# Multipole expansions: test on a function $\zeta=\zeta(t)$

Considering the potential $\psi(r)$ of a sphere of mass $M$ with density $\rho(\mathbf r')$, connected by a small volume positioned in the $P'$ point as shown in the figure:

The $\psi(r)$ is:

\begin{align} \psi(r)=-G\iiint_{\mathcal{V}} \frac{\rho(\mathbf r')dx'dy'dz'}{|\mathbf r - \mathbf r'|}= -G\iiint_{\mathcal{V}}\frac{\rho(\mathbf r')dV'}{\sqrt{{r}^2+{r'}^{2}-2rr'\cos \theta}}=& \nonumber \\ =-\frac Gr\iiint_{\mathcal{V}}\frac{\rho(\mathbf r')dV'}{\sqrt{1+(r'/r)^2-2 (r'/r)\cos \theta}} \end{align}

If we want multipoles expansions of the gravitational potential, we need the $O$ origin of the reference system to be close to the sphere so that $r\gg r'$. Putting $r'/r=\beta$, $\cos \theta=t$, with $t$ parameter where $|t|\le 1$, the the potenzial $\psi(r)$ can be written:

$$f(x)=\frac{1}{\sqrt{1+x}}$$ with $x=\beta^2-2\beta t$. The idea of a multipole expansion of the gravitational potential of a mass distribution, valid for distant points, is to write it in a series of powers of the type $1/r$, where we indicated with $r$ the distance from the point $P$ in question. We know that if $g$ is a function of the type $g(x)= (1+x)^\alpha$ true for $1+x>0$ and $\alpha\in\mathbb R$ then $g(x)$ is written as a binomial series in the way:

$$(1+x)^\alpha =\sum_{n=1}^\infty \binom{\alpha}{n-1} x^{n-1}$$

If the binomial series is convergent then $|x|<1$. We may therefore also consider cases where $x=\pm 1$ but it all depends to $\alpha$. We wonder then when it will happen that $$|x|=|\beta^2-2\beta t|<1$$ This condition is equivalent to resolving the following system:

$$\begin{cases} \beta^2-2\beta t-1<0& (\star) \\ \beta^2-2\beta t+1>0& (\star \star) \end{cases}$$

The $(\star \star)$ is always true for every $t\in\mathbb R$. In fact being $\Delta/4=t^2-1\le 0$ then $|t|\le 1$. Consider the $(\star)$. Solving it we find the solutions

$$t-\sqrt{t^2+1}\le \beta \le t+\sqrt{t^2+1}\quad \tag{*}$$

But $\beta=r'/r>0$ so the ($^*$) will be:

$$0\le \beta \le t+\sqrt{t^2+1}\quad \tag{**}$$

Let's consider the function $$\zeta(t)=t+\sqrt{t^2+1} \quad \tag{***}$$ we must take the minimum. But

$$\zeta'(t)=1+\frac{2t}{2\sqrt{t^2+1}}=\frac{\sqrt{t^2+1}+t}{\sqrt{t^2+1}}>0$$

Hence $\zeta(t)$ is always increase in $\mathbb R$, without minimum or maximum point. But if I plot the function:

$$h(\beta)=h(\cos \theta)=\cos \theta + \sqrt{\cos^2 \theta +1} \tag{§}$$

it have severals points of mininum (with periodicity $2\pi$) as shown in the second figure:

My questions are:

1. Why $\sqrt{t^2+1}>t$, $\forall |t|<1$.

2. If I can find the minimum point $(-1, -1+\sqrt 2)$ from the function $$h(\beta), \tag{§}$$ why I have needed of the function $\zeta=\zeta(t)$?

3. Why $\zeta(-1)=-1+\sqrt 2$; therefore the $(-1, -1+\sqrt 2)$ point is a

minimum for the $\zeta$ function and therefore

$$0\le \beta=\frac{r'}r\le \sqrt 2-1\quad ?$$

• According to your rationale the condition $\beta\leq \sqrt{2}-1$ comes from $|x|<1$, however, in $f(x) = \frac{1}{\sqrt{1+x}}$ $x$ can be as large as you want, there is no limit. The limit comes only in if one wants to use the development in the binomial series. – Frederic Thomas Jul 16 '18 at 8:37
• Of course x cannot be smaller than -1 or equal -1.. – Frederic Thomas Jul 16 '18 at 8:53