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Short electrostatics question. Given that a dipole moment of a charge distribution is defined as $$P = \int r' \rho(r')d \tau '$$ where $r'$ is the position vector of the source charge, and $\rho$ is the charge density. Why does it follow that if we displace the origin by an amount $\vec{a}$ then the new dipole moment $\bar{P}$ is $$\bar{P} = \int \bar{r}' \rho(r')d \tau ' = \int (r'-\vec{a})\rho(r')d \tau'$$ rather than $$\bar{P} = \int \bar{r}' \rho(\bar{r}')d \tau '?$$

I might be missing something simple? Give me a hint if possible, thanks.

Alex
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3 Answers3

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The dipole moment of a continuos charge distribution is dependent (unless your total charge is 0) by the pole you choose. If we set our pole equal to the origin of our system then we have: $$\vec{p}=\int_V \rho (\vec{r}')\vec{r}' d\tau$$ where $\vec{r}'$ is the vector position from your origin to the point you consider during integration(if you have a discrete charge distribution, then $\vec{r_i}'$ is the position of the charge $q_i$ with respect to the origin, $i$ is the index of the sum that replace the integral), and remember that you are integrating upon $d\tau=dxdydz$ which is expressed with respect to the origin. Now let's say we choose another point $A=(x_A,y_A,z_A)$. Reasoning with discrete distribution we say that we are searching the contribution of the $q_i$ charge that is still at the point $P'=\vec{r_i}'$ (we didn't change the origin) to the moment but with respect to another point $A$ which is in turn in the position $\vec{r}_A$ (thus we have the term $(\vec{r_i}'-\vec{r}_A)$), but the volume upon we are integrating is the same $d\tau$ and the point $P'$ is expressed by the same $\vec{r_i}'$ because we didn't change the origin, but only the pole. It follows that the moment with respect to the point $A$ is: $$\vec{p}_A=\int_V \rho(P'=\vec{r}')(\vec{r}'-\vec{r}_A)d\tau$$ where $V$ and $d\tau=dxdydz$ are the same of before.

pier94
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  • Thanks for your response. What is your definition of 'pole' that you are using? I think there are some typos in your answer. You start using 'momentum'? – Alex Aug 26 '16 at 14:08
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    Yes, sorry for the typos but i'm answering from my phone with italian dictionary inserted so sometimes it makes typos; of course I meant 'moment' and not momentum; i will edit my answer. The pole is the point with respect to you are calculating your quantity, that, it may be, but is not necessarily, the origin. Think about the moment of inertia, that is completely analogous. You can calculate it with respect to the origin, or any another point you are interested in, and generally the values of it will vary in function of the point (the pole) you choose (it's a polar quantity). – pier94 Aug 26 '16 at 14:23
  • Yes, I think I am starting to understand what you are saying. Just to confirm: So for the discrete case. The dipole moment with respect to the origin is $$P = \sum_{i=1}^{n}q_ir_i$$ and with respect to a point $\vec{a}$ is $$P = \sum_{i=1}^{n}q_i(r_i-\vec{a})?$$ – Alex Aug 26 '16 at 14:42
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    @Alex yes! In general you can cast the spacial density of charge of a discrete charge distribution like $$\rho(\vec{r})=\sum\limits_{i=1}^n q_i\delta(\vec{r}-\vec{r}i)$$ where $\vec{a}$ is the position of the pole, and $\delta(x)$ is the Dirac's delta. You have $$\vec{p}=\sum\limits{i=1}^n q_i\int_V \delta(\vec{r}'-\vec{r}i)(\vec{r}'-\vec{a})d\tau=\sum\limits{i=1}^n q_i(\vec{r}_i-\vec{a})$$ due to delta's properties, and remember the specifications we made about $d\tau$. – pier94 Aug 26 '16 at 15:12
  • @pier94 What you describe is different from say shifting the whole coordinate system down a few units? But it seems that the dipole moment also remains unchanged if the system is neutral when this shifting is done. Do you know if there is a general result which states that the dipole moment is independent of a change of coordinates iff the system is neutral? –  Sep 05 '16 at 15:08
  • @LucioD if i've understood what you mean, if you bring on the calculations you have $$\vec{p}_A=\int_V \rho(\vec{r}')(\vec{r}'-\vec{r}_A)d\tau=\int_V \rho(\vec{r}')\vec{r}'d\tau-\vec{r}_A\int_V\rho(\vec{r}')d\tau$$ since $\int_V\rho(\vec{r}')d\tau=0$ you have only the term $$\int_V \rho(\vec{r}')\vec{r}'d\tau$$ thus your $\vec{p}_A$ does not depend on the pole you choose. If you change the frame of reference simply translating i would make the following reasoning. – pier94 Sep 05 '16 at 23:42
  • If I choose as new origin the previous point $P(\vec{a})\equiv O$ you have the formula $$\vec{p}{O}=\int{V(O)} \rho(\vec{r})\vec{r}d\tau$$ where $\vec{r}$ is the same vector $(\vec{r}'-\vec{a})$ of before (you can draw the vectors for persuade yourself). For what concerns $\rho$ you must think that is strictly connected to the space and your charge distribution and not to your frame, so it gives the same contribution for the same point. Looking now that this integral is similar to the one with the old origin but with pole $\vec{a}$ it's easy to see that $\vec{p}$ is the same. – pier94 Sep 05 '16 at 23:59
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The dipole moment of a charge distribution $\rho(\mathbf r)$, which is always defined as $$ \mathbf d=\int\mathbf r\rho(\mathbf r) \: \mathrm d\mathbf r $$ with respect to some coordinate origin, is independent of said coordinate origin if and only if the system is neutral. Otherwise, as you note, it transforms to a new origin $\mathbf r_0$ as $$ \mathbf d' =\int(\mathbf r-\mathbf r_0)\rho(\mathbf r) \: \mathrm d\mathbf r =\mathbf d-\mathbf r_0\int\rho(\mathbf r) \: \mathrm d\mathbf r =\mathbf d-Q\mathbf r_0 $$ for $Q$ the total charge of the system.

(This transformation law comes from the fact that in our original equation, $\mathbf r$ actually plays two roles, that of a position and that of an arrow. In this calculation of $\mathbf d'$ I'm keeping this calculation in the old reference frame, which keeps the $\rho(\mathbf r)$ intact, but I need to change the arrow to $\mathbf r-\mathbf r_0$ so that it now points from the new coordinate origin, $\mathbf r_0$, to the position $\mathbf r$.)

For a charged system, the dipole moment takes a backseat role, but it is still a useful quantity - for example, it is still the coefficient in the $1/r^2$ term of the multipole expansion for the electrostatic potential far away from the system, $$ V(\mathbf r) = \frac{Q}{r} + \frac{\mathbf d\cdot\mathbf r}{r^3}+\cdots, $$ with quadrupole terms after that. However, it is always important to keep in mind this dependence on the coordinate origin, and to clearly state this origin whenever the dipole moment is reported for a charged system.

Emilio Pisanty
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  • You mention here that you are keeping the old reference frame. So the coordinate origin and the origin are defined as two different things the way you are using it? Would I be right in stating that if we changed coordinates (say shifted all axis $x$, $y$ and $z$ down tow units) then the result would still hold that the dipole moment is independent of the new coordinates iff the system is neutral? –  Sep 05 '16 at 15:02
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You are absolutely correct . In the new reference frame dipole moment will be P¯= ∫r¯′ρ(r¯′)dτ′ And in old reference frame dipole moment will be P = ∫r′ρ(r′)dτ′

But you will have to understand what this term ρ(r′)dτ′ is . Its small charge dq which is at any given point lets say point A (Mark it red in your mind) in the charge distribution. For the same charge dq at A when observed from new frame we will have to write it as ρ(r¯′)dτ′. isn't it ?

So you can see that we can use ρ(r′)dτ′ and ρ(r¯′)dτ′ interchangeably because it points to same charge dq from different reference frame.

And thus your observation P¯= ∫r¯′ρ(r¯′)dτ′ is same as P¯=∫r¯′ρ(r′)dτ′ given in books.

Hope you got my point .

dτ′ term will be same in both reference frame . Its just dxdydz

Om Aditya
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