Past exam of the short options course of the University of Oxford / 2015 / 8 / b / i / Solution

Now use $(ct,x{)}^T$ and ${\eta }_2=\mathrm{diag}(1,-1)$. WritePreservation of the Minkowski metric says that the first column has squared norm $1$, the second has squared norm $-1$, and the columns are orthogonal for the Minkowski inner product:Orthochronousness gives $a\ge 1$. From the orthogonality equation, $b=de/a$; substituting in the second norm identity gives $e^2=a^2$. Hence $(b,e)=\epsilon (d,a)$ with $\epsilon =\pm 1$. The determinant is then $\epsilon (a^2-d^2)=\epsilon$, so properness forces $\epsilon =1$. Thus $b=d$ and $e=a$.

Set $\beta =d/a$. The first norm identity implies $|\beta |<1$ and $a=(1-{\beta }^2{)}^{-1/2}$. With $u=c\beta$, this givesConversely, the displayed matrix has ${\gamma }^2-(\gamma u/c{)}^2=1$, equal off-diagonal entries, determinant one, and positive time-time entry. Direct multiplication therefore gives $L^T{\eta }_{2}L={\eta }_2$, proving both directions. The plus signs are the Lorentz boost convention used in the paper; reversing the boost direction replaces $u$ by $-u$.