Relativistic electrodynamics: transformation of electric dipole moment

[lys04]

Homework Statement

^^As title suggests

Relevant Equations

Lorentz transformation matrix, current 4-vector

A charge distribution stationary in its own frame S’ has a static charge density $\rho ’$, a total charge of 0 and a net electric dipole $\vec{p'}$. An observer in frame S sees the charge distribution moving with constant velocity $\vec{v}=c \vec{\beta}$ What is the magnetic dipole moment in S?

Split the electric dipole moment into components perpendicular and parallel $\vec{\beta}$, i.e $\vec{p'}_\parallel$ and $\vec{p'}_\perp$

Then the current density in frame S is given by $\vec{j} = c \gamma \rho ' \vec{\beta}$

Now using the definition of magnetic dipole moment: $\vec{m} = \frac{1}{2} \int_V \vec{x} \times \vec{j} dV$

Split $\vec{x}$ into components perpendicular and parallel to $\vec{\beta}$, i.e $\vec{x}_\parallel$ and $\vec{x}_\perp$

$\vec{m} = \frac{c \gamma}{2} \int_V ( \vec{x}_\parallel + \vec{x}_\perp) \times (\rho ' \vec{\beta}) dV$

$\vec{m} = \frac{c \gamma}{2} \int_V \rho ' ( \vec{x}_\parallel + \vec{x}_\perp) \times \vec{\beta} dV$

$\vec{m} = \frac{c \gamma}{2} \int_V \rho ' \vec{x}_\perp \times \vec{\beta} dV$

$\vec{m} = \frac{\gamma}{2} \int_V \rho ' \vec{x}_\perp \times \vec{v} dV$

$\vec{m} = \frac{\gamma}{2} \int_V \rho ' \vec{x}_\perp dV \times \vec{v}$

$\implies \vec{m} = \frac{\gamma}{2} \vec{p'}_\perp \times \vec{v}$

Is this correct?

 

[QuarkyMeson]

This doesn't look correct to me.

What about the volume element, how does that/ should that change? (I just did this in emag2 and we blew through this at the end, so.... I could be wrong.)

 

[lys04]

This doesn't look correct to me.

What about the volume element, how does that/ should that change? (I just did this in emag2 and we blew through this at the end, so.... I could be wrong.)

Yeah I thought about that, I’m not sure how to do it tho since the direction is not specified but instead in some arbitrary direction.

 

[QuarkyMeson]

Arbitary should be fine, infact you don't even need to pick a basis for this.

So dV = dLdA, you can now choose dL so that it is parallel and dA so that its perpendicular. How would dL change under the transformation? Then go back to dV' and you're basically done.

 

[lys04]

Arbitary should be fine, infact you don't even need to pick a basis for this.

So dV = dLdA, you can now choose dL so that it is parallel and dA so that it’s perpendicular. How would dL change under the transformation? Then go back to dV' and you're basically done.

Sorry for the late reply.
Hm so by length contraction for the component of dl parallel to the direction of v it gets contracted? dl’=dl/gamma?

 

[QuarkyMeson]

Yeah, this is how I would I do it.

 

[lys04]

Yeah, this is how I would I do it.

So then $dV'=dA dl' = \frac{dA dl}{\gamma} = \frac{dV}{\gamma}$?
This is dV' when the component of dl is parallel to the direction of v. How do I deal with the component that's perpendicular? It doesn't change right? (But is the perpendicular component really needed anyways?)
So then if I plug that back into my integral I get the same answer but without the gamma factor? $\frac{1}{2} \vec{p'}_\perp \times \vec{v}$?

 

[QuarkyMeson]

Yes, the perpendicular component doesn't change. $dV = \frac{dV'}{\gamma}$, so you plug that back in and you get $\frac{1}{2}\vec{p'}_{\perp} \times \vec{v}$.. which obviously matches the non relativistic expression. (This isn't a great check here since what you wrote before also matches the non relativistic expression when $\gamma \approx 1$.)

 

[lys04]

plug that back in and you get $\frac{1}{2}\vec{p'}_{\perp} \times \vec{v}$.. which obviously matches the non relativistic expression.

Yeah and also only the perpendicular component contribute to the magnetic dipole moment which kinda surprises me since the parallel component is the bit that's moving?

And also this annoys me but what do I do with the dl that's perpendicular to the direction of v? it seems like that's being ignored..

 

[JimWhoKnew]

So then $dV'=dA dl' = \frac{dA dl}{\gamma} = \frac{dV}{\gamma}$?
This is dV' when the component of dl is parallel to the direction of v. How do I deal with the component that's perpendicular? It doesn't change right? (But is the perpendicular component really needed anyways?)
So then if I plug that back into my integral I get the same answer but without the gamma factor? $\frac{1}{2} \vec{p'}_\perp \times \vec{v}$?

Note that $\frac{1}{2} \vec{p'}_\perp \times \vec{v}=\frac{1}{2} \vec{p'}\times \vec{v}$ .

Although the result seems OK (up to a possible power of $c$), I think the replacement of the time-dependent $\rho$ by the time-independent $\rho'/\gamma$ can have a better justification (hint: $\rho(t,\vec{x}$ ) should be expressed as a function of the S coordinates.

Split $\vec{x}$ into components perpendicular and parallel to $\vec{\beta}$, i.e $\vec{x}_\parallel$ and $\vec{x}_\perp$

$\vec{m} = \frac{c \gamma}{2} \int_V ( \vec{x}_\parallel + \vec{x}_\perp) \times (\rho ' \vec{\beta}) dV$

$\vec{m} = \frac{c \gamma}{2} \int_V \rho ' ( \vec{x}_\parallel + \vec{x}_\perp) \times \vec{\beta} dV$

The charge distribution is moving in S, so $\vec{x}_\parallel$ should be replaced here by $(\vec{x}_\parallel+\vec{v}t)/\gamma$ . Since both vanish after the $\times$ product, this omission doesn't affect the later results.

 

[QuarkyMeson]

Yeah and also only the perpendicular component contribute to the magnetic dipole moment which kinda surprises me since the parallel component is the bit that's moving?

And also this annoys me but what do I do with the dl that's perpendicular to the direction of v? it seems like that's being ignored..

For the first bit, It's just the nature of what a magnetic dipole moment is and comes from the cross product. For the second but, it's not being ignored, it's actually the relevant part.

I wanted to give a better answer, so I've been reading up and found this article: MDM

This is for the transformation of a magnetic dipole moment, not electrical dipole, but I think it should be analogous.

I would say what you have is correct enough now for a homework problem, but I'm not sure its complete now. If that makes sense.