The relative velocity is $$v_{rel}=\sqrt2v_{mean}$$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$$$v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$$ because the reduced mass of two of the same particles is half the mass of one. In general: $$v_{rel}=(\frac{8KT}{\pi \mu})^{\frac12}$$

However, my question is: what is the need for reduced mass? If $v_{rel}=\sqrt2v_{mean}$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$. Wouldn't that just leave you with: $v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ Where does reduced mass come into it?


1 Answer 1


Your equation $v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$ is incorrect.

$v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ (your last equation) is also incorrect.

Instead $v_{rel}=(\frac{16KT}{\pi m})^{\frac12}$ and $v_{rel}=(\frac{8KT}{\pi \frac m2})^{\frac12}$ would both be correct.

The need for reduced mass is for when the particles are not the same mass, usually in the context of considering collisions between two different molecules or atoms.

$\mu = m_1m_2/(m_1 + m_2)$ by definition. If $m_1 = m_2$ then $\mu = m/2$

When there are two different masses, there are two different mean velocities.

To see where the equation with the reduced mass comes from, look at appendix 1.4 of Kinetic Theory of Gases, along with pages 1-30 to 1-34.

  • $\begingroup$ Thank you. The reduced mass must also be used for when the particles are the same (i.e use $\frac m2$) to get the required equation rather than the one I wrote at the end of my question with the 16. $\endgroup$
    – RobChem
    Dec 24, 2014 at 15:00
  • $\begingroup$ Is reduced mass just always used for two-body problems? $\endgroup$
    – RobChem
    Dec 24, 2014 at 15:07
  • 1
    $\begingroup$ @RobChem I added to my answer, see if it is clear now. $\endgroup$
    – DavePhD
    Dec 24, 2014 at 16:13
  • $\begingroup$ Perfect, thanks. Sorry to be a pain but would you please take a look at chemistry.stackexchange.com/questions/22037/… $\endgroup$
    – RobChem
    Dec 24, 2014 at 16:48

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