1. ## Maxwell-Boltzmann law-urgent

According to the Maxwell-Boltzmann law of theoretical physics, the probability density of V, the velocity of a gas molecule, is
f(v)=kv^2*e^(- βv^2), for v>0; =0, elsewhere.
Where β depends on its mass and the absolute temperature and k is an appropriate constant. Show that the kinetic energy E=(1/2)mV^2, where m the mass of the molecule is a random variable having a gamma distribution.

2. Originally Posted by Yan
According to the Maxwell-Boltzmann law of theoretical physics, the probability density of V, the velocity of a gas molecule, is
f(v)=kv^2*e^(- βv^2), for v>0; =0, elsewhere.
Where β depends on its mass and the absolute temperature and k is an appropriate constant. Show that the kinetic energy E=(1/2)mV^2, where m the mass of the molecule is a random variable having a gamma distribution.

If you still have trouble after reading this, please show your working and where you get stuck.

3. Originally Posted by mr fantastic

If you still have trouble after reading this, please show your working and where you get stuck.
The problem is i don't understand what is the question ask for..
can you explain to me? and step by step???

Originally Posted by mr fantastic

If you still have trouble after reading this, please show your working and where you get stuck.
In this problem, i'm confuse about the sentence<where m the mass of the molecule is a random variable having a gamma distribution.>what is the relationship between this sentence and the problem?

4. Originally Posted by Yan
The problem is i don't understand what is the question ask for..
can you explain to me? and step by step???
You're given the pdf for V and asked to find the pdf of $\displaystyle \frac{m}{2} V^2$.

5. Originally Posted by mr fantastic
You're given the pdf for V and asked to find the pdf of $\displaystyle \frac{m}{2} V^2$.

G(y) = P (Y<y)
= P ($\displaystyle \frac{m}{2} V^2$<y)
= 2 P(V<$\displaystyle \sqrt{\frac{2x}{M}}$)
= 2* ....

I just confuse about the m, because in the problem, it says that m is a random variable having a gamma distribution. what is this information use for???

6. Originally Posted by Yan
G(y) = P (Y<y)
= P ($\displaystyle \frac{m}{2} V^2$<y)
= 2 P(V<$\displaystyle \sqrt{\frac{2x}{M}}$)
= 2* ....

I just confuse about the m, because in the problem, it says that m is a random variable having a gamma distribution. what is this information use for???
There is a comma missing after molecule, it should read:

"Show that the kinetic energy E=(1/2)mV^2, where m the mass of the molecule, is a random variable having a gamma distribution."

m is not a random variable, E (the kinetic energy) is the random variable in question.

CB