Past exam of the symmetry and relativity course of the University of Oxford / 2021 / 4

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.