Relativistic Dynamics

Learning Objectives¶

By the end of this chapter, you should be able to:

• Explain why the Newtonian definition of momentum, $p = mu$, is not conserved in all inertial frames once the Lorentz transformation replaces the Galilean transformation, and derive the corrected, relativistic definition from a symmetric elastic-collision thought experiment.

• Apply the relativistic expressions for momentum, kinetic energy, and total energy.

• Use the mass–energy relation $E = mc^2$ and the energy–momentum invariant $E^2 = (pc)^2 + (mc^2)^2$.

• Distinguish rest mass, relativistic momentum, kinetic energy, and total energy, and apply these to problems involving massive particles and photons.

• Represent energy and momentum as components of a single four-vector, and use conservation of this four-vector to analyze collisions, decays, and annihilation events.

• Transform to the center-of-momentum frame of a collision and compute threshold energies for particle-production reactions.

• Recognize the low-speed (Newtonian) and high-speed (ultrarelativistic) limits of the relativistic energy and momentum expressions.

• Explain why the combination of momentum blowing up as $u \to c$ and the causality argument of Chapter 2 together rule out faster-than-light material particles.

Introduction¶

Chapter 2 replaced the Galilean transformation with the Lorentz transformation, because only the latter is consistent with the invariance of the speed of light. But momentum and energy in Newtonian mechanics are defined and conserved using Galilean kinematics: $p = mu$, conserved because Newton’s third law and Galilean-invariant forces guarantee it in every Galilean frame. Once the underlying kinematics changes, the old definitions of momentum and energy no longer transform consistently between inertial frames, and a collision that conserves Newtonian momentum in one frame will not, in general, conserve it in another frame related by a Lorentz transformation. This chapter derives the corrected definitions — relativistic momentum and relativistic energy — that are conserved in every inertial frame, assembles them into a single four-component object whose conservation law captures both at once, and works out their most important consequences: mass and energy are, up to a conversion factor $c^2$, the same quantity, and a whole new class of practical calculations — threshold energies for particle production — becomes possible.

3.1Relativistic Momentum¶

Deriving Relativistic Momentum¶

Rather than simply asserting the corrected formula, it is worth seeing exactly where it comes from, because the argument is a direct descendant of the light-clock argument used to derive time dilation in Chapter 2, and it makes clear that the correction is forced on us by the Lorentz transformation, not chosen for convenience.

Consider two identical particles, $A$ and $B$, each of rest mass $m$, in an idealized elastic scattering event. Choose the incoming and outgoing velocities so that the $x$ components are unchanged while each particle’s $y$ component reverses sign. In frame $S$, let $B$’s transverse speed be $w_0$. Let $S'$ move at speed $v$ along the $x$-axis relative to $S$, and choose $S'$ so that $A$ has no $x'$ motion. In $S'$, let $A$’s transverse speed be the same $w_0$. Because $\Delta y = \Delta y'$ while $\Delta t = \gamma(v)\left(\Delta t' + v\Delta x'/c^2\right) = \gamma(v)\Delta t'$ for $A$ with $\Delta x' = 0$, the transverse speed measured in $S$ is reduced by a factor of $\gamma(v)$:

In $S$, $A$ therefore has an $x$ component $v$ and a $y$ component $-w_0/\gamma(v)$ before the bounce, while $B$ has a $y$ component $+w_0$. After the bounce, both $y$ components reverse sign. Writing the momentum of a particle moving at speed $u$ as $p = m(u)\,u$, for some as-yet-undetermined function $m(u)$ that reduces to the particle’s ordinary mass at $u=0$, conservation of $y$ momentum says that the incoming and outgoing transverse momenta are equal. Since the outgoing total is the negative of the incoming total, each must be zero; hence

where $u(A)$ and $u(B)$ denote the total (not just transverse) speeds of $A$ and $B$ in frame $S$. Substituting the magnitudes $|u_y(A)| = w_0/\gamma(v)$ and $|u_y(B)| = w_0$, and cancelling the common factor of $w_0$,

Finally, take the limit $w_0 \to 0$: in this limit, $B$’s transverse motion vanishes entirely, so $u(B) \to 0$ and $m(u(B)) \to m(0) \equiv m$, the particle’s ordinary rest mass. Meanwhile $A$’s transverse velocity component also vanishes, leaving $A$ moving purely along $x$ at the frame’s relative velocity, so $u(A) \to v$. The boxed relation becomes

for arbitrary $v$, since $v$ was the (otherwise unconstrained) relative speed of the two frames. In other words, momentum conservation across inertial frames, together with the Lorentz transformation’s effect on transverse velocities, forces the “effective mass” $m(u)$ appearing in $p = m(u)u$ to be exactly $\gamma(u)m$ — not simply $m$, as Newtonian mechanics assumed. This is the relativistic momentum derived below; unlike the Newtonian formula, it is exactly the definition needed to make momentum conservation frame-independent, consistent with the Lorentz transformation of Chapter 2.

Relativistic Momentum¶

The result of the argument above is

where $m$ is the particle’s rest mass — an intrinsic, frame-independent property of the particle, equal to the mass measured by an observer at rest relative to it — and $u$ is the particle’s speed in the frame in question. For $u \ll c$, $\gamma \to 1$ and this reduces to the Newtonian $\vec p = m\vec u$. As $u \to c$, however, $\gamma \to \infty$, so $p \to \infty$ as well: a finite force acting for a finite time supplies only a finite impulse, whereas reaching the speed of light would require an unbounded impulse and work. This is the precise dynamical reason no massive object can reach or exceed $c$, complementing the kinematic argument of Chapter 2 (relativistic velocity addition never produces $u \ge c$ from sub-light inputs).

Worked Example: Testing $p = \gamma m u$ Against Data¶

The formula $\vec p = \gamma m \vec u$ is not merely a theoretical nicety; it has been tested directly by measuring the momentum and speed of fast electrons and protons independently — momentum from the radius of curvature in a known magnetic field ($p = qBr$), speed from time-of-flight over a measured distance. Measurements of this kind confirm $p/(mu) = \gamma(u)$ across a wide range of speeds and rule out the Newtonian prediction $p/(mu) = 1$ at appreciable fractions of $c$. For a proton with $u = 0.60c$, for instance, $\gamma = 1/\sqrt{1-0.36} = 1.25$, so the relativistic momentum is $25\%$ larger than the Newtonian formula predicts.

3.2Relativistic Energy¶

A parallel argument — demanding that the work-energy theorem, $dK = \vec F\cdot d\vec x$, hold with the relativistic force $\vec F = d\vec p/dt$ — leads to the relativistic kinetic energy

It is useful to expand this for $u \ll c$ using the binomial approximation $\gamma \approx 1 + \tfrac{1}{2}u^2/c^2 + \cdots$:

recovering the familiar Newtonian kinetic energy as the low-speed limit — a necessary consistency check, since Newtonian mechanics is extremely well tested at everyday speeds.

The kinetic energy expression separates naturally into two terms: $\gamma mc^2$, and a constant $mc^2$ subtracted off. Einstein’s insight was to take both terms seriously as energy, not just their difference. Define the total energy

and the rest energy

Then $K = E - E_0$: kinetic energy is the energy above and beyond the energy $mc^2$ a particle possesses simply by virtue of having rest mass $m$, even at rest ($u=0$, $\gamma=1$). This is the celebrated mass–energy equivalence: rest mass is a form of energy, convertible (in principle and, in nuclear and particle processes, routinely in practice) into other forms of energy, and vice versa. The conversion factor $c^2 \approx 9\times 10^{16}\ \text{m}^2/\text{s}^2$ is enormous, which is why converting even a small amount of rest mass releases a very large amount of energy — the physical basis of the energy released in nuclear fission and fusion, examined in Chapter 13.

3.3Four-Vectors and Threshold Energies¶

The Energy–Momentum Four-Vector¶

Momentum and energy are not independent; eliminating $u$ and $\gamma$ between $\vec p = \gamma m\vec u$ and $E = \gamma mc^2$ gives the energy–momentum invariant,

a relation that holds for every particle, in every inertial frame.

This is directly analogous to the invariant spacetime interval $(\Delta s)^2 = c^2\Delta t^2 - \Delta x^2$ of Chapter 2, and the analogy is not superficial: just as $(ct, x, y, z)$ can be assembled into a single spacetime-displacement four-vector that transforms under a Lorentz boost according to the Lorentz transformation, the quadruple $(E/c, p_x, p_y, p_z)$ can be assembled into a single energy–momentum four-vector,

which transforms from one inertial frame to another by exactly the same Lorentz transformation rule used for $(ct,x,y,z)$ in Chapter 2 (with $ct \to E/c$ and $x \to p_x$). Its invariant magnitude is

which is precisely the energy–momentum relation in Equation (3.14), rearranged; the rest mass $m$ plays the same role for the energy–momentum four-vector that the invariant interval plays for the spacetime-displacement four-vector — a quantity every observer computes to be the same, regardless of the frame in which $E$ and $\vec p$ individually are measured. For a massive particle, the energy–momentum four-vector is proportional to its four-velocity. Its component ratio therefore gives the ordinary velocity, $v_x = p_xc^2/E$ (and similarly for the other components). A slower particle has a larger ratio of $E/c$ to its momentum, as this expression requires.

The relation can be read geometrically as a right triangle, as shown in Figure 3.1: the total-energy term is the hypotenuse, while rest energy and momentum provide the two legs.

Two limits of the energy–momentum relation are worth committing to memory:

• Massive particle at rest ($p = 0$): $E = mc^2$, the rest energy alone.

• Massless particle ($m = 0$), such as a photon: $E = pc$. Massless particles carry momentum and energy but no rest energy, and — consistent with the momentum argument above, which forbids a massive particle from reaching $u=c$ — they travel at exactly $c$ in every inertial frame.

• Ultrarelativistic massive particle ($pc \gg mc^2$, equivalently $K \gg mc^2$): $E \approx pc$ and $u \approx c$. The massless relation is therefore also an excellent approximation for a sufficiently energetic massive particle.

The energy–momentum relation is often more convenient than working with $u$ and $\gamma$ directly, particularly for high-energy particles and for photons, where speed is fixed at $c$ and carries no information about energy.

Why the Four-Vector Formalism Earns Its Keep¶

The payoff of packaging $(E/c, \vec p)$ as a single object is that the energy–momentum four-vector is exactly conserved in every collision or decay, in every inertial frame, component by component — precisely because the underlying, separately-conserved quantities (in every frame) are energy and the three components of momentum. This means a conservation calculation can be carried out entirely by four-vector addition: add up the four-vectors of everything going into a collision, add up the four-vectors of everything coming out, and set the two sums equal, four components at a time. Because the four-vector’s magnitude, $(E/c)^2 - p^2 = (mc)^2$, is invariant, this magnitude can be computed in whichever frame is most convenient — often a frame in which one particle is initially at rest, or the frame in which the total three-momentum is zero (the subject of the next section) — and the resulting relation between energies and masses will hold in every other frame as well.

Electron–positron annihilation, revisited: consider an electron and a positron (each of mass $m$), both essentially at rest, annihilating. Charge conservation alone would permit $e^- + e^+ \to \gamma$ (a single photon), but the four-vector of the initial state is $(2mc, 0,0,0)$ (both particles at rest, energies $mc^2$ each, zero total momentum), while any single photon’s four-vector must satisfy $E=pc$, i.e., have equal, nonzero energy and momentum magnitude — it cannot have zero momentum unless its energy is also zero. A single outgoing photon is therefore impossible; conservation of the energy–momentum four-vector, not merely of energy or momentum separately, forces at least two photons, emitted back-to-back so that their momenta cancel, each carrying energy $mc^2$. This is the same conclusion reached by separate energy and momentum arguments in Problem 3 below, but the four-vector language makes clear that both conservation laws are really a single, unified statement, and this is exactly the physical process (positron annihilation, producing two back-to-back $511\ \text{keV}$ gamma rays) exploited in medical positron-emission tomography (PET) scans to locate a radioactive tracer inside the body.

Center-of-Momentum Frame and Threshold Energies¶

Many practical problems in nuclear and particle physics — will one particle collision produce a new particle, or not? — are most easily solved by transforming to the center-of-momentum (CM) frame: the unique inertial frame in which the total three-momentum of a system is zero. Because $(E/c)^2 - p^2$ is Lorentz-invariant, the total invariant mass of a system of particles,

is the same number whether computed in the lab frame or in the CM frame — but in the CM frame, where $p_{\text{total}} = 0$ by definition, it simplifies to $Mc^2 = E_{\text{total,CM}}$: the total CM-frame energy alone. This invariant $M$ determines whether a reaction is kinematically possible: by conservation of the energy-momentum four-vector, a reaction that produces a set of final-state particles with total rest mass $\sum m_f$ is only possible if the available energy in the CM frame is at least $\left(\sum m_f\right)c^2$ — i.e., if $M \ge \sum m_f$. Other conservation laws and interaction probabilities can still forbid or suppress a reaction. The threshold condition is $M = \sum m_f$ exactly, corresponding, for massive final-state particles, to all of them being at rest relative to the CM frame (and hence relative to each other), with no leftover kinetic energy to spare.

The CM frame is not a relativistic invention, and it is worth recovering the non-relativistic intuition before leaning on it. Figure 3.2 runs elastic and inelastic collisions in one and two dimensions, with the center of mass drawn on the screen and a momentum diagram beside it. Whatever the pucks do, that marker glides on at constant velocity — the collision cannot touch it, because the internal forces cancel in pairs — which is what makes its rest frame a natural place to do the bookkeeping. Choose the masses and velocities so that the marker stands still, and the momentum diagram shows what has been bought: two arrows equal and opposite before the collision, two arrows equal and opposite after it, however much kinetic energy was lost in between. The relativistic version below keeps that structure exactly, replacing $m\vec u$ by $\gamma m \vec u$ and the total mass by the invariant $M$; what changes is that $M$ is no longer the sum of the parts.

Worked Example: The Threshold for Antiproton Production¶

In 1955, the Bevatron at Berkeley was built specifically to search for the antiproton, via the reaction

a proton beam striking a stationary proton (in a hydrogen target), producing an additional proton–antiproton pair. (Baryon number and charge are each automatically conserved by this reaction; nothing prevents it kinematically once enough energy is available.) What is the minimum, or threshold, kinetic energy the beam proton must have, in the lab frame, where the target proton is at rest?

Let $m$ be the proton (and antiproton) rest mass, $m_1 = m_2 = m$ for beam and target, and let $M = 4m$ be the total rest mass of the four final-state particles, all momentarily at rest in the CM frame at threshold. The invariant $M^2c^4 = E_{\text{total}}^2 - (p_{\text{total}}c)^2$ can be computed in the lab frame, where the target is at rest ($E_2 = mc^2$, $p_2 = 0$) and the beam proton has energy $E_1$ and momentum $p_1$:

Using $E_1^2 - (p_1c)^2 = (mc^2)^2$ (the beam proton’s own invariant), this simplifies to

The threshold kinetic energy is $K_1 = E_1 - mc^2 = 6mc^2$. With $mc^2 = 938\ \text{MeV}$ for the proton, this is $K_1 = 6(938\ \text{MeV}) = 5.6\ \text{GeV}$ — dramatically larger than the naive $2mc^2 = 2(938\ \text{MeV}) = 1.9\ \text{GeV}$ one might have guessed from simply counting the rest-mass energy of the new particle pair. The extra factor of three arises because, in the lab frame, the newly created particles must all share the same velocity as the CM frame itself (since at threshold they are at rest in the CM frame, which is itself moving relative to the lab), so a substantial fraction of the beam’s kinetic energy is unavoidably “wasted” maintaining the overall forward motion of the collision products rather than being converted into new rest mass. The Bevatron was deliberately designed to reach a beam energy of $6.2\ \text{GeV}$, comfortably above this threshold, and the antiproton was discovered there later that same year by Owen Chamberlain, Emilio Segrè, and collaborators.

This “wasted energy” problem is exactly why modern particle physics favors colliders, in which two beams travel toward each other and collide head-on. If the beam and target in the reaction above were replaced by two protons of equal and opposite momentum (so that the lab frame is the CM frame), the threshold condition becomes simply $2E_1 = 4mc^2$, i.e., $K_1 = mc^2$ per beam — six times less kinetic energy required per proton than the fixed-target case, precisely because no energy needs to be spent maintaining a net forward CM velocity.

The contrast between the two arrangements is summarized in Figure 3.3.

The historical setting of this threshold calculation is shown in Figure 3.4: the Bevatron was built in the 1950s to reach the energies needed to discover the antiproton.

Aside: Why There Are No Faster-Than-Light Massive Particles¶

Chapter 2 argued, from causality alone, that no signal or influence can travel faster than $c$ without permitting effects to precede their causes in some valid inertial frame. This chapter’s momentum formula, $p = \gamma m u$, gives an independent, purely dynamical reason a massive particle in particular can never reach or exceed $c$: as $u \to c^-$, $\gamma \to \infty$, so accelerating a massive particle arbitrarily close to $c$ requires arbitrarily large — and, at $u=c$ itself, literally infinite — momentum and energy. No finite amount of work can supply this, so $c$ is a strict, unreachable asymptote for any object with $m>0$, approached but never attained no matter how long or how powerfully it is accelerated.

It is sometimes asked whether a hypothetical particle might simply be born moving faster than $c$, without ever having to accelerate through $c$ — such a hypothetical particle is called a tachyon. Formally, applying the energy–momentum relation at $u>c$ gives a spacelike four-momentum, so the invariant mass-squared would be negative (often described informally as an “imaginary rest mass”). This is not the rest mass of an ordinary particle, and no such particle has been observed. More importantly, a tachyon would connect spacelike-separated events (Chapter 2) — and the causality argument there shows that some inertial observer would measure any spacelike-connecting signal to travel backward in time, arriving before it was sent. Thus the mass–energy relation excludes faster-than-light motion for ordinary massive particles, while the light-cone argument gives a separate causal reason that controllable faster-than-light signals cannot exist.

Worked Example: An Electron Accelerated Through a Potential Difference¶

An electron (rest energy $m c^2 = 0.511\ \text{MeV}$) is accelerated from rest through a potential difference of $2.00\ \text{MV}$, gaining kinetic energy $K = qV = 2.00\ \text{MeV}$.

Total energy: $E = K + mc^2 = 2.00\ \text{MeV} + 0.511\ \text{MeV} = 2.511\ \text{MeV}$.

Momentum: from $E^2 = (pc)^2 + (mc^2)^2$,

so $p = 2.458\ \text{MeV}/c$.

Speed: from $E = \gamma mc^2$, $\gamma = E/mc^2 = (2.511\ \text{MeV})/(0.511\ \text{MeV}) = 4.914$, and $u = c\sqrt{1 - 1/\gamma^2} = 0.979c$.

Note that a Newtonian calculation of the speed from $K = \tfrac12 mu^2$ would give $u = c\sqrt{2K/mc^2} = c\sqrt{2(2.00\ \text{MeV})/(0.511\ \text{MeV})} \approx 2.8c$ — an unphysical result exceeding $c$, and a sharp reminder that the Newtonian kinetic-energy formula must not be used once $K$ is comparable to or larger than $mc^2$.

3.4Summary¶

• Because momentum must be conserved in every inertial frame under the Lorentz transformation, the correct definition is $\vec p = \gamma m\vec u$, not the Newtonian $m\vec u$. This can be derived by analyzing a symmetric elastic collision in two frames related by a boost: requiring conservation of transverse momentum, together with the Lorentz-transformation rule for transverse velocities, forces a moving particle’s effective inertia to scale as $\gamma(u)m$.

• As $u \to c$, $p \to \infty$, which is why no massive particle can reach the speed of light; this is confirmed directly by measurements of $p/(mu)$ for fast electrons and protons, which match $\gamma(u)$ and rule out the Newtonian prediction of 1.

• Relativistic kinetic energy is $K = (\gamma - 1)mc^2$, which reduces to $\tfrac12 mu^2$ for $u \ll c$.

• Total energy $E = \gamma mc^2$ and rest energy $E_0 = mc^2$ satisfy $K = E - E_0$: mass is a form of energy (mass–energy equivalence), convertible to and from other forms of energy.

• The energy–momentum invariant $E^2 = (pc)^2 + (mc^2)^2$ holds for all particles in all frames; it reduces to $E = mc^2$ for a particle at rest and to $E = pc$ for a massless particle such as a photon, which necessarily travels at $c$.

• For an ultrarelativistic massive particle with $pc \gg mc^2$, the same invariant gives $E \approx pc$ and $u \approx c$.

• Energy and momentum combine into a single energy–momentum four-vector, $p^\mu = (E/c, p_x, p_y, p_z)$, which transforms like $(ct,x,y,z)$ under a Lorentz boost and whose invariant magnitude is the rest mass, $(E/c)^2 - p^2 = (mc)^2$. Conservation of energy and momentum together is equivalent to conservation of this single four-vector.

• The center-of-momentum (CM) frame is the frame in which total momentum is zero; the invariant total mass $M$ of a system, computed from $M^2c^4 = E_{\text{total}}^2 - (p_{\text{total}}c)^2$, sets the threshold condition, $M = \sum m_f$, for whether a reaction producing a given set of final-state particles is energetically possible, and fixed-target reactions require substantially more beam kinetic energy than head-on collider reactions to reach the same threshold.

• Combining the divergence of $p=\gamma m u$ as $u\to c$ with the causality argument of Chapter 2 rules out faster-than-light motion for ordinary massive particles and shows why controllable faster-than-light signals would create causal paradoxes.

• Newtonian expressions for momentum and kinetic energy are the $u \ll c$ (equivalently $K \ll mc^2$) limit of the relativistic expressions and must not be used when this condition fails.

3.5Problems¶

Solution to Exercise 3.3 #

Initially, both particles are essentially at rest, so their total energy is their combined rest energy:

$$E_i=(0.511\ \text{MeV})+(0.511\ \text{MeV})=1.022\ \text{MeV}.$$

(3.30)

If the two photons have equal energy $E_\gamma$, energy conservation gives

$$2E_\gamma=1.022\ \text{MeV}, \qquad E_\gamma=0.511\ \text{MeV}.$$

(3.31)

The initial total momentum is zero. Each photon has momentum magnitude $p_\gamma=E_\gamma/c$, so two equal nonzero momentum vectors can sum to zero only if they point in opposite directions. Therefore, the annihilation produces two photons of energy $0.511\ \text{MeV}$ each, traveling in exactly opposite directions.

Solution to Exercise 3.9 #

At threshold, the final particles are at rest relative to one another in the CM frame, so their total rest mass is

$$M=2m_p+m_X=2m_p+10m_p=12m_p.$$

(3.51)

For a stationary target, the initial invariant is

$$M^2c^4=(E_b+m_pc^2)^2-(p_bc)^2 =2m_p^2c^4+2E_bm_pc^2,$$

(3.52)

where $E_b^2-(p_bc)^2=m_p^2c^4$ was used. Set this equal to $(12m_pc^2)^2$:

$$144m_p^2c^4=2m_p^2c^4+2E_bm_pc^2,$$

(3.53)

so

$$E_b=71m_pc^2, \qquad K_b=E_b-m_pc^2=70m_pc^2.$$

(3.54)

For equal and opposite collider beams, the lab is the CM frame, and threshold requires

$$2E_b=12m_pc^2, \qquad E_b=6m_pc^2, \qquad K_b=5m_pc^2$$

(3.55)

per proton. The ratio is

$$\frac{70m_pc^2}{5m_pc^2}=14.$$

(3.56)

Therefore, the fixed-target threshold is $70m_pc^2$ of kinetic energy, the collider threshold is $5m_pc^2$ per beam proton, and colliders require fourteen times less kinetic energy per incident proton because the CM has no net forward motion.