Part II · Special Relativity — Chapter 2.5

Relativistic Dynamics

Conservation laws have to be statements about four-vectors, or they are not laws. Everything else in this chapter is the bill for that sentence.

Tools you'll need  — Chapter 0.3 §2: the binomial series $(1+x)^{\alpha}$ for arbitrary real $\alpha$, and its worked example 2, which already expanded $\gamma mc^{2}$ without saying what it was. Chapter 0.6 §5: the multivariable chain rule, used constantly below to convert $\dd/\dd\tau$ into $\gamma\,\dd/\dd t$. Chapter 1.2 §3.5 and §8.1: the Euler–Lagrange equation, and the table of actions this book will write down. Section 5 here supplies the second entry in that table. Chapter 1.4 §3: Noether's theorem, in particular that invariance under time translation gives a conserved energy. We use it in §5 as an independent check. Chapter 2.3 §5 (proper time as the length of a worldline), §6 (straight worldlines maximise it), §4 (the causal classification of four-vectors), and §7 (four-velocity, sketched there and built properly here). Chapter 2.4 §4 (raising and lowering), §5 (the operations preserving tensor character, and the quotient theorem), and above all §6, the invariance theorem, which is the load-bearing wall of this entire chapter.

1 · Velocity is not a four-vector, and the repair

Start with the object we would most like to keep. A particle traces a worldline through spacetime, and the obvious thing to call its velocity is the rate at which its coordinates change with time:

That is four numbers, built from a four-vector, containing exactly the information you want. It is also not a four-vector. The reason is worth stating precisely rather than gesturing at, so let's do that first.

1.1 · Exactly what goes wrong

Being a four-vector means one thing and one thing only, by Chapter 2.4 §2: the four components must transform as $A'^{\mu}=\Lambda^{\mu}{}_{\nu}A^{\nu}$. So let's test (2.5.1) against that requirement.

The numerator is fine. $\dd x^{\mu}$ is the prototype four-vector, so $\dd x'^{\mu}=\Lambda^{\mu}{}_{\nu}\dd x^{\nu}$. The denominator is the problem. $t$ is not a scalar. It is the zeroth component of a four-vector, and under a boost along $x$ it becomes

where the second form divides out $\dd t$ and uses $\dd x/\dd t=v_{x}$. Now put the transformed numerator over that transformed denominator and see what we are left with:

Let's look at what that line is actually saying. $V^{\mu}$ transforms almost correctly. It picks up the right matrix $\Lambda$, and then it is spoiled by an extra scalar factor out front.

Worse, that factor depends on the particle's own velocity $v_{x}$. Two different particles observed in the same boost are therefore multiplied by two different numbers, and no rescaling of the definition can repair that.

This single line is also the whole reason Chapter 2.2's velocity-addition law came out as an awkward quotient rather than a matrix acting on a column. The awkward denominator in $u'=(u+v)/(1+uv/c^{2})$ is the factor in (2.5.3).

Now state the diagnosis in a form that tells you the cure. We divided a four-vector by a frame-dependent quantity. So divide by a frame-independent one instead. Chapter 2.3 §5 supplied exactly one natural candidate along a worldline: the proper time $\tau$, which every observer computes and every observer agrees on.

1.2 · The bridge: $\dd t/\dd\tau=\gamma$

Before we can use $\tau$ we need to know how it relates to coordinate time. This is a two-line recomputation of a result from Chapter 2.3 §5. It is worth having in front of you, because we will use it perhaps thirty times.

Proper time along a worldline is defined by $c^{2}\dd\tau^{2}=\dd s^{2}$. So our first job is to evaluate $\dd s^{2}$ between two neighbouring events on the worldline, and then to pull a factor of $\dd t^{2}$ out to the front:

The bracket is $1-v^{2}/c^{2}=1/\gamma^{2}$, so $c^{2}\dd\tau^{2}=c^{2}\dd t^{2}/\gamma^{2}$. Take the positive root, since both $\tau$ and $t$ increase toward the future, and we have the relation we came for:

Two remarks here will save confusion later.

First, $\gamma$ in this formula is the particle's own $\gamma$, evaluated with its instantaneous speed in whatever frame you are using. It is a function of $t$, and it has nothing to do with the $\gamma$ of any boost between frames.

Second, (2.5.5) is the reason $\dd/\dd\tau$ is computable at all. Whenever a $\tau$-derivative turns up, you convert it into the $t$-derivative you already know how to take:

which is nothing but the chain rule of Chapter 0.6 §5 applied to a one-parameter reparametrisation.

1.3 · The four-velocity

Now define what we were forced to define:

This is a four-vector, and the proof is one sentence. The numerator transforms with $\Lambda$ and the denominator is a scalar, so $u'^{\mu}=\Lambda^{\mu}{}_{\nu}u^{\nu}$ with no leftover factor. That is the entire content of the definition, and it is why relativity keeps replacing $\dd/\dd t$ by $\dd/\dd\tau$ everywhere.

Next we want the components of $u^{\mu}$ in terms of quantities a laboratory can measure, so apply the conversion rule (2.5.6) to the definition:

Note what happened, because it is the pattern of the whole chapter. The repaired object is the old object times $\gamma$.

At low speed $\gamma\to1$ and $u^{\mu}\to(c,\vv v)$, so nothing familiar is lost. At high speed the components run off to infinity. That is exactly the behaviour that lets a fixed-length object point in ever more extreme directions without ever exceeding the speed limit.

Which brings us to the property that makes $u^{\mu}$ special.

1.4 · $u\cdot u=c^{2}$, by explicit contraction

We want the Minkowski length of $u^{\mu}$, so contract it with itself using the metric. Chapter 2.4 §4.3 tells us how the bookkeeping goes. The time component keeps its sign and the spatial components flip:

The answer is $c^{2}$, identically. Not for special particles, and not in special frames. It holds for every particle, at every instant, in every frame, whatever that particle happens to be doing. The four-velocity is a four-vector of fixed Minkowski length $c$, and all it can do is point in different future-timelike directions.

There is a second proof, shorter, which explains why rather than merely confirming. It is the one to remember. Write the two $\tau$-derivatives as a single ratio of differentials and the answer falls out of the definition of $\tau$:

So the constraint is not a fact about particles at all. It is the definition of $\tau$, rearranged.

We chose to parametrise the worldline by its own arc length, and a curve parametrised by arc length has a unit tangent vector. Here "unit" means length $c$, because we measured arc length in seconds and tangent vectors in metres per second. Everything in this chapter that looks like a miracle is this sentence in disguise.

1.5 · Four-acceleration, and an orthogonality you get for free

Differentiate again with respect to the invariant:

Same argument as before, run twice over: $a^{\mu}$ is a four-vector.

Now do something that costs one line and pays for the rest of the chapter. We want to know what the fixed length of $u^{\mu}$ forces on $a^{\mu}$, so differentiate the identity (2.5.9) with respect to $\tau$. Its right-hand side is the constant $c^{2}$, so whatever the left-hand side works out to has to vanish:

where the two terms merged because $\eta_{\mu\nu}$ is a constant array and is symmetric, so the second sum is the first with the dummy names swapped. Divide by the 2 and what remains is

and it holds always, for every worldline. In Minkowski language, four-acceleration is orthogonal to four-velocity. The word "orthogonal" is doing honest work here. It is the same bilinear form, and the same definition of perpendicularity, that Chapter 0.5 gave for Euclidean space, with one sign changed.

Grind box — components of $a^{\mu}$, and $u\cdot a=0$ checked the hard way

We will need the components in §6, and computing them is a good drill in (2.5.6). First we need the derivative of $\gamma$. With $\gamma=(1-v^{2}/c^{2})^{-1/2}$ and $v^{2}=\vv v\cdot\vv v$, the chain rule gives

$$\dv{\gamma}{t} = -\frac12\left(1-\frac{v^{2}}{c^{2}}\right)^{-3/2}\cdot\left(-\frac{1}{c^{2}}\dv{(\vv v\cdot\vv v)}{t}\right) = \frac{\gamma^{3}}{c^{2}}\,\vv v\cdot\vv a,$$

using $\dd(\vv v\cdot\vv v)/\dd t = 2\vv v\cdot\vv a$, with $\vv a\equiv\dd\vv v/\dd t$ the ordinary three-acceleration. Write $\dot\gamma$ for this. Now apply (2.5.6) to (2.5.8):

$$a^{\mu} = \gamma\dv{}{t}\big(\gamma c,\ \gamma\vv v\big) = \Big(\gamma\dot\gamma c,\ \ \gamma\dot\gamma\,\vv v + \gamma^{2}\vv a\Big).$$

Now contract, carefully, keeping every term:

$$\begin{aligned} u\cdot a &= (\gamma c)(\gamma\dot\gamma c) - (\gamma\vv v)\cdot\big(\gamma\dot\gamma\vv v+\gamma^{2}\vv a\big)\\[3pt] &= \gamma^{2}\dot\gamma c^{2} - \gamma^{2}\dot\gamma v^{2} - \gamma^{3}\,\vv v\cdot\vv a\\[3pt] &= \gamma^{2}\dot\gamma c^{2}\left(1-\frac{v^{2}}{c^{2}}\right) - \gamma^{3}\,\vv v\cdot\vv a\\[3pt] &= \dot\gamma c^{2} - \gamma^{3}\,\vv v\cdot\vv a. \end{aligned}$$

Substituting $\dot\gamma=\gamma^{3}(\vv v\cdot\vv a)/c^{2}$ from the first line makes the two terms cancel identically. ✓ The abstract one-liner (2.5.12) and the brute-force component calculation agree, as they must. The point of doing both once is to see how much labour the index notation is saving.

One consequence worth extracting now. Go to the particle's instantaneous rest frame, where $\vv v=0$ and $u^{\mu}=(c,\vv 0)$. Then $u\cdot a = c\,a^{0}=0$ forces $a^{0}=0$, so in that frame $a^{\mu}=(0,\vv a_{0})$, where $\vv a_{0}$ is the acceleration the particle itself feels. Call it the proper acceleration. Its invariant square is

$$a\cdot a = 0 - \abs{\vv a_{0}}^{2} = -\,\alpha^{2}, \qquad \alpha \equiv \abs{\vv a_{0}},$$

so four-acceleration is always a spacelike vector, and its invariant magnitude is the proper acceleration, meaning what an accelerometer bolted to the particle reads. That number is frame-independent. This is why "the rocket burns at $1g$" is a meaningful statement, while "the rocket's acceleration is $9.8\,\mathrm{m\,s^{-2}}$" is a statement about somebody's coordinates. Problem 2 uses this.

2 · Four-momentum, and the argument this chapter turns on

We have two ingredients on the table. One is a four-vector that describes a particle's motion. The other is a scalar that describes the particle itself, namely its rest mass $m$, which by the conventions of this Part always means the rest mass and never anything else.

Multiplying a four-vector by a scalar gives a four-vector (2.4 §5.1, operation 2), so we are entitled to define the four-momentum

Its spatial part is $\gamma m\vv v$, which reduces to the Newtonian $m\vv v$ when $\gamma\to1$. So we have at least a candidate for "momentum", and we will call it $\vv p\equiv\gamma m\vv v$.

The time component $p^{0}=\gamma mc$ is, at this stage, an unnamed number that came along for the ride. We are not going to name it until §3, and we are not going to name it by guessing. The name will be forced on us.

Everything now rests on a single question: why should this be the conserved quantity? The answer is the intellectual centre of the chapter, so we take it slowly.

2.1 · What a conservation law has to be

A conservation law is a claim about a physical process. Some quantity computed before the process equals the same quantity computed after it. Consider a collision, with particles coming in and particles going out, and the assertion

Now ask the question that Chapter 2.4 taught us to ask about every equation. Who is asserting this, and does the assertion survive translation into another observer's language?

The question is not pedantry. If the assertion does not survive, then (2.5.15) is not a law of nature. It is a report from one laboratory, of no more fundamental standing than "the train is moving".

Chapter 2.4 §6 gave the tool for deciding. If the two sides of an equation are tensors of the same type, then the equation holds in every frame as soon as it holds in one. If they are not, all bets are off.

Chapter 2.4's closing warning was blunt about which side of that line statements like $\vv p=0$ fall on. They are statements about components, and therefore about a frame, and they must never be carried across a boost unexamined.

So the demand is clear. Whatever is conserved must be a tensor, and the conservation statement must be an equation between tensors of the same type. Three numbers are not a tensor under the Lorentz group. Four numbers that transform with $\Lambda$ are.

2.2 · Newtonian momentum conservation does not survive a boost

That is an argument from principle. Now let's see the failure in arithmetic, in a collision you can hold in your head.

Two identical lumps of putty, each of rest mass $m$, approach each other along the $x$-axis in frame $S$ with speeds $+u$ and $-u$. They collide and stick. By symmetry the composite is at rest in $S$, and we will call its rest mass $M$, whatever that turns out to be.

Start by checking Newtonian momentum conservation in $S$, where the arithmetic is easiest:

Perfect. But note how little that check actually tested. By symmetry it would have passed for any definition of momentum of the form $\vv p = m f(v)\,\vv v$, with $f$ any function whatever of the speed, because the two terms cancel no matter what $f$ is.

A symmetric collision in a single frame is almost completely uninformative. The information is in the boost, so let's boost.

Move to a frame $S'$ travelling at $-v$ along $x$ relative to $S$, so that in $S'$ everything is drifting in the $+x$ direction. Chapter 2.2's velocity-addition law gives the new velocities of the two lumps:

and the composite, at rest in $S$, moves at $v$ in $S'$. Our next goal is the incoming Newtonian momentum in $S'$, so add those two velocities. Write $a\equiv uv/c^{2}$ for compactness and put the two fractions over the common denominator $1-a^{2}$:

where the last step used $au = u^{2}v/c^{2}$, so that $2v-2au = 2v(1-u^{2}/c^{2})$.

Meanwhile the Newtonian momentum after the collision is $Mv$, and Newtonian mechanics also insists that mass is additive, $M=2m$. So here are the two sides of the conservation law as $S'$ reads them:

These are equal only if $u=0$ or $v=0$, which is to say only if there was no collision or no boost. Put in numbers. For $u=v=0.6c$ the fraction is $(1-0.36)/(1-0.1296)=0.64/0.8704=0.7353$, so an observer in $S'$ finds the outgoing momentum 36% larger than the incoming momentum. The books balance in $S$ and fail badly in $S'$.

2.3 · The four-vector version cannot fail

Now run the same collision with $p^{\mu}$ and watch the disease vanish for structural reasons.

Suppose that in frame $S$ the total four-momentum is conserved, meaning $\sum_{\text{in}}p^{\mu} = \sum_{\text{out}}p^{\mu}$ for all four components. To test that claim in another frame we want a single object to transform, so define the shortfall

Each $p^{\mu}$ is a four-vector, and sums and differences of tensors of the same type are tensors of that type (2.4 §5.1, operation 1). So $\Delta P^{\mu}$ is itself a four-vector.

Our hypothesis is now the tensor equation $\Delta P^{\mu}=0$, and 2.4 §6 says a tensor equation true in one frame is true in all of them. Written out explicitly, that reads

That is the whole proof, and it is deliberately anticlimactic. The transformation law is linear and homogeneous, so it maps zero to zero, and there is nowhere for a violation to come from. If four-momentum is conserved for one observer, it is conserved for every observer, automatically, with no further conditions.

Grind box — the same collision, done relativistically, term by term

The abstract argument is airtight, but you should see the numbers come out once. We need one identity, and it is worth having permanently.

The velocity-addition identity for $\gamma$. If $u'=(u+v)/(1+uv/c^{2})$, then

$$1-\frac{u'^{2}}{c^{2}} = \frac{\left(1+\dfrac{uv}{c^{2}}\right)^{2}-\dfrac{(u+v)^{2}}{c^{2}}}{\left(1+\dfrac{uv}{c^{2}}\right)^{2}}.$$

Expand the numerator, being careful with every term:

$$\begin{aligned} \text{num} &= 1 + \frac{2uv}{c^{2}} + \frac{u^{2}v^{2}}{c^{4}} - \frac{u^{2}}{c^{2}} - \frac{2uv}{c^{2}} - \frac{v^{2}}{c^{2}}\\[3pt] &= 1 - \frac{u^{2}}{c^{2}} - \frac{v^{2}}{c^{2}} + \frac{u^{2}v^{2}}{c^{4}} \;=\; \left(1-\frac{u^{2}}{c^{2}}\right)\left(1-\frac{v^{2}}{c^{2}}\right). \end{aligned}$$

The $2uv/c^{2}$ terms cancel and what remains factorises. Taking reciprocals and square roots,

$$\boxed{\;\gamma(u') = \gamma(u)\,\gamma(v)\left(1+\frac{uv}{c^{2}}\right).\;}$$

This is just the statement that rapidities add, $\cosh(\phi_{u}+\phi_{v}) = \cosh\phi_{u}\cosh\phi_{v} +\sinh\phi_{u}\sinh\phi_{v}$, written in velocities. Chapter 2.2 §5 proved the rapidity version.

Momentum before, in $S'$. Using the identity with $u\to u$ and then $u\to-u$:

$$\gamma(u_{1}')\,u_{1}' = \gamma(u)\gamma(v)\left(1+\frac{uv}{c^{2}}\right)\cdot\frac{u+v}{1+uv/c^{2}} = \gamma(u)\gamma(v)\,(u+v),$$ $$\gamma(u_{2}')\,u_{2}' = \gamma(u)\gamma(v)\left(1-\frac{uv}{c^{2}}\right)\cdot\frac{v-u}{1-uv/c^{2}} = \gamma(u)\gamma(v)\,(v-u).$$

The messy denominators cancel exactly, which is the identity earning its keep. Adding the two and multiplying by $m$:

$$\sum_{\text{before}} \gamma m\,u' \;=\; m\gamma(u)\gamma(v)\big[(u+v)+(v-u)\big] \;=\; 2m\,\gamma(u)\gamma(v)\,v.$$

Momentum after, in $S'$. The composite has rest mass $M$ and moves at $v$, so its momentum is $\gamma(v)Mv$. Equality demands

$$\gamma(v)\,M\,v = 2m\,\gamma(u)\gamma(v)\,v \qquad\Longrightarrow\qquad M = 2\gamma(u)\,m.$$

Read that twice. Relativistic momentum conservation in the boosted frame does not merely permit the composite's rest mass to exceed $2m$. It requires it, and fixes the excess exactly. The lump of putty is heavier than its parts, by a factor $\gamma(u)$, and nothing in the calculation gave us a choice. Section 9 is about that number.

The time components, for free. Do the same with $p^{0}=\gamma mc$:

$$\sum_{\text{before}}\gamma mc = mc\,\gamma(u)\gamma(v)\left[\left(1+\frac{uv}{c^{2}}\right)+\left(1-\frac{uv}{c^{2}}\right)\right] = 2mc\,\gamma(u)\gamma(v),$$

while after the collision $p^{0}=\gamma(v)Mc = 2\gamma(u)\gamma(v)mc$, using the $M$ just derived. Equal ✓, and note that we did not impose this. It came out. That is (2.5.21) being true in coordinates.

And in $S$ itself. Before the collision $p^{0}=2\gamma(u)mc$, and after it $p^{0}=Mc=2\gamma(u)mc$ ✓, while the spatial parts vanish on both sides by symmetry. So the four-vector statement holds in $S$, holds in $S'$, and by (2.5.21) holds in every frame there is.

2.4 · The part you cannot refuse: momentum forces energy

So far we have shown that four-momentum conservation is self-consistent across frames. Now comes the sharper statement, and it is the one that makes this chapter inevitable.

Suppose you are a minimalist. You accept that the conserved object must be built from the four-vector $p^{\mu}$, since you have seen §2.2 and you are not going back. But you want to conserve only the three spatial components, because those are the ones you have experimental evidence for. You want the law "$\Delta\vv P=0$ in every frame", and you decline to say anything at all about $\Delta P^{0}$.

You cannot have it. Here is why, in three lines.

Look at what that argument did. It used no dynamics, no experiment, and no assumption about forces or collisions. It used only the transformation law.

And what it says is this. A universe in which momentum is conserved for every observer is a universe in which $p^{0}$ is conserved too. You are not permitted to conserve three components of a four-vector. The boost mixes the fourth into the first three and drags it into the law whether you invited it or not.

That is the whole chapter in one paragraph. The demand that conservation laws be frame-independent selects $p^{\mu}$ out of all the candidate momenta. Having selected it, you are stuck with its time component, a fourth conserved quantity you did not ask for, did not derive from any experiment, and cannot discard. Section 3 asks what on earth it is.

3 · What the time component is

We have an unnamed conserved quantity $c\,p^{0}=\gamma mc^{2}$. Its dimensions are those of energy, since it is a mass times a velocity squared. Dimensions are suggestive and prove nothing, so let's find out what it actually is by the only honest route available. We expand it at low speed and see what it turns into.

3.1 · The expansion

Apply the binomial series of Chapter 0.3 §2, which for arbitrary real exponent $\alpha$ reads $(1+x)^{\alpha}=\sum_{k\ge0}\binom{\alpha}{k}x^{k}$ with $\binom{\alpha}{k}=\alpha(\alpha-1)\cdots(\alpha-k+1)/k!$. Take $\alpha=-\half$ and $x=-\beta^{2}$, where $\beta=v/c$, and the first few coefficients come out as

Substituting $x=-\beta^{2}$ flips the sign of every odd power, so all the signs become positive:

That is the expansion of $\gamma$ alone. The quantity we actually want is $\gamma mc^{2}$, so multiply through by $mc^{2}$ and restore $\beta=v/c$, keeping an eye on what each term becomes:

3.2 · The identification, made carefully

Now the part that deserves care, because it is usually skated over. What (2.5.24) shows is a mathematical fact about a function, not a physical identification. It says that the conserved quantity $\gamma mc^{2}$ equals a constant, plus the Newtonian kinetic energy, plus terms that vanish as $v/c\to0$.

Getting from there to "$\gamma mc^{2}$ is the energy" takes one further argument, and here it is.

Newtonian mechanics has its own conserved quantity for elastic collisions, $\sum_{i}\half m_{i} v_{i}^{2}$, verified to exhaustion at low speeds. Our new law says instead that $\sum_{i}\gamma_{i}m_{i}c^{2}$ is conserved. We want to see how the two claims are related, so expand the new one particle by particle at low speed:

Now suppose the collision does not change what the particles are, so the same species go in as come out. That is the only kind of collision Newtonian mechanics ever contemplated.

Then the sum $(\ast)$ is the same before and after, and it cancels out of the conservation statement entirely. What is left is exactly Newtonian kinetic-energy conservation.

So the new law contains the old one, and the quantity it conserves must be what the old theory called energy. That, and only that, is what licenses the name:

The four-momentum is the energy and the momentum, welded into one object by the boost that mixes them. From here on $p^{\mu}$ is called the energy–momentum four-vector, and the two conservation laws you learned separately are one law with four components.

3.3 · The first correction, and where you can see it

The $\tfrac38 mv^{4}/c^{2}$ term in (2.5.24) is not decoration. To judge its size we need something to judge it against, so compare it with the Newtonian term:

The correction is quadratically small in $\beta$. That is why Newtonian mechanics survived so long, and it is also why the correction becomes visible the moment you have a system with a characteristic speed and a spectroscope.

The electron in a hydrogen atom has $\beta\approx\alpha=1/137$. Chapter 0.3 §5 got that from dimensional analysis alone, since the natural speed $k_{e}/\hbar$ divided by $c$ is exactly $\alpha$. Putting that number into the ratio above gives

a relative shift of four parts in $10^{5}$ in the atom's energy levels. On a $13.6\,\mathrm{eV}$ scale that is of order $\alpha^{2}\times13.6\,\mathrm{eV}\approx7\times10^{-4}\,\mathrm{eV}$, which is comfortably resolvable.

It is one of the three contributions to the fine structure of hydrogen that Chapter 0.3's worked example 2 flagged and could not yet explain. Chapter 4.16 computes the full splitting. The point here is only that the leading piece of it is a term in (2.5.24) and nothing more exotic.

Grind box — the same correction in momentum, and why the coefficient changes sign

Open a quantum mechanics text and you will find the relativistic correction to the hydrogen Hamiltonian written as $-\hat p^{4}/8m^{3}c^{2}$, a negative term with coefficient $\tfrac18$. We just derived a positive term with coefficient $\tfrac38$. Both are correct. Chapter 0.3's worked example 2 noticed the clash and named the culprit. Now that we have actually derived $\vv p=\gamma m\vv v$ and the mass shell, we can close it arithmetically.

The kinetic energy in terms of $p$. Anticipating §4's mass shell, $E=\sqrt{p^{2}c^{2}+m^{2}c^{4}}$. Factor out $mc^{2}$ and expand with the binomial series again, now with $x=p^{2}/m^{2}c^{2}$ and $\alpha=+\tfrac12$, for which $\binom{1/2}{1}=\tfrac12$ and $\binom{1/2}{2}=\frac{(\frac12)(-\frac12)}{2}=-\tfrac18$:

$$E = mc^{2}\left(1+\frac{p^{2}}{m^{2}c^{2}}\right)^{1/2} = mc^{2}\left[1+\frac{p^{2}}{2m^{2}c^{2}} - \frac{p^{4}}{8m^{4}c^{4}}+\cdots\right] = mc^{2}+\frac{p^{2}}{2m}-\frac{p^{4}}{8m^{3}c^{2}}+\cdots$$

There it is, sign and coefficient. Quantum mechanics uses this form because $\hat p$ is the natural operator, not $v$.

Why the two agree. The variable is different, because $p=\gamma mv$ rather than $mv$. So expand $\gamma$ to the order needed,

$$p^{2} = \gamma^{2}m^{2}v^{2} = m^{2}v^{2}\left(1-\frac{v^{2}}{c^{2}}\right)^{-1} = m^{2}v^{2}\left(1+\frac{v^{2}}{c^{2}}+\cdots\right),$$

so the two terms become

$$\frac{p^{2}}{2m} = \frac{mv^{2}}{2}+\frac{mv^{4}}{2c^{2}}+\cdots, \qquad -\frac{p^{4}}{8m^{3}c^{2}} = -\frac{m^{4}v^{4}}{8m^{3}c^{2}}+\cdots = -\frac{mv^{4}}{8c^{2}}+\cdots$$

Add them: $\tfrac12 mv^{2} + \left(\tfrac12-\tfrac18\right)mv^{4}/c^{2} = \tfrac12 mv^{2}+\tfrac38 mv^{4}/c^{2}$ ✓, which is (2.5.24). The apparent discrepancy was bookkeeping. A $\tfrac12$ of the quartic term comes from $p$ not being $mv$, and a $-\tfrac18$ comes from the square root, and $\tfrac12-\tfrac18=\tfrac38$.

The moral is one you will need again in Part V: a perturbative coefficient is meaningless until you say which variable is being held fixed. Terms move between orders when you change variables. Only the physical prediction is invariant, which here means the size of the fine-structure shift, and that is $\sim\alpha^{2}$ either way.

4 · The mass shell

Every four-vector has an invariant square, and the invariant square of $p^{\mu}$ turns out to contain everything.

4.1 · $p\cdot p=m^{2}c^{2}$, and what it says

Contract $p^{\mu}=mu^{\mu}$ with itself. The mass is a scalar, so it comes straight out of the contraction, and (2.5.9) finishes the job:

That is the answer in terms of the mass. We also want it in terms of the things a detector actually reports, so compute the same quantity from the components in (2.5.26), remembering that lowering flips the sign of the spatial parts:

Two expressions for one quantity. Set them equal and multiply through by $c^{2}$:

This is the mass-shell relation, and it is the most useful equation in particle physics. Three observations before we use it.

It is an invariant statement. The left side involves $E$ and the right involves $p$. Both change under a boost, and the particular combination $E^{2}-p^{2}c^{2}$ does not. Different observers assign different energies and momenta to the same electron, and all of them compute the same $m$.

So mass is a Lorentz invariant. It is not a quantity that grows with speed. It is the fixed label attached to the four-vector, exactly as the interval is the fixed label attached to a pair of events.

It is Chapter 2.3's geometry, in different variables. Plot $E$ against $p$ and (2.5.31) is a hyperbola with asymptote $E=pc$. Compare 2.3 §3.2, where the orbits of a boost in the $(ct,x)$ plane were the hyperbolae $c^{2}t^{2}-x^{2}=\text{const}$ with asymptotes at $45^{\circ}$.

Same curves, and for the same reason. A boost is a hyperbolic rotation, and it slides any four-vector along the level surface of its own invariant square. Momentum space is spacetime with the axes relabelled.

It has no reference to $\tau$. That matters more than it looks, and §4.3 collects the debt.

4.2 · Velocity from the four-momentum

Before we take the massless case we need one more result, and the order matters, because this result is what makes the massless case work at all. We want the particle's velocity written in terms of $E$ and $\vv p$. Both components of $p^{\mu}$ carry the same factor $\gamma m$, so dividing one by the other kills it:

Read it as a recipe. Given a particle's energy and momentum, its velocity is the ratio. No $\gamma$, no $m$, no square roots.

Two checks. At low speed, $E\approx mc^{2}$ and $\vv p\approx m\vv v$, so the ratio returns $\vv v$ ✓. And in general $E\ge pc$ by (2.5.31) whenever $m^{2}\ge0$, so the formula automatically gives $v\le c$. The speed limit is not imposed on (2.5.32). It falls out of it.

4.3 · Massless particles

Set $m=0$ in (2.5.31). Taking the positive root, since energies of real particles are positive:

That fixes the relation between energy and momentum. To get the speed, feed it into the recipe (2.5.32) and watch the momentum cancel:

Exactly $c$, for every energy, with no limiting process and no approximation. A massless particle does not travel near the speed of light. It travels at it, always, and it has no rest frame in which to be at rest.

rest mass m

momentum p

hide Newtonian curve

let m → 0

a proton

find the 1% point

m = 1.000 p = 0.620 E = √(p²+m²) = 1.1766

v/c = pc²/E = 0.5269 γ = E/mc² = 1.1766

Newtonian m + p²/2m = 1.1922 error = 1.325 %

Newtonian error passes 1% at p = 0.571 (v/c = 0.496, γ = 1.152)

Grind box — why the Newtonian curve is so good for so long

The figure claims the Newtonian energy is within 1% of the truth up to $v\approx0.5c$, which is surprising if you expect errors of order $\beta^{2}$. Here is the reason, and it is a nice piece of Chapter 0.3 bookkeeping.

Work in units of $mc$ and $mc^{2}$. Write $x=p/mc$, so the exact energy is $E/mc^{2}=\sqrt{1+x^{2}}$ and the Newtonian one is $1+x^{2}/2$. The fractional error is

$$\varepsilon(x) \;=\; \frac{1+x^{2}/2}{\sqrt{1+x^{2}}} - 1.$$

Expand with the binomial series, $(1+x^{2})^{-1/2}=1-\tfrac12x^{2}+\tfrac38x^{4}-\cdots$:

$$\left(1+\frac{x^{2}}{2}\right)\left(1-\frac{x^{2}}{2}+\frac{3x^{4}}{8}\right) = 1 - \frac{x^{2}}{2} + \frac{3x^{4}}{8} + \frac{x^{2}}{2} - \frac{x^{4}}{4} + O(x^{6}) = 1 + \frac{x^{4}}{8}+O(x^{6}).$$

The $x^{2}$ terms cancel exactly. That cancellation is not luck. The Newtonian formula was constructed to reproduce the first two terms of (2.5.24), so the leading error is necessarily the third. Hence $\varepsilon\approx x^{4}/8$, which is quartic rather than quadratic, and a quartic is very small for a while and then is not.

Numbers. Setting $\varepsilon=0.01$ exactly, numerically rather than from the expansion, gives $x=0.5715$, whence $v/c=x/\sqrt{1+x^{2}}=0.4962$ and $\gamma=1.152$. Some more, exact:

$p/mc$	$v/c$	$E/mc^{2}$	Newtonian	error
$0.5$	$0.447$	$1.1180$	$1.1250$	$0.62\%$
$1$	$0.707$	$1.4142$	$1.5000$	$6.07\%$
$2$	$0.894$	$2.2361$	$3.0000$	$34.2\%$
$3$	$0.949$	$3.1623$	$5.5000$	$73.9\%$
$5$	$0.981$	$5.0990$	$13.500$	$165\%$

This is the quantitative answer to "why didn't anyone notice?". Everything in nineteenth-century mechanics ran at $\beta\lt10^{-4}$, where $\varepsilon\lt10^{-17}$. You do not find relativity by doing careful mechanics. You find it by doing electromagnetism, which is Chapter 2.1's story, or by building something that goes fast.

5 · The action is proper time

We have the repaired momentum $\vv p=\gamma m\vv v$, and we expect the equation of motion to read $\dd\vv p/\dd t=\vv F$. That is Newton's second law in the form he actually wrote it, with the repaired momentum substituted for the old one. (Section 6 shows this is the spatial part of a proper tensor equation. For now, take it as the target.)

Chapter 1.2 established that respectable dynamics comes from an action principle. So the question in front of us is this: what Lagrangian produces this?

We are not going to guess. We impose the answer and solve for $L$. That is the honest way round, and it is exactly how Chapter 1.2 §4 checked that $L=T-V$ reproduces Newton, run backwards.

5.1 · Solving for $L$

Take a particle in a potential, $L=L_{0}(\vv v)-V(\vv x)$, with the free part depending only on the velocity. We need the Euler–Lagrange equation for the coordinate $x^{i}$, which gives one equation per degree of freedom, $\dv{}{t}\big(\partial L/\partial\dot q^{i}\big)=\partial L/\partial q^{i}$, from Chapter 1.2 §3.5. Here it reads

We want the left-hand side to be $\dd p_{i}/\dd t$ with $p_{i}=\gamma m v^{i}$. Matching the two demands

Now solve this for $L_{0}$. The right-hand side is $v^{i}$ times a function of the speed alone, so $L_{0}$ can depend on $\vv v$ only through $v=\abs{\vv v}$. The chain rule then gives $\partial L_{0}/\partial v^{i} = L_{0}'(v)\,\partial v/\partial v^{i} = L_{0}'(v)\,v^{i}/v$. Comparing with (2.5.36), the direction $v^{i}$ cancels and we are left with a single ordinary differential equation:

All that remains is to integrate it. The substitution that clears the square root is $s=1-v^{2}/c^{2}$, for which $\dd s = -2v\,\dd v/c^{2}$ and therefore $v\,\dd v = -\tfrac{c^{2}}{2}\dd s$:

The constant is genuinely free. Chapter 1.2's Problem 4 showed that adding a constant to $L$ adds a term linear in $t$ to $S$, which is a total derivative and changes no equation of motion. Set it to zero, for a reason that will be apparent in ten lines:

Check the limit before going on. Expanding the square root with the binomial series, $\sqrt{1-\beta^{2}} = 1-\tfrac12\beta^{2}-\tfrac18\beta^{4}-\cdots$, so

which is Chapter 1.2's $T-V$ plus the irrelevant constant $-mc^{2}$ plus small corrections ✓.

Note in passing that $L$ is emphatically not $T-V$ any more. The quantity $-mc^{2}/\gamma$ is not the kinetic energy of anything. Chapter 1.2 §5 warned that "$T-V$" is a special case rather than a definition, and here is the first place that warning cashes out.

5.2 · The rewrite that makes it obvious

Now form the action and use (2.5.5) in the direction we have not yet used it. Until now we have converted $\tau$-derivatives into $t$-derivatives. This time we go the other way and turn $\dd t$ into $\dd\tau$. For a free particle ($V=0$),

Stop here. This is the most important equation in the chapter after (2.5.31), and it deserves to be read as a sentence rather than a formula.

Grind box — three independent checks on $L$, including Noether's

Check 1: the canonical momentum. Chapter 1.2 §6 defines the momentum conjugate to $x^{i}$ as $\partial L/\partial\dot x^{i}$. From (2.5.39), with $\dot x^{i}=v^{i}$,

$$\pdv{L}{\dot x^{i}} = -mc^{2}\cdot\frac{1}{2}\big(1-\beta^{2}\big)^{-1/2}\cdot\left(-\frac{2v^{i}}{c^{2}}\right) = \frac{m v^{i}}{\sqrt{1-\beta^{2}}} = \gamma m v^{i} = p_{i}. \ \checkmark$$

Which it had better, since (2.5.36) is what we solved. But it is reassuring that the canonical momentum of Lagrangian mechanics and the spatial part of the four-momentum are literally the same object. They did not have to be.

Check 2: Noether's energy. Chapter 1.4 §3 showed that if $L$ has no explicit time dependence, the conserved Noether charge is $H=\sum_{i}\dot x^{i}\,\partial L/\partial\dot x^{i} - L$. Compute it for the free case:

$$\begin{aligned} H &= \vv v\cdot(\gamma m\vv v) + mc^{2}\sqrt{1-\beta^{2}}\\[3pt] &= \gamma m v^{2} + \frac{mc^{2}}{\gamma}\\[3pt] &= \gamma m\left(v^{2} + \frac{c^{2}}{\gamma^{2}}\right) \;=\; \gamma m\big(v^{2} + c^{2}-v^{2}\big) \;=\; \gamma mc^{2}. \end{aligned}$$

where the third line used $c^{2}/\gamma^{2}=c^{2}(1-\beta^{2})=c^{2}-v^{2}$. So the Noether charge of time-translation invariance is exactly $E=\gamma mc^{2}$, the quantity §3 identified by an entirely different argument, namely matching a Taylor expansion to Newtonian kinetic energy. Two independent derivations, one answer. ✓

Check 3: the equation of motion, in full. Confirm that (2.5.39) really does give the promised law and not something that merely resembles it. With $V=0$, Euler–Lagrange says $\dd(\gamma mv^{i})/\dd t = 0$, so $\gamma m\vv v$ is constant, so $\vv v$ is constant, since $\gamma$ is a monotonic function of $v$ and fixing $\gamma m v$ therefore fixes $v$. The path is a straight line traversed uniformly. Free particles move in straight lines at constant speed, which is Newton's first law, recovered from an action whose Lagrangian is nothing like $\half mv^{2}$. ✓

An aside on manifest covariance. (2.5.41) is manifestly a scalar, since $m$ is a scalar and $c$ is a scalar and $\tau$ is a scalar. So the action is frame-independent by inspection, in the sense of Chapter 2.4 §6. The Lagrangian (2.5.39) is not, and neither is $\dd t$. Only their product is. This is the standard situation in relativistic mechanics, and it is why the action formulation is preferred. The object with the good transformation properties is the integral, not the integrand-in-$t$.

6 · Force, and the death of relativistic mass

Newton's second law is the one piece of the old mechanics we have not yet replaced. Chapter 2.4 §6.2 already announced the verdict, which is that $\vv F=m\vv a$ is not a tensor equation while $\dd p^{\mu}/\dd\tau=f^{\mu}$ is. Announcing is not deriving. Here is the object, and here is what it costs.

6.1 · The covariant force

Define the four-force by the only construction available: differentiate the four-momentum by the invariant.

where the last step holds for particles of constant rest mass. Both sides are $(1,0)$ tensors, so by Chapter 2.4 §6 this is a legitimate law. Written once, it holds in every frame.

It also inherits a constraint at no cost. We know $u\cdot a=0$ from (2.5.13), so contract the definition with $u$ and let the mass ride along:

So a four-force has only three independent components. Whatever three you specify, the fourth is determined. That is not a defect. It is a theorem with a name you already know, as we are about to see.

6.2 · The three-force, and the work–energy theorem for free

Connect to the laboratory. Define the ordinary three-force as the rate of change of the ordinary momentum with respect to ordinary time, $\vv F\equiv\dd\vv p/\dd t$. That is Newton's second law in its correct form, the form he actually wrote. Using (2.5.6) on (2.5.42), component by component:

Now we cash in the constraint. We have $f^{\mu}$ in laboratory quantities, and (2.5.43) says its contraction with $u^{\mu}$ vanishes. So write out that contraction, using $u^{\mu}=\gamma(c,\vv v)$, and see what it forces:

and since $\gamma^{2}\neq0$, the bracket itself must vanish, which leaves

That is the work–energy theorem. The rate of change of energy is the power delivered by the force. We did not assume it, define it, or import it. It is the time component of a four-vector equation, forced by the geometric identity $u\cdot a=0$.

In Newtonian mechanics the work–energy theorem is a separate derivation. Here it is the statement that the four-force is orthogonal to the four-velocity, which is the statement that the four-velocity has fixed length, which is the definition of proper time. Everything is the same thing.

6.3 · $\vv F$ and $\vv a$ are not parallel

Now the damage. Expand $\vv F=\dd(\gamma m\vv v)/\dd t$ with the product rule, using $\dot\gamma=\gamma^{3}(\vv v\cdot\vv a)/c^{2}$ from §1.5's grind box:

The bracket contains a piece along $\vv a$ and a piece along $\vv v$. Unless those two directions coincide, or unless the second term vanishes, the force and the acceleration point in different directions.

Push a fast particle sideways and it does not accelerate sideways. It accelerates partly forward, or partly backward, depending on the sign of $\vv v\cdot\vv a$. Nothing in Newtonian mechanics prepares you for this, and it is not a small effect.

The two special cases where they are parallel are worth having explicitly.

Longitudinal, meaning a push along the motion. Then $\vv a\parallel\vv v$, so $\vv v\cdot\vv a=va$ and the bracket is $a\big(1+\gamma^{2}\beta^{2}\big)\hat v$. Use the identity $\gamma^{2}-\gamma^{2}\beta^{2}=\gamma^{2}(1-\beta^{2})=1$, that is $1+\gamma^{2}\beta^{2}=\gamma^{2}$, and the whole factor collapses:

Transverse, meaning a push across the motion. Then $\vv v\cdot\vv a=0$, the second term dies, and what is left is

Two different laws, differing by $\gamma^{2}$, for the same particle at the same instant. At $\gamma=10$ a longitudinal push is a hundred times less effective at producing acceleration than a transverse one of the same magnitude.

This is not an exotic regime. It is routine at any electron accelerator, and it is why the transverse focusing of a beam and its longitudinal acceleration are engineered as separate problems with separate hardware.

Grind box — inverting the relation: $\vv a$ in terms of $\vv F$

(2.5.47) gives $\vv F$ from $\vv a$. For solving actual problems you want the other direction, and inverting it is a small exercise in projection that is worth doing once, because the same trick appears in Chapter 3.3.

Step 1, get the component along $\vv v$. Dot (2.5.47) with $\vv v$:

$$\vv F\cdot\vv v = \gamma m\left[\vv v\cdot\vv a + \frac{\gamma^{2}}{c^{2}}(\vv v\cdot\vv a)v^{2}\right] = \gamma m\,(\vv v\cdot\vv a)\left[1+\gamma^{2}\beta^{2}\right] = \gamma^{3}m\,(\vv v\cdot\vv a),$$

using $1+\gamma^{2}\beta^{2}=\gamma^{2}$ again. So

$$\vv v\cdot\vv a = \frac{\vv F\cdot\vv v}{\gamma^{3}m}.$$

Step 2, substitute back. Put that into (2.5.47) and solve for $\vv a$:

$$\vv F = \gamma m\,\vv a + \gamma m\cdot\frac{\gamma^{2}}{c^{2}}\cdot\frac{\vv F\cdot\vv v}{\gamma^{3}m}\,\vv v = \gamma m\,\vv a + \frac{(\vv F\cdot\vv v)}{c^{2}}\,\vv v,$$

hence

$$\boxed{\;\vv a = \frac{1}{\gamma m}\left[\vv F - \frac{(\vv F\cdot\vv v)}{c^{2}}\,\vv v\right].\;}$$

Check it. For $\vv F\parallel\vv v$ the bracket is $\vv F(1-\beta^{2})=\vv F/\gamma^{2}$, so $a=F/\gamma^{3}m$ ✓ matching (2.5.48). For $\vv F\perp\vv v$ the second term vanishes, so $a=F/\gamma m$ ✓ matching (2.5.49).

What the formula says. The bracket removes from $\vv F$ the piece $(\vv F\cdot\hat v)\beta^{2}\hat v$, a projection along the motion weighted by $\beta^{2}$. So the acceleration is a distorted image of the force, squashed along the direction of travel.

At $\beta\to1$ the longitudinal response is suppressed entirely. You can push a nearly-light-speed particle as hard as you like along its motion and it barely speeds up, even though its energy rises steadily at the rate (2.5.46) says it must. The energy has to go somewhere, and it goes into $\gamma$ rather than into $v$.

One consequence worth naming. Set $\vv F$ constant, along $x$, starting from rest. Then $\dd(\gamma mv)/\dd t=F$ integrates immediately to $\gamma mv=Ft$, so

$$v(t) = \frac{Ft/m}{\sqrt{1+(Ft/mc)^{2}}} \;\longrightarrow\; c \quad\text{as } t\to\infty,$$

approaching but never reaching $c$, whereas the Newtonian $v=Ft/m$ sails past it at $t=mc/F$. Same constant force, same duration, entirely different destination. And the relativistic answer required no new physical assumption beyond $\vv p=\gamma m\vv v$.

6.4 · Why this book will not say "relativistic mass"

There is a tempting way to make (2.5.48) and (2.5.49) look like $\vv F=m\vv a$, which is to absorb the $\gamma$'s into the mass. Define $m_{\text{rel}}=\gamma m$ and you can write $\vv p=m_{\text{rel}}\vv v$ and $E=m_{\text{rel}}c^{2}$, and the formulas look Newtonian again. Textbooks did this for sixty years. It is a mistake, and the calculation above is the cleanest way to see why.

It would have to be two different numbers at once. To keep $\vv F=m\vv a$ you need $m_{\parallel}=\gamma^{3}m$ for a longitudinal push and $m_{\perp}=\gamma m$ for a transverse one. A "mass" that depends on which way you shove is not a property of the particle. It is a badly chosen name for the components of a tensor relation.

And for a general push, by (2.5.47), no scalar whatsoever works, because $\vv F$ and $\vv a$ are not even parallel. You would need a matrix, and calling a matrix "the mass" abandons the only thing the word was supposed to mean.

It makes $E=mc^{2}$ empty. With $m_{\text{rel}}\equiv E/c^{2}$, the celebrated equation $E=m_{\text{rel}}c^{2}$ says $E=E$. All the physical content lives in the rest mass and is invisible in the relativistic one: that a particle at rest still has energy $mc^{2}$, that binding changes rest mass, and that mass is not additive.

It hides the invariant. The real structure is (2.5.31). Here $m$ is what you get by contracting $p^{\mu}$ with itself, a Lorentz scalar, the same for all observers, in exactly the way the interval is the same for all observers.

Redefining "mass" to mean $E/c^{2}$ replaces an invariant with a component and throws away the geometry. It is the momentum-space version of saying that a metre stick "really" gets shorter when it moves.

So $m$ means rest mass, always, everywhere in this book, exactly as the conventions for Part II state. Energy is $E=\gamma mc^{2}$ and momentum is $\vv p=\gamma m\vv v$, and the $\gamma$'s stay where they are, visible, attached to the motion rather than smuggled into the particle. The phrase "relativistic mass" does not appear again.

7 · Light: the wave four-vector, and Doppler

Everything so far has been about particles. Waves need one more four-vector, and it comes from an argument so simple it is easy to undervalue.

7.1 · Why $k^{\mu}$ has to be a four-vector

A plane wave has a phase

with $\vv k$ the wave vector ($\abs{\vv k}=2\pi/\lambda$, pointing along propagation) and $\omega$ the angular frequency. Here is the physical claim, and it is the whole argument:

Now use the scalar to manufacture the four-vector. We want (2.5.50) written as a contraction, since a contraction is the shape the quotient theorem knows how to act on. With $x^{\mu}=(ct,x,y,z)$ and the metric lowering the spatial indices,

So $\Phi=-k_{\mu}x^{\mu}$. We know $x^{\mu}$ is a four-vector, and we have just argued that $\Phi$ is a scalar for every event $x^{\mu}$, not merely for one.

That is precisely the hypothesis of the quotient theorem (Chapter 2.4 §5.1's grind box). If $k_{\mu}A^{\mu}$ is a scalar for every four-vector $A^{\mu}$, then $k_{\mu}$ is a covector, and raising with $\eta$ makes $k^{\mu}$ a four-vector. Hence

For light in vacuum the dispersion relation is $\omega=c\abs{\vv k}$, which Chapter 2.1 §2 derived from Maxwell. That says precisely $k\cdot k = \omega^{2}/c^{2}-\abs{\vv k}^{2}=0$, so the wave four-vector of light is null. Set that beside $p\cdot p=0$ from §4.3 and the shape of Chapter 4.1's punchline is already visible. We come back to it below.

7.2 · Longitudinal Doppler

Everything about the Doppler effect is now one matrix multiplication. Let a light wave travel in the $+x$ direction in frame $S$, so

the spatial part having magnitude $\omega/c$ as the null condition requires. Now boost to $S'$ moving at $+\beta c$ along $x$, which is an observer running away from the source, in the same direction the light is going. Chapter 2.4 §2.1's boost matrix acts on any four-vector, so it acts on this one:

Both components pick up the same factor, as they must, since $k'^{\mu}$ has to be null too. The frequency we want is $\omega'=ck'^{0}$, so read that off and simplify with $\gamma=1/\sqrt{(1-\beta)(1+\beta)}$:

For a receding observer, $\beta\gt0$ and $\omega'\lt\omega$, which is a redshift. For an approaching one, replace $\beta\to-\beta$ and get a blueshift. At $\beta\ll1$ this reduces to $\omega'\approx\omega(1-\beta)$, the classical result, so nothing familiar is lost. What is new is the exactness, and the second-order term, which is where the physics is.

7.3 · Transverse Doppler: time dilation, seen directly

Now a case with no classical counterpart at all. Let a source move at $+\beta c$ along $x$ past a detector, and consider the light that reaches the detector travelling in the $+y$ direction in the detector's frame. That is the light emitted, as the lab sees it, at the moment of closest approach, so that the source's velocity is entirely perpendicular to the line of sight. Classically there is no Doppler shift whatsoever in that configuration, because the distance is momentarily not changing.

In the lab frame $S$, the received wave has

What we want to compare it with is the frequency the source emits, so transform to the source's rest frame $S'$, which moves at $+\beta c$ along $x$. Only the time component is needed:

But $ck'^{0}$ is the frequency in the source's own frame, which is the frequency the source actually emits. Call it $\omega_{0}$, a property of the atom. Then $\omega_{0}=\gamma\omega$, which rearranges to

A redshift, of exactly the factor $\gamma$, in a geometry where the classical prediction is precisely nothing. And its content is unmistakable. The factor $1/\gamma$ is the time-dilation factor of Chapter 2.2. The source is a clock, the clock is moving, so it runs slow, and you see its emission slowed.

The transverse Doppler shift is time dilation with no admixture of anything else, which is exactly why it was worth measuring.

Grind box — the general Doppler formula, aberration, and what Ives and Stilwell actually measured

Arbitrary angle. Let the light travel at angle $\theta$ to the $x$-axis in $S$, so $k^{\mu}=(\omega/c)(1,\cos\theta,\sin\theta,0)$. Boosting along $x$ by $\beta$:

$$\begin{aligned} \omega' &= \gamma\omega\big(1-\beta\cos\theta\big),\\[2pt] k'^{1} &= \gamma\frac{\omega}{c}\big(\cos\theta-\beta\big), \qquad k'^{2} = \frac{\omega}{c}\sin\theta. \end{aligned}$$

The first line is the general Doppler formula. Check the special cases. Setting $\theta=0$ gives $\omega'=\gamma\omega(1-\beta)$, which is (2.5.55) ✓. Setting $\theta=\pi/2$ gives $\omega'=\gamma\omega$, which is (2.5.58) ✓, with $\omega'=\omega_{0}$.

Aberration. Divide the second line by the first to get the direction in $S'$:

$$\cos\theta' = \frac{k'^{1}}{k'^{0}} = \frac{\cos\theta-\beta}{1-\beta\cos\theta},$$

which is the relativistic aberration of light. It is the same formula that turns the rain on your windscreen into a forward-slanting streak, but exact, and with the crucial difference that the aberration of light does not depend on any medium. This is Problem 3, where it becomes the headlight effect.

⚑ Ives and Stilwell, 1938: quoted experiment, derived analysis. The transverse configuration is hard to arrange, because "transverse" must be specified in a definite frame and a tiny angular error contaminates the result with a first-order longitudinal shift, which is $\gamma-1\approx\beta^{2}/2$ times larger in the relevant sense. Ives and Stilwell got around this by measuring longitudinally in both directions at once. Let a source recede with $\beta$ and approach with $\beta$, so that in wavelengths $\lambda_{\pm}=\lambda_{0}\sqrt{(1\pm\beta)/(1\mp\beta)}$. Average them:

$$\frac{\lambda_{+}+\lambda_{-}}{2} = \frac{\lambda_{0}}{2}\left[\sqrt{\frac{1+\beta}{1-\beta}}+\sqrt{\frac{1-\beta}{1+\beta}}\right] = \frac{\lambda_{0}}{2}\cdot\frac{(1+\beta)+(1-\beta)}{\sqrt{(1-\beta)(1+\beta)}} = \gamma\,\lambda_{0}.$$

The first-order shifts cancel exactly and what survives is pure $\gamma$, which is the transverse effect extracted from a longitudinal measurement. Ives and Stilwell ran hydrogen canal rays at $\beta\approx5\times10^{-3}$, where $\gamma-1\approx1.25\times10^{-5}$, and found the predicted second-order displacement of the mean. It remains one of the cleanest direct tests of time dilation, and the derivation above is the entire theory of the experiment. We quote the experiment, and the algebra is ours.

8 · Collisions: invariant mass, and why colliders exist

The whole practical value of §2 is that it turns collision problems into linear algebra. You write down the four-momenta, you add them, you square whatever combination kills the unknowns, and you are done. This section builds the three tools and then uses them to say something quantitative and slightly outrageous about accelerator design.

8.1 · The invariant mass of a system

Take any collection of particles, whatever they are doing, and add up their four-momenta:

A sum of four-vectors is a four-vector, so $P^{\mu}$ transforms properly and has an invariant square. That square is the object we want, so give it a name. Define the system's invariant mass $M$ by

Everyone agrees on $M$. And the system's total four-momentum is conserved, by §2, so $M$ is conserved as well. That is a strong constraint on what a system may turn into.

8.2 · The centre-of-momentum frame

Suppose $P^{\mu}$ is timelike, $P\cdot P\gt0$. Then by exactly the construction of Chapter 2.3 §6.1 there is a frame in which its spatial part vanishes. Boost with $\vv\beta_{\text{cm}}=\vv P\big/P^{0} = c\sum_{i}\vv p_{i}\big/\sum_{i}E_{i}$, which has magnitude less than $1$ precisely because $P^{\mu}$ is timelike. In that frame, called the centre-of-momentum frame or CM frame,

So the invariant mass of a system is its total energy in the frame where it is collectively at rest, divided by $c^{2}$. Notice that this is the direct generalisation of $E_{0}=mc^{2}$ from one particle to many.

Notice also that $\vv\beta_{\text{cm}}$ is exactly (2.5.32) applied to the system's four-momentum, which is a small piece of evidence that we defined things sensibly.

8.3 · Mandelstam $s$

For a two-body collision the invariant mass of the initial state gets its own name. Define

(The factor of $c^{2}$ is bookkeeping so that $\sqrt{s}$ is an energy in SI units. In the natural units of Part V, where $c=1$, the definition is simply $s=(p_{1}+p_{2})^{2}$ and this parenthesis is unnecessary. Chapter 5.9 introduces the companions $t$ and $u$. Here we need only $s$.)

Our goal is to see what $s$ depends on, so expand the product once, using $p_{i}\cdot p_{i}=m_{i}^{2}c^{2}$ on the two square terms:

Everything now depends on the single cross term, and $p_{1}\cdot p_{2}$ is an invariant we can evaluate in whichever frame is convenient. Two configurations matter.

Fixed target. Particle 1 has energy $E$ and hits particle 2 at rest, $p_{2}^{\mu}=(m_{2}c,\vv 0)$. Then $p_{1}\cdot p_{2}=(E/c)(m_{2}c) - \vv p_{1}\cdot\vv 0 = m_{2}E$, so

Collider. Two beams of energy $E$ meet head-on with equal and opposite momenta. Then $\sum\vv p_{i}=0$, so you are already in the CM frame and

8.4 · Threshold energies, and the antiproton

Here is the technique that makes $s$ worth defining. A reaction can proceed only if the initial state carries enough invariant mass to build the final state.

The minimum case is unambiguous. At threshold the products have no kinetic energy left over in the CM frame, so they are all at rest there and moving together as one. The final invariant mass is then just the sum of the final rest masses. Since $s$ is conserved and invariant,

Take the reaction that mattered historically, which is producing an antiproton by smashing a proton beam into a hydrogen target,

Charge and baryon number both force the extra proton to accompany the antiproton, since you cannot make $\bar p$ alone, so four particles of mass $m_{p}$ must emerge. (Antiparticles have the same rest mass as their partners. ⚑ We quote that here, and Chapter 5.5 derives it, where it falls out of the Dirac equation.) Now set $\sqrt{s}=4m_{p}c^{2}$ in (2.5.64) with $m_{1}=m_{2}=m_{p}$ and solve for the beam energy:

That $E$ is the projectile's total energy, and what an accelerator is rated by is the kinetic energy it delivers. So subtract the rest energy the proton had before anyone switched the machine on:

Six proton rest energies to make two particles' worth of new rest mass. The factor of three overhead is exactly the energy locked up in the forward motion of the centre of mass, which (2.5.64) says you cannot spend. ⚑ Historically: the Bevatron at Berkeley was designed to reach $6.2\ \mathrm{GeV}$, comfortably above this threshold and for this reason, and the antiproton was found there in 1955. The number in (2.5.69) is a piece of accelerator engineering that came out of four lines of four-vector algebra.

Grind box — the general threshold formula, and a second worked case

Do the algebra once in general so you never have to repeat it. A projectile of mass $m_{1}$ and total energy $E$ strikes a stationary target of mass $m_{2}$ and produces final particles of total rest mass $\mathcal{M}\equiv\sum_{f}m_{f}$. Equate (2.5.64) with (2.5.66):

$$\big(m_{1}^{2}+m_{2}^{2}\big)c^{4} + 2m_{2}c^{2}E \;=\; \mathcal{M}^{2}c^{4},$$ $$\boxed{\;E_{\text{threshold}} \;=\; \frac{\big[\mathcal{M}^{2}-m_{1}^{2}-m_{2}^{2}\big]c^{2}}{2m_{2}}.\;}$$

Check against (2.5.68): $\mathcal{M}=4m_{p}$, $m_{1}=m_{2}=m_{p}$ gives $E=(16-1-1)m_{p}c^{2}/2=7m_{p}c^{2}$ ✓.

A second case, pion photoproduction, $\gamma+p\to p+\pi^{0}$. Now $m_{1}=0$ for the photon, $m_{2}=m_{p}=938.27\ \mathrm{MeV}/c^{2}$, and $\mathcal{M}=m_{p}+m_{\pi^{0}}$ with $m_{\pi^{0}}c^{2}=134.98\ \mathrm{MeV}$ (⚑ quoted mass). Then

$$E_{\gamma} = \frac{(m_{p}+m_{\pi})^{2}-m_{p}^{2}}{2m_{p}}c^{2} = \frac{2m_{p}m_{\pi}+m_{\pi}^{2}}{2m_{p}}c^{2} = m_{\pi}c^{2}\left(1+\frac{m_{\pi}}{2m_{p}}\right).$$

Numerically $134.98\times(1+0.0719)=144.7\ \mathrm{MeV}$. The photon must supply about 7% more than the pion's rest energy, the surplus being the recoil the proton is obliged to take. Note how the formula degrades gracefully. For a very heavy target, $m_{2}\to\infty$ and the overhead vanishes, which is why a nucleus makes a better anvil than a single proton.

The same physics, from the CM frame. There is a shortcut worth knowing. The overhead in fixed-target work is the kinetic energy of the centre of mass, which by (2.5.32) moves at $\beta_{\text{cm}}=p_{1}c/(E+m_{2}c^{2})$. Everything above can be re-derived by boosting to that frame and demanding the products be at rest, but the invariant route is shorter precisely because it never needs $\beta_{\text{cm}}$ at all. That is the habit to acquire: find the invariant, evaluate it in the easy frame, use it in the hard one.

9 · Mass is not additive

Section 2.3's grind box already produced the result that makes this section necessary. Two lumps of putty of rest mass $m$ each, colliding and sticking, form a body of rest mass $2\gamma m$, not $2m$. The kinetic energy did not disappear. It became rest mass. Run that backwards and you have the most consequential fact in applied physics.

9.1 · Binding energy and the mass defect

Consider a bound system: a nucleus, an atom, a planet. Assemble it from constituents that start far apart and at rest, and let it settle by radiating away the excess. The total four-momentum is conserved throughout, so in the frame where the final object sits at rest,

where $B\gt0$ is the energy carried away, called the binding energy. The bound system is lighter than its parts, by exactly the energy you would have to put back to take it apart, over $c^{2}$.

That difference is the mass defect. It is not a small correction to a picture in which mass is additive. It is a demonstration that mass was never additive.

Real numbers, since the whole point is that this is measurable. ⚑ The masses below are quoted from standard tables, and everything done with them is ours.

System	Constituents ($\mathrm{MeV}/c^{2}$)	Bound mass	$B$	$B/\sum m c^{2}$
Deuteron $\ ^{2}$H	$938.272 + 939.565 = 1877.837$	$1875.613$	$2.224\ \mathrm{MeV}$	$0.118\%$
Helium-4	$2(938.272)+2(939.565)=3755.675$	$3727.379$	$28.30\ \mathrm{MeV}$	$0.753\%$
Hydrogen atom	$938.272+0.511=938.783$	$938.783$	$13.6\ \mathrm{eV}$	$1.4\times10^{-8}$

Read the last column. Chemistry runs at $10^{-8}$ or $10^{-10}$ of the rest energy. That is why Lavoisier could weigh reactants and products and conclude that mass is conserved. His balance was many orders of magnitude too coarse to see the defect, and he was right to the precision available.

Nuclear binding runs at $10^{-3}$, which is five to seven orders of magnitude larger. That gap is the entire reason nuclear energy is a different kind of thing rather than a better kind of chemistry.

Grind box — the arithmetic, and a fusion reaction end to end

Deuteron. A proton and a neutron bound together. The constituent rest energies are $938.272+939.565=1877.837\ \mathrm{MeV}$, and the measured deuteron rest energy is $1875.613\ \mathrm{MeV}$. Difference:

$$B = 1877.837 - 1875.613 = 2.224\ \mathrm{MeV}, \qquad \frac{B}{\sum m c^{2}} = \frac{2.224}{1877.837} = 1.184\times10^{-3}.$$

So a deuteron weighs about $0.12\%$ less than the parts you made it from. That is a fractional mass difference roughly $10^{5}$ times larger than any chemical bond, and comfortably measurable with a mass spectrometer.

Helium-4. The constituents come to $2\times938.272 + 2\times939.565 = 3755.675$ and the measured mass to $3727.379$, so $B=28.296\ \mathrm{MeV}$. That is $7.07\ \mathrm{MeV}$ per nucleon and $0.753\%$ of the total. Helium-4 is unusually tightly bound for its size, which is why $\alpha$ particles exist as a distinct thing and why stellar nucleosynthesis piles up at helium.

A complete reaction: D–T fusion. $\ ^{2}\mathrm{H} + {}^{3}\mathrm{H} \to {}^{4}\mathrm{He} + n$. Using nuclear rest energies in MeV:

$$\begin{aligned} \text{in:}\quad & 1875.613 + 2808.921 = 4684.534\\ \text{out:}\quad & 3727.379 + 939.565 = 4666.944\\[3pt] \text{released:}\quad & Q = 4684.534 - 4666.944 = 17.59\ \mathrm{MeV}. \end{aligned}$$

As a fraction of the input rest energy, $17.59/4684.534 = 3.755\times10^{-3}$, which is $0.375\%$. Per kilogram of fuel that is $0.00375\,c^{2}=3.4\times10^{14}\ \mathrm{J\,kg^{-1}}$, against roughly $5\times10^{7}\ \mathrm{J\,kg^{-1}}$ for burning hydrocarbons. The ratio is about seven million.

The lesson to extract. Every one of these is the same calculation: add the rest energies before, add them after, and the difference is kinetic energy. Nothing was "converted into energy" in a way that requires new physics. The bookkeeping quantity $\sum_{i}E_{i}$ was conserved throughout, and all that changed was how much of it was sitting in the rest-mass column.

9.2 · Where your mass actually comes from

Push the argument one step further than nuclear physics does and it stops being a correction and becomes the main effect.

A proton has rest energy $938\ \mathrm{MeV}$. It is made of three light quarks whose rest energies sum to roughly $9\ \mathrm{MeV}$. ⚑ That figure is quoted from Chapter 6.5's tables, and note that even defining a quark mass takes care, since quarks are never found alone. It accounts for about $1\%$ of the proton.

The other $99\%$ is exactly what §8.2's warning described. It is the energy of quarks moving relativistically inside a small volume, plus the energy of the gluon field binding them, all of it appearing as invariant mass because the constituent four-momenta point in many different directions and their spatial parts cancel while their energies add.

So the mass of ordinary matter is, to better than $99\%$, not a property of its ingredients. It is confined kinetic and field energy, measured in the frame where the total momentum vanishes. That is (2.5.60), applied to a bound state of a strongly coupled field theory.

Chapter 6.5 does that calculation, or rather explains why it can only be done numerically. The structural statement is available now, and it belongs to this chapter rather than to quantum chromodynamics.

10 · Worked examples

11 · Your turn