# Understanding of dipole moment and its vector property

I have a trouble understanding the electric dipole moment.

The electric dipole moment formula is $${\bf p}= \int {\bf r}' \rho({\bf r}')d\tau '$$ I'm interested in the coordinate, the origin of which is changed into $\bf a$. $${\bf r}' = {\bar {\bf r}}' + {\bf a}$$ Now calculate dipole moment in the new coordinate \begin{align} {\bar {\bf p}} &= \int {\bar {\bf r}}' \rho ({\bar {\bf r}}') d\bar\tau' \\ &= \int ({\bf r}'-{\bf a}) \rho ({\bar {\bf r}'}) d\bar\tau' \end{align} In Griffiths, it says $$\rho ({\bar {\bf r}}') = \rho ({{\bf r}}')$$ so that yields ${\bar {\bf p}}= {\bf p} - Q {\bf a}$.

I don't understand how to verify $\rho ({\bar {\bf r}}') = \rho ({{\bf r}}')$.

• I think, to get $\bar{\bf{p}}$, you need to work with $\bar{\rho}(\bf{r})=\rho(\bf{r}-\bf{a})$. Then you substitute $\bar{\bf{r}}=\bf{r}-\bf{a}$ in order to do the integral. Just using your old $\rho$ and shifting the integration variable will give you $\bf{p}$ again rather than $\bar{\bf{p}}$. – Photon Jan 20 '16 at 11:26

I think you are confused with the new coordinates system.

You sould correct the above equations like this:

\begin{align} \bar{\mathbf p} &= \int \bar{\mathbf r} \bar\rho(\bar{\mathbf r})d\bar\tau \\ &=\int ({\mathbf r}-\mathbf a) \rho(\mathbf r)d\tau\\ &=\mathbf p - \mathbf a\int\rho(\mathbf r)d\tau = \mathbf p - Q\mathbf a \end{align}

In other words, you should consider a new density function $\bar\rho$.