# Homework Help: Electric dipole EM field using Lorentz Transformation

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1. Nov 26, 2017

### leo.

1. The problem statement, all variables and given/known data
An electric dipole instantaneously at rest at the origin in the frame $K'$ has potentials $\Phi'=\mathbf{p}\cdot\mathbf{r}'/r'^3$ and $\mathbf{A}'=0$ (and thus only an electric field). The frame $K'$ moves with uniform velocity $\mathbf{v}=\vec{\beta }c$ in the frame $K$.

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

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

3. The attempt at a solution
Since frame $K'$ moves with respect to $K$ with velocity $\mathbf{v}$, we know that the Lorentz transformation from $K'$ to $K$ is just $\Lambda(-\vec{\beta})$. Using the exponential form it is $\Lambda(-\vec{\beta})=e^{\hat{\beta}\cdot \mathbf{K}\tanh^{-1}\beta}$.

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

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

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

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 $\beta$ I was able to derive that $$\dfrac{\partial \xi_i}{\partial \beta_j}\approx \delta_{ij}+\delta_{ij}\dfrac{\beta^3}{3}-\dfrac{\beta_i\beta_j\beta}{3}-\beta_i\beta_j\left(\dfrac{1}{\beta^2}-\dfrac{1}{1-\beta^2}\right).$$

But I'm stuck. I believe only the $\delta_{ij}$ should remain when taking $\beta\to 0$. 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.

2. Nov 26, 2017

### TSny

To express this to first order in $\beta$, you can start by writing $\tanh^{-1}\beta$ to first order in $\beta$.
$\hat{\beta}$ is a fixed unit vector.

3. Nov 26, 2017

### leo.

I believe I got it, expanding $\tanh^{-1}\beta \approx \beta$ I have $\Lambda(-\vec{\beta})=e^{\vec{\beta}\cdot \mathbf{K}}$. Then expanding the exponential I get $\Lambda(-\vec{\beta})\approx 1 + \vec{\beta}\cdot \mathbf{K}$ which is first order in $\beta$.

With this I can derive that $A^{\mu}(x) = \Lambda^{\mu}_\nu(-\vec{\beta}){A'}^\nu(x')$ gives $\Phi(x) = \Phi'(x')+\vec{\beta}\cdot \mathbf{A}'(x')$ and $\mathbf{A}(x)=\vec{\beta}\Phi'(x')+\mathbf{A}'(x')$. Considering that $\Lambda(-\vec{\beta})$ maps $K'$ to $K$ the inverse matrix $\Lambda(\vec{\beta})$ maps $K$ to $K'$ and thus we can write $x' = \Lambda(\vec{\beta}) x$.

Now inputing the information of the problem we have that $$\Phi(x) = \Phi'(x'),\quad \mathbf{A}(x) = \vec{\beta}\Phi'(x').$$ But we have ${x'}^0=x^0 - \vec{\beta}\cdot \mathbf{x}$ and $\mathbf{x}'=\mathbf{x}-x^0\vec{\beta}$. Thus since $x^0 = ct$ we get $t' = t - \mathbf{v}\cdot\mathbf{x}$ and $\mathbf{x}'=\mathbf{x}- \mathbf{v}t$.

With this data, setting $\mathbf{R} = \mathbf{x}- \mathbf{x}_0(t)$ which is $\mathbf{x}_0(t) = \mathbf{v}t$ since $\Phi$ doesn't depend on time we get $$\Phi(t,\mathbf{x}) = \dfrac{\mathbf{p}\cdot\mathbf{R}}{R^3},\quad \mathbf{A}(t,\mathbf{x})=\vec{\beta} \dfrac{\mathbf{p}\cdot \mathbf{R}}{R^3}$$

Is part (1) correct? The only thing I'm unsure above is that $\mathbf{x}_0(t) = \mathbf{v}t$. Jackson just says that $\mathbf{v} = d\mathbf{x}_0/dt$ at time $t$, I believe there is something here about the analysis being based on the dipole instantaneously at rest on $K'$ but I'm unsure how this is formalized.

4. Nov 26, 2017

### TSny

I believe your work is correct, except I don't think you are meant to assume $\mathbf x_0(t) = \mathbf v t$.

I think the problem is assuming that the dipole is moving along an arbitrary trajectory $\mathbf x_0(t)$ with respect to frame K. At some instant of time $t$ you would like to know the values of $\Phi (t, \mathbf x)$ and $\mathbf A (t, \mathbf x)$ in frame K at some field point $\mathbf x$.

The dipole has velocity $\mathbf v = \frac{d \mathbf x_0(t)}{dt}$ at time $t$.

The frame K' (not shown above) is an inertial frame that is instantaneously comoving with the dipole at time $t$.

(One minor thing I noticed is that you wrote $t' = t - \mathbf{v}\cdot\mathbf{x}$ which I believe is missing some factors of $c$. But I don't think this affects the result.)

5. Nov 28, 2017

### leo.

But then I can't see why the $\mathbf{R}$ appears on the result. Notice that by my derivation we have (upon writing explicitly $\mathbf{x}'=\mathbf{x}-\mathbf{v}t$ that comes from the Lorentz transformation) $$\Phi(t,\mathbf{x})=\dfrac{\mathbf{p}\cdot(\mathbf{x}-\mathbf{v}t)}{|\mathbf{x}-\mathbf{v}t|^3},\quad \mathbf{A}(t,\mathbf{x})=\vec{\beta}\Phi(t,\mathbf{x})$$

Now, reading your post, I agree that $\mathbf{x}_0(t)\neq \mathbf{v}t$. The dipole has velocity $\mathbf{v}$ at time $t$ but its trajectory need not to be a straight line. Either way, the next idea would be to expand $\mathbf{x}_0(t+\delta t)\approx \mathbf{x}_0(t) + \mathbf{v}\delta t$ since we are looking at this instantaneously, but this doesn't help. It doesn't give $\mathbf{v}t$, just $\mathbf{v}\delta t$ and expanding around zero doesn't seem to be right, since $\mathbf{v}$ is the derivative at $t$ and not zero.

How then we end up replacing $\mathbf{x}-\mathbf{v}t$ on the equations by $\mathbf{R}(t,\mathbf{x})=\mathbf{x}-\mathbf{x}_0(t)$?

6. Nov 28, 2017

### TSny

Let's go back to your post #3 where you wrote

I agree with this.

This isn't necessarily the correct transformation of the spacetime coordinates of an event $(t, \mathbf x)$ between the $K$ and $K'$ frames since these transformation equations assume that the origins of the two frames coincide at $t = t' = 0$. But, this might not be true for the frames $K$ and $K'$ used here. However, you can use these Lorentz transformation equations to relate space and time intervals between the two frames.

That is, if $a$ and $b$ are two events with coordinates $(ct_a,\mathbf x_a)$ and $(ct_b, \mathbf x_b)$ in frame $K$ and coordinates $(ct_a', \mathbf x_a')$ and $(ct_b', \mathbf x_b')$ in frame $K'$, then $(\Delta ct,\Delta \mathbf x) \equiv (ct_b - ct_a, \mathbf x_b - \mathbf x_a)$ and $(\Delta ct',\Delta \mathbf x') \equiv (ct_b' - ct_a', \mathbf x_b' - \mathbf x_a')$ will be related by the Lorentz transformation equations.

You can let event $a$ correspond to the event where the dipole is located at position $\mathbf x_0(t_a)$ in frame $K$. Let event $b$ be the observation of the potentials at the field point $\mathbf x_b$ in frame $K$ at the instant $t_b$ that is simultaneous with event $a$ according to frame $K$. That is, $t_b = t_a$ for frame $K$.

In frame $K$ the space interval between the two events is $\Delta \mathbf x = \mathbf x_b - \mathbf x_a = \mathbf x_b - \mathbf x_0(t_a) = \mathbf R(t_a)$.

In frame $K'$, the space interval between the two events is $\Delta \mathbf x' = \mathbf x_b' - \mathbf x_a' = \mathbf r'$.
The Lorentz transformation to order $\beta$ then gives $\mathbf r' = \mathbf x_b - \mathbf x_0(t_a) = \mathbf R(t_a)$

There is a complication due to the fact that events $a$ and $b$ don't occur simultaneously in frame $K'$. That is, according to frame $K'$, event $b$ doesn't occur at the same moment that the dipole is at rest at the origin of frame $K'$. But I think you can argue that to order $\beta$ this time difference does not matter when calculating the potential in frame $K'$. That is, $\Phi'(\mathbf r', t_b') \approx \Phi'(\mathbf r', t_a')$ to order $\beta$.

7. Nov 28, 2017

### leo.

I believe I got your point. The Lorentz transformation's actually take place on the tangent spaces of spacetime relating tangent vectors, they are not transformations on the spacetime manifold itself. So what I did is fine for the four-potential because it is actually a tangent vector, but for the events themselves, in the case of Minkowski spacetime I need to include the translation relating the origins.

I understand also that the separations can be related by Lorentz transformations. So in frame K the separation between the events "dipole at $\mathbf{x}_0(t)$" and "observation of the potentials at $\mathbf{x}$" is $(0,\mathbf{x}-\mathbf{x}_0(t))$. The corresponding separation in K' is thus $(-\vec{\beta}\cdot\mathbf{x}+\mathbf{x}_0(t)\cdot \vec{\beta}, \mathbf{x}-\mathbf{x}_0(t))$.

Now I need the coordinates of the event $x'$ which are the coordinates in K' of the event "observation of the potentials". Since I know the separation I just need the coordinates in frame K' of the event corresponding to "dipole at $\mathbf{x}$ in K. But since the dipole is at the origin, I would have coordinates $(0,0)$ so that the coordinates of $x'$ needed on the potentials is just the separation itself. Is that the idea?

There are just two things: first, that the first event (dipole at $\mathbf{x}$ on K) maps to spatial coordinate zero on K' there is no doubt since on K' the dipole is at the origin. But am I right to set the corresponding time to zero? If so how to justify it? Because I feel this is the case but I don't know how to justify.

Apart from that, what you say about the time, the potentials in K' are time independent, so time wouldn't even enter expression, right?

8. Nov 28, 2017

### TSny

Yes, that all sounds good to me.

Yes.

Right.
Well, the choice of origin of time ($t' = 0$) in frame K' is not important for this problem. But I see no reason why you couldn't make the choice that event $a$ has time $t_a' = 0$ in K'. The only thing that might be of significance would be the time interval between events $a$ and $b$ in frame K': $\delta t' = t_b' - t_a'$, which you have calculated.

I think so. When the dipole is instantaneously at rest in frame K', the potentials are not changing in time at that instant in the vicinity of the origin of K'. This assumes that we can neglect retardation effects. For points "far" from the origin, the potential would depend on the behavior of the dipole at the retarded time. So, I think that there is an implicit assumption that the field point is not so far from the dipole that you would need to use retarded times.

Anyway, neglecting such effects, I think the potential $\Phi'(t_b', \mathbf x_b')$ in frame K' at event $b$ could be approximated as

$\Phi'(t_b', \mathbf x_b') \equiv \Phi'(t_b', \mathbf r') \approx \Phi'(t_a', \mathbf r') + \frac {\partial \Phi'}{\partial t'} \left(t_a', \mathbf r' \right) \delta t' +$ (higher order terms in $\delta t'$)

The second term on the right can be neglected since the rate of change of the potential in K' at the field point at the instant $t_a'$ is essentially zero. Since $\delta t'$ is of order $\beta$, the higher order terms on the right side would be of higher order in $\beta$.

Thus, $\Phi'(t_b', \mathbf r') \approx \Phi'(t_a', \mathbf r')$ to order $\beta$.

9. Dec 1, 2017

### leo.

Thanks for the help, it is solved now. The other items were just a matter of performing computations with what was obtained on the first one.

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