Mass, Energy, Linear Momentum; The Law of Motion

Three results that look unrelated — what mass is, why a receding source reddens, and why F = ma is wrong — turn out to be one statement, and none of them needs a new assumption.

🎯 Why this matters

A detector measures energies and momenta in whatever frame it happens to sit in, and can still report a mass. No part of the apparatus needs to know how fast anything was moving, and that is what makes results from different experiments comparable at all.

§1.1 built a transformation for the coordinates of an event. The payoff is that the same transformation, with the same γ, applies to energy and momentum — and the invariant it leaves behind is what we call mass.

The energy–momentum 4-vector

\htmlClass{t-pxp}{p'_x} = \htmlClass{t-g}{\gamma}\left(\htmlClass{t-px}{p_x} - \htmlClass{t-b}{\beta}\frac{\htmlClass{t-E}{E}}{c}\right), \qquad p'_y = p_y, \quad p'_z = p_z, \qquad \frac{\htmlClass{t-Ep}{E'}}{c} = \htmlClass{t-g}{\gamma}\left(\frac{\htmlClass{t-E}{E}}{c} - \htmlClass{t-b}{\beta}\,\htmlClass{t-px}{p_x}\right)

(1.27)

Compare with Eq. (1.3) on the previous page. It is the same matrix — (E/c, p) transforms exactly as (ct, r) does.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

⚙️ Engineer’s bridge

A 4-vector is a length-4 array with one rule attached: under a change of frame it gets multiplied by a fixed matrix — the same matrix for every 4-vector in the theory. Position, energy–momentum, current density, spin: all the same matrix.

The matrix is the one from §1.1, and it is not orthogonal. The quantity it preserves uses a flipped sign:

\text{Euclidean: } x^2 + y^2 + z^2 \qquad\text{vs}\qquad \text{Minkowski: } E^2 - p_x^2 - p_y^2 - p_z^2 .

So: a boost is a matrix multiply, and the Minkowski norm is a checksum that survives the transform. Two frames that disagree about every single component still compute the same checksum — and the checksum is the particle’s mass.

Where it breaks: a checksum in the usual sense is a redundant function of the data, deliberately added. This one is not added — it is the only frame-independent number the data contains. Everything else is a matter of who is looking.

What the invariant is

\htmlClass{t-m}{m^2}c^4 = \htmlClass{t-E}{E}^2 - \htmlClass{t-p}{p}^2c^2 \qquad\longrightarrow\qquad \htmlClass{t-m}{m^2} = \htmlClass{t-E}{E}^2 - \htmlClass{t-p}{p}^2 \;\;(\hbar = c = 1)

(1.28)

This is a definition, not a derived result: the norm of the energy–momentum 4-vector IS what mass means.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — mass is the length of a 4-vector, not an amount of matter

Mass is not “how much matter there is”. It is the length of a 4-vector, in a geometry with one flipped sign. Two consequences the book insists on:

A massless particle has $E = pc$ exactly — energy and momentum are the same number. That is not an approximation for the photon; it is exact.
The energy of a particle can be anything from $m$ upwards, depending on who is looking, but never below $m$ . Slide the boost in the widget and watch $E'$ bottom out at exactly the mass.

🧮 4-vector under a boost — everything moves except the checksum

mass m (GeV)0.9383pₓ (GeV)3.000p_y (GeV)0.000

boost β along xβ = 0.600 · γ = 1.2500

component	frame S	frame S′
E	3.1433	1.6791
pₓ	3.0000	1.3925
p_y	0	0
p_z	0	0
E² − p² → m²	0.88035	0.88035

The boost, written out, is a 2×2 matrix acting on (E, pₓ) while p_y and p_z ride along untouched:

( E′ )   (   1.2500   -0.7500 ) ( E  )
( pₓ′)   (  -0.7500    1.2500 ) ( pₓ )   det = 1.000000

Discrepancy in the invariant: 1.33e-15 GeV² — floating point only. Algebraically it is exactly zero, for every β.

E′(β)
pₓ′(β)
the invariant m — flat, by construction

E′ has its minimum exactly at the rest frame, and the minimum value is m. No boost can make a particle's energy smaller than its mass — which is what 'mass is the invariant' means, drawn.

Two words this book refuses to use

Aside — “relativistic mass” and “rest mass”

You will meet both outside this book. Bettini calls them “useless and misleading”, and he is right for a concrete reason:

“Relativistic mass” means $m\gamma$ . But $m\gamma$ is just $E/c^2$ — the fourth component of a 4-vector, which changes from frame to frame. Calling it a mass invites you to substitute it into $F = ma$ , which is wrong (see below), and it has no meaning at all for a massless particle.
“Rest mass” is a retronym for the only thing that was ever mass. The adjective is redundant: mass is invariant, so there is nothing else it could be.

The site follows the book: mass means the invariant, always, and $E$ is called energy.

The three relations you actually use, all of which follow from (1.28):

Relation	Book eq.	Holds for	Reading
$\mathbf{p} = (E/c^2)\,\mathbf{v}$	(1.29)	everything, massive or not	the most general link between the three
$pc = E$	(1.30)	massless only	photon momentum and energy are one number
$E = m\gamma c^2$	(1.31)	$m \neq 0$	energy diverges as $\beta \to 1$ : the speed limit, enforced by economics
$\mathbf{p} = m\gamma\mathbf{v}$	(1.32)	$m \neq 0$	the momentum that goes into Newton’s law below

⚠️ A caveat you will need in Chapter 8

Mass is defined for eigenstates of the free Hamiltonian — just as only a monochromatic wave has a well-defined frequency. Chapter 8 meets two-state systems (

K^0_S

B^0,\ \bar B^0

B_s^0,\ \bar B_s^0

D^0,\ \bar D^0

) that are produced in states which are not stationary. For those it is improper to speak of “the” mass or “the” lifetime at all. Flagged now so it does not come as a shock later.

🔢 Worked example — the Doppler effect falls out for free (Example 1.1)

A source emits a photon of energy

E_0

forwards, along

x

. The source moves towards you at

\beta

. What energy do you measure?

For a photon $p_x = E_0/c$ , so the inverse of the last line of (1.27) gives

\frac{E}{c} = \gamma\left(\frac{E_0}{c} + \beta p'_x\right) = \gamma\frac{E_0}{c}(1+\beta) \qquad\Longrightarrow\qquad \frac{E}{E_0} = \gamma(1+\beta) = \sqrt{\frac{1+\beta}{1-\beta}} .

At $\beta = 0.5$ that is $\sqrt{3} = 1.732$ : the photon arrives 73 % more energetic.

And this is exactly the relativistic Doppler formula for frequency, $\nu/\nu_0 = \sqrt{(1+\beta)/(1-\beta)}$ — which is no coincidence, because $E = h\nu$ . The Doppler effect and the transformation of a 4-vector are the same statement. One line of kinematics reproduced a result usually derived from wave theory.

§1.3 The law of motion — where Newton survives and where he does not

Newton’s law is fine at any speed in the form he actually wrote it:

\mathbf{F} = \frac{d\mathbf{p}}{dt}, \qquad \mathbf{p} = m\gamma\mathbf{v} .

The form $\mathbf{F} = m\mathbf{a}$ — which Einstein used in 1905 — is simply wrong. Here is what you get instead.

🪜 Why force and acceleration are not parallel

Step 1 of 5 — Differentiate the momentum(1.33)

\mathbf{F} = \frac{d\mathbf{p}}{dt} = m\gamma\,\mathbf{a} + m\,\frac{d\gamma}{dt}\,\mathbf{v}

Why you may do this: Product rule on p = mγv. Newtonian mechanics has no second term because there γ ≡ 1 and its derivative vanishes.

Bettini pp. 10–11. The whole difficulty is that γ depends on the speed, so it carries its own time derivative.

force ⊥ velocity: a = F/(mγ)
force ∥ velocity: a = F/(mγ³)
Newton: a = F/m

The same force, applied two ways. Pushing along the motion becomes futile far faster than pushing across it — the two differ by γ², which is 5.3 at β = 0.9 and 26 at β = 0.99. This is why every accelerator separates the jobs: RF cavities push forwards, magnets push sideways.

⚙️ Engineer’s bridge — inertia becomes anisotropic

In Newtonian mechanics inertia is one scalar: one number relating any force to the acceleration it produces. Relativistically it is direction-dependent — the response along the motion is

\gamma^3

times stiffer than the response across it. If you insisted on writing

F_i = M_{ij}a_j

, the “mass” would have to be a tensor, not a scalar, with eigenvalues

m\gamma^3

(longitudinal) and

m\gamma

(twice, transverse).

That is a familiar situation — an anisotropic medium, a crystal with direction-dependent stiffness, an inductance matrix with off-diagonal terms. And it is precisely why the historical “longitudinal mass” and “transverse mass” terminology was invented, and then abandoned: the honest statement is that $\mathbf{F} = d\mathbf{p}/dt$ still holds and $\mathbf{F} = m\mathbf{a}$ never did.

Where it breaks: an anisotropic material really is stiffer along one axis — the anisotropy belongs to the object and travels with it. Here it belongs to nothing: the “longitudinal” and “transverse” masses are properties of the relation between force and acceleration in one frame, they change if you change frame, and the particle has exactly one mass in all of them. That is precisely why the terminology was invented and then abandoned. The useful residue is the warning, not the model: $\mathbf F = m\mathbf a$ was never the law, and the places it appears to fail are the places it was never entitled to hold.

Reproduce it

import numpy as np

m, px = 0.938272, 3.0                      # GeV, natural units (c = 1)
E = np.hypot(m, px)
print(f"proton p=3 GeV:  E = {E:.6f} GeV, beta = {px/E:.6f}, gamma = {E/m:.6f}")

def boost(E, px, b):                       # the 2x2 matrix of Eq. (1.27)
    g = 1 / np.sqrt(1 - b*b)
    return g*(E - b*px), g*(px - b*E)

for b in (0.0, 0.6, px/E, -0.9):
    Ep, pxp = boost(E, px, b)
    print(f"  beta={b:+.5f}: E'={Ep:9.6f}  px'={pxp:+9.6f}  m'={np.sqrt(Ep**2 - pxp**2):.9f}")

err = max(abs(np.sqrt(np.subtract(*np.square(boost(E, px, b)))) - m)
          for b in np.linspace(-0.999, 0.999, 100000))
print(f"max |m' - m| over 10^5 boosts = {err:.2e} GeV  (floating point only)")

b = 0.5
print(f"Doppler at beta=0.5:  E/E0 = {np.sqrt((1+b)/(1-b)):.6f}")
g = 1 / np.sqrt(1 - 0.9**2)
print(f"at beta=0.9: a_parallel = {1/g**3:.5f} F/m, a_perp = {1/g:.5f} F/m, "
      f"ratio = gamma^2 = {g**2:.4f}")

prints

proton p=3 GeV:  E = 3.143303 GeV, beta = 0.954410, gamma = 3.350098
beta=+0.00000: E'= 3.143303  px'=+3.000000  m'=0.938272000
beta=+0.60000: E'= 1.679129  px'=+1.392523  m'=0.938272000
beta=+0.95441: E'= 0.938272  px'=+0.000000  m'=0.938272000
beta=-0.90000: E'=13.405457  px'=+13.372581  m'=0.938272000
max |m' - m| over 10^5 boosts = 2.64e-12 GeV  (floating point only)
Doppler at beta=0.5:  E/E0 = 1.732051
at beta=0.9: a_parallel = 0.08282 F/m, a_perp = 0.43589 F/m, ratio = gamma^2 = 5.2632

🔑 If you remember only three things

Newton’s second law never failed; one of its two forms did. Force as the rate of change of momentum survives at any speed; force as mass times acceleration does not.
Mass is defined here, not discovered. Eq. (1.28) says what the word means, and every later use — systems, resonances, missing mass — inherits that definition rather than extending it.
Two frames disagree about every component and agree about one number. Which components you measure is a fact about your apparatus; the number they share is a fact about the particle.

Where this goes next

§1.4–1.5 applies the invariant to systems of particles — where it stops being the sum of the parts and starts being the thing every collision is analysed with.
§1.6–1.7 puts numbers on all of this and introduces the cross-section.
§2.5 The Dirac equation is where $E^2 = p^2 + m^2$ becomes a wave equation and the negative-energy root stops being a nuisance and becomes antimatter.

✅ Check yourself — mass, energy and the law of motion

0/5 answered · 0 correct

1.In the 4-vector widget, load the proton with $p_x = 3$ GeV and sweep the boost. Which quantity does not change?
2.Why does this book refuse the term "relativistic mass" for $m\gamma$ ?
3.A source moving towards you at $\beta = 0.5$ emits a photon of energy $E_0$ forwards. What do you measure, and what is the deeper point?
4.At $\beta = 0.9$ , the same force produces about 5.3 times more acceleration when applied across the motion than along it. Where does that factor come from, and what is it used for?
5.If you insisted on writing $F_i = M_{ij}a_j$ relativistically, what would $M$ have to be?