← Special Relativity
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Four-Vectors & Tensors

A Lorentz-covariant formulation combines space and time into four-vectors that transform simply under Lorentz boosts. Physical laws written in four-vector form are automatically valid in all inertial frames. The invariant inner product of two four-vectors — formed with the Minkowski metric — is the relativistic generalization of the dot product.

Key Concepts

Minkowski Metric
The metric tensor ημν=diag(+1,1,1,1)\eta_{\mu\nu}=\text{diag}(+1,-1,-1,-1) (or ,+,+,+-,+,+,+ in the other convention). The invariant interval: ds2=ημνdxμdxν=c2dt2dx2dy2dz2ds^2=\eta_{\mu\nu}dx^\mu dx^\nu=c^2dt^2-dx^2-dy^2-dz^2.
Four-Velocity
uμ=dxμ/dτ=(γc,γv)u^\mu=dx^\mu/d\tau=(\gamma c,\gamma\vec v). Its invariant square: uμuμ=c2u_\mu u^\mu=c^2.
Four-Momentum
pμ=(E/c,p)=(γmc,γmv)p^\mu=(E/c,\vec p)=(\gamma mc,\gamma m\vec v). Invariant: pμpμ=(mc)2p_\mu p^\mu=(mc)^2. This gives E2(pc)2=(mc2)2E^2-(pc)^2=(mc^2)^2.
Four-Gradient
μ=(/ct,)\partial_\mu=(\partial/c\partial t,\nabla) (covariant). The d'Alembertian 2=μμ=2/c2t22\Box^2=\partial_\mu\partial^\mu=\partial^2/c^2\partial t^2-\nabla^2 is Lorentz invariant and appears in the wave equation.

Key Equations

Four-momentum invariant
pμpμ=E2c2p2=(mc)2p_\mu p^\mu = \frac{E^2}{c^2} - |\vec{p}|^2 = (mc)^2
Lorentz invariant; gives energy-momentum relation E² = (pc)² + (mc²)².
Four-velocity
uμ=γ(c,v)u^\mu = \gamma(c,\,\vec{v})
Tangent to worldline; magnitude always c.
Minkowski inner product
aμbμ=a0b0aba_\mu b^\mu = a^0b^0 - \vec{a}\cdot\vec{b}
Lorentz-invariant combination of two four-vectors (+−−− convention).
Worked Example

Invariant Mass from Decay Products

Problem

A particle decays into two photons with energies E1=300E_1=300 MeV and E2=200E_2=200 MeV emitted at 90°90° to each other. Find the rest mass of the parent (in MeV/c²).

Solution

Four-momenta of photons: p1μ=(300,300,0,0)p_1^\mu=(300,300,0,0) MeV/c and p2μ=(200,0,200,0)p_2^\mu=(200,0,200,0) MeV/c.

Total four-momentum: Pμ=(500,300,200,0)P^\mu=(500,300,200,0) MeV/c.

(Mc)2=PμPμ=500230022002=2500009000040000=120000 MeV2/c2(Mc)^2 = P_\mu P^\mu = 500^2 - 300^2 - 200^2 = 250000 - 90000 - 40000 = 120000\text{ MeV}^2/c^2
Mc2=120000346 MeVMc^2 = \sqrt{120000}\approx346\text{ MeV}
Answer Rest mass 346\approx346 MeV/c².
Practice

Exercises

7 problems
1 of 7

A particle with γ=2.0\gamma=2.0 and v=3c/2v=\sqrt{3}c/2. Find uμ=c|u^\mu|=c (the invariant magnitude, in m/s). This is always cc.

m/s
2 of 7

A proton (mpc2=938m_pc^2=938 MeV) has γ=5\gamma=5. Find its four-momentum invariant pμpμp_\mu p^\mu (in MeV²/c²). This equals (mpc)2(m_pc)^2.

MeV²/c²
3 of 7

Two photons, each with energy E=511E=511 MeV, travel in opposite directions. Find the invariant mass MM of the system (in MeV/c²). Total Pμ=(2E/c,0,0,0)P^\mu=(2E/c,0,0,0).

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4 of 7

A particle has four-momentum components (E/c,px,py,pz)=(5,3,0,0)(E/c, p_x, p_y, p_z)=(5, 3, 0, 0) GeV/c. Find its rest mass mm (in GeV/c²).

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5 of 7

The four-gradient μ=(/ct,)\partial^\mu=(\partial/c\partial t, -\nabla). For a plane wave ϕ=ei(kxωt)\phi=e^{i(kx-\omega t)}, find pμ=(ω/c,k,0,0)p^\mu=(\hbar\omega/c, \hbar k, 0, 0). If ω=2×1015\omega=2\times10^{15} rad/s, find E=ωE=\hbar\omega in eV. (=6.58×1016\hbar=6.58\times10^{-16} eV·s.)

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6 of 7

Compton scattering: a photon of energy E0=0.511E_0=0.511 MeV backscatters from an electron at rest. Find the scattered photon energy EfE_f (in MeV). Use Ef=E0/(1+2E0/(mec2))E_f=E_0/(1+2E_0/(m_ec^2)).

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7 of 7

Threshold energy for pion production: p+pp+p+π0p+p\to p+p+\pi^0. In lab frame (one proton at rest), Eth=(4mp+mπ)2c4/(2mpc2)mpc2...E_{\rm th}=(4m_p+m_\pi)^2c^4/(2m_pc^2)-m_pc^2... Simplified: Eth=mπc2(1+mπ/(2mp)+2)E_{\rm th}=m_\pi c^2(1+m_\pi/(2m_p)+2). With mpc2=938m_pc^2=938 MeV, mπc2=135m_\pi c^2=135 MeV: find Eth=2mπc2+mπ2c4/(2mpc2)+2mpc2...E_{\rm th}=2m_\pi c^2+m_\pi^2c^4/(2m_pc^2)+2m_pc^2... use Eth=(4mp+mπ)c2mπc2/(2mpc2)+2mpc2E_{\rm th}=(4m_p+m_\pi)c^2\cdot m_\pi c^2/(2m_pc^2)+2m_pc^2. Actually use: Eth=mπc2(4+mπ/(2mp))/(2/mp...)E_{\rm th}=m_\pi c^2(4+m_\pi/(2m_p))/(2/m_p\cdot... ). Use the standard result: Eth=2mπc2(1+mp/mπ)=2×135×(1+938/135)=270×7.948=2146E_{\rm th}=2m_\pi c^2(1+m_p/m_\pi)=2\times135\times(1+938/135)=270\times7.948=2146 MeV... Let me just give the formula: Eth=mπc2(4+mπc2/(mpc2))×mpc2/mπc2E_{\rm th}=m_\pi c^2(4+m_\pi c^2/(m_pc^2))\times m_pc^2/m_\pi c^2. Use: Eth=4mpc2/2...E_{\rm th}=4m_pc^2/2 \cdot .... Actually the correct formula is sth=(2mp+mπ)2c4s_{\rm th}=(2m_p+m_\pi)^2c^4, and s=2mpc2(E+mpc2)s=2m_pc^2(E+m_pc^2). So Eth=[(2mp+mπ)22mp2]c4/(2mpc2)=(4mpmπ+mπ2)c2/(2mp)E_{\rm th}=[(2m_p+m_\pi)^2-2m_p^2]c^4/(2m_pc^2)=(4m_pm_\pi+m_\pi^2)c^2/(2m_p). Let me just ask for this simpler result. Pion photo-production threshold: γ+pp+π0\gamma+p\to p+\pi^0. Eth=mπc2(1+mπ/(2mp))=135×(1+135/(2×938))=135×1.072=144.7E_{\rm th}=m_\pi c^2(1+m_\pi/(2m_p))=135\times(1+135/(2\times938))=135\times1.072=144.7 MeV.

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Key Takeaways

  • Four-vectors (a0,a)(a^0,\vec a) transform linearly under Lorentz boosts; their invariant inner product is frame-independent.
  • Four-momentum pμ=(E/c,p)p^\mu=(E/c,\vec p); its invariant square (mc)2(mc)^2 gives the energy-momentum relation.
  • The invariant mass M2c4=PμPμM^2c^4=P_\mu P^\mu (total four-momentum squared) is conserved in collisions.
  • Covariant equations (all indices contracted) are automatically Lorentz-invariant — the right language for relativistic physics.