Chapter 2: Relativistic Mechanics Introduction We know that Newton’s laws of motion hold and are invariant in inertial frames of reference that are related by the Galilean transformation. But it is the Lorentz trans- formation rather than the Galilean transformation that is correct. That means that the laws of classical mechanics cannot be correct, and we must find new laws of relativistic mechanics. In doing so, we will be guided by 3 principles: 1. A correct relativistic law must hold in all inertial frames, i.e., it must be invariant under the Lorentz transformation. 2. Relativistic definitions and laws must reduce to their nonrelativistic counterparts when applied to systems moving much slower than the speed of light. 3. Our relativistic laws must agree with experiment. Mass How do we define mass m? We know that at slow speeds, mass can be defined classically by m = F/a. In relativity we will measure the mass m of an object in its rest frame. This is called the rest mass or the proper mass. Observers in different rest frames could measure the mass of an object by bringing the object to rest in their frame of reference, and then measuring its mass. The mass they would measure would be the same in all inertial reference frames. Relativisitic Momentum Classically, momentum is defined by ~p = m~u (1) where m and ~u the mass and velocity of an object. What is the correct relativistic definition of momentum? We have the freedom to define momentum any way we like, but to be useful, we would like to define it such that total momentum is conserved in all inertial frames. Recall the law of momentum conservation. If there are n bodies with momenta ~p1, ..., ~pn, then, in the absence of external forces, the total momentum is given by ~p = ~p1 + ... + ~pn (2) X cannot change. We would like to define momentum relativistically to preserve conserva- tion of momentum. As your book describes in section 2.3, we can’t use the classical definition Eq. (1) because of the way velocity transforms between reference frames. For example, the way ′ uy transforms in going from S to S depends on ux. The classical definition of momentum is d~r ~p = m~u = m (3) dt u Classical c Relativistic p Figure 1: Velocity versus momentum. The problem is that both ~r and t are subject to the Lorentz transformation and that makes things messy. The relativistic way to define momentum is d~r ~p = m (4) dt0 where t0 is the proper time, i.e., the time measured in the rest frame of the particle or object. Notice that with this definition, dy py = m (5) dt0 which is invariant when one goes between reference frames that are moving relative to ′ one another along the x−axis. That is because t0 is invariant and y = y . Similarly for pz. 2 2 Since dt = γdt0 where γ =1/ 1 − u /c , we can write q m~u ~p = = γm~u (6) 1 − u2/c2 q A constant force changes the momentum and hence the velocity. Notice that as u → c, γ and hence p increase without limit, but the speed u never reaches the speed of light (see plot of u versus p in Fig 2.2 of your book). Thus no object can go faster than the speed of light. We can define a relativistic mass M(u) by M M(u)= = Mγ (7) 1 − u2/c2 q 2 Your book calls this a variable mass mvar, but it doesn’t want to use this. However, it does make it easy to write the momentum in the traditional form ~p = M(u)~u (8) Notice from Eq. (7) that if the speed u of an object with nonzero mass is equal to the speed of light, the relativistic mass is infinity. If the object’s speed exceeds the speed of light, then the relativistic mass M(u) is imaginary. Clearly, this is absurd and so, once again, the velocity of an object cannot exceed the speed of light. If the object has mass, it cannot travel at the speed of light; its speed must be less than the speed of light. Relativistic Energy Now we need to define a relativistic energy. We will use 2 criteria: 1. When applied to slowly moving bodies, the new definition reduces to the classical definition. 2. The total energy E of an isolated system should be conserved in all reference frames. P The relativistic energy that satisfies these requirements turns out to be mc2 E = = γmc2 (9) 1 − u2/c2 q This applies to any single body, no matter how big or how small. Notice that the units are right. (It’s always important to check units.) γ is dimensionless, and mc2 has units of energy. (Recall that kinetic energy is (1/2)mv2.) Now let’s take the nonrelativistic (slow) limit u ≪ c. Then −1 2 u2 / 1 u2 γ = 1 − 2 ≈ 1+ 2 (10) c ! 2 c Therefore, when u ≪ c, 2 2 1 u 2 2 1 2 E = γmc ≈ 1+ 2 mc = mc + mu (11) 2 c ! 2 The first term is a constant independent of the speed u, and we can always add a constant to the energy since we can set the zero of the energy (E = 0) anywhere. The second term is just the usual classical expression for the kinetic energy. So our definition of the relativistic energy satisfies the criterion of reducing to the classical kinetic energy in the nonrelativistic (slow) limit. Now let’s look at that constant mc2. If the object is at rest, then u = 0, γ = 1, and Eq. (9) reduces to the famous equation E = mc2 (12) 3 This is called the rest energy of the mass m. If we could convert a mass m completely into energy, then this is the amount of energy we would get. It’s a lot of energy because the speed of light c is so large, and c2 is even larger. For example, the rest energy of a 1 kg lump of metal is E = mc2 = (1 kg) × (3 × 108 m/s)2 =9 × 1016 joules (13) which is about the energy generated by a large power plant in one year. E = mc2 is the basic equation behind nuclear power, and atomic bombs. These are powered by nuclear fission in which, typically, uranium 235 nuclei are split apart, converting about 1/1000 of the rest energy into heat. Thus 1 kg of 235U can yield a fantastic amount of heat (9×1013 J). When an object is not at rest, its energy E = γmc2 is the sum of its rest energy mc2 plus its kinetic energy K =(E − mc2): E = mc2 + K (14) where K = E − mc2 =(γ − 1)mc2 (15) 1 2 In the nonrelativistic limit K ≈ 2 mu , the classical kinetic energy. However, in the relativistic limit of u → c, we get something quite different. Because γ can approach infinity, K can approach infinity even though the speed u can never reach the speed of light c. Notice that K, like its classical counterpart, is always positive (K ≥ 0). Two Useful Relations Let’s derive 2 useful relations starting from ~p = γm~u (16) and E = γmc2 (17) If we divide Eq. (16) by Eq. (17), we obtain ~p ~u = (18) E c2 or ~u ~pc β~ = = (19) c E which give the dimensionless velocity β~ = ~u/c in terms of ~p and E. Notice that ~pc has dimensions of energy, so the ratio ~pc/E is dimensionless, just like β~. Eq. (19) is one useful relation. The other is 2 2 E2 =(pc) + mc2 (20) 4 This has the same form as the Pythagorean relation for a right triangle with sides pc and mc2 and hypotenuse E, though there isn’t any deep meaning to this. To show that this is correct, we can plug in E = γmc2 and p = γmv, solve for γ2 and show that this reduces to the definition γ2 = (1 − v2/c2)−1. Units Since we are dealing with small things, we should use appropriate units. For example, if I ask how tall you are, you would tell me so many feet and so many inches. You could give me your height in miles or light years, but those are not appropriate units. Similarly, joules is not an appropriate unit for atomic and subatomic particles like electrons. The commonly used unit is the electron volt or eV. This is defined as the amount of work needed to move an electron (q = −e = −1.60 × 10−19 Coulomb) through a voltage drop of 1 volt (∆V = −1 volt); thus 1eV= q∆V = −1.60 × 10−19 C × (−1 Volt) = 1.60 × 10−19 J (21) Typical atomic energies are on the order of 1 eV or so, while those in nuclear physics are on the order of 106 eV = 1 MeV. Masses of particles are often given in eV with the understanding that they are referring to the rest energy mc2. Techically, the mass is properly in units of eV/c2. So, for example, the “mass” of the electron is given as 0.511 MeV, meaning mc2 = 0.511 MeV. Similarly, when momentum p is given in units of MeV, they really mean the quantity pc, or the units of [p] is eV/c. If you look up a table for the mass of atoms, the mass is often given in atomic mass units (denoted u). The conversion is 1u=1.66 × 10−27 kg (22) Conversion of Mass to Energy The famous equation E = mc2 implies that if you can convert mass into energy, then you would get a lot of energy because the speed of light squared is so big.