A New Look at Momentum — Question 8

Section 37-10 · Physics pages 6226–6227

Question 8 A reflecting particle beam exerts pressure through momentum flux A dilute beam of identical particles has laboratory number density n and speed v = 0.60c normal to a stationary wall. Each particle reverses its velocity elastically. The density refers to the incoming beam alone. Use p = γmv, γ = (1 −v2/c2)−1/2, and F = dp/dt. Here m is invariant mass. reflected Schematic; labels, not drawing lengths, specify the data. (a) Find the wall pressure from the arrival rate and impulse per particle. (b) Compare with the pressure predicted by classical momentum at the same laboratory density and speed.
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Question 8 - Solution A reflecting particle beam exerts pressure through momentum flux reflected (a) Method and calculation ˙N A = nv, |∆pparticle| = 2γmv. P = 2nγmv2 . (b) Interpretation and further calculation β = .60 : P = .900 nmc2 . Pcl = 2nmv2 = .720 nmc2, P/Pcl = 1.25 . Physical check The particle impulse points away from the wall; the equal and opposite impulse on the wall creates the positive pressure.

Question 8

A reflecting particle beam exerts pressure through momentum flux

A dilute beam of identical particles has laboratory number density n and speed v=0.60cv=0.60c normal to a stationary wall. Each particle reverses its velocity elastically. The density refers to the incoming beam alone. Use 𝒑=γm𝒗\mathbf p=\gamma m\mathbf v, γ=(1−v2/c2)−1/2\gamma=(1-v^2/c^2)^{-1/2}, and 𝑭=d𝒑/dt\mathbf F=d\mathbf p/dt. Here m is invariant mass.

Diagram shown in the original SVG worksheet.

  1. Find the wall pressure from the arrival rate and impulse per particle.

  2. Compare with the pressure predicted by classical momentum at the same laboratory density and speed.

Question 8 - Solution

A reflecting particle beam exerts pressure through momentum flux

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ṄA=nv,|Δpparticle|=2γmv.\frac{\dot N}{A}=nv,\qquad |\Delta p_{\mathrm{particle}}|=2\gamma mv.P=2nγmv2.\boxed{P=2n\gamma mv^2}.

(b) Interpretation and further calculation

β=.60:P=.900nmc2.\beta=.60:\quad\boxed{P=.900\,nmc^2}.Pcl=2nmv2=.720nmc2,P/Pcl=1.25.P_{\mathrm{cl}}=2nmv^2=.720\,nmc^2,\quad\boxed{P/P_{\mathrm{cl}}=1.25}.

Physical check

The particle impulse points away from the wall; the equal and opposite impulse on the wall creates the positive pressure.