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For an atomic form factor $F(\textbf q)$, the corresponding charge density distribution is given by $$ \rho(\textbf r) = \frac{1}{(2\pi)^3}\int\text{d}^3 \textbf q \,F(\textbf q)\,\text{e}^{-\text{i}\textbf q\cdot \textbf r}.\tag{1} $$ where $\textbf q$ denotes the momentum transfer in an elastic scattering process. What if the form factor is only known as a function of the four-momentum squared, $Q^2=-q^2$: $F(Q^2)$? How would one calculate the charge density? Do I integrate over $\text{d}^4q$? Do I leave $Q^0$ constant and get the result as a function of $\textbf r$ and $Q^0$?

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Maybe it makes sense to transform the equation, and reduce the ammount of information you are trying to extract? When you say you only know as a function of four momentum, what do you mean?

Do you mean that instead of $F=F\left(\mathbf{q}\right)$ you only know $K(Q)=\oint d^2\Omega_\mathbf{\hat{q}} F\left(\mathbf{\hat{q}}\sqrt{Q}\right)$, where $\oint d^2\Omega_\mathbf{\hat{q}}$ is the integral over the solid angle?

Why not try to reduce the spatial information?

$\bar{\rho}\left(r\right)=\oint d^2 \Omega_\mathbf{\hat{r}} r\rho\left(\mathbf{\hat{r}}r\right)=\frac{1}{\left(2\pi\right)^3}\int d^3\mathbf{q}F\left(\mathbf{q}\right)\oint d^2\Omega_\mathbf{\hat{r}} \exp\left(-iqr\left(\mathbf{\hat{q}.\hat{r}}\right)\right)=\frac{1}{\left(2\pi\right)^3}\int d^3\mathbf{q}F\left(\mathbf{q}\right) \frac{-4\pi }{qr}\sin\left(qr\right)$

Where $\oint d^2\Omega_\mathbf{\hat{r}}=\int^{2\pi}_0 d\phi\int^\pi_0 \sin\theta d\theta$

So:

$\bar{\rho}\left(r\right)=\frac{-1}{2\pi^2 r}\int^\infty_0 \frac{q^2 dq}{q} \sin\left(qr\right) \oint d^2\Omega_\mathbf{\hat{q}} F\left(\mathbf{\hat{q}}q\right)=\frac{-1}{2\pi^2 r}\int^\infty_0 q\, dq\, \sin\left(qr\right)\, K\left(q^2\right)$

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  • $\begingroup$ I mean the four momentum $Q^\mu$, such that $Q^2 = (Q^0)^2-\textbf{Q}^2$. $\endgroup$ – Stephan Apr 27 '19 at 8:17
  • $\begingroup$ Hmm. You could do the same thing I did in four-dimensions. However, if you want to work in four dimensions shouldn't you work with four-current density and the 4d fourier transform? Also, it would help to define the quantities are working with. For example, based on you initial post I can define $F\left(\mathbf{q}\right)=\int d^3 r \rho\left(\mathbf{r}\right)\exp\left(i\mathbf{q.r}\right)$. How do you define $F$ as a function of the magnitude of $Q^2$? $\endgroup$ – Cryo Apr 27 '19 at 21:37
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Using the Breit frame (which, for elastic scattering, is the center of mass frame), we have $Q^2 = Q^\mu Q_\mu =-q^2 = -(\nu^2 - \textbf q^2) = \textbf q^2$. This is because the Breit frame is defined to have no energy transfer: $\nu\equiv 0$.

Since the density distribution is a Lorentz scalar, we can choose any frame and the result should be the same.

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