Past exam of the symmetry and relativity course of the University of Oxford / 2017 / 1

For a massive particle on a timelike world line, the four-velocity is $U^{\mu }=d{X}^{\mu }/d\tau =\gamma (c,\mathbf{v})$, with $U^2=c^2$. Its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$ and its four-force is $F^{\mu }=d{P}^{\mu }/d\tau$. Here $\tau$ is proper time. A massless trajectory does not admit this massive-particle proper-time parametrization.

The asserted classification of arbitrary orthogonal vectors is false. With the chosen Minkowski metric, $D=(0,1,0,0)$ and $G=(0,0,1,0)$ are nonzero, orthogonal and both spacelike vectors. Also $D=(1,1,0,0)$ and $G=2D$ are orthogonal null vectors. The valid statement is that a nonzero vector orthogonal to a timelike vector is spacelike. To prove it, choose the rest frame of the timelike vector, $D=(d,0,0,0)$; then $D\cdot G=d{G}^0=0$ forces $G^0=0$, so $G^2=-|\mathbf{G}{|}^2<0$.

The relativistic energy-momentum relation givesTherefore $\boxed{P^2=0⟹m=0}$. A photon is an example: $E=c|\mathbf{p}|$. A collection of photons need not itself have zero invariant mass.

Along a massive particle's world line, proper time satisfies $d\tau =dt\sqrt{1-v^2/c^2}$. A pure force preserves its rest mass. Differentiating $P=mU$ and $U^2=c^2$ givesso $\boxed{F\cdot U=0}$ for a pure force. This is a restriction on rest-mass transfer, not a statement that its work in every frame vanishes.

Write $\mathbf{a}=d\mathbf{v}/dt$. Since $\dot{\gamma }={\gamma }^3\mathbf{v}\cdot \mathbf{a}/c^2$, the four-acceleration isIf $\mathbf{v}\cdot \mathbf{a}=0$, this reduces to $A=(0,{\gamma }^2\mathbf{a})$, whenceHere $\alpha$ is the proper acceleration. The positive expression printed in the source is $A^2$ in the opposite metric signature, or the squared acceleration magnitude in ours. For example, at rest with nonzero acceleration, $A=(0,\mathbf{a})$ has $A^2=-a^2$ in our convention. Thus one must not combine the printed positive sign with a mostly-minus norm.

There is no nonzero, literally constant four-force that is a pure force throughout this motion. Initially $U=(c,0,0,0)$, so purity forces $F^0=0$. If the full four-vector were constant, $U(\tau )=U(0)+F\tau /M_0$; its square would be $c^2+F^2{\tau }^2/M_0^2$, which cannot remain $c^2$ for a nonzero initially spatial $F$. This is a genuine ambiguity in the source wording.

The usual consistent interpretation is constant laboratory relativistic force $f=d{p}_x/dt$, equivalently constant one-dimensional proper acceleration $f/M_0$. Take $f>0$ and let $s=ft/(M_{0}c)$. Integrating momentum conservation with the applied impulse gives $p_x=ft$, soThe corresponding four-force has components $F^{\mu }=\gamma f(\beta ,1,0,0)$, which are not constant, although its invariant magnitude is $F^2=-f^2$. Its temporal component is indeed $F^0=(\gamma /c)dE/dt$. The speed rises from zero and approaches $c$ asymptotically; $\gamma$ starts at one, initially rises quadratically and becomes linear in $t$ at late times.

If instead only the spatial four-force component $F^x=f_x$ is held constant, the distinct consistent answer is most simply parametrized by $u=f_x\tau /(M_{0}c)$:This follows by integrating $d{U}^x/d\tau =f_x/M_0$ and $dt/d\tau =U^0/c$. The temporal component must still vary. These two interpretations must not be conflated with a constant four-vector.

The equal and opposite incoming momenta make the laboratory the centre-of-mass frame. The initial total four-momentum is timelike. One outgoing photon would have null four-momentum, so an isolated two-body annihilation cannot produce exactly one photon. Two photons are kinematically possible and must have opposite momenta and equal energies:The quoted $0.511\mathrm{MeV}$ is the rest energy $m_e{c}^2$, not a mass in SI units. The energy balance includes both rest energies as well as the $1\mathrm{GeV}$ total kinetic energy.

This minimum concerns the complete final state. A detector with incomplete acceptance could register only one of the two photons, or none; conservation laws do not impose a minimum detected count without an ideal-detection assumption. Additional outgoing photons are also possible.

The electron moves in the direction for which the electric field does positive work. Its kinetic energy gain is the magnitude $eEL=30\mathrm{MeV}$. HenceThis uses the relativistic energy-momentum relation, rather than the nonrelativistic kinetic-energy formula. Acceleration is opposite the field direction for a negatively charged electron, although the energy gain is positive along its actual displacement.