Relativistic Bullet & Bell — Two Observers


Relativistic Bullet Striking a Bell: Frame Analysis

Events in the bell’s rest frame S:

  • A: bullet strikes bell at (t=0,\;x=0).
  • B: bell begins emitting a pressure (sound) wave at (t=\tau_e,\;x=0), with \tau_e > 0 a (possibly tiny) response time.
Timelike separation (same place, later time):

    \[ \Delta s^2 \;=\; c^2(\Delta t)^2 - (\Delta x)^2 \;=\; c^2\tau_e^2 \;>\; 0. \]

Consequences: For timelike-separated events, all inertial observers agree on the order. Thus A occurs before B in every frame—no frame can render them simultaneous.

Observer 1 (on the bell; frame S)

She is at rest with the bell, so:

    \[ \Delta t_{(1)} \;=\; t_B - t_A \;=\; \tau_e \;>\; 0, \qquad \Delta x_{(1)} \;=\; 0. \]

Not simultaneous. The strike precedes the onset of emission by \tau_e.

Observer 2 (receding from the bell along the bullet’s direction)

Let Observer 2 move at constant speed u in +x. Their inertial frame is S' with Lorentz factor \gamma = 1/\sqrt{1-u^2/c^2}. For the emission event B (which has x=0 in S):

    \[ \Delta t'_{A\to B} = \gamma\!\left(\Delta t - \frac{u\,\Delta x}{c^2}\right) = \gamma\,\tau_e \;>\; \tau_e. \]

Thus the time gap between strike and start of emission is even longer in S' (time dilation). Still not simultaneous.

When does Observer 2 actually hear the sound?

Let sound propagate in the air rest frame S at speed v_s \ll c. Suppose at t=0 Observer 2 is at x=x_0>0 and recedes at speed u. The sound front, launched at t=\tau_e, obeys:

    \[ x_{\text{sound}}(t) \;=\; v_s\,(t-\tau_e), \qquad t \ge \tau_e, \]

    \[ x_{2}(t) \;=\; x_0 + u\,t. \]

The reception time t_R solves x_{\text{sound}}(t_R)=x_2(t_R):

    \[ v_s\,(t_R-\tau_e) \;=\; x_0 + u\,t_R \;\;\Longrightarrow\;\; t_R \;=\; \frac{x_0 + v_s\,\tau_e}{\,v_s - u\,}. \]

  • If u < v_s: t_R is finite and positive ⇒ Observer 2 eventually hears the bell.
  • If u \ge v_s: the sound never catches up ⇒ Observer 2 never hears the bell (in the medium’s rest frame).

The reception event R:(t_R,\;x_R=x_0+u t_R) is also timelike relative to A (any sub-c signal yields c^2\Delta t^2 - \Delta x^2 > 0). Hence the ordering A \rightarrow R is invariant across frames. In S', the elapsed time from strike to reception is:

    \[ \Delta t'_{A\to R} = \gamma\!\left(t_R - \frac{u\,x_R}{c^2}\right) = \gamma\!\left(t_R - \frac{u(x_0+u t_R)}{c^2}\right) = \gamma\!\left( t_R\!\left(1-\frac{u^2}{c^2}\right) - \frac{u x_0}{c^2}\right) = \frac{t_R}{\gamma} \;-\; \gamma\,\frac{u x_0}{c^2}. \]

You can insert the expression for t_R to see explicitly how x_0, u, v_s, and \tau_e shape the delay in S'.

Key Takeaways

  • Never simultaneous in any frame: A (strike) precedes B (emission start) for all observers since the separation is timelike.
  • Moving observer sees a larger gap: the interval between strike and emission start is \gamma \tau_e in the receding frame.
  • Hearing depends on outrunning sound: if u \ge v_s, the receding observer never hears the bell.
  • Causal order is invariant: strike \to emit \to (possibly) receive is preserved in all inertial frames; only the measured time intervals differ.