Past exam of the symmetry and relativity course of the University of Oxford / 2021 (A13481W1)

Source: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

Source bundle pages 70–73; the cover identifies Trinity Term 2021 despite the attachment filename. Use coordinates $x^{\mu }=(ct,x,y,z)$, metric signature $(-,+,+,+)$, and $A^{\mu }=(\varphi /c,\mathbf{A})$. For the electromagnetic field tensor use $F^{0i}=E_i/c$ and $F^{ij}={ϵ}_{ijk}{B}_k$. Prescribed external fields are treated classically, neglecting radiation reaction in the trajectories; radiation losses are computed separately where requested.

Let $e>0$ be the magnitude of the electron's electric charge, $m=m_e$, $F=eE>0$, and $\ell =m{c}^2/F$. The Lorentz force is $d{p}_x/dt=qE$, so the electron and Positron acquire opposite momenta:Using the relativistic energy-momentum relation, $u_x=p{c}^2/\mathcal{E}$, and integrating from rest gives the world linesFor $t<0$, the world lines are vertical at $x=0,d$ on a spacetime diagram. They join smoothly to the accelerated branches with zero initial velocity. At small $t$, $x_e\simeq -F{t}^2/(2m)$ and $x_p-d\simeq F{t}^2/(2m)$; at large $t$, their velocities approach $-c,+c$, respectively. Each accelerated world line is a branch of a hyperbola, since $(x_e-\ell {)}^2-c^2{t}^2={\ell }^2$ and $(x_p-d+\ell {)}^2-c^2{t}^2={\ell }^2$. The electric field separates the opposite electric charges in this orientation. The asymptotes are lightlike, while every finite-time tangent is timelike.

Write $\Gamma =(1-v^2/c^2{)}^{-1/2}$ for the Lorentz factor of the inertial frame $S^{\prime }$, distinguishing it from the particles' changing Lorentz factors. Apply the Lorentz transformationto each world line in a. Both particles initially have velocity $-v$ in $S^{\prime }$. The switching events have coordinatesThus the Positron starts accelerating earlier in $S^{\prime }$. Its velocity first becomes less negative, vanishes when its laboratory velocity reaches $v$, and then becomes positive. The electron continues at $-v$ until $t^{\prime }=0$, then accelerates toward $-c$. The relativistic velocity-addition formula makes this precise:The electric field parallel to the Lorentz boost has unchanged magnitude, but its switching surface is $t^{\prime }+v{x}^{\prime }/c^2=0$, rather than $t^{\prime }=0$ everywhere.

In the laboratory, the total three-momentum is zero for all $t$: the two Lorentz forces cancel at equal laboratory times. In $S^{\prime }$, each active force is still $qE$, since this is one-dimensional motion with a parallel electric field. Define $t_p^{\prime }=-\Gamma vd/c^2$. For $v>0$ the total three-momentum, evaluated at equal $t^{\prime }$, isThis follows by integrating $d{p}_p^{\prime }/d{t}^{\prime }=F$ from its switching event and $d{p}_e^{\prime }/d{t}^{\prime }=-F$ from its later switching event. During the interval between them only the Positron feels the prescribed electric field, so momentum conservation for the particle pair fails. Once both forces act, the total three-momentum is constant again.

The equal-time momentum of an externally driven particle pair illustrates a distinction. The sum of particle four-momenta at a fixed pair of events is always a four-vector. However, a fresh sum at equal time in each inertial frame generally uses different events. Here transforming an equal-$t$ laboratory sum would give $-2\Gamma v\sqrt{m^2{c}^4+F^2{c}^2{t}^2}/c^2$, which varies with $t$; this differs from the equal-$t^{\prime }$ answer above. For example, with $c=m=F=d=1$ and $v=0.6$, the late-time equal-$t^{\prime }$ sum is $-0.75$, whereas transforming the equal-$t=0$ sum gives $-1.5$. This is a concrete failure of treating the pair's equal-time sums as components of a single frame-independent four-vector.

A closed system's total four-momentum, including the electromagnetic field and the apparatus supplying it, is conserved and independent of the choice of spacelike integration surface under the usual isolated-system boundary conditions. The particles alone exchange energy and momentum with an external system, so neither that independence nor universal particle-only momentum conservation is guaranteed.

With the reversed electric field, reflection about $x=d/2$ exchanges the electron and Positron. They therefore meet there with equal and opposite momenta. Each travels $d/2$, and the work-energy theorem gives its kinetic energy:Their combined kinetic energy is $1.000\mathrm{MeV}$, while each total relativistic energy is $\mathcal{E}=0.511+0.500=1.011\mathrm{MeV}$. The laboratory is the center-of-momentum frame of the colliding pair.

Local four-momentum conservation in the electron-positron annihilation excludes one real photon: the incoming total four-momentum is timelike, with zero spatial component and positive energy, while a single photon's four-momentum is null and has $|\mathbf{p}|=\mathcal{E}/c>0$. Two photons can balance their momenta by travelling in opposite directions, soThe equal photon energies follow from opposite equal momenta, not merely from sharing the incoming energy arbitrarily. Here $hc\simeq 1.23984\mathrm{MeV}\mathrm{pm}$. This is the free local annihilation idealization implicit in the question; significant momentum transfer to the external apparatus would change the kinematic system being considered.

The laboratory remains the center-of-momentum frame. At the relativistic collision mass threshold, the two final W bosons can have arbitrarily small opposite momenta. The available relativistic energy is $2m_e{c}^2+eEd$, soEach incoming particle must acquire about $80.378489\mathrm{GeV}$ of kinetic energy, travelling about $16.08\mathrm{km}$. Equality is the ideal kinematic threshold; a nonzero production rate needs energy above it. This estimate neglects acceleration losses and treats the W bosons with the supplied nominal rest mass.

A linear particle accelerator avoids repeatedly bending high-energy electrons and Positrons. This greatly reduces the bending synchrotron radiation loss, which at fixed bending radius grows approximately as the fourth power of the relativistic energy for a given particle species. Its useful collision energy can therefore be increased without the severe circular electron collider radiation penalty. The price is long structures or high accelerating gradients: each bunch normally passes once, requiring substantial beam power, fresh bunches, and very tight focusing for high luminosity. A circular particle accelerator reuses its accelerating structures and circulates bunches through many collisions, but needs a large ring and substantial radiofrequency power to replace synchrotron radiation losses. Heavier particles such as protons radiate far less at a given energy and radius, although a proton's constituent structure makes the hard-collision energy less sharply specified than in an electron-positron annihilation experiment.

Let $z^{\mu }(\tau )$ be the point electric charge's world line, with proper time $\tau$ and four-velocity $U^{\mu }=d{z}^{\mu }/d\tau =\gamma (c,\mathbf{u})$. For the field event $x^{\mu }$, choose the retarded source event: $R^{\mu }=x^{\mu }-z^{\mu }({\tau }_{\mathrm{ret}})$ is future-directed and null. Write $R^{\mu }=(R,R\mathbf{n})$, where $R=|\mathbf{x}-\mathbf{z}(t_{\mathrm{ret}})|=c(t-t_{\mathrm{ret}})$. In the instantaneous rest inertial frame of that source event,Consequently the covariant expression matching this Coulomb potential isThe denominator is a Lorentz scalar, so the right-hand side is a four-vector with the required instantaneous-rest components. This describes the retarded potential in Lorenz gauge, with no added free electromagnetic wave. Lorentz covariance alone does not establish the boundary conditions or exclude additional terms depending on four-acceleration. The physical input is the retarded point-source solution of Maxwell equations: the potential at the retarded event depends on the instantaneous four-current, while its derivatives generate the acceleration field. Equivalently, evaluating the retarded source integral gives a Jacobian $1/(1-\mathbf{n}\cdot \mathbf{\beta })$ from the light-cone constraint; this fixes that input for arbitrary smooth subluminal motion.^[ref]

In any inertial frame, put $\mathbf{\beta }=\mathbf{u}/c$ and $\kappa =1-\mathbf{n}\cdot \mathbf{\beta }$. Then $-R\cdot U=\gamma cR\kappa$, and the Liénard–Wiechert potential isDifferentiate using $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$, including the position and time dependence of the retarded time. For example, $\partial t_{\mathrm{ret}}/\partial t=1/\kappa$ and $\nabla t_{\mathrm{ret}}=-\mathbf{n}/(c\kappa )$. The resulting Liénard-Wiechert field isHere the dot denotes a derivative with respect to source coordinate time. The first term is the velocity or non-radiative field: it falls as $R^{-2}$ and exists even without acceleration. The second is the acceleration or radiation field, falling as $R^{-1}$ and transverse to $\mathbf{n}$. Its Poynting flux falls as $R^{-2}$, so its outward power integrated over a large sphere stays finite as its radius tends to infinity. The first term's squared-field flux instead vanishes in that limit. This large-distance distinction defines radiation; a moving near field can carry local energy without constituting radiation escaping to infinity.

For uniform circular motion, the transverse Lorentz force supplies the change in direction of three-momentum:The magnetic rigidity relation gives $pc=0.299792458B[\mathrm{T}]r[\mathrm{m}]\mathrm{GeV}=2.99792458\mathrm{GeV}$. Using $m_e{c}^2=0.511\mathrm{MeV}$,The coordinate centripetal acceleration is $a=u^2/r$. For transverse acceleration, the magnitude of the four-acceleration, or proper acceleration, is $a_0={\gamma }^{2}a$. The supplied Larmor formula, in its proper acceleration form, therefore givesThe emitted power here is also the laboratory total emitted power: in the instantaneous rest inertial frame the dipole radiation pattern has zero net radiated three-momentum, so both emitted energy and elapsed coordinate time gain the same Lorentz factor on transformation. This does not assert that a directional detector flux is invariant.

Numerically $\mathcal{T}\simeq 1.048\times 10^{-5}\mathrm{s}$, $P\simeq 2.18\times 10^{-10}\mathrm{W}$ per electron, andThe synchrotron's radiofrequency accelerating system must replace this lost energy to keep $r$ and magnetic rigidity fixed over many revolutions. The fraction lost per turn is about $4.77\times 10^{-6}$, so a fixed-orbit calculation over one turn is self-consistent.^[ref]

For an ultrarelativistic particle, relativistic beaming concentrates the synchrotron radiation within an angle of order $1/\gamma$ around its instantaneous velocity. A fixed viewing direction in the orbital plane receives an intense signal only when the tangent sweeps through that direction. One electron, or one sufficiently short bunch, therefore produces one pulse per revolution, separated by $\mathcal{T}\simeq 10.48\mu \mathrm{s}$. Several bunches produce the corresponding additional pulses per revolution.

The time during which the velocity points within that cone is of order $r/(\gamma c)$, but this is the emission interval. If $t_a$ is the arrival time at a distant detector,Near alignment, $1-\beta \cos \theta \simeq ({\gamma }^{-2}+{\theta }^2)/2=O({\gamma }^{-2})$. Thus synchrotron pulse compression reduces the observed duration by a further factor of order ${\gamma }^2$:A more explicit geometric estimate chooses the cone edges $\theta =\pm 1/\gamma$. With $\omega =u/r$, $x=r\sin \theta$ along the line of sight, the difference in arrival-time compression for a moving source isThe order-one coefficient depends on how pulse width is defined; the robust scale is $r/(c{\gamma }^3)$. The usual central-cone estimate $r/(2c{\gamma }^3)$ gives $4.1\times 10^{-18}\mathrm{s}$. These estimates agree on the attosecond scale and must not be confused with $r/(\gamma c)\simeq 2.84\times 10^{-10}\mathrm{s}$.^[ref]

Bunch qualification. The source says only that the bunches are short, without their longitudinal length $L_b$. Different electrons pass the same emitting tangent at times separated by up to about $L_b/u$, with essentially the same path length from that tangent to the detector. The bunch-length broadening of synchrotron radiation means that the measured bunch envelope is the single-electron pulse convolved with the bunch's passage-time profile; its scale is generally at least of order $\max \{L_b/u,r/(c{\gamma }^3)\}$, with additional detector bandwidth effects. For example, a $1\mathrm{mm}$ bunch, very short compared with this ring, gives an approximately $3.3\mathrm{ps}$ envelope, much longer than the single-electron duration. No unique numerical bunch-envelope width follows from the supplied data. The pulse argument also presumes that the accepted $+x$ direction is tangent to the orbit; a direction far outside the orbital plane receives no comparable intense forward pulse.

Assume $E,B>0$ for the directions and parametrizations; the boost formulas retain their signed meaning for reversed fields. A Lorentz boost with velocity $w\hat{\mathbf{z}}$ along $\mathbf{E}\times \mathbf{B}$ transforms the transverse electric field and magnetic field asIn the electric-dominated case chooseThe inequality ensures $|w|<c$, so this is an inertial frame. The field-removing boosts for perpendicular electric and magnetic fields are governed by the electromagnetic field invariants $\mathbf{E}\cdot \mathbf{B}=0$ and $E^2-c^2{B}^2>0$ explain why a purely electric inertial frame exists and a purely magnetic one does not.

The electron is not initially at rest in this convenient inertial frame: its initial velocity is $-w\hat{\mathbf{z}}$ and its initial three-momentum is $p_z^{\prime }=-{\gamma }_{w}mw$. Thereafter $p_x^{\prime }=-e{E}^{\prime }{t}^{\prime }$, while $p_z^{\prime }$ stays constant. The electric field accelerates it toward $-x$; as its relativistic energy increases, its negative transverse velocity decreases in magnitude. Defining $a=e{E}^{\prime }/m>0$ and $s=a\tau /c$ gives a useful parametric world line:Here $\tau$ is the proper time. These formulas follow by integrating $U^{\prime 0}={\gamma }_{w}c\cosh s$, $U^{\prime x}=-{\gamma }_{w}c\sinh s$, and constant $U^{\prime z}=-{\gamma }_{w}w$. Its projection onto the $t^{\prime },x^{\prime }$ plane is a hyperbola, but the spatial trajectory includes the initial transverse motion and is not a straight line.

Transforming back with $z={\gamma }_w(z^{\prime }+w{t}^{\prime })$, $t={\gamma }_w(t^{\prime }+w{z}^{\prime }/c^2)$ givesThe laboratory trajectory starts from rest toward $-x$, then bends toward $+z$. It is unbounded in $-x$ and ultimately has $u_z\to w=c^{2}B/E$ and $u_x\to -c/{\gamma }_w$. Thus the late velocity has magnitude $c$ as a limit, consistent with the relativistic energy-momentum relation, rather than undergoing a closed cyclotron orbit.

In the magnetic-dominated case the suitable Lorentz boost instead hasThe hypothesis gives $|w|<c$. The electromagnetic field invariants now have $E^2-c^2{B}^2<0$, allowing a purely magnetic inertial frame. The initial velocity is again $-w\hat{\mathbf{z}}$, perpendicular to ${\mathbf{B}}^{\prime }$. Since a magnetic field does no work, the electron's relativistic energy and speed in this inertial frame remain constant. Its trajectory is a circle in the $x^{\prime }{z}^{\prime }$ plane with Lorentz factor ${\gamma }_w$ and cyclotron frequency $\Omega =e{B}^{\prime }/({\gamma }_{w}m)$.

For $E,B>0$, set $\theta =\Omega t^{\prime }$ and $\rho =w/\Omega$. A parametrization with the correct initial velocity and Lorentz force isTransforming the world line back to the laboratory givesThe laboratory trajectory is a stretched cycloid, with cusps at $\theta =2\pi n$ where the electron is momentarily at rest. It first accelerates toward $-x$, then curves toward $+z$, returning to its original $x$ coordinate each cycle while advancing in $z$. Its mean drift is $\Delta z/\Delta t=w=E/B$, the E-cross-B drift. Its laboratory kinetic energy oscillates as the electric field does positive and then negative work; it does not grow indefinitely. The strict inequalities in the two branches exclude the null case $|E/B|=c$, where neither field can be removed by a finite Lorentz boost.

Use the electromagnetic field tensor convention stated above, and lower indices with the Minkowski metric $g=\mathrm{diag}(-1,1,1,1)$. Lowering a single time index changes its sign. Summing both orders of each antisymmetric index pair gives the Lorentz scalarFor $T^{00}$, $F_j{}^0=F^{j0}=-E_j/c$ and $g^{00}=-1$. Direct substitution in the printed electromagnetic stress-energy tensor yieldsFor a spatial index $i$, $g^{0i}=0$ and $F_j{}^i=F^{ji}=-{ϵ}_{ijk}{B}_k$. HenceThus the full top row isHere $u_{\mathrm{EM}}$ is the electromagnetic energy density and $\mathbf{S}$ the Poynting vector, using ${ϵ}_0{\mu }_0{c}^2=1$. By the symmetry $T^{0i}=T^{i0}$, these components describe both energy flux divided by $c$ and $c$ times the electromagnetic momentum density $\mathbf{g}=\mathbf{S}/c^2$. Keeping the factor $c$ is essential because the time coordinate is $ct$.

Let starred quantities denote the capacitor's rest inertial frame, $u_0={ϵ}_0{E}^2/2$, $\beta =v/c$, and $\gamma =(1-{\beta }^2{)}^{-1/2}$. Work in the uniform interplate region of an ideal parallel-plate capacitor, neglecting its fringing field of a capacitor. To transform from that rest inertial frame to a laboratory in which the capacitor moves along $+x$, use the positive-sign Lorentz transformationFor a purely electric field, the spatial block of the electromagnetic stress-energy tensor isThis is the negative of the Maxwell stress tensor: it has $-u_0$ along the electric field and $+u_0$ perpendicular to it. The stress contribution to energy transport in a moving capacitor is essential to the Lorentz transformation of the energy density and flux.

Plates parallel to $xy$. Choose ${\mathbf{E}}_{\ast }=E\hat{\mathbf{z}}$ and ${\mathbf{B}}_{\ast }=0$. The electromagnetic field tensor and electromagnetic stress-energy tensor areThe transformed nonzero independent electromagnetic field tensor components are $F^{03}=\gamma E/c$ and $F^{13}=\gamma \beta E/c$. Thus $\mathbf{E}=\gamma E\hat{\mathbf{z}}$ and $\mathbf{B}=-\gamma vE\hat{\mathbf{y}}/c^2$, consistent with $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$. Transforming the electromagnetic stress-energy tensor givesIn particular, the laboratory Poynting vector is $\mathbf{S}=2u_0{\gamma }^{2}v\hat{\mathbf{x}}$, and the electromagnetic energy density is $u=u_0{\gamma }^2(1+{\beta }^2)$. The ratio $S_x/u=2v/(1+{\beta }^2)$ is subluminal for $v<c$, but differs from the capacitor's velocity. The difference from a simple convective flux isThe extra term is $v{T}_{\ast }^{xx}$, the rest-frame stress multiplied by the velocity. This shows why treating stored field energy as an unstressed parcel moving rigidly with the plates misses part of the energy transport.

Plates parallel to $yz$. Now ${\mathbf{E}}_{\ast }=E\hat{\mathbf{x}}$. The rest tensors areThe electric field is parallel to the Lorentz boost, so $F^{01}=E/c$ remains unchanged and $\mathbf{B}=0$. The cancellation between $T_{\ast }^{00}$ and $T_{\ast }^{11}$ givesThere is no Poynting flux in the ideal interior in this orientation, although the region containing the electric field moves. The longitudinal negative stress contribution cancels the convective term: $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})=0$. Field energy is removed at the receding boundary and supplied at the advancing boundary by work on the charged plates; it need not flow through the interior to accompany the moving region. At edges, omitted fringing fields provide additional local structure.

For a finite idealized interplate rest volume $V_0$, length contraction gives $V=V_0/\gamma$. The field-only integrated laboratory energies would consequently be $U_{xy}=\gamma (1+{\beta }^2)U_0$ and $U_{yz}=U_0/\gamma$, with $U_0=u_0{V}_0$. Their different transformation laws are not paradoxical: the field alone is not a closed system. Include the plates and the supports holding them apart, whose mechanical stresses balance the Maxwell stress tensor. The total closed-system four-momentum then has the ordinary Lorentz transformation law. This is the same distinction between a subsystem and a conserved whole that appears in b.

A symmetry is a transformation of a physical system that preserves its laws: it maps admissible solutions and consistently transformed data to admissible solutions. The central symmetry of special relativity is Lorentz invariance, expressing equivalence of inertial frames and preservation of the Minkowski interval and the universal speed of light. With spacetime translations included, the spacetime symmetry group is the Poincaré group.

A practical sufficient check is to express an equation as an equality between Lorentz tensors of the same type, with every repeated upper/lower pair correctly contracted and the same free indices on both sides. Every ingredient, including sources and background fields, must have its stated Lorentz transformation rule. For example, $p^{\mu }{p}_{\mu }=-m^2{c}^2$ is a Lorentz scalar equality, and $d{p}^{\mu }/d\tau =q{F}^{\mu }{}_{\nu }{U}^{\nu }$ equates four-vectors. Such manifest Lorentz covariance makes the form of the equation the same in every inertial frame.

This is a sufficient test when the transformation rules are genuine physical ones, not a guarantee from writing index symbols. A fixed preferred four-vector could make an equation formally covariant if it too were transformed, while its chosen physical background breaks Lorentz invariance. Conversely, an equation written in three-dimensional components can be fully consistent with special relativity, as the Lorentz force is, even when the covariance is not manifest.

For an active clockwise rotation of column vectors in the usual oriented $xy$ plane, takeDirect matrix multiplication givesThus the rotation matrix is an orthogonal matrix, preserving the Euclidean inner product, with determinant $+1$.

For a passive one-dimensional Lorentz boost, use rapidity $\eta$, $\beta =\tanh \eta$:This is not $I$ unless $\eta =0$. Hence a nontrivial Lorentz boost is not an orthogonal matrix for the Euclidean inner product. Instead it preserves the Minkowski metric:by ${\cosh }^2\eta -{\sinh }^2\eta =1$. Thus it is pseudo-orthogonal. The zero-rapidity identity is the explicit exception to the assertion that a Lorentz boost is not orthogonal.

The supplied matrix obeysSeparate the even and odd powers in the matrix exponential to obtainIt fixes the $z$ component and is a counterclockwise active rotation viewed from $+z$. This sign differs from the clockwise convention in b, because the source fixes the sign of both $J_3$ and the exponential. Replacing $\theta$ by $-\theta$ yields the clockwise rotation.

With $(J_k{)}_{ij}=-i{ϵ}_{kij}$, the generators of rotations obeyThe commutator measures the leading difference between performing two infinitesimal rotations in opposite orders. For example, with $X=-i\alpha J_1$ and $Y=-i\beta J_2$,The failure to commute is itself an infinitesimal rotation about the perpendicular axis; the Lie algebra closes on the same three generators of rotations.

To pass from this local statement to the full group property, put $A_k=-i{J}_k$. These are real antisymmetric matrices, with $[A_i,A_j]={ϵ}_{ijk}{A}_k$. The Baker--Campbell--Hausdorff formula shows locally that the logarithm of a product of their exponentials remains a linear combination of the $A_k$, because all nested commutators do. Globally, each exponential has $R^{T}R=I$ and $\det R=e^{\mathrm{tr}A}=1$. If $R_1,R_2$ have these properties, thenEvery real three-dimensional orthogonal matrix of determinant $+1$ has an eigenvector with eigenvalue $+1$: its nonreal eigenvalues come in conjugate pairs, and the product of all three is $+1$. On the perpendicular plane it acts as a two-dimensional rotation. It therefore has axis-angle form and can be written $e^{-i\theta {\hat{n}}^k{J}_k}$. Thus the product is again a rotation. The identity is $\theta =0$, the inverse reverses $\theta$, and associativity comes from matrix multiplication. This proves that all these rotations form the special orthogonal group $SO(3)$. The commutator supplies local closure, while the orthogonality and axis-angle argument supplies the necessary global step.

Let $K_i$ generate Lorentz boosts with the same exponential convention as the rotation generators of rotations $J_i$. The Lie algebra of the Lorentz group isConsequently non-collinear Lorentz boosts do not close among themselves: the product contains a rotation, the Wigner rotation. For an explicit counterexample, in coordinates $(ct,x,y)$ write $c_a=\cosh a$, $s_a=\sinh a$ and similarly for $b$. ThenA pure rotationless Lorentz boost has a symmetric matrix, whereas this product has spatial entries $(xy)=0$ and $(yx)=s_a{s}_b\ne 0$ for $a,b\ne 0$. It is therefore a concrete product of two pure Lorentz boosts that is not a pure Lorentz boost. Collinear Lorentz boosts are the special case that do form a one-parameter group, with additive rapidities.

Allowing all spatial directions and closure generates both the $K_i$ and, through their commutators, all $J_i$. The smallest connected group containing all these Lorentz boosts is the proper orthochronous Lorentz groupIt contains arbitrary compositions of spatial rotations and Lorentz boosts continuously connected to the identity. It has six continuous parameters. Here “smallest” refers to the subgroup generated by all continuous pure Lorentz boosts, not to an arbitrary collection already including discrete reversals.

The full Lorentz group $O(1,3)$ has four connected components, distinguished by the sign of the determinant and the direction of time. Representatives are the identity, parity $P=\mathrm{diag}(1,-1,-1,-1)$, time reversal $\mathsf{T}=\mathrm{diag}(-1,1,1,1)$, and $P\mathsf{T}=-I$. ThusMultiplication by these discrete transformations relates the components algebraically; no continuous path within the Lorentz group joins different components. Translations are not required for closure here; adding them gives the larger Poincaré group.