1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maxwell's speed distribution law

  1. Mar 9, 2007 #1
    In a non-traditional type of derivation of Maxwell's speed distribution for gases,I happen to face the following problem:
    They say since P(v_x),P(v_y),P(v_z) are independent,so the combined probability wil be P=P(v_x)P(v_y)P(v_z).
    This much is OK.Then they say the only function having the property f(a+b+c)=f(a)f(b)f(c)
    is an exponential function.So, consider the P(v_x) as to have exponential dependence P(v_x)=K exp[-L*(v_x)^2].This makes me uncomfortable.Did we have P=P(v_x+P_y+P_z)?I am a bit new to statistical ideas,so really cannot be sure when we said the joint probability is P,it means P=P(v_x+P_y+P_z).
    Please help.
  2. jcsd
  3. Mar 9, 2007 #2
    Do you what a partition function is?
  4. Mar 9, 2007 #3
    OK, I do not know.Is P acting as a partition function here?How do we know that?And after all,what does it do?
  5. Mar 9, 2007 #4


    User Avatar
    Gold Member

    I think the argument holds because

    [tex]V^2 = V_x^2 + V_y^2 + V_z^2[/tex]
    [tex]P(V^2) = P( V_x^2 + V_y^2 + V_z^2 )[/tex]
    and the next step follows from the independence of V_x, V_y and V_z.
  6. Mar 9, 2007 #5
    Thank you all very much
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Maxwell's speed distribution law