Supersonic Speeds, Shock Waves — Question 7

Question 7

The cone marks the merger of two retarded emission times

A source has moved for all relevant past times along $(vt_e,h)$ with $v>c$. A stationary observer at the origin receives a signal at $t=t_e+\sqrt{v^2t_e^2+h^2}/c$. Consider positive reception times and retain the physical condition $t-t_e\ge0$.

1. Derive a quadratic equation for $t_e$ and find the earliest positive reception time from its discriminant.

2. Explain the number of physical emission times just before, at, and after that threshold. Why is a divergent linear amplitude near the envelope not a complete shock theory?

Question 7 - Solution

The cone marks the merger of two retarded emission times

(a) Method and calculation

Squaring $c(t-t_e)=\sqrt{v^2t_e^2+h^2}$ and rearranging gives$$(v^2-c^2)t_e^2+2c^2tt_e+h^2-c^2t^2=0.$$Its discriminant is$$D=4[c^2v^2t^2-(v^2-c^2)h^2].$$For positive $t$, $D\ge0$ requires $t\ge t_b=h\sqrt{v^2-c^2}/(cv)=h\sqrt{M^2-1}/v$.

(b) Interpretation and further calculation

Before $t_b$ there is no real emission time. At $t_b$ the two roots merge into one tangent event. After it there are two real roots, both satisfying $t_e<t$ for positive $t$; two different earlier emissions reach the same event. Concentration of wavefronts explains the envelope, but finite-amplitude nonlinear response, dissipation and source size determine an actual shock profile.

Physical check

Negative reception times admitted by the squared discriminant must still pass the unsquared propagation condition; the positive-time restriction prevents that spurious interpretation here.