- #1

deuteron

- 57

- 13

- Homework Statement
- find the potential caused by the rotationally symmetric charge distribution ##\rho_{\vec r_q}\equiv \rho_{r_q}##

- Relevant Equations
- ##\phi_{e;\vec r} = \iiint\limits_{\R^3} d^3r\ \rho_{\vec r_q} \frac 1 {|\vec r-\vec r_q|}

First, we rewrite the term ##|\vec r-\vec r_q|## in the following way:

$$|\vec r-\vec r_q|= \sqrt{(\vec r-\vec r_q)^2} = \sqrt{\vec r^2 + \vec r_q^2 -2\vec r\cdot\vec r_q} = \sqrt{r^2 + r_q^2 -2rr_q\cos\theta}$$

Due to rotational symmetry, we go to spherical coordinates:

$$\phi_{e;\vec r_q} = \int\limits_0^{2\pi}d\varphi \int\limits_0^\pi d\theta \int\limits_0^\infty dr_q\ r_q^2 \sin^2\theta\ \rho_{r_q} \frac 1 {\sqrt{ r^2 + r_q^2 - 2rr_q\cos\theta}}= 2\pi \int\limits_0^\pi d\theta \int\limits_0^\infty dr_q\ r_q^2 \sin^2\theta\ \rho_{r_q} \frac 1 {\sqrt{ r^2 + r_q^2 - 2rr_q\cos\theta}}$$

For the ##\theta## integral, we do the substitution ##\theta\mapsto\cos\theta,\ d\theta \mapsto (-\frac 1 {\sin\theta})d\cos\theta##, and we get:

$$= 2\pi \int\limits_{1}^{-1} d\cos\theta \int\limits_0^\infty dr_q \ (-\frac 1 {\sin\theta})\ r_q^2 \sin^2\theta\ \rho_{\vec r_q} \frac 1{\sqrt{r^2 + r_q^2 -2rr_q\cos\theta}}\\ = 2\pi \int\limits_{-1}^1 d\cos\theta \int\limits_0^\infty dr_q\ r_q^2 \sin\theta \ \rho_{\vec r_q} \sqrt{r^2 + r_q^2 -2rr_q\cos\theta}^{-1} \\ = 2\pi \int\limits_0^\infty dr_q\ r_q^2 \rho_{\vec r_q} [\frac 1 {rr_q} (r^2 + r_q^2 -2rr_q\cos\theta)^{\frac 12}]_{\cos\theta=-1}^{\cos\theta=1},$$

which leads us to the right answer, at least this is what I have in my solution sheet.

What I don't understand is, when we do the ##\theta##-substitution, why don't we write for the ##\cos\theta## inside the square root as ##\cos(\cos\theta)##, since I would expect the substitution to affect the ##\theta## inside another function too.

$$|\vec r-\vec r_q|= \sqrt{(\vec r-\vec r_q)^2} = \sqrt{\vec r^2 + \vec r_q^2 -2\vec r\cdot\vec r_q} = \sqrt{r^2 + r_q^2 -2rr_q\cos\theta}$$

Due to rotational symmetry, we go to spherical coordinates:

$$\phi_{e;\vec r_q} = \int\limits_0^{2\pi}d\varphi \int\limits_0^\pi d\theta \int\limits_0^\infty dr_q\ r_q^2 \sin^2\theta\ \rho_{r_q} \frac 1 {\sqrt{ r^2 + r_q^2 - 2rr_q\cos\theta}}= 2\pi \int\limits_0^\pi d\theta \int\limits_0^\infty dr_q\ r_q^2 \sin^2\theta\ \rho_{r_q} \frac 1 {\sqrt{ r^2 + r_q^2 - 2rr_q\cos\theta}}$$

For the ##\theta## integral, we do the substitution ##\theta\mapsto\cos\theta,\ d\theta \mapsto (-\frac 1 {\sin\theta})d\cos\theta##, and we get:

$$= 2\pi \int\limits_{1}^{-1} d\cos\theta \int\limits_0^\infty dr_q \ (-\frac 1 {\sin\theta})\ r_q^2 \sin^2\theta\ \rho_{\vec r_q} \frac 1{\sqrt{r^2 + r_q^2 -2rr_q\cos\theta}}\\ = 2\pi \int\limits_{-1}^1 d\cos\theta \int\limits_0^\infty dr_q\ r_q^2 \sin\theta \ \rho_{\vec r_q} \sqrt{r^2 + r_q^2 -2rr_q\cos\theta}^{-1} \\ = 2\pi \int\limits_0^\infty dr_q\ r_q^2 \rho_{\vec r_q} [\frac 1 {rr_q} (r^2 + r_q^2 -2rr_q\cos\theta)^{\frac 12}]_{\cos\theta=-1}^{\cos\theta=1},$$

which leads us to the right answer, at least this is what I have in my solution sheet.

What I don't understand is, when we do the ##\theta##-substitution, why don't we write for the ##\cos\theta## inside the square root as ##\cos(\cos\theta)##, since I would expect the substitution to affect the ##\theta## inside another function too.