What should the normalization condition be for an energy distribution function?

[LuisVela]

Hello everybody.
I have the following problem:
I am given a Speed distribution function, f(v). This distribution is normalized to the total number of particles there are, that's equivalent to:

$$\int_0^\infty f(v) dv = n_0$$

(particles traveling away from my target are not considered.)Now. I know the following equation should hold:

$$\int_w^\infty v f(v) dv = \alpha$$

where $\alpha$ is a constant, and $w=\sqrt{\frac{2e\epsilon}{m}}$. Here e is the charge of the electron, and $\epsilon$ is another constant.

All this is part of the theory I am trying to simulate on my computer program.
The thing is, when I started my program, I decided to work with energies rather than velocities, so I have to express both, the normalization condition, and the equation itself in terms of my energy distribution function f(E).

Since energy and velocity are related as follow:

$$\frac{1}{2}mv^2 = eE$$

(the energy is in eV)

Then,

$$dV=\sqrt{\frac{e}{2mE}}dE$$

If you replace all the v's with E's and realize that $f(v)=f(v(E))=f(E)$, then you have:

$$\frac{e}{m}\int_{\epsilon}^\infty f(E) dE = \alpha$$

I think until now I have got it correct.
Im just a little confused about what should $\int_0^\infty f(E) dE$ be normalized to. I mean...either:

a.) Since f(E) is also a distribution function, it should also be normalized to the total number of particles. $$\int_0^\infty f(E) dE = n_0$$

b.) If you perform the substitution $\frac{1}{2}mv^2 = eE$ to the normalization condition for f(V) you get the following:

$$\int_0^\infty \sqrt{\frac{e}{2mE}}f(E)dE=n_0$$

which is not precisely what's written under option a.)...

Do you understand my dilemma?

Can anybody help me out ?

 

[vanhees71]

I do not exactly understand what you are up to, but your energy formula cannot be right. On the left-hand side you write a scalar quantity, nonrelativistic kinetic energy, on the right-hand side a vector. That cannot match! For an electrostatic situation you have

$$E=\frac{m}{2}\vec{v}^2+e \Phi,$$

where $\Phi$ is the potential of the electric field,

$$\vec{E}=-\vec{\nabla} \Phi.$$

 

[Petr Mugver]

Be careful, f(v(E)) is not equal to f(E), because your function v(E) is not the identity transformation. Rather you shuld define g(E) = f(v(E)) and replace all your f(E)'s with g(E)'s.
This way the alpha-condition you wrote is correct, and the correct normalization condition is a).

 

[LuisVela]

@ Vanhees 71:
Hey!, thanks for your reply. But I still don't understand your comment.
$$\vec{v}^2=\vec{v}\cdot\vec{v}\in R$$
which is also a scalar.

I guess you misunderstood my no0tation (sorry for thta XD, my mistake). E is not the electric field. E is the total mechanical energy of the system. e, on the other side, is the charge of the electron.


@P. Mugver

Thanks for your comment.
f(v)=f(v(E)) is easy to understand. Now, it seems obvious to me that from there, f(v(E))=f(E) where f(E) is simply the velocity distribution function, expressed in terms of the energy.

If now, there would exist g(E) (energy distribution function function) then I am pretty sure a.) will hold. However, what I am not sure of is:

Does g(E) = f(v(E))?

Where g(E) is the energy distribution function, normalized to n_0, and f(v(E)) is simple the velocity distribution function expressed in terms of the energy.

 

[Petr Mugver]

g(E)=f(v(E)) is a definition, not the energy distribution, which is, instead g(E)dv/dE, and the normalization is the same.