Form factor - spherically symmetric

[Max Eilerson]

1) Use the fact that the form factor, F(q), is the Fourier transform of the normalised charge distribution p(r), which in the spherically symmetric case gives
,

$$F(q) = \int \frac{4\pi\hbar r}{q}p(r)sin(\frac{qr}{\hbar}) dr$$

to find an expression for F(q) for a simple model of the proton considered as a uniform spherical charge distribution of radius R.

This just means I can use coulomb's law as an expression for p(r). $$p(r) = \frac{q}{4\pi\epsilon_0R}$$ ?

 

[OlderDan]

1) Use the fact that the form factor, F(q), is the Fourier transform of the normalised charge distribution p(r), which in the spherically symmetric case gives
,

$$F(q) = \int \frac{4\pi\hbar r}{q}p(r)sin(\frac{qr}{\hbar}) dr$$

to find an expression for F(q) for a simple model of the proton considered as a uniform spherical charge distribution of radius R.

This just means I can use coulomb's law as an expression for p(r). $$p(r) = \frac{q}{4\pi\epsilon_0R}$$ ?

I take this to mean

$$\rho(r) = \frac{q}{V} = \frac{3q}{4 \pi R^3}$$ for r < R and zero elsewhere.

 

[Max Eilerson]

Thanks :). That's is what I was thinking after I posted. Don't see many $$\epsilon_0$$ floating around in these things.

 

[Meir Achuz]

This just means I can use coulomb's law as an expression for p(r). $$p(r) = \frac{q}{4\pi\epsilon_0R}$$ ?

You know it has nothing to do with Coulomb's law.

 

[Max Eilerson]

So I evaulated the integral with $$\rho(r) = \frac{3e}{4 \pi R^3}$$
q = momentum transfer, e = proton charge.

$$\frac{3e\hbar^2(\hbar\sin[\frac{qr}{\hbar}] - qr\cos[\frac{qr}{\hbar}])}{q^3 R^3} + C$$
Evaluated between R and 0, (F(q) = 0 between R and infinity since p(r) = 0.)


$$F(q) = \frac{3e\hbar^2(\hbar\sin[\frac{qR}{\hbar}] - qR\cos[\frac{qR}{\hbar}])}{q^3 R^3}$$

It asks me to show that for $$\fracq{qR}{\hbar} << 1$$ the form factor reduces to 1, I'm not seeing how the charge e can disappear here

 

[OlderDan]

So I evaulated the integral with $$\rho(r) = \frac{3e}{4 \pi R^3}$$
q = momentum transfer, e = proton charge.

$$\frac{3e\hbar^2(\hbar\sin[\frac{qr}{\hbar}] - qr\cos[\frac{qr}{\hbar}])}{q^3 R^3} + C$$
Evaluated between R and 0, (F(q) = 0 between R and infinity since p(r) = 0.)


$$F(q) = \frac{3e\hbar^2(\hbar\sin[\frac{qR}{\hbar}] - qR\cos[\frac{qR}{\hbar}])}{q^3 R^3}$$

It asks me to show that for $$\fracq{qR}{\hbar} << 1$$ the form factor reduces to 1, I'm not seeing how the charge e can disappear here

What does normalised charge distribution mean?

I wonder if that means you should be dividing by a factor something like

$$\int \rho(r) dr$$

which would remove the quantity of charge from the calculation

 

[Max Eilerson]

Just found this page which says normalised means you can omit the e http://www.physics.rutgers.edu/ugrad/418/FormFactors.pdf (page 3)

The series expansions give me one in that limit with the e omitted:).

 

[Max Eilerson]

What does normalised charge distribution mean?

I wonder if that means you should be dividing by a factor something like

$$\int \rho(r) dr$$

which would remove the quantity of charge from the calculation

$$1 = \int \rho(r) d^3r$$

 

[OlderDan]

$$1 = \int \rho(r) d^3r$$

I like that even better. I even wrote my integral that way at first and changed it because yours just had the r integral.