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Take $V=|\mathbf{V}|$ to be the test body's instantaneous speed relative to the nonstreaming background, rather than an ensemble root mean square speed. Use a local isotropic galactic distribution function $f(v)$ normalized as a number distribution, and assume independent encounters and a large, approximately constant Coulomb logarithm in stellar dynamics.

For a background particle with relative incoming velocity $\mathbf{w}=\mathbf{V}-\mathbf{v}$, a gravitational scattering angle $\chi$ obeysIn the centre-of-momentum frame, the test body's share of the relative velocity is $m/(M+m)$. The azimuthal average cancels transverse changes, leaving a change parallel to the incoming relative velocity,The number of encounters per unit time in $d^{3}v$ and impact interval $db$ is $f(v)d^{3}vw2\pi bdb$. Integrating the longitudinal change from $b=0$ to $b_m$ givesAt leading logarithmic order replace the logarithm by $2\ln \Lambda$, evaluated at a representative speed. For isotropic $f$, the velocity-shell identity for dynamical friction isOne proof takes the divergence with respect to $\mathbf{V}$: the inverse-square vector field has divergence $4\pi {\delta }^{(3)}(\mathbf{V}-\mathbf{v})$, whose velocity integral is $4\pi f(V)$, and rotational symmetry plus the divergence theorem determines the radial field. Exterior velocity shells give zero field. Consequently the systematic Chandrasekhar dynamical friction is antiparallel to motion, withIn the massive-test-body approximation $M\gg m$, this becomesThe finite-test-mass dynamical friction coefficient shows the needed qualification: $M>m$ alone does not make $M+m=M$. The quoted expression also neglects diffusion and the velocity dependence of the logarithmic cutoff. For the lowered Maxwellian velocity distribution, set $f=0$ for $v>v_e$; continuing the printed subtraction beyond escape would give a negative distribution. If $V>v_e$, the integral stops at $v_e$.