Form factor - spherically symmetric

In summary: Thanks for catching that! In summary, 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 which can be simplified to \frac{3e\hbar^2}{q^3} for r < R and zero elsewhere.
  • #1
Max Eilerson
121
1
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
,

[tex] F(q) = \int \frac{4\pi\hbar r}{q}p(r)sin(\frac{qr}{\hbar}) dr [/tex]

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). [tex] p(r) = \frac{q}{4\pi\epsilon_0R} [/tex] ?
 
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  • #2
Max Eilerson said:
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
,

[tex] F(q) = \int \frac{4\pi\hbar r}{q}p(r)sin(\frac{qr}{\hbar}) dr [/tex]

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). [tex] p(r) = \frac{q}{4\pi\epsilon_0R} [/tex] ?
I take this to mean

[tex] \rho(r) = \frac{q}{V} = \frac{3q}{4 \pi R^3} [/tex] for r < R and zero elsewhere.
 
  • #3
Thanks :). That's is what I was thinking after I posted. Don't see many [tex]\epsilon_0 [/tex] floating around in these things.
 
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  • #4
Max Eilerson said:
This just means I can use coulomb's law as an expression for p(r). [tex] p(r) = \frac{q}{4\pi\epsilon_0R} [/tex] ?
You know it has nothing to do with Coulomb's law.
 
  • #5
So I evaulated the integral with [tex] \rho(r) = \frac{3e}{4 \pi R^3} [/tex]
q = momentum transfer, e = proton charge.

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


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

It asks me to show that for [tex] \fracq{qR}{\hbar} << 1 [/tex] the form factor reduces to 1, I'm not seeing how the charge e can disappear here
 
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  • #6
Max Eilerson said:
So I evaulated the integral with [tex] \rho(r) = \frac{3e}{4 \pi R^3} [/tex]
q = momentum transfer, e = proton charge.

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


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

It asks me to show that for [tex] \fracq{qR}{\hbar} << 1 [/tex] 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

[tex] \int \rho(r) dr [/tex]

which would remove the quantity of charge from the calculation
 
  • #8
OlderDan said:
What does normalised charge distribution mean?

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

[tex] \int \rho(r) dr [/tex]

which would remove the quantity of charge from the calculation


[tex] 1 = \int \rho(r) d^3r [/tex]
 
  • #9
Max Eilerson said:
[tex] 1 = \int \rho(r) d^3r [/tex]

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

Related to Form factor - spherically symmetric

What is the definition of form factor - spherically symmetric?

The form factor - spherically symmetric refers to a mathematical function that describes the shape and size of a spherical object. It is often used in physics and engineering to calculate the scattering of particles off of a spherical target.

How is the form factor - spherically symmetric calculated?

The form factor - spherically symmetric is calculated by taking the integral of the scattering amplitude over all angles. This integral takes into account the size and shape of the spherical object, as well as the properties of the particles being scattered.

What is the importance of the form factor - spherically symmetric in physics?

The form factor - spherically symmetric is important in understanding the scattering of particles off of spherical targets. It is also used in the study of nuclear physics, as it can provide information about the structure of atomic nuclei.

Can the form factor - spherically symmetric be applied to non-spherical objects?

No, the form factor - spherically symmetric is specifically designed for spherical objects. However, there are other form factors that can be used for non-spherical objects, such as the form factor - elliptically symmetric.

How is the form factor - spherically symmetric used in experiments?

The form factor - spherically symmetric is often used in experiments involving particle scattering, such as in electron microscopy and X-ray diffraction. It can also be used in experiments involving the study of atomic nuclei and their structure.

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