- #1
LuisVela
- 33
- 0
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:
[tex]\int_0^\infty f(v) dv = n_0 [/tex]
(particles traveling away from my target are not considered.)Now. I know the following equation should hold:
[tex] \int_w^\infty v f(v) dv = \alpha [/tex]
where [itex] \alpha [/itex] is a constant, and [itex] w=\sqrt{\frac{2e\epsilon}{m}} [/itex]. Here e is the charge of the electron, and [itex] \epsilon [/itex] 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:
[tex] \frac{1}{2}mv^2 = eE[/tex]
(the energy is in eV)
Then,
[tex] dV=\sqrt{\frac{e}{2mE}}dE [/tex]
If you replace all the v's with E's and realize that [itex] f(v)=f(v(E))=f(E) [/itex], then you have:
[tex] \frac{e}{m}\int_{\epsilon}^\infty f(E) dE = \alpha[/tex]
I think until now I have got it correct.
Im just a little confused about what should [itex] \int_0^\infty f(E) dE [/itex] 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. [tex]\int_0^\infty f(E) dE = n_0 [/tex]
b.) If you perform the substitution [itex] \frac{1}{2}mv^2 = eE[/itex] to the normalization condition for f(V) you get the following:
[tex] \int_0^\infty \sqrt{\frac{e}{2mE}}f(E)dE=n_0[/tex]
which is not precisely what's written under option a.)...
Do you understand my dilemma?
Can anybody help me out ?
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:
[tex]\int_0^\infty f(v) dv = n_0 [/tex]
(particles traveling away from my target are not considered.)Now. I know the following equation should hold:
[tex] \int_w^\infty v f(v) dv = \alpha [/tex]
where [itex] \alpha [/itex] is a constant, and [itex] w=\sqrt{\frac{2e\epsilon}{m}} [/itex]. Here e is the charge of the electron, and [itex] \epsilon [/itex] 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:
[tex] \frac{1}{2}mv^2 = eE[/tex]
(the energy is in eV)
Then,
[tex] dV=\sqrt{\frac{e}{2mE}}dE [/tex]
If you replace all the v's with E's and realize that [itex] f(v)=f(v(E))=f(E) [/itex], then you have:
[tex] \frac{e}{m}\int_{\epsilon}^\infty f(E) dE = \alpha[/tex]
I think until now I have got it correct.
Im just a little confused about what should [itex] \int_0^\infty f(E) dE [/itex] 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. [tex]\int_0^\infty f(E) dE = n_0 [/tex]
b.) If you perform the substitution [itex] \frac{1}{2}mv^2 = eE[/itex] to the normalization condition for f(V) you get the following:
[tex] \int_0^\infty \sqrt{\frac{e}{2mE}}f(E)dE=n_0[/tex]
which is not precisely what's written under option a.)...
Do you understand my dilemma?
Can anybody help me out ?
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