Past exam of the symmetry and relativity course of the University of Oxford / 2015 / 2 / c

For the source's definition $s^{\alpha }=\gamma (\mathbf{s}\cdot \mathbf{v}/c,\mathbf{s})$, direct contraction with $u^{\alpha }=\gamma (c,\mathbf{v})$ givesIt remains zero under the torque-free transport law. Differentiate the contraction, using the product rule and $u^2=-c^2$:Thus an initially spatial spin in the instantaneous rest frame remains orthogonal to the velocity. This is the orthogonal-spin specialization of Fermi-Walker transport. The sign is tied to the negative timelike norm; using the opposite metric sign without changing the transport formula would spoil the cancellation.

Take the dot here to mean $d/d\tau$, as in the source. Differentiate $u^{\alpha }=\gamma (c,\mathbf{v})$ and the given spin components. Their contraction simplifies because the terms involving $\dot{\gamma }$ cancel:The spatial part of the transport equation consequently readsUse $\dot{\gamma }/\gamma ={\gamma }^2\mathbf{v}\cdot \dot{\mathbf{v}}/c^2$ and divide by $\gamma$. The result isNo assumption of acceleration perpendicular to the velocity was made. In particular the second term cannot be dropped for a particle speeding up along a straight line. This $\mathbf{s}$ is the three-vector in the source's particular spin four-vector parametrization; its Euclidean magnitude need not be the invariant proper-spin magnitude.

Let the fixed direction of motion be the unit vector $\mathbf{n}$ and decompose $\mathbf{s}=s_{\parallel }\mathbf{n}+{\mathbf{s}}_{\perp }$. In ii the two parallel contributions cancel, whereas the transverse equation isFor the stated speed law, $1-v^2/c^2=e^{-2\Gamma \tau }$, hence $\gamma =e^{\Gamma \tau }$. Since the particle is initially at rest, $\gamma (0)=1$, andThe corresponding spin four-vector has components $s^0=\gamma \beta s_{\parallel }(0)$, $s_{\parallel }^{\mathrm{spatial}}=\gamma s_{\parallel }(0)$, and ${\mathbf{s}}_{\perp }^{\mathrm{spatial}}={\mathbf{s}}_{\perp }(0)$. Its squared norm is therefore $s_{\parallel }(0{)}^2+|{\mathbf{s}}_{\perp }(0){|}^2$, independent of time. The transverse decay of the source's laboratory three-vector is a boost-normalization effect, not a loss of proper angular momentum or a torque.

For a real increasing speed take $\Gamma >0$ and $\tau \ge 0$. The prescribed profile has a singular initial acceleration: $v\sim c\sqrt{2\Gamma \tau }$ near zero. The differential transport calculation applies for $\tau >0$ and the solution has the continuous initial limit displayed above. This trajectory should not be silently described as motion at constant proper acceleration.