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Maxwell's speed distribution law 
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#1
Mar907, 01:18 PM

P: 140

In a nontraditional 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
Mar907, 01:31 PM

P: 728

Do you what a partition function is?



#3
Mar907, 01:34 PM

P: 140

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?



#4
Mar907, 06:57 PM

PF Gold
P: 4,087

Maxwell's speed distribution law
I think the argument holds because
[tex]V^2 = V_x^2 + V_y^2 + V_z^2[/tex] therefore [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. 


#5
Mar907, 09:36 PM

P: 140

Thank you all very much



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