Potential of a rotationally symmetric charge distribution

[deuteron]

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.

 

[Orodruin]

The substitution is $u = \cos\theta$, implying that $du = d\cos\theta = -\sin\theta\, d\theta$. It is usually much clearer to introduce a simple new variable rather than using a function of the original variable to represemt the new one. $\theta \to \cos\theta$ does not mean you replace $\theta$ by $\cos \theta$, it means you use $\cos\theta$ as your integration variable instead of $\theta$.

The 3D volume element also contains only a single $\sin\theta$.

 

[deuteron]

The substitution is $u = \cos\theta$, implying that $du = d\cos\theta = -\sin\theta\, d\theta$. It is usually much clearer to introduce a simple new variable rather than using a function of the original variable to represemt the new one. $\theta \to \cos\theta$ does not mean you replace $\theta$ by $\cos \theta$, it means you use $\cos\theta$ as your integration variable instead of $\theta$.

The 3D volume element also contains only a single $\sin\theta$.

Thank you! But I am still confused about why we are not changing the $\theta$ that is already inside the $\cos$. For example, a substitution like $x\mapsto \sin \theta$ would look like the following:

$$ \int \sqrt{1-x^2}\ dx \stackrel{x\mapsto \sin \theta} {=} \int \sqrt{1-\sin ^2\theta } \cos\theta\ d\theta = \int\cos^2\theta\ d\theta$$

where we substitute in the function $f(x)=\sqrt{1-x^2}$

however, in the above example, if we let $f(x)=r_q^2\ \sin\ \frac1 {\sqrt{r^2 + r_q^2 + \cos\ }}$, I would expect the substitution $\theta\mapsto \cos\phi$ to affect the $\theta$ in the $\sin$ and $\cos$ too, which would look something like:

$$\int\limits_0^\pi d\theta \ r_q^2 \sin\theta \frac 1 {\sqrt{r^2 + r_q^2 + \cos\theta}}\stackrel{\theta\mapsto \cos\phi}{=} \int\limits_{1}^{-1} d\cos\phi \ (-\frac 1{\sin\phi}) r_q^2 \sin(\sin\phi) \frac 1 {\sqrt{r^2 + r_q^2 + \cos(\sin\phi)}}$$

I still haven't understand why we don't change the variables **in** the $\cos$ and $\sin$ when we do a substitution

 

[Orodruin]

But I am still confused about why we are not changing the θ that is already inside the cos.

You are changing it. If $u = \cos\theta$ then $\theta = \acos u$ so $\cos\theta = \cos(\acos u) = u$.

 

[deuteron]

Thank you very much!!