Past exam of the physics 1 paper of the University of Oxford / 2015 / 11

Use the tunnel frame $S$, with the entrance at $x=0$, the far end at $x=L$, and $t=0$ at both runner entry and photon emission. The runner has $x_R=vt$, while the approaching photon has $x_{\gamma }=L-ct$. At detection,For $0<v<c$, $0<f<1/2$: the photon meets the runner before the runner reaches the tunnel midpoint.

Let $S^{\prime }$ be the runner's inertial frame, with origin at the entry event. The Lorentz transformation isThe emission event $E=(L,0)$ has coordinates $x_E^{\prime }=\gamma L$, $t_E^{\prime }=-\gamma vL/c^2$. It is earlier than entry in $S^{\prime }$: entry and far-end emission are simultaneous only in the tunnel frame. This is relativity of simultaneity, not an extra delay of the light source.

The entrance clock is at $x=0$. Let $G$ be the event on that clock simultaneous with emission in the runner's frame. Since $t_G^{\prime }=\gamma t_G=t_E^{\prime }$, its reading isThis is the tunnel clock's reading, not the runner-frame coordinate time $t_E^{\prime }=-\gamma vL/c^2$. It is also not a calculation of what a photograph of that distant clock would show, which would require an additional light-travel condition.

At that simultaneous entrance-clock event, $x_G=0$ and $t_G=-vL/c^2$. Its runner-frame position isThe runner is at $x^{\prime }=0$, so the entrance is still ahead by distance $\gamma {\beta }^{2}L$ at the runner-frame emission time. The simultaneously measured distance between the two tunnel ends is $x_E^{\prime }-x_G^{\prime }=\gamma L(1-{\beta }^2)=L/\gamma$, consistent with length contraction.

The photon starts at $x_E^{\prime }=\gamma L$ and travels left to the runner at the fixed position $x^{\prime }=0$. ThusThis distance is not the simultaneous contracted tunnel length $L/\gamma$, because the runner is not at the entrance when the photon is emitted in $S^{\prime }$. Adding the entrance-to-runner gap from part (ii) to $L/\gamma$ indeed gives $\gamma L$.

During the photon's runner-frame flight, the tunnel entrance moves left by $v\Delta t_{\gamma }^{\prime }=\gamma \beta L$. It began ahead of the runner by $\gamma {\beta }^{2}L$, so at detection it lies behind byDivide this distance into the tunnel by the simultaneously measured tunnel length $L/\gamma$:The runner is stationary in $S^{\prime }$, but their location relative to the moving tunnel still has this fraction. Omitting the initial entrance gap would incorrectly count the runner's progress from an emission event at which they had not yet entered in this frame. These simultaneous positions give the tunnel photon timing in two inertial frames.

The photon propagates along $-x$ in the tunnel frame and has energy $E_{\gamma }=hc/{\lambda }_0$ and momentum $p_x=-E_{\gamma }/c$. A Lorentz transformation into the approaching runner's frame givesSince $E_{\gamma }^{\prime }=hc/{\lambda }^{\prime }$, the approaching-source relativistic Doppler effect isIt is a blueshift. With $q={\lambda }^{\prime }/{\lambda }_0=500.0/589.1$, solving $q^2=(1-\beta )/(1+\beta )$ givesThis shift depends on the relative velocity along the light ray, not on the tunnel length or the location of detection.