Four-Vectors & Relativistic Doppler

A proton in the LHC carries 7 TeV of energy. Fire it at a bottle of liquid hydrogen sitting on a bench and the collision has about 115 GeV to spend on making new particles: less than two percent of what you paid for. Turn the machine around so two 7 TeV beams meet head-on and suddenly you have 14 TeV to spend. Same protons, same energy per proton, a hundred-fold difference in what the collision can create.

That factor is not a detail of accelerator engineering. It falls out of one number you can compute in a line, once you stop treating energy and momentum as two separate bookkeeping problems and start treating them as one object. This lesson is about that object, about the single invariant it carries, and about what the same idea says when the particle you are tracking is a photon and the thing you measure is its colour.

Energy and momentum are one thing

The boundary-value methods you just finished worked because you chose coordinates that matched the symmetry of the problem. Relativity rewards the same instinct, except the symmetry is between frames.

You already know that time and space mix when you change frame. A clock reading $t$ and a position $x$ in one frame become a different pair in another, tangled by the Lorentz transformation. Energy and momentum do exactly the same thing, with exactly the same coefficients. That is not a coincidence; it is forced on you by the fact that momentum is mass times a rate of change of position, and the position and the time both transform.

So stack them. Write

$P = (E/c,\; p_x,\; p_y,\; p_z)$

and call it the energy-momentum four-vector. $E$ is total energy including rest energy, and $p_x, p_y, p_z$ are the ordinary relativistic momentum components, $\vec{p} = \gamma m \vec{v}$, where $\gamma = 1/\sqrt{1-\beta^2}$ and $\beta = v/c$. Dividing $E$ by $c$ just puts all four entries in the same units so they can mix.

The payoff is immediate. Conservation of energy and conservation of momentum stop being four equations you have to track separately. They become one statement:

$\sum P_{\text{in}} = \sum P_{\text{out}}$

and because both sides are four-vectors, that equation holds component by component in every frame at once.

The invariant is the whole trick

An ordinary vector has a length that does not care how you rotate your axes. A four-vector has an analogue, but with a minus sign in front of the space part:

$P \cdot P = \frac{E^2}{c^2} - |\vec{p}\,|^2$

Work that out for a single particle of mass $m$. Its energy is $E = \gamma mc^2$ and its momentum is $p = \gamma m v$, so

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

since $\gamma^2(1-\beta^2) = 1$ exactly. Every $\gamma$ cancelled. What is left contains nothing but the rest mass, which is a property of the particle and not of your motion, so every observer computes the same number. Rearranged, that is the relation you will use more than any other:

$E^2 = (pc)^2 + (mc^2)^2$

Read it as a right triangle. The hypotenuse is the total energy, one leg is $pc$, the other is the rest energy $mc^2$. Grow the momentum and the energy grows, but never below $mc^2$, which is the floor a massive particle can never get under. Set $m = 0$ and the triangle collapses: a photon has $E = pc$ and no floor at all.

Working in MeV saves you from carrying factors of $c$. Quote masses as $mc^2$ in MeV, momenta as $pc$ in MeV, and the triangle becomes a statement about three numbers in the same unit.

Example 1. A proton has total energy $10.00$ GeV. Find $pc$. Take $m_pc^2 = 938.3$ MeV $= 0.9383$ GeV.

$(pc)^2 = E^2 - (mc^2)^2 = (10.00)^2 - (0.9383)^2 = 100.00 - 0.8804 = 99.120\ \text{GeV}^2$

So $pc = \sqrt{99.120} = 9.956$ GeV. The proton's momentum is within half a percent of its energy: at ten times the rest energy, it is already almost a photon as far as the triangle is concerned.

The mass of a system

Now do the same squaring for a collection of particles. Add up all their four-vectors into a total $P_{\text{tot}}$, and square:

$s \equiv P_{\text{tot}} \cdot P_{\text{tot}} \, c^2 = \left(\sum E\right)^2 - \left(\sum \vec{p}c\right)^2$

This is called the Mandelstam $s$, and $\sqrt{s}$ has units of energy. Because it came from squaring a four-vector, it is the same in every frame. And because the total four-vector is conserved, it is also the same before and after the collision.

$\sqrt{s}$ has a physical meaning worth holding onto. Go to the frame where the total momentum is zero: the centre-of-momentum frame, the natural one for any collision. There $\sum \vec{p} = 0$, so $\sqrt{s}$ is nothing but the total energy in that frame. That is the energy genuinely available to make new things.

Example 2. A 7.00 TeV proton hits a proton at rest. Find $\sqrt{s}$.

Before the collision the beam has $E_1 = 7000$ GeV and $p_1c = \sqrt{E_1^2 - m^2c^4}$. The target has $E_2 = mc^2$ and $p_2 = 0$. So

$s = (E_1 + mc^2)^2 - (p_1c)^2 = E_1^2 + 2E_1 mc^2 + (mc^2)^2 - E_1^2 + (mc^2)^2$

$s = 2(mc^2)^2 + 2E_1mc^2 = 2mc^2\left(E_1 + mc^2\right)$

That two-line result is worth memorising for fixed-target problems. Numerically, $s = 2(0.9383)(7000 + 0.9383) = 13135\ \text{GeV}^2$, so $\sqrt{s} = 114.6$ GeV.

Compare the collider: two 7 TeV beams head-on have $\sum \vec{p} = 0$ already, so $\sqrt{s} = 14000$ GeV. The fixed target wastes almost everything because most of the beam energy goes into carrying the debris forward, and momentum conservation will not let it stop. The $\sqrt{s}$ of a fixed target grows only as $\sqrt{E_1}$, which is why every high-energy machine built since the 1970s is a collider.

Thresholds

Here is where the invariant earns its keep. A threshold problem asks for the minimum beam energy to produce some set of products. In the lab that final state is several particles with unknown momenta, which looks hopeless. But you never have to touch the lab-frame final state.

At threshold, the products have just barely enough energy to exist and none left over for motion: in the centre-of-momentum frame they all come out at rest, moving together. So in that frame $\sqrt{s}$ after the collision is exactly the sum of the product rest energies. Set that equal to $\sqrt{s}$ computed before the collision in the lab, where you know everything.

Example 3. Protons strike protons at rest to make an antiproton: $p + p \to p + p + p + \bar{p}$. Find the minimum beam kinetic energy.

Four particles come out, each of rest energy $938.3$ MeV, so at threshold $\sqrt{s} = 4m_pc^2$. From Example 2, the lab side gives $s = 2m_pc^2(E_1 + m_pc^2)$. Equate:

$2m_pc^2\left(E_1 + m_pc^2\right) = 16(m_pc^2)^2$

$E_1 + m_pc^2 = 8m_pc^2 \quad\Longrightarrow\quad E_1 = 7m_pc^2$

That is total energy, so the kinetic energy is $K = E_1 - m_pc^2 = 6m_pc^2 = 6(938.3) = 5630$ MeV, or $5.63$ GeV. You need six proton masses of kinetic energy to make one antiproton, and five of those six are wasted on forward motion.

The same three lines handle every threshold. Only two things change: what $\sqrt{s}$ is before, and what the products weigh. If the incoming particle is a photon, put $E_\gamma$ and $p_\gamma c = E_\gamma$ into the sum. If the two incoming particles collide head-on rather than one being at rest, the momenta subtract instead of vanishing, and for a head-on collision of a high-energy proton with a low-energy photon you get $s \approx (m_pc^2)^2 + 4E_pE_\gamma$, which is the whole of the GZK cutoff problem.

Decays, which are thresholds run backwards

A particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$. The parent has $E = Mc^2$ and $\vec{p} = 0$, so the products carry equal and opposite momenta of size $p$. Write energy conservation and use the triangle on each product:

$Mc^2 = \sqrt{(pc)^2 + (m_1c^2)^2} + \sqrt{(pc)^2 + (m_2c^2)^2}$

Solving that for $p$ is messy directly. The clean route is to isolate one particle. Move particle 2's four-vector to the left, $P_2 = P_M - P_1$, and square:

$(m_2c^2)^2 = (Mc^2)^2 + (m_1c^2)^2 - 2Mc^2E_1$

because the cross term is $2 P_M \cdot P_1 c^2 = 2Mc^2E_1$ when the parent is at rest. Rearranged,

$E_1 = \frac{(Mc^2)^2 + (m_1c^2)^2 - (m_2c^2)^2}{2Mc^2}$

Example 4. A pion at rest decays, $\pi^+ \to \mu^+ + \nu_\mu$, with $m_\pi c^2 = 139.6$ MeV, $m_\mu c^2 = 105.7$ MeV, and a massless neutrino. The muon's energy is

$E_\mu = \frac{(139.6)^2 + (105.7)^2 - 0}{2(139.6)} = \frac{19488 + 11172}{279.2} = \frac{30660}{279.2} = 109.8\ \text{MeV}$

The neutrino takes the rest: $139.6 - 109.8 = 29.8$ MeV. Since it is massless, its momentum is $p_\nu c = 29.8$ MeV, and the muon must carry the same $29.8$ MeV of momentum in the opposite direction. Check the triangle: $\sqrt{29.8^2 + 105.7^2} = \sqrt{888 + 11173} = 109.8$ MeV. It closes.

Photons: Compton and Doppler

A photon's four-vector is $(E/c,\, \vec{p}\,)$ with $|\vec{p}\,|c = E$. Feed that into the same machinery and two famous results drop out.

For Compton scattering, a photon of energy $E$ hits an electron at rest and leaves at angle $\theta$ with energy $E'$. Conservation says $P_\gamma + P_e = P'_\gamma + P'_e$. You do not care about the recoiling electron, so isolate it and square, which kills its unknown direction:

$\frac{1}{E'} - \frac{1}{E} = \frac{1 - \cos\theta}{m_ec^2}$

In wavelength this is the familiar $\Delta\lambda = (h/m_ec)(1-\cos\theta)$, with $h/m_ec = 2.43$ pm. A $1.000$ MeV photon at $60.0^\circ$: $1/E' = 1/1.000 + (1-0.500)/0.511 = 1 + 0.9785 = 1.9785$, so $E' = 0.5054$ MeV. The largest possible loss is at $\theta = 180^\circ$, which is the Compton edge.

For the Doppler shift, transform the photon's four-vector instead. A source at rest emits frequency $f_0$, so its photon has $E_0 = hf_0$. Boost to the frame where the source moves at $\beta$ and the energy transforms like any time-component: $E = \gamma(E_0 + \beta p'_x c)$. For a photon emitted at angle $\theta$ from the motion in the observer's frame the result is

$\frac{f}{f_0} = \frac{1}{\gamma\left(1 - \beta\cos\theta\right)}$

Three cases matter. Straight toward you, $\theta = 0$, and this simplifies to $f = f_0\sqrt{(1+\beta)/(1-\beta)}$: a blueshift. Straight away, flip the sign of $\beta$: a redshift. And sideways, $\theta = 90^\circ$, where the classical prediction is no shift at all but you get $f = f_0/\gamma$. That transverse shift is pure time dilation, and measuring it is direct evidence that moving clocks run slow.

Redshift is usually quoted as $z = \Delta\lambda/\lambda_0$, so $1 + z = \lambda/\lambda_0 = f_0/f$. For a receding source, $1+z = \sqrt{(1+\beta)/(1-\beta)}$. A quasar at $z = 2.00$ gives $3 = \sqrt{(1+\beta)/(1-\beta)}$, so $9(1-\beta) = 1+\beta$ and $\beta = 0.800$.

The same $\gamma(1-\beta\cos\theta)$ factor bends directions as well as frequencies. Light emitted at $90^\circ$ in the source frame arrives at $\cos\theta = \beta$ in the lab, so a fast source beams half its light into a narrow forward cone of half-angle $\arccos\beta \approx 1/\gamma$. That is the headlight effect, and running it with $\beta$ tiny gives stellar aberration, $\Delta\theta \approx \beta$ radians.

Recognising the method

Reach for four-vectors and invariants when you see any of these:

"Find the minimum energy to produce...": a threshold. Compute $s$ in the lab before, set $\sqrt{s}$ equal to the sum of product masses after, solve.

"A particle at rest decays into...": isolate one product, square, use the two-body energy formula. The answer depends only on the three masses.

A final state with more than two unknown momenta. Component equations will drown you. Squaring a four-vector deletes exactly the unknowns you do not care about, which is why you always isolate the particle you were not asked about before squaring.

Any question about a photon's measured colour, frequency, wavelength or redshift when something is moving: Doppler, with $\theta$ measured in the observer's frame. Check whether the problem means approach, recession or transverse before you pick a sign.

"As measured in the rest frame of..." with two velocities given: either a Lorentz transformation of a four-velocity, or the invariant $P_1 \cdot P_2 = \gamma_{\text{rel}} m_1m_2c^2$, which hands you the relative $\gamma$ in one step.

The habit that makes all of this fast: before writing a single number, write the conservation statement as four-vectors, decide which particle you want to eliminate, move it to one side alone, and square. The frame you evaluate in is then yours to choose, and you should always choose the lazy one.

That closes the course. Everything here rests on the same instinct you have been building since conservation of energy: find the quantity that does not change, and let it do the work.