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

Use coordinates $X=(ct,x,y,z{)}^T$ and Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. A real $4\times 4$ matrix $L$ is a Lorentz transformation precisely when it preserves the Minkowski interval for every $X$, equivalentlyIt is proper when $\boxed{\det L=+1}$, and orthochronous when $\boxed{L^0{}_0>0}$, equivalently $L^0{}_0\ge 1$. The latter selects the orthochronous Lorentz group, which preserves future-directed timelike four-vectors. Indeed the time-time entry of the preservation identity gives $(L^0{}_0{)}^2-{\sum }_{i=1}^3(L^i{}_0{)}^2=1$, so $|L^0{}_0|\ge 1$. Imposing both conditions gives a proper orthochronous Lorentz transformation.

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 $-u$.

The signed rapidity corresponding to velocity $u$ isEquivalently $u/c=\tanh \varphi$, $\gamma =\cosh \varphi$, and $\gamma u/c=\sinh \varphi$. Thus the Lorentz boost from part (i) becomesMultiplying two such matrices gives diagonal entries $\cosh \varphi \cosh \psi +\sinh \varphi \sinh \psi =\cosh (\varphi +\psi )$ and off-diagonal entries $\cosh \varphi \sinh \psi +\sinh \varphi \cosh \psi =\sinh (\varphi +\psi )$, by the addition identities for hyperbolic functions. HenceThe resulting rapidity is $\varphi +\psi$. In velocity variables this is the relativistic velocity-addition formula $u\oplus v=(u+v)/(1+uv/c^2)$.

For $N\ge 2$, each adjacent pair of walkways has relative rapidity ${\varphi }_N=\mathrm{artanh}(1/N)$. There are $N$ such steps between $P_0$ and $P_N$, so the equal-rapidity chain of collinear Lorentz boosts has total rapidity $N{\varphi }_N$. Thus the actual velocity isIt is strictly below the speed of light for every finite $N\ge 2$. Adding the adjacent velocities as ordinary numbers ignores the relativistic velocity-addition formula.

For large $N$, the Taylor series of the inverse hyperbolic tangent givesConsequentlyMore precisely, $v_N/c=\tanh 1+{\mathrm{sech}}^2(1)/(3N^2)+O(N^{-4})$, so the limit is approached from above. The case $N=1$ would prescribe an adjacent relative velocity equal to the speed of light; it does not define an admissible walkway inertial frame.