# Maxwell Boltzmann speed distribution: why isn't speed element integrated when converting from velocity distribution? [duplicate]

Maxwell Boltzmann velocity distribution is given by $$f_{\vec v}(v_x,v_y,v_z)=A^{3/2}\exp{[B(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})]}$$

To convert the velocity distribution into speed distribution, spherical coordinates were used, with $$v^2=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}$$, and volume element $$v^2 \sin\theta dv d\theta d\phi$$.

The integral is $$f(v)=A^{3/2} \exp{[Bv^2]} \int_{0}^{2\pi} \int_{0}^{2\pi} v^2 \sin\theta dv d\theta d\phi$$

, which becomes $$f(v)=A^{3/2} \exp{[Bv^2]} 4\pi v^2$$

Why isn't speed element integrated here?

• Feb 3 at 19:29

'Integrating out' the angular components leads to a result that depends only on the magnitude of the velocity vector (i.e. the speed). That is what you want: a function that describes the distribution of speeds. If you integrated wrt to the speed, then you would not have a function of speed anymore (you would get instead the probability of finding a particle with speed between $$v$$ and $$v + dv$$).