- #1
jeebs
- 325
- 4
Here is the problem. I've been messing around with it for a while but I'm not sure if what I'm trying to do is the right way to approach this.
The form factor F(q) = [tex]\int\rho(\vec{r})e^{i\vec{q}.\vec{r}/\hbar}d^{3}\vec{r}[/tex] is the 3D Fourier Transform of the normalised charge distribution [tex]\rho[/tex]([tex]\vec{r}[/tex]).
For a simplified model of a proton's charge distribution, [tex]\rho(r)\propto[/tex] (e[tex]^{-r/R}[/tex])/r.
R can be considered as some characteristic "size" of the proton, setting the rate at which the charge dies away, but does not constitute a hard edge to the proton.
i) Find the constant of proportionality required to normalise [tex]\rho[/tex] correctly.
ii) something else that presumably needs the answer to i) first.
I am new to all this particle physics business, so I am in unfamiliar territory and I'm not sure how to approach this question. I've so far just aimlessly waded into this and ended up with a couple of sides of mindless mathematical messing around. This could be a simple question or a complicated one for all I know, so I thought i'd post it here before I bother my busy lecturer...
Anyway, my closest attempted solution:
-what I thought was that I should assume the proton has its highest charge density at its centre and it gradually fades away, uniformly in all directions.
-I'm also thinking that r must be the distance from the centre of the proton, so that as r tends to infinity, the charge density [tex]\rho(r)[/tex] approaches zero.
-I'm trying to find some constant of proportionality here, let's call it A, so that [tex]\rho(r) =[/tex] A(e[tex]^{-r/R}[/tex])/r.
-I'm thinking that if I do [tex]\int \rho(r) dV = 1[/tex] then I can solve for A, but this is as far as I have got, I'm struggling with how to take this integral any further.
Am I on the right lines, has anyone got any suggestions that would make my life easier?
thanks.
The form factor F(q) = [tex]\int\rho(\vec{r})e^{i\vec{q}.\vec{r}/\hbar}d^{3}\vec{r}[/tex] is the 3D Fourier Transform of the normalised charge distribution [tex]\rho[/tex]([tex]\vec{r}[/tex]).
For a simplified model of a proton's charge distribution, [tex]\rho(r)\propto[/tex] (e[tex]^{-r/R}[/tex])/r.
R can be considered as some characteristic "size" of the proton, setting the rate at which the charge dies away, but does not constitute a hard edge to the proton.
i) Find the constant of proportionality required to normalise [tex]\rho[/tex] correctly.
ii) something else that presumably needs the answer to i) first.
I am new to all this particle physics business, so I am in unfamiliar territory and I'm not sure how to approach this question. I've so far just aimlessly waded into this and ended up with a couple of sides of mindless mathematical messing around. This could be a simple question or a complicated one for all I know, so I thought i'd post it here before I bother my busy lecturer...
Anyway, my closest attempted solution:
-what I thought was that I should assume the proton has its highest charge density at its centre and it gradually fades away, uniformly in all directions.
-I'm also thinking that r must be the distance from the centre of the proton, so that as r tends to infinity, the charge density [tex]\rho(r)[/tex] approaches zero.
-I'm trying to find some constant of proportionality here, let's call it A, so that [tex]\rho(r) =[/tex] A(e[tex]^{-r/R}[/tex])/r.
-I'm thinking that if I do [tex]\int \rho(r) dV = 1[/tex] then I can solve for A, but this is as far as I have got, I'm struggling with how to take this integral any further.
Am I on the right lines, has anyone got any suggestions that would make my life easier?
thanks.