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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


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    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
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