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## Homework Statement

An electric dipole instantaneously at rest at the origin in the frame [itex]K'[/itex] has potentials [itex]\Phi'=\mathbf{p}\cdot\mathbf{r}'/r'^3[/itex] and [itex]\mathbf{A}'=0[/itex] (and thus only an electric field). The frame [itex]K'[/itex] moves with uniform velocity [itex]\mathbf{v}=\vec{\beta }c[/itex] in the frame [itex]K[/itex].

- Show that in frame [itex]K[/itex] to first order in [itex]\beta[/itex] the potentials are [tex]\Phi = \dfrac{\mathbf{p}\cdot \mathbf{R}}{R^3},\quad\mathbf{A}=\vec{\beta }\dfrac{(\mathbf{p}\cdot\mathbf{R})}{R^3}[/tex] where [itex]\mathbf{R}=\mathbf{x}-\mathbf{x}_0(t)[/itex] with [itex]\mathbf{v} = \mathbf{x}_0'(t)[/itex].
- Show explicitly that the potentials in [itex]K[itex] satisfy the Lorentz condition.
- Show that to first order in [itex]\beta[/itex] the electric field [itex]\mathbf{E}[/itex] in [itex]K[/itex] is just the electric dipole field centered at [itex]\mathbf{x}_0[/itex] or a dipole field plus time-dependent higher multipoles, if viewed from a fixed origin, and the magnetic field is [itex]\mathbf{B}=\vec{\beta}\times \mathbf{E}[/itex]. Where is the effective dipole moment of Problem 6.21 or 11.27a?

## Homework Equations

The transformation law for vector fields [itex]{A'}^{\mu}(x')=\Lambda^{\mu}_\nu A^{\nu}(x)[/itex] and the expression for boosts in terms of the generators as [itex]\Lambda(\vec{\beta})=e^{-\hat{\beta}\cdot\mathbf{K} \tanh^{-1}\beta}[/itex] where [itex]K_i[/itex] are the boost generators.

## The Attempt at a Solution

Since frame [itex]K'[/itex] moves with respect to [itex]K[/itex] with velocity [itex]\mathbf{v}[/itex], we know that the Lorentz transformation from [itex]K'[/itex] to [itex]K[/itex] is just [itex]\Lambda(-\vec{\beta})[/itex]. Using the exponential form it is [itex]\Lambda(-\vec{\beta})=e^{\hat{\beta}\cdot \mathbf{K}\tanh^{-1}\beta}[/itex].

In that case, since we know the four potential at [itex]K'[/itex] and this transformation takes [itex]K'[/itex] to [itex]K[/itex] we can compute the potentials at [itex]K[/itex] as [tex]A^{\mu}(x)=\Lambda^{\mu}_\nu(-\vec{\beta}){A'}^{\nu}(x').[/tex]

The thing now is expand this matrix to first order in [itex]\beta[/itex]. My idea would be to expand the exponent first. Setting [itex]\xi_i(\vec{\beta})=\hat{\beta}_i \tanh^{-1}(\beta)[/itex] the exponent is [itex]\sum_i K_i\xi_i(\vec{\beta})[/itex]. So expanding to first order in [itex]\beta[/itex] we get [tex]\xi_i(\vec{\beta})=\xi_i(0)+\sum_j \beta_j\dfrac{\partial \xi_i}{\partial \beta_j}\bigg|_{\beta=0}[/tex]

The derivative is then given by (using [itex]\beta = |\vec{\beta}|[/itex]) [tex]\dfrac{\partial \xi_i}{\partial \beta_j}=\dfrac{\partial}{\partial \beta_j} \left(\dfrac{\beta_i}{\beta}\tanh^{-1}\beta \right)=\dfrac{\partial}{\partial \beta_j}\left(\dfrac{\beta_i}{\beta}\right)\tanh^{-1}\beta + \dfrac{\beta_i}{\beta} \dfrac{1}{1-\beta^2}\dfrac{\beta_j}{\beta}=\dfrac{\delta_{ij}\beta-\frac{\beta_i\beta_j}{\beta}}{\beta^2}\tanh^{-1}\beta + \dfrac{\beta_i\beta_j}{1-\beta^2}.[/tex]

Now I'm stuck. The closer I got was to think that since I need this derivative at zero, I can expand the inverse hyperbolic tangent. Expanding to third order in [itex]\beta[/itex] I was able to derive that [tex]\dfrac{\partial \xi_i}{\partial \beta_j}\approx \delta_{ij}+\delta_{ij}\dfrac{\beta^3}{3}-\dfrac{\beta_i\beta_j\beta}{3}-\beta_i\beta_j\left(\dfrac{1}{\beta^2}-\dfrac{1}{1-\beta^2}\right).[/tex]

But I'm stuck. I believe only the [itex]\delta_{ij}[/itex] should remain when taking [itex]\beta\to 0[/itex]. With this I believe part (1) follows immediately. Part (2) seems to be just a matter of computing the Lorentz condition. Then part (3) up to now I had no idea what to do.