Week 1: “Natural” Dimensions (Graduate)

In particle and nuclear physics, ﬁnd it choosing energy as the basic dimension and deﬁning the speed of light and Planck’s constant c =h ¯ ≡ 1 and dimensionless is often convenient Applying these more or less arbitary choices1 to physical laws determine a system of dimensions. This system is typically referred to as “natural.”2 Designate the distance dimension L and the time dimension T . Then, for light (photon) motion, L and T are related by L = cT . Since c is dimensionless, L and T have the same natural dimension. The photon has energy, E = ¯hω = 2π¯hc/λ, where ω is the angular velocity, which has dimension 1/T , and λ is the wavelength, dimension L. From these, T = L = 1/[E], that is, the natural dimension of both time and distance is inverse energy.3 Since velocity is the ratio of a distance and a time interval, velocity must be (as c is) dimensionless, and always less than or equal to c. In general, nuclear and particle physicists report velocity as a fraction of c, or β = v/c, from Special Relativity. But because c = 1 and dimensionless, β = v. Because the ratio p/E = β = v is dimensionless, momentum’s natural dimension is also energy. Because E = mc2, the natural dimension of mass is energy, as well. Since the (integral of) the (dot- )product between force and distance, R F · ds, gives (potential or kinetic) energy, force has the natural dimension of energy-squared. And so forth.

1. What is the natural dimension of acceleration?

2. What is the natural dimension of cross-section? 3. What is the natural dimension of charge?

Theoretical nuclear and particle physicists employ only natural dimensions in their calculations. Phenomenologists and experimentalists need to make con- tact with the everyday world and so often employ “hybrid” units, in which most quantities that in natural dimensions have dimension of energy (e.g., mass and momentum) retain an energy dimension, but those that have dimensions not equal to energy to the ﬁrst power (e.g., time and length) have everyday

1c and h are physical constants that characterize Special Relativity and Quantum Mechan- ics, and energy is arguably the most important quantity computed and measured in particle and nuclear physics 2Not much work is necessary to make objects of very small mass go very fast. For example, 500 kV (the voltage in some U.S. transmission lines) will accelerate an electron to 10% of light speed. Considering that 1. the typical kinematics of elementary particles is relativistic, 2. zero velocity is arbitrary (reference frame-dependent), and 3. light speed, c, is measured to have the same magnitude in all reference frames, designating light speed as the reference speed is a natural choice. Even more convenient is to measure speed relative to light speed, so that light speed has the dimensionless value 1 and all other speeds are dimensionless fractions between 0 and 1. 3High-energy projectiles probe small distances and short time scales.

1 dimensions. Hybrid systems retain the dimensionless value of one forh ¯ and c in expressions that might involve these in everyday dimensions, but hybrid systems employ mixed (hybrid) units of these parameters to translate other quantities from natural to everyday units. In everyday dimension systems, ¯h has dimensions energy × time, [E] × T , and c has dimensions length / time, L/T . The producthc ¯ then has dimensions energy × length, [E] × L. The natural unit of energy is the electronvolt: 1 eV = 1.6 × 10−19 J (recall that Joule = kg-m2/s2). In hybrid units ¯h ≈ 6.6 × 10−16 ev-s, c ≈ 3 × 108 m/s, and ¯hc ≈ 2 × 10−7 eV-m). With these, natural dimensions and units can be converted into everyday dimensions and units, and vice versa. In everyday units, the (charge) radius of a proton is approximately rp ≈ 0.8 × 10−15 m (or about 1 fm–pronounced “fermi”). In natural units, this is −9 −1 −1 rp/¯hc ≈ 4 × 10 eV or 4 GeV . This implies that investigating the internal structure of the proton requires an energy scale greater than 108 MeV, or roughly 1 GeV, which happens to be −27 approximately the mass of the proton in natural units: mp ≈ 1.7 × 10 kg 2 −27 16 −10 −10 −19 9 ⇒ mpc ≈ 1.7×10 ×9×10 ≈ 1.5×10 J ⇒ 1.5×10 /1.6×10 ≈ 10 eV = 1 GeV. CERN’s Large Hadron Collider (LHC) operates at a center-of-mass energy of 13 TeV = 13 × 1012, which suggests that, in principle, new particles of mass ∼3 TeV could be produced, if they existed, and structures of linear size of about 3 × 10−11 eV−1 could be investigated. That is, masses of 3 × 1012 × 1.6 × −19 16 −24 10 /9 × 10 ≈ 5 × 10 kg (about 5000 × mp) could be created, or distances −11 −7 −18 of 3 × 10 × 2 × 10 = 6 × 10 m (about rp/1000) could be explored. 4. Show how to convert momentum p in everyday units into momentum p in natural units? 5. In natural dimensions, an object’s kinetic energy is K = E − m. Convert this equation into everyday dimensions. 6. The everyday dimensions of the universal gravitational constant, G, are length-cubed / (mass times time-squared), L3/(MT2). What is its natural dimension? Its value in Standard Interna- tional (SI) units is 6.674 × 10-11 m3/ (kg-s2). What is its value in natural units? 7. In SI units, Coulomb’s Law is

1 Q1Q2 F = 2 4πε0 r

(a) Argue that the ratio Q1Q2/ε0 is dimensionless in natural dimensions. (b) Recall that c2 = 1 and that c ≡ 1 and dimensionless in ε0µ0 natural dimensions. Then, ε0 = µ0 = 1 and dimensionless. Argue that charge is dimensionless in natural dimensions.

2 (c) Recall that the ﬁne structure constant,

e2 1 α = = 4πε0¯hc 137 in all systems of units. Calculate the charge of the proton, e, in natural units.

8. Consider two nearly touching protons. Compare the strengths of their gravitational and Coulomb interactions. Such a comparison is especially straightforward in natural units. 9. The Standard Model of particle physics integrates three of the four known interaction–strong, electromagnetic, and weak–but ignores the gravitational. Explain why ignoring gravity doesn’t impact the model’s predictions. 10. Recall that the number of scattering interactions in a thin target, Ni = σNbntz, where Nb is the number of projectiles, nt is the number of target entities per unit volume, z is the thickness, and σ is the cross section (effective scattering area of the target entities). The number of particles, Nf , that make it through the target unscattered then is Nf = Nb − Ni. In a direct total cross section measurement, the number in and the number out are counted to determine the number scattered, Ni = Nb − Nf . For scattering through a not-so-thin target, it is necessary to integrate over a differential thickness, dz. With each differential thickness traversed, the number of unscattered particles, N, decreases, and so Nf at one thickness becomes Nb at the next, and so forth. For such an experiment, then, with a not-so-thin target, the differential number of particles scattered at any given differential thickness will be dNi = −dN = σNntdz. By integrat- ing dN from an initial Nb to a final Nf over the entire thickness from 0 to Z, find an expression for the number of particles that go unscattered, Nf . In terms of this result and the total number of projectiles, Nb, find an expression for the total number scattered, Ni.