Past exam of the symmetry and relativity course of the University of Oxford / 2016 / 1

For a massive particle on a timelike world line, proper time satisfies $d\tau =dt\sqrt{1-v^2/c^2}=dt/\gamma$. Its four-velocity and four-acceleration are $U=dX/d\tau$ and $A=dU/d\tau$. Differentiating $U=\gamma (c,\mathbf{v})$ givesThe invariant Minkowski inner products areFor the explicitly requested contraction,Thus four-acceleration is orthogonal to four-velocity. Equivalently this follows by differentiating the constant Lorentz scalar $U^2=c^2$ with respect to proper time.

The event separation is timelike precisely whenA simultaneous pair in an inertial frame would instead have $(D-B{)}^2=-|\Delta {\mathbf{x}}^{\prime }{|}^2\le 0$. Since a Lorentz transformation preserves this interval, no such simultaneous frame exists. A timelike pair can be brought to the same spatial position: the required velocity is $\mathbf{u}=\Delta \mathbf{x}/\Delta t$, whose magnitude is less than $c$.

The separate condition $D\cdot B=0$ says that the two event-position four-vectors, measured from the chosen spacetime origin, are orthogonal in the Minkowski inner product. It is invariant under Lorentz transformations about that origin but not under arbitrary translations. It is not a general criterion for their causal separation.

For a concrete example, put $B=(cT,0,0,0)$ and $D=(0,R,0,0)$. Their inner product vanishes, but their separation has squared interval $c^2{T}^2-R^2$, which can be positive, zero or negative. If $B$ is timelike, a rest frame for $B$ makes $D^0=0$; this means that $D$ and the chosen origin are simultaneous in that frame, not that $D$ and $B$ are simultaneous.

The proper acceleration is $\alpha =\sqrt{-A^2}$, the magnitude of the instantaneous rest-frame three-acceleration. A pure force is a four-force preserving rest mass, or equivalently satisfying $F\cdot U=0$. Indeed $F=d(mU)/d\tau$ gives $F\cdot U=c^{2}dm/d\tau$.

For the right-hand branch of the hyperbola, use $c=1$, $x>0$ and $x=\sqrt{L^2+t^2}$, taking $L>0$. Direct differentiation givesThe sign of $v$ follows the sign of $t$, whereas the coordinate acceleration on this branch is positive. ConsequentlyThe proper-time parameterization $t=L\sinh (\tau /L)$, $x=L\cosh (\tau /L)$ independently gives constant proper acceleration. Restoring dimensions, a hyperbola $x^2-c^2{t}^2=L^2$ has $\alpha =c^2/L$. Constant proper acceleration does not mean that all inertial-frame components of the four-force are constant.

Let $f=|\mathbf{f}|$ and $\hat{\mathbf{f}}=\mathbf{f}/f$. A constant laboratory three-force means $d\mathbf{p}/dt=\mathbf{f}$, hence $\mathbf{p}=\mathbf{f}t$ from rest. Using $E^2=m^2{c}^4+p^2{c}^2$ givesThe dimensionless time $s=ft/(mc)$ makes the shapes transparent: $\gamma =\sqrt{1+s^2}$ starts at one with zero slope and becomes asymptotically linear; $\beta =s/\sqrt{1+s^2}$ starts linearly and approaches one without reaching it at finite time. The acceleration decreases as $f/(m{\gamma }^3)$.

Integrating the velocity gives the distance along the force,Thus a constant three-force produces hyperbolic motion with constant proper acceleration $f/m$, although its velocity is never Newtonian at large time. The following original curves supply the requested sketches.

The magnitude of the electric force is $f=e\mathcal{E}$, directed so that the electron gains energy. The work across the gap is $e\mathcal{E}L=50\mathrm{MeV}$. With the source precision $m{c}^2=0.511\mathrm{MeV}$,Reducing the field by 20 percent gives kinetic energy $40\mathrm{MeV}$ andA large fractional energy change therefore produces only a small speed change, since the original electron is already highly relativistic.

For the transit time, combine the distance formula from solution with ${\gamma }^2-1=[ft/(mc){]}^2$:This is slightly longer than $L/c$, since the electron starts from rest. For the reduced field it becomes $3.37798\times 10^{-8}\mathrm{s}$.