This one is part b) of the problem I asked a day ago. Special thanks to MarkMitchison and sbp on the first part. I finally got it. ().   Alright, back to topic.      Let $\rho(r)$ be an arbitary charge distribution now. Under what circumstances does the dipole moment of the displaced charge distribution $\rho '(\vec{r})=\rho(\vec{r}-\vec{b})$ differ from the one not displaced at all.    Now this one is even nastier (excuse my language) than the one before. In hindsight, the one before was easy because I could express the integral in spherical coordinates.   But how do I approach this one? Am I supposed to solve $p=\int r'\cos\theta' \rho d^3r'$ in carthesian coordinates? And what would the boundaries be like?
a) There's a radially symmetric charge density $\rho(r)$ centered around the origin. Determine the dipolemoment of that charge density. b) Let $\rho(r)$ be an arbitrary charge density now. Under what circumstances does the dipole moment of the displaced charge density  $\rho '(\vec{r}) = \rho (\vec{r}-\vec{b})$ differ from the one not displaced at all.  Here were my ideas so far: a) Just thinking about the situation it has to be zero, right? I mean, since there's no real dipole. But how do I show that mathematically? I was thinking of just going like this (it may be wrong): Let the charge density be $\rho (r)=kr$, then we can get the charge q by integrating: $$q=4\pi \int_0^R kr\cdot r^2dr=\pi k R^4$$ I'm looking at the charge distribution as a spherical electron cloud with radius $R$. Then, since $p=qd$ and $d$ is zero because there are no two different charges the dipole moment is zero. Is that sufficient as an answer? b) I don't know how to approach this one. My guess is that its dipole moment is also zero because we're only looking at a displacement here. Anyone got any idea? I would appreciate any advice on this.
What influences findability on a post?   Upvotes Views Asked date Last active time   How is this done in the Stack Exchange internal search engine? And how does Google do it?  Why is it that when searching on Google for 'iPhone development stackoverflow' the fourth result is ? ii) Show that $u$ is not a real part of the function which analytic on $\mathbb{C} \backslash \lbrace 0 \rbrace$ Suppose $u(x,y)=\ln(x^2+y^2)$  i) Show that $u$ is harmonic on $\mathbb{C} \backslash \lbrace 0 \rbrace$  ii) Show that $u$ is not the real part of a function which analytic on  $\mathbb{C} \backslash \lbrace 0 \rbrace$  I manage to show the first part. For the second part, note that $u(x,y)=\ln(x^2+y^2)=\ln(|z|^2)=2 \Re \log(z)$  But this only show that $u$ is not a real part of $\log(z)$. I don know how to show $u$ cannot be the real part of a function which analytic on $\mathbb{C} \backslash \lbrace 0 \rbrace$  Can anyone guide me? 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It is clear that if $card(U) = n$ for some $n \in \omega$ then we only have a finite set of ordinals $\{\alpha_1, \ldots, \alpha_n\}=U$ so $\varphi$ is constant for $\alpha &gt; \alpha_n.$  If $U$ is countably infinite, then $U = \{\alpha_n\}_{n \in \omega}$ and since $[0, \omega_1)$ is sequentially compact and the $\alpha_n$'s are countable ordinals  we obtain a convergent subsequence $\{\alpha_{n_k}\}$ of $U$ to some countable ordinal $\beta \in [0, \omega_1)$ $[\beta = \cup_{n_k \in \omega}\alpha_{n_k}???],$ hence $\varphi$ is constant for all $ \alpha &gt; \beta.$     But what if $U$ is uncountable?    Idea: If $U = [0, \omega_1),$ then since $\varphi \rightarrow x$, for every $n \in \omega$ there is $\alpha_n \in U$ such that $\alpha \geq \alpha_n \implies \varphi(\alpha) \in B_{\frac{1}{n}}(x).$ This is a countable sequence, so since $[0, \omega_1)$ is sequentially compact, there is a subsequence converging to a countable ordinal $\beta \in [0, \omega_1)$. 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