Suppose I have a charge situated at rest in one inertial coordinate system. The field of that drops inversely with the square of the distance. If I perform a Lorentz boost where $$v^2« c^2$$ (not necessarily $v«c$), should I expect the magnetic field to vary with the inverse square of the distance calculated in the first frame, or should I expect it to vary inversely with the square of the "retarded distance" in the second frame? The "retarded" distance depends on the velocity of the charge and the radial vector to the charge and the dependency is of order $\frac vc$. So at some points, the charge should appear closer due to retardation of the field, while at other points towards the "antipode", the charge should appear farther due to retardation of the field. This would be in addition to the length contraction which makes the charge appear closer at both ends, but that effect is of order $$\frac{v^2}{ c^2}$$ which we could neglect here.

  • $\begingroup$ This is a very standard calculation done in just about every undergrad E&M text... have you seen it and do you have a more specific question about it? Because yes at one level a distance is calculated in the stationary frame to find an expression for the scalar and vector potentials; but no at another level we are evaluating position and time derivatives of those potentials along the time and space axes in the moving frame to find the fields. $\endgroup$
    – CR Drost
    Apr 22, 2019 at 5:07
  • $\begingroup$ I'm not sure how your "yes" and "no" map to my question. So yes to, "If I perform a Lorentz boost where v^2«c^2 (not necessarily v«c), should I expect the magnetic field to vary with the inverse square of the distance caculated in the first frame" and no to, "should I expect it to vary inversely with the square of the 'retarded distance' in the second frame" ? Whether fields or potentials, they will be retarded. So my thought was "no" to the first one and "yes" to the second one. I've seen low velocity approximations where v«c such that the first one is "yes". But here my focus is on v^2«c^2. $\endgroup$ Apr 23, 2019 at 15:43
  • $\begingroup$ Let's suppose for the first n observers in the first frame where the charge is at rest that they are positioned equidistant from the charge at rest. They have synchronized clocks. Now consider a "flock" of n observers whose trajectory through space-time intersects these first observers' trajectories, where the first observers each report encountering a member of the "flock" at time t moving at velocity v, with the n(n-1) messages of these n events being recieved at different times between the first observers. The messages include info by the "flock" of what they saw at each point of encounter. $\endgroup$ Apr 23, 2019 at 16:04

1 Answer 1


A moving point charge $q$ with velocity $c~\vec \beta$ at position $\vec r = r~\hat r,$ $\|\hat r\| = 1,$ a time $t=0$ is known to have relativistic fields at the spacetime origin of (SI units) $$ \vec E = -\frac{q~\vec r}{4\pi \epsilon_0 r^3}~\frac{\gamma}{\big(1 + (\gamma~\vec \beta \cdot \hat r)^2\big)^{3/2}},\\ \vec B = \frac{1}{c^2} \vec u \times \vec E, $$where $\gamma=1/\sqrt{1 - \beta^2}$. See for example Richard Fitzpatrick's derivation although this is a staple of most undergrad electromagnetism textbook, as they all want to eventually excite their students with relativistic calculations.

The field points directly at the charge but is attenuated by this strange $\alpha = \gamma ~\vec \beta \cdot \hat r$ term which can certainly be merged into the $r^3$ term to be a field strength going like $\gamma r/(r')^3$ where $r'$ is the proper distance between the charged particle's world line and the field point at the spacetime origin.

It is not immediately clear to me what connection this has to the delayed distance $\ell_-.$ In cylindrical coordinates the position $\vec r$ can be treated as being at some $(\rho\cos\theta,\rho\sin\theta,z)$ position and choosing the $z$-direction parallel to $\vec \beta$ the delayed distance should be the positive solution of $$\ell_-^2 = \rho^2 +(z + \beta ~\ell_-)^2,$$or,$$(1-\beta^2)\ell_-^2 - 2\beta z\ell_- - \rho^2 - z^2 = 0,$$ so $$\ell_- = \gamma^2\beta z + \gamma^2\sqrt{\beta^2z^2 + (\rho^2 + z^2)(1-\beta^2)}.$$ With a bit more manipulation then, we have $\rho^2 + z^2 = r^2$ while $\beta z=\vec \beta \cdot \vec r$ and this looks like $$\ell_- = \gamma~r~\left(\gamma \vec \beta \cdot \hat r + \sqrt{1 + (\gamma ~\vec \beta\cdot \hat r)^2}\right). $$ This connection between the term I above called $\alpha$ and $\ell_-$ can be written, in other words, as $$\ell_-/(\gamma~r) = \alpha + \sqrt{1 + \alpha^2}$$ which we could even solve for $ \alpha$. Haphazardly working through it, I think I get that $$\vec E = -\frac{q~\vec r}{4\pi\epsilon_0 r^3}~\frac{\gamma}{\left(\frac{\ell_-}{2\gamma r} + \frac{\gamma r}{2 \ell_-}\right)^3},$$ which seems simple enough but it is not really “obvious” in any way that I think you are hoping it would be obvious.


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