Lecture Three: Theory of Relativity

3.1 Relativistic Kinematics

In the previous lecture I introduced the four-velocity of a particle U. Now, let us suppose that the particle has a certain property, its rest-mass m_{0}, which we use to define a fourmomentum, P:

\begin{aligned}P & = & m_{0}U = m_{0}(γc, γu) \\ & = & (mc, p). \end{aligned}

The components of the four-momentum have the following interpretation:

m = γ(u)m_{0} ← relativistic mass
p = mu  ← relativistic three − momentum

In analogy with conservation of momentum in Newtonian mechanics, we can introduce a new principle into Special Relativity, namely that four-momentum is conserved in interactions.

Suppose that some particles go into a little box, where they experience some sort of interaction. Particles (not necessarily the same ones) come out of the box leaving nothing behind. Conservation of four-momentum requires

∑_{i} P_{i}^{in} = ∑_{i} P_{i}^{out},

where the sums extend over all particles. Each component of the four-momentum must satisfy this equation separately. Thus the relativistic masses before and after must balance, as must the relativistic three-momenta:

∑_{i} m_{i}^{in} = ∑_{i} m_{i}^{out},  (1)
∑_{i} p_{i}^{in} = ∑_{i} p_{i}^{out}.  (2)

Now let us look at the first of these equations. It expresses the conservation of the quantity

m = m_{0}(1 − u^{2}/c^{2})^{−1/2}.

For speeds much smaller than the speed of light, u ≪ c, we can expand the term in brackets

m ≈ m_{0} + \frac{1}{2}m_{0}u^{2}/c^{2} + . . . . . .

↗

classical kinetic energy Evidently, our conservation law is equivalent to the conservation of energy in the nonrelativistic limit. This suggests that we identify mc^{2} with the energy of a particle leading to the most famous equation in physics:

E = mc^{2}

Furthermore, this identification tells us that a particle has a rest energy

E_{0} = m_{0}c^{2}.  (3)

Equation (1) expresses the conservation of energy in special relativity. The conclusion that particles have a rest energy (equation 3) is profound and follows from equation (1) since rest energy can be converted into kinetic energy and vice-versa (depending on what happens in the interaction). This conclusion follows from the requirement that the laws of nature have a tensorial form. (We generalized the principle of conservation of momentum to apply to the four-momentum P).

The identification E = mc^{2} allows us to write the four-momentum as

P = (E/c, p)

and note that P^{2} is invariant

P^{2} = E^{2}/c^{2} − p^{2} = m^{2}c^{2} − p^{2},

or equivalently

E^{2} = p^{2}c^{2} + m_{0}^{2}c^{4}.  (4)

(Think about how to get equation (4) from the previous equation without having to substitute for the relativistic mass and three-momentum.)

3.2 Photons and Planck’s Hypothesis

Equation (4) applies to particles with a rest mass m_{0}. But what if m_{0} = 0? Equation (4) tells us that the energy of the particle is

E = pc.

Suppose, as before, that we have two inertial frames S and S′, with S′ moving at speed v relative to S as in Figure 1.

Let the energy of a zero mass particle travelling in the negative x′-direction in S′ be E_{0}, then, transforming the components of the four-momentum to S (remembering that they transform just like the coordinates)

E = γ(E_{0} + vp′_{x})

Figure 1:

But, according to equation (4), for a zero mass particle p′_{x} = −E_{0}/c (the minus sign comes about because the particle is travelling in the negative x direction), so

E = γE_{0}(1 − v/c) = E_{0} \begin{matrix}(1 − v/c) & 1/2 \\ (1 + v/c) & 1/2 \end{matrix} .  (5)

Now, if we identify the zero mass particle with light of frequency ν_{0} in S′, then we know that its frequency in S will differ because of the relativistic radial Doppler effect,

\frac{ν_{0}}{ν} = (\frac{1 + v/c}{1 − v/c})^{1/2} .  (6)

Comparing equations (5) and (6) we find the startling result that

\frac{E}{E_{0}} = \frac{ν}{ν_{0}} .

In other words, the energy of the particle is proportional to its frequency,

E = hν,

with a constant of proportionality h that we now know as Planck’s constant.

As I am sure many of you know, Einstein published a quantum theory of the photo-electric effect in 1905, the same year in which he published his fundamental paper on Special Relativity. In explaining the photo-electric effect, Einstein introduced the notion of the photon, a particle with zero mass and energy given by Planck’s formula E = hν. In fact, Einstein was awarded the Nobel prize for his explanation of the photo-electric effect, not for developing the theory of relativity.

In various books that I have read, Einstein’s introduction of the photon and Special Relativity are portrayed as completely disconnected theories. Yet, the arguments given in this section show that Special Relativity requires Planck’s hypothesis. My guess is that Einstein must have known about this, and that it gave him added confidence in his concept of the photon. Special Relativity may have played a key role in Einstein’s Nobel prize after all.

3.3 Two examples of kinematic calculations

In my experience, students often have trouble in doing kinematic calculations in Special Relativity. In fact most of these calculations are quite easy. The key principle is to choose inertial frames in which the problem is simplified and in which invariant quantities take obvious values. This principle saves lots of time and algebra. If you follow it, you will find that kinematic problems can be done very easily. I will illustrate what I mean by working through two examples:

3.3.1 Example I: Threshold Energies

P_{1}  P_{2}
• − − − − − − − ⟶ •

↑

target particle at rest

Consider the collision of two particles to produce a third. Particle 1 (with four-momentum P_{1}) is moving towards particle 2 (four-momentum P_{2}) which is at rest as shown in the Figure above. Let the post-collision momentum be P. Conservation of four-momentum requires that

P_{1} + P_{2} = P.

Now square this equation,

P_{1}^{2} + P_{2}^{2} + 2P_{1} · P_{2} = P^{2}.

Now we can easily write down what P_{1}^{2} is,

P_{1}^{2} = m^{2}c^{2} − p^{2} = m_{0_{1}}^{2}c^{2},

where m_{0_{1}} is the rest mass of particle 1. You see, it doesn’t matter that the particle is moving, P_{1}^{2} is an invariant and so we can evaluate it in the particles rest frame. Likewise,

P_{2}^{2} = m_{0_{2}}^{2}c^{2}.

The dot product P_{1} · P_{2} can be written

P_{1} · P_{2} = c^{2}m_{0_{1}}m_{2} = c^{2}m_{1}m_{0_{2}}

= c^{2}γ(v)m_{0_{1}}m_{0_{2}} where v is the relative speed between the two particles. So, we can write the post-collision momentum as

\begin{aligned}P^{2} & = & P_{1}^{2} + P_{2}^{2} + 2P_{1} · P_{2} \\ & = & m_{0_{1}}^{2} + m_{0_{2}}^{2} + 2m_{0_{1}}m_{0_{2}}γ(v). \end{aligned}

Let us now pose a specific problem. Suppose, for example, that we have two protons that interact to produce a pion:

p + p → p + p + π^{0}

what is the threshold energy for the production of a pion? In this example, the relevant rest masses are

\begin{aligned}m_{0_{1}} & = & m_{p} & = & M, \\ m_{0_{2}} & = & m_{p} & = & M, \\ m_{0_{3}} & = & m_{π} & = & m. \end{aligned}

As derived above, conservation of four-momentum requires

M^{2} + M^{2} + 2Mmγ(v) = m̄_{cm}^{2},

where m̄_{cm}^{2} is the sum of the post-collision relativistic masses in the post-collision center of momentum frame (i.e. the frame in which the post-collision four-momentum is P = (m̄_{cm}c, 0). Now at the threshold for the production of a pion

m̄_{cm} = (M + M + m),

(the three post-collision particles are at rest with respect to each other). Hence we can calculate the threshold value of γ for the production of a pion,

M^{2} + M^{2} + 2M^{2}γ(v) = 4M^{2} + 4Mm + m^{2},

i.e.

γ(v) ≥ 1 + \frac{2m}{M} + \frac{m^{2}}{2M^{2}} .

If we use a particle accelerator to accelerate the proton towards a stationary target, we therefore need to supply an energy of

E_{Accelerator} = γMc^{2} − Mc^{2}
= (γ − 1)Mc^{2}

with γ set to the threshold value if we are to create a particle of mass m. If we want to create a very massive particle m ≫ M, we need to supply an energy

E_{Accelerator} ≥ (\frac{m}{2M} ) mc^{2},

i.e. much greater than the rest mass energy of the particle that we want to create. This is why particle physicists have developed colliders which accelarate oppositely charged particles (e.g. e^{+} and e^{−}) towards each other.

Figure 2: The upper figure shows how the collision appears in the laboratory frame. The lower figure shows the collision in the centre-of-momentum frame.

3.3.2 Example II: Elastic collisions

In our second example, we will consider an elastic collision between two particles of equal mass, one of which is at rest (see Figure 2).

Our goal is to relate the two angles θ and ϕ of the scattered particles in the laboratory frame to the speed V of the incoming particle.

Now the problem looks much more symmetrical in the centre-of-momentum (com) frame. In the com frame, the particles (each moving at speed v) approach each other and scatter at some angle θ′. Their post collision speed is v, because the collision is assumed to be elastic. To relate the situation in the laboratory frame to the simpler situation in the com frame, we must apply a Lorentz transformation between the two frames.

Let us first look at the following simple problem. A particle is moving with a velocity u in the x − y plane in the inertial frame S. What is its velocity in frame S′ moving at speed v relative to S in the x-direction (see Figure 3). We have already solved this problem. The components of u′ in S′ are given by the velocity addition formulae:

u′_{x} = \frac{u_{x} − v}{(1 − u_{x}v/c^{2})},
u′_{y} = \frac{u_{y}}{γ(v) (1 − u_{x}v/c^{2})}.

Figure 3:

The angle α′ of the velocity u′ with respect to the x′ axis in frame S′, is therefore given by

tan α′ = \frac{u′_{y}}{u′_{x}} = \frac{1}{γ(v)} \frac{sin α}{(cos α − v/u) .}  (1)

This is sometimes called the particle abberation formula.

How is this calculation related to our problem of elastic collisions? Identify S′ with the laboratory frame and S with the com frame. We now want the inverse of the particle aberation formula to relate the angles θ and θ′ (replace v by −v in equation (1)),

tan θ = \frac{sin θ′}{γ(v) [cos θ′ + 1]}

Likewise for the angle ϕ′,

tan ϕ = \frac{sin ϕ′}{γ(v) [cos ϕ′ + 1]}.

But since in the com frame ϕ′ = π − θ′ (see Figure 2), the second of these equations becomes

tan ϕ = \frac{sin θ′}{γ(v) [− cos θ′ + 1]}.

Combining the equations for tan θ and tan ϕ gives,

tan θ tan ϕ = \frac{sin^{2} θ′}{γ^{2}(v) [1 − cos^{2} θ′]}  = \frac{1}{γ^{2}(v)}

= \frac{2}{γ(V) + 1}.

The last line comes from the γ relations in the velocity addition law. Thus, we have solved our problem and derived a relation between the scattering angles θ and ϕ that must apply if the collision is elastic (and so can be used to check whether the collision is indeed elastic).

3.4 The three-force and four-force

In non-relativistic mechanics, Newton’s second law is

f = ma = \frac{dp}{dt}.

This suggests that in Special Relativity we write

F = m_{0} \frac{dU}{dτ} = \frac{dP}{dτ}

defining a four-force F. Now since the four-momentum is

P = m_{0}(γc, γu)

we can write the equation of motion of a particle as

F = \frac{dP}{dτ} = γ \frac{dP}{dt} = γ (c\frac{dm}{dt}, \frac{dp}{dt})
= γ (\frac{1}{c} \frac{dE}{dt} , \frac{dp}{dt})
= γ (\frac{1}{c} \frac{dE}{dt} , f) .

where f is the relativistic three-force. Now, from our definitions of the four-force and four-velocity:

Four − force  F = γ (c\frac{dm}{dt}, f)
Four − velocity U = γ (c, u)

we can form the dot product,

U · F = g_{μν}U^{μ}F^{ν}
= γ^{2}c^{2} \frac{dm}{dt} − γ^{2}f · u.

U · F is an invariant under any coordinate transformation and so we can evaluate it in the instantaneous rest frame of the particle (in which m = m_{0} and u = 0). Thus,

U · F = c^{2} \frac{dm_{o}}{dτ} .

We see that in Special Relativity, the action of a force can alter the rest mass of a particle. A force that preserves the rest mass of the particle must have

U · F = 0.

Such forces are called pure forces.

So far, our discussion of Special Relativity has been entirely kinematical. To discuss forces we need dynamics, and this requires new physics. At the time that Special Relativity was invented only two forces were known, electromagnetism and gravity. Electromagnetism, as I mentioned in the Introduction, actually led to the development of Special Relativity. I will therefore discuss electromagnetism in some detail in the next lecture. That will complete our discussion of Special Relativity and we will then turn our attention to gravity. We will see that understanding gravity requires a generalization of the ideas of relativity theory, culminating in the full dynamical theory of General Relativity.