PHY6200 W07

Chapter 15: Special Relativity


In Taylor, read sections 15.13 - 15.17 for this week.

u = γ[(dx/dt), c(dt/dt)] = γ(v, c)
p = mu = (γmv, γmc).

Example: GZK Limit

High energy collisions of two photons can yield electron-positrons pairs. Due to the cosmic microwave background radiation (black body spectrum with T = 2.725K), this process sets an upper limit to the energy of photons that can freely propagate across galactic distances. Find the threshold energy (in eV) for photons to scatter (inelastically) into an electron-positron pair in the CM frame. Determine the threshold energy for a photon to scatter with a micrwave background photon (peak energy of the spectrum) and produce an electron-positron pair (this is the lab frame).

A positron is the anti-particle of an electron; it has the same mass (mec² = 511keV) but opposite charge. The energy of the scattered particles (electron and positron) is the sum of their rest energies plus kinetic energies, Ef = 2mec² + Te + Tp. The threshold energy is the smallest possible energy for the process to proceed, and this occurs when the kinetic energies are zero, Ef = 2mec². The photon is a massless particle, so it's four-vector is p = (p3, p3), where p3 is the 3-momentum. In the CM frame, the two photons have equal and opposite 3-momenta, so their total energy (fourth component) is 2p3c. Setting this equal to the threshold energy yields the photon energy of Eγ = p3c = mec².

The energy of a photon at the peak of a blackbody spectrum is Eμ = hc/λmax = 2π(hbar)cT/b = 2π(197.3×10-9 eV m) (2.725 K) / (2.898×10-3 m K) = 1.165×10-3eV. To determine the threshold energy for a colliding photon to produce an electron-positron pair, note that in the CM, the sum of the photon four-momenta is P = (0, 0, 0, 2mec;), and its square is an invariant, P² = -4(mec)². For scattering off microwave photons, P = pγ + pμ = (pγ - pμ, 0, 0, pγ + pμ), and equating the squares, we find (pγ - pμ)² - (pγ + pμ)² = -4pγpμ = -4(mec)². Therefore the threshold energy of the photon is Eth = pγc = me²c4 / Eμ = (5.11×105 eV)² / (1.165×10-3eV) = 2.24×1014eV.

Relativistic Dynamics

Up to know everything we've done has been kinematics -- describing motions without explaiing the underlying cause. This second step is dynamics, and this leads us to look at how to modify Newton's laws and the Lagrangian for a correct relativistic description of motion. A complication in relativistic dynamics is that a composite object (a nucleus, an atom, or a composite particle) can have internal degrees of freedom. When a force is applied, the internal energy of the object can be changed, and due to the connection of mass and energy, this causes the rest mass of the object to change. For the purposes of this course, we assume that this doesn't happen, and take the rest mass of an object as a fixed constant.


One of the most important forces in special relativity is the Lorentz force on a charge moving in electric and magnetic fields. Classically, this force is given by

F = q(E + v×B)

This form also holds relativisically if we make one (small) change to Newton's second law. (Note that charge is a relativistic invariant.) We must write

F = dp/dt = (d/dt)(γmv)

where the relativistic form of three-momentum is used. Note that the derivative is a regular time derivative, not a derivative with respect to proper time! This definition of force and Newton's second law isn't manifestly Lorentz invariant -- that is, it isn't written in terms of scalars and four-vectors -- but it does give correct results, and can be used to prove the generalization of the work-KE theorem in relativity (see Sec. 15.15 of Taylor).

Motion with a Constant Force

Find the motion of an object of fixed rest mass m acted on by a constant force F. Such a force occurs for a charge in a constant electric field, F = qE. Assume the object is released from rest at x=0, at t=0, and find its position and velocity for later times.

Start from F = dp/dt, and since the force is constant, both sides can be integrated with respect to time, yielding p = Ft, where I've used the initial condition to see that p(t=0) = 0. From p = γβmc, and the fact that γ²β² + 1 = γ² we have γ = √(1 + (p/mc)²) = √(1 + (Ft/mc)²). Therefore

v(t) = p/mγ = (Ft/m) / √(1 + (Ft/mc)²).

In the non-relativistic limit (c going to infinity, or small values of t), we recover the non-relativistic result v = Ft/m. In the limit of t going to infinity (which must give a relativistic result for any finite constant force), we obtain v ≈ (Ft/m)(mc/Ft) = cFhat meaning that the velocity approches the speed of light, in the direction of the force. Careful consideration shows that the speed never quite reaches the speed of light, consistent with the result that the speed of light is an upper limit to the speed of a material particle.

We find the position as a function of time by integrating the velocity yielding

x(t) = ∫ vdt = (F/m) (mc/F)² [√(1 + (Ft/mc)²) - 1].

In the non-relativistic limit, (Ft/mc) small compared to 1, we can expand the square root in a Taylor series to find x(t) ≈ (F/m) (mc/F)² ½(Ft/mc)² = ½(F/m)t², in agreement with non-relativisitic physics.


The three-force is the force that can be measured by test masses or particles at rest in the frame of reference. That is, it is the force we are used to dealing with in mechanics, and can be calculated from a potential energy in the normal way. But it is only valid in that frame of reference; it isn't a four-vector and doesn't transform following the Lorentz transformation. (It is curious to note that the Lorentz transformation comes from Maxwell's equations, and, as mentioned in the previous example, a constant electric field will produce a three-force, but this three-force doesn't Lorentz transform! Read the last two sections of chapter 15 to see how this is fixed.)

There can be cases where it is useful to have a force that Lorentz transforms. The definition of the three-force can be adapted to the relativisitic world by defining it as the derivative of the four-momentum by the proper time, τ, a scalar:

K = dp/dτ = (1/γ)[dp/dt, dE/cdt] = (1/γ)[F, vF/c]

We won't make use of the four-force here.

Relativistic Lagrangian

The Lagrangian is very useful in classical mechanics. It is also useful in relativistic mechanics (especially field theory), but has some funny properties. Namely, although the Lagrangian is a single valued function, it isn't manifestly Lorentz invariant, and isn't a relativistic scalar! This arises due to the definition of the Lagrangian as the function integrated over the path of the motion to give the action:

S = ∫ L dt

Since the action is the quantity that is minimized for the correct path, it should be relativistically invariant -- we want a process to proceed by one path. The Lagrangian must be integrated over a variable that parameterizes the path, so time is an obvious choice. But then, since t is not relativistically invariant, the Lagrangian can't be either, in fact, it must be invariant in a way opposite to t, so that the combination is invariant. (Why not integrate with repect to proper time? Well ...)

What about the definition of the Lagrangian, classically, L = T - U. It is rather obvious though that it must be modified for use in relativisitic mechanics. We start this process by seeking the relativisitic analogy of the kinetic energy term in L to get pi = ∂L/∂vi = γmvi = mvi / √(1 - v²/c²). Remember that v² = ∑vi². This can be integrated to give

L = -mc²√(1 - β²) - U(x) + C

where C is a constant of integration, and I've shown the potential energy with explicit dependence on position. Note: when taking the partial derivative, β must be replaced by v/c!

This is clearly not the same as T - U, even allowing for the relativisitic form of kinetic energy, T = (γ - 1)mc².

It is interesting to see what the Hamiltonian is for this Lagrangian. Recall the definition of the Hamiltonian, H = ∑ qidot pi - L. Using vi for qidot the Hamiltonian for a single particle becomes

H = ∑ vipi - L = ∑ (pi²/γm) + mc²/γ + U = (p²c² + m²c4)/(γmc²) + U = Ep²/Ep + U = Ep + U,

where Ep is the relativistic energy of a particle. So this form for the Lagrangian results in the Hamiltonian being the total energy of a particle. This is a satisfying result, since the Hamiltonian is the total energy of a system in the non-relativistic case.

© 2007 Robert Harr