Thermal Physics · Problem 1

Kinetic Theory — Charles Kittel & Herbert Kroemer

Physics — Statistical Mechanics · mean-speeds-maxwellian-distribution

Problem

Mean speeds in a Maxwellian distribution. (a) Show that the root mean square velocity vrmsv_{\text{rms}} is vrms=⟨v2⟩1/2=(3τ/M)1/2.(118)v_{\text{rms}} = \langle v^2 \rangle^{1/2} = (3\tau/M)^{1/2}. \tag{118} Because ⟨v2⟩=⟨vx2⟩+⟨vy2⟩+⟨vz2⟩\langle v^2 \rangle = \langle v_x^2 \rangle + \langle v_y^2 \rangle + \langle v_z^2 \rangle and ⟨vx2⟩=⟨vy2⟩=⟨vz2⟩\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle, it follows that ⟨vx2⟩1/2=(τ/M)1/2=vrms/31/2.(119)\langle v_x^2 \rangle^{1/2} = (\tau/M)^{1/2} = v_{\text{rms}}/3^{1/2}. \tag{119} The results can also be obtained directly from the expression in Chapter 3 for the average kinetic energy of an ideal gas. (b) Show that the most probable value of the speed vmpv_{\text{mp}} is vmp=(2τ/M)1/2.(120)v_{\text{mp}} = (2\tau/M)^{1/2}. \tag{120} By most probable value of the speed we mean the maximum of the Maxwell distribution as a function of vv. Notice that vmp<vrmsv_{\text{mp}} < v_{\text{rms}}. (c) Show that the mean speed cˉ\bar{c} is cˉ=∫0∞dv vP(v)=(8τ/πM)1/2.(121)\bar{c} = \int_0^\infty dv\, v P(v) = (8\tau/\pi M)^{1/2}. \tag{121} The mean speed may also be written as ⟨∣v∣⟩\langle |v| \rangle. The ratio vrms/cˉ=1.086.(122)v_{\text{rms}}/\bar{c} = 1.086. \tag{122} (d) Show that cˉz\bar{c}_z, the mean of the absolute value of the zz component of the velocity of an atom, is cˉz≡⟨∣vz∣⟩=12cˉ=(2τ/πM)1/2.(123)\bar{c}_z \equiv \langle |v_z| \rangle = \tfrac{1}{2}\bar{c} = (2\tau/\pi M)^{1/2}. \tag{123}

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