=c= 1; since [.I = (LI[TI-I (where the symbol [ ] means the dimension of ), we have [L] = [TI, and since E2 = p2 + M2c4 we find [El = [MI = [PI = [kl, where p = hk. Because [h] = [M][LI2[T]-, setting ti= c= 1 yields (A.1) [MI = [LJ- = [TI-. (A4 Hence [MI can be chosen as the single independent dimension of our set of natural units. For electromagnetic interactions it is convenient to introduce a set of units in which the MKS constants 0 and po are set to unity, and factors of 47r are expunged from the fields, appearing in the forces instead. This is done in the rationalized Heaviside-Lorentz system of units, which is common in high energy physics. Further, with natural units c = 1, so the c s would not appear in the equations of electrodynamics. The Maxwell equations in these units are given by eqns. (2.112). The fine structure constant is a dimensionless ratio of the electrostatic repulsion between two electrons separated by one Compton wavelength and the electron rest mass: 1 e2 1 47rhc 137 (1/137 corresponds to the asymptotic value of the running coupling constant; see ). Therefore, e = (47r~)~ ~ in natural units. 511

3 512 Appendix A Example: The pion Compton wavelength in natural units is n 1 A, = --t - 21 (140 MeV)-' (-4.3) Mxc Mx We may convert this to conventional units by multiplying with a combination of Ti and c to give a distance unit. Since iic = MeV- fm (with 1 frn e 1 x cm), in natural units we have 1fm=- MeV-' = GeV-' fm-' = MeV 1 GeV = fm-'. Hence (197.3 MeV.fm) = 1.41 fm. For cross sections we have the dimensions [u] = [LI2 = [ My, and from (A.4), 1 fm2 = 25.7 GeV-2 in natural units. But 1 b = 100 fm2 and 1 mb = 0.1 fm2, so cm2 = 1 GeV-2 = mb 1 mb = GeT2 1 fm2 = 10 mb. Example: A typical hadronic cross section is of order u _N AT 2 1 T -- MeV-' M: - (140)2 If u is to be in fm2, we must multiply by a combination of ti and c with units MeV2. fm2. This is just h2c2 = ( 197.3)2 MeV2. fm2 and ii2c2 N 2 fm2 N 20 mb. = (140)2 MeV2 Example: In natural units the mean lifetime for the decay Co 3 A + y is (Sakurai, 1967, 84.1) TN MA + M c)~ e2 E," Since [MI = [El, this has dimension [MI-' and we must multiply the right side by ti = 6.58 x MeV. sec to make it dimensionally correct. Using

4 Natural Units 513 experimental values (MA + ME) = 2307 MeV, E7 = 74.5 MeV, and n/e2 = 1/40 = 137/4, 7-=- 137 (2307 MeV)2 (6.58 x 4 (74.5 MeV)3 MeVasec) x 1O-l sec. It is important to know the dimensions of various field operators in these natural units. Assume d = 4 spacetime dimensions; the Lagrangian has the dimension of mass [Lagrangian] = [MI, (A.5) and the Lagrangian density then has units The Hamiltonian and Hamiltonian density have the same dimensions as the Lagrangian and Lagrangian density, respectively, and the action is dimensionless in natural units. From the free field equations (see $2.3), we can then infer the dimensions of various fields. For a spinor field for scalar fields 141 = [MI, and for the photon and massive vector fields Likewise, for derivative or covariant derivative operators [a ] = [D ] = [MI. (A.lO) Using these dimensions and (A.6), we can find the dimension of any coupling constant appearing in the interaction Lagrangian density (see Exercise 6.4). Example: Consider the Skyrme Lagrangian density (13.94), for which E~ is dimensionless and [fr] = [MI. From the above considerations, [o] = [MI (scalar field) [%I = [MI. Therefore U and Ut are dimensionless, [L,] = [MI, and [k] = [MI4, as befits a Lagrangian density. The preceding equations assume spacetime has d = 4 dimensions. Examples where d # 4 (see ) are considered in Exercises

5 514 Appendix A In cosmology one often sets fi = c = kg = 1, where kg is the Boltzmann constant. Then, 1 GeV = 1.2 x 1013 K, (A.11) and the gravitational constant G is (A.12) where the Planck mass is Mp = 1.2 x lo1 GeV (A.13) From (A.4) the corresponding Planck length is 1 Cp = - = 1.6 x MP cm. (A.14) Multiplying by 1/c gives the corresponding Planck time tp = 5.4 x sec, (A.15) and using (A.ll) in (A.13) gives the Planck temperature Tp = 1.4 x K. (A.16)