# maxwell's speed distribution law

by Kolahal Bhattacharya
Tags: distribution, maxwell, speed
 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.
 P: 725 Do you what a partition function is?
 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?
PF Patron
P: 3,924

## maxwell's speed distribution law

I think the argument holds because

$$V^2 = V_x^2 + V_y^2 + V_z^2$$
therefore
$$P(V^2) = P( V_x^2 + V_y^2 + V_z^2 )$$
and the next step follows from the independence of V_x, V_y and V_z.
 P: 140 Thank you all very much

 Related Discussions Introductory Physics Homework 8 Introductory Physics Homework 9 Advanced Physics Homework 2 Classical Physics 3 Advanced Physics Homework 5