# Ideal gas

#### ehrenfest

1. The problem statement, all variables and given/known data
The following is a probability density distribution for an ideal gas:
$$p(v_x,v_y,v_z) = (2 \pi \sigma ^2)^{-3/2} \exp\left({\frac{v_x^2 +v_y^2+v_z^2}{2 \sigma^2}}\right)$$

$$E = 0.5 m |\vec{v}|^2$$

Find p(E).

2. Relevant equations

3. The attempt at a solution
I want to just plug in what E but that gives the wrong answer and I cannot figure out why.
$$p(E) = (2 \pi \sigma ^2)^{-3/2} \exp\left({\frac{E}{m \sigma^2}}\right)$$
Why is that wrong? It makes no sense, but I just plugged in what E was!! This is really bothering me.

#### kdv

1. The problem statement, all variables and given/known data
The following is a probability density distribution for an ideal gas:
$$p(v_x,v_y,v_z) = (2 \pi \sigma ^2)^{-3/2} \exp\left({\frac{v_x^2 +v_y^2+v_z^2}{2 \sigma^2}}\right)$$

$$E = 0.5 m |\vec{v}|^2$$

Find p(E).

2. Relevant equations

3. The attempt at a solution
I want to just plug in what E but that gives the wrong answer and I cannot figure out why.
$$p(E) = (2 \pi \sigma ^2)^{-3/2} \exp\left({\frac{E}{m \sigma^2}}\right)$$
Why is that wrong? It makes no sense, but I just plugged in what E was!! This is really bothering me.

What is the correct answer? It's a probability *density* so there is also a factor $$dv_x dv_y dv_z$$ that must be multiplied to get a probability. If you want the probability density in terms of the energy you wil have to relate the above factors to the infinitesimal $$dE$$

EDIT: it's a density in speed so the factor to convert is probably dv

#### ehrenfest

Interesting. BTW, is it generally true that if y=y(x) and x=x(z), then y=y(x(z))? That is, you can write y as a function of z just by plugging in x as a function of z. That rule seems to fail here. But in general, when is it true?

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#### kdv

Interesting. BTW, is it generally true that if y=y(x) and x=x(z), then y=y(x(z))? That is, you can right y as a function of z just by plugging in x as a function of z. That rule seems to fail here. But in general, when is it true?
It is true in general, yes.

The problem here is that what you are looking for is *not* simply y(x(z))!

what you are looking for is the probability density in energy so the probability of measuring an energy between E and E+dE.. If you just plug in v in terms of energy, you are still giving the probability density in *speed*, so you are giving th eprobability of measuring the speed between v and v+dv. This is not what you are asked to find!

#### ehrenfest

So, back to the original problem. $p(v_x, v_y, v_z)dv_x dv_y dv_z[/tex] gives the probability of finding the particle's velocity in the box with side lengths [itex]dv_x, dv_y, dv_z$ with the one corner at $(v_x, v_y, v_z)$.

And I want to find p(E) such that $p(E)dE[/tex] gives the probability of finding the particle's energy between E and E+dE. We know that [itex]dE = 1/2m(2v_x dv_x + 2v_y dv_y+2v_z dv_z)$, but I still don't really know how to proceed. Do express each of dv_x dv_y dv_z as a function of dE somehow?

You said something about just using dv, but I am not sure what you meant. Can you elaborate?

#### kdv

So, back to the original problem. $p(v_x, v_y, v_z)dv_x dv_y dv_z[/tex] gives the probability of finding the particle's velocity in the box with side lengths [itex]dv_x, dv_y, dv_z$ with the one corner at $(v_x, v_y, v_z)$.

And I want to find p(E) such that $p(E)dE[/tex] gives the probability of finding the particle's energy between E and E+dE. We know that [itex]dE = 1/2m(2v_x dv_x + 2v_y dv_y+2v_z dv_z)$, but I still don't really know how to proceed. Do express each of dv_x dv_y dv_z as a function of dE somehow?

You said something about just using dv, but I am not sure what you meant. Can you elaborate?
You can't directly get from the distribution in v_x, v_y an dv_z to the density in Energy since there is more information in the first than in the second (since E depends only on v^2). So you need to find the probability density in the speed first. So you integrate over all directions using $$dv_x dv_y dv_z = v^2 dv d\Omega$$. But the result of the angular integral is simply $$4 \pi v^2 dv$$ times the probability density in v_x, v_y and v_z so

$$\rho(v) dv = 4 \pi \rho(v_x,v_y,v_z) v^2 dv$$
where it is understood that in the sceon term you rewrite everything in terms of the speed v.

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