maxwell's speed distribution law


by Kolahal Bhattacharya
Tags: distribution, maxwell, speed
Kolahal Bhattacharya
Kolahal Bhattacharya is offline
#1
Mar9-07, 01:18 PM
P: 140
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.
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Manchot
Manchot is offline
#2
Mar9-07, 01:31 PM
P: 728
Do you what a partition function is?
Kolahal Bhattacharya
Kolahal Bhattacharya is offline
#3
Mar9-07, 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?

Mentz114
Mentz114 is offline
#4
Mar9-07, 06:57 PM
PF Gold
P: 4,081

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.
Kolahal Bhattacharya
Kolahal Bhattacharya is offline
#5
Mar9-07, 09:36 PM
P: 140
Thank you all very much


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