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Lorentz Transformations

A Lorentz transformation is a linear change of inertial coordinates that preserves the spacetime interval. Rotations change spatial directions; boosts mix space and time. Their composition governs how relativistic wave equations and their solutions compare between inertial observers. This page constructs the transformations; Four-Vectors explains the physical objects on which they act.

Required background. Metric and Units supplies the metric and index notation.

For x′μ=Λμνxνx'^\mu=\Lambda^\mu{}_{\nu}x^\nu, preserving xμημνxνx^\mu\eta_{\mu\nu}x^\nu for every xx requires

ΛTηΛ=η,Λ−1=η−1ΛTη.\Lambda^{\mathsf T}\eta\Lambda=\eta, \qquad \Lambda^{-1}=\eta^{-1}\Lambda^{\mathsf T}\eta.

Taking determinants gives (det⁡Λ)2=1(\det\Lambda)^2=1. A Lorentz matrix is generally not orthogonal with respect to the Euclidean inner product; replacing Λ−1\Lambda^{-1} by ΛT\Lambda^{\mathsf T} is incorrect for boosts.

A spatial rotation has the block form diag⁡(1,R)\operatorname{diag}(1,R) with RTR=I3R^{\mathsf T}R=I_3. Proper spatial rotations also obey det⁡R=1\det R=1. They preserve time components and rotate spatial components in the usual way.

Use a passive convention: the primed frame moves at velocity vx^v\hat{\mathbf x} relative to the unprimed frame, and the origins coincide at t=t′=0t=t'=0. Then

(ct′x′)=(γ−γβ−γβγ)(ctx),y′=y,z′=z,\begin{pmatrix}ct'\\x'\end{pmatrix} = \begin{pmatrix}\gamma&-\gamma\beta\\-\gamma\beta&\gamma\end{pmatrix} \begin{pmatrix}ct\\x\end{pmatrix}, \qquad y'=y,\quad z'=z,

where β=v/c\beta=v/c and γ=(1−β2)−1/2\gamma=(1-\beta^2)^{-1/2}. The worldline x=vtx=vt indeed gives x′=0x'=0, which checks the physical meaning of the signs. The inverse transformation sends β\beta to −β-\beta.

Define the rapidity ξ\xi by

β=tanh⁡ξ,γ=cosh⁡ξ,γβ=sinh⁡ξ.\beta=\tanh\xi,\qquad \gamma=\cosh\xi,\qquad \gamma\beta=\sinh\xi.

The boost is a hyperbolic rotation,

Bx(ξ)=(cosh⁡ξ−sinh⁡ξ−sinh⁡ξcosh⁡ξ),B_x(\xi)= \begin{pmatrix}\cosh\xi&-\sinh\xi\\-\sinh\xi&\cosh\xi\end{pmatrix},

in its ct,xct,x block. Multiplying two such blocks and using the hyperbolic addition formulas gives

Bx(ξ2)Bx(ξ1)=Bx(ξ1+ξ2),β12=β1+β21+β1β2.B_x(\xi_2)B_x(\xi_1)=B_x(\xi_1+\xi_2), \qquad \beta_{12}=\frac{\beta_1+\beta_2}{1+\beta_1\beta_2}.

Rapidity, rather than velocity, adds for collinear boosts. Two successive boosts with β1=β2=3/5\beta_1=\beta_2=3/5 have combined β12=15/17\beta_{12}=15/17, less than one; their combined Lorentz factor is 17/817/8.

For a frame velocity βc\boldsymbol\beta c in an arbitrary direction, resolve the displacement parallel and perpendicular to that direction:

ct′=γ(ct−β⋅x),x′=x+γ−1β2(β⋅x)β−γβct.\begin{aligned} ct'&=\gamma(ct-\boldsymbol\beta\cdot\mathbf x),\\ \mathbf x'&=\mathbf x+ \frac{\gamma-1}{\beta^2} (\boldsymbol\beta\cdot\mathbf x)\boldsymbol\beta -\gamma\boldsymbol\beta ct. \end{aligned}

The continuous β→0\beta\to0 limit is the identity. This form makes it explicit that the perpendicular component is unaffected by a pure boost.

Which transformations preserve the future?

Section titled “Which transformations preserve the future?”

The Lorentz condition implies ∣Λ00∣≥1|\Lambda^0{}_{0}|\geq1. Thus two discrete choices distinguish four connected components: det⁡Λ=±1\det\Lambda=\pm1 and the sign of Λ00\Lambda^0{}_{0}. The identity component is SO+(1,3)SO^+(1,3), the proper orthochronous Lorentz group. It contains ordinary rotations and finite boosts connected continuously to the identity.

Parity P=diag⁡(1,−1,−1,−1)P=\operatorname{diag}(1,-1,-1,-1) reverses spatial orientation. Time reflection T=diag⁡(−1,1,1,1)T=\operatorname{diag}(-1,1,1,1) reverses the time orientation. Neither can be reached continuously from the identity within the Lorentz group. Their action on quantum states requires additional structure; a classical time-reflection matrix does not itself specify the antiunitary quantum time-reversal operator.

For a future-directed on-shell momentum, E>0E>0 and E≥c∣p∣E\geq c|\mathbf p| (with E>c∣p∣E>c|\mathbf p| for positive mass), a finite boost gives

E′=γ(E−v⋅p)>0.E'=\gamma(E-\mathbf v\cdot\mathbf p)>0.

Hence a change between proper orthochronous frames cannot turn a positive energy into a negative energy. The two sheets of the massive mass shell are not two ordinary observers’ descriptions of one positive-energy particle.

Transforming both pp and xx with the same Lorentz matrix leaves p⋅xp\cdot x unchanged, so the phase e−ip⋅x/ℏe^{-ip\cdot x/\hbar} is invariant. For a photon moving along +x+x, an xx boost gives

E′=Eγ(1−β)=Ee−ξ.E'=E\gamma(1-\beta)=Ee^{-\xi}.

The Doppler shift changes the separate frequency and wavelength, while the phase assigned to the same event remains unchanged. A scalar amplitude satisfies ϕ′(x′)=ϕ(x)\phi'(x')=\phi(x); spinors have an additional transformation on their components, developed in the covariant Dirac equation.

Collinear rapidity addition does not extend to arbitrary boost directions. For an infinitesimal example, let bib_i be the real four-vector matrix with entries (bi)0i=(bi)i0=−1(b_i)^0{}_i=(b_i)^i{}_0=-1, and all other entries zero. Then [bx,by][b_x,b_y] has spatial entries (x,y)=1(x,y)=1 and (y,x)=−1(y,x)=-1: it generates a rotation. Non-collinear boost composition consequently contains a rotation, which is the kinematic source of Wigner rotations. These real matrices should not be confused with Hermitian generators acting on a quantum Hilbert space.

  1. Show directly that the ct,xct,x boost preserves (ct)2−x2(ct)^2-x^2.
Solution

Expanding the difference of squares cancels the mixed terms and leaves γ2(1−β2)[(ct)2−x2]=(ct)2−x2\gamma^2(1-\beta^2)[(ct)^2-x^2]=(ct)^2-x^2.

  1. A primed observer measures a velocity ux′u'_x. Derive its unprimed value.
Solution

Differentiate the inverse boost: dx=γ(dx′+vdt′)dx=\gamma(dx'+vdt') and dt=γ(dt′+v dx′/c2)dt=\gamma(dt'+v\,dx'/c^2). Their ratio gives ux=(ux′+v)/(1+vux′/c2)u_x=(u'_x+v)/(1+vu'_x/c^2). In particular ux′=cu'_x=c implies ux=cu_x=c.

  1. For a spacelike displacement with Δx>cΔt>0\Delta x>c\Delta t>0, find a boost for which the two events are simultaneous. Can a larger subluminal boost reverse their time ordering?
Solution

Δt′=γ(Δt−vΔx/c2)\Delta t'=\gamma(\Delta t-v\Delta x/c^2) vanishes at v=c2Δt/Δx<cv=c^2\Delta t/\Delta x<c. Any vv between this value and cc makes Δt′<0\Delta t'<0. No causal signal can connect such spacelike events.

  • J. D. Jackson, Classical Electrodynamics, 3rd ed., Wiley, 1998, chapter 11 — boosts and the Lorentz group.
  • W. Rindler, Introduction to Special Relativity, 2nd ed., Oxford University Press, 1991 — rapidity and relativistic kinematics.
  • S. Weinberg, The Quantum Theory of Fields, Volume I: Foundations, Cambridge University Press, 1995, section 2.3 — Lorentz transformations and quantum representations.