Past exam of the symmetry and relativity course of the University of Oxford / 2019 / 3

An invariant has the same value when the same physical event or state is described in different inertial frames. A conserved quantity has the same value as the physical system evolves, subject to its equations and isolation conditions. These are different comparisons. The invariant mass of a heated object can change with time, while an isolated system's conserved energy has different numerical values in boosted frames.

Under a Lorentz transformation $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$, a Lorentz four-vector transforms asA covector transforms by the inverse matrix, $W_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{W}_{\nu }$. A general tensor has one corresponding factor for each index; for example,Contraction with the Minkowski metric produces Lorentz scalars; it does not by itself make those scalars conserved.

To prove the vanishing component theorem for a Lorentz four-vector, first suppose $Y^0=0$ in every inertial frame. An infinitesimal boost along spatial direction $i$ gives $Y^{\prime 0}=Y^0-\beta Y^i+O({\beta }^2)$. It too vanishes for every $\beta$, so each $Y^i=0$. If instead a specified spatial component vanishes in every frame, spatial rotations first imply that all three spatial components vanish in a chosen frame, and a boost in the specified direction then forces $Y^0=0$. ThereforeA physical consequence is that conservation of energy in every inertial frame implies full four-momentum conservation: apply the result to the difference between outgoing and incoming total four-momenta.

The statement fails for general tensors. A nonzero antisymmetric rank-two tensor has $F^{00}=0$ in every frame, because antisymmetry is preserved. For an off-diagonal component, the nonzero Minkowski metric is another counterexample: ${\eta }^{01}=0$ in all orthonormal Lorentz frames, while $\eta$ is certainly not zero. The transformation law can preserve structural zero components without forcing the entire tensor to vanish.

Use $c=1$ and the mostly-plus Minkowski metric fixed in part a. The electromagnetic four-potential and its covariant form areDefine the electromagnetic field tensor by $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$. Since $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$,Raising both indices changes the signs of its time-space entries. The lower-index signs agree with ^[ref]; the signature and index placement must be kept consistent.

For a passive boost to a frame moving with $v\hat{\mathbf{x}}$, transform the tensor asFor example, $F^{\prime 02}=\gamma F^{02}-\gamma v{F}^{12}=\gamma (E_y-v{B}_z)$ and $F^{\prime 12}=\gamma (B_z-v{E}_y)$. Reading off all entries givesFor an arbitrary boost velocity $\mathbf{v}$, rotational covariance gives the Lorentz transformation of electromagnetic fieldsHere all velocities are in $c=1$ units. Equivalently the parallel components are unchanged and the perpendicular components have the cross-product expressions multiplied by $\gamma$.

The scalar contraction of the electromagnetic field tensor isWith orientation ${ϵ}^{0123}=+1$, define the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. Direct substitution givesThe first is a Lorentz scalar; the second is a pseudoscalar, invariant under proper Lorentz transformations such as boosts and rotations. Under spatial inversion $\mathbf{E}$ changes sign while $\mathbf{B}$ does not, so $\mathbf{E}\cdot \mathbf{B}$ reverses sign. For example, $\mathbf{E}=\mathbf{B}=\hat{\mathbf{z}}$ has product $+1$ before inversion and $-1$ afterwards. Thus the unrestricted statement including improper transformations needs this qualification.

For the spacetime integral, assume a globally regular electromagnetic four-potential on ordinary Minkowski space and sufficient boundary decay. Using $\nabla \cdot \mathbf{B}=0$ and ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$,ConsequentlyThe magnetic helicity boundary term controls the integral. If its surface flux vanishes and the magnetic helicity $H_B(t)=\int \mathbf{A}\cdot \mathbf{B}{d}^{3}x$ vanishes at both temporal ends, thenCompactly supported smooth potentials are one sufficient version of the intended falloff assumption. The local identity is exact, but vanishing fields at spatial infinity alone is insufficient. For a concrete counterexample, take $\varphi =0$ and, in dimensionless coordinates,where $a$ changes smoothly from zero to one and $\dot{a}$ has compact temporal support. Both electric fields and magnetic fields vanish rapidly at spatial infinity, butThe counterexample leaves a final magnetic field and therefore fails the extra temporal boundary condition. It shows exactly why the printed integral cannot follow from a merely spatial reading of “at infinity”; vanishing of the full boundary term is the needed hypothesis.