Magnetic field of a point charge moving uniformly

  • Thread starter Hiero
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  • #1
Hiero
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Everywhere I look online I see the formula for the magnetic field of a uniformly moving charge is,
$$\frac{\mu_0 q \vec v \times \vec r}{4\pi r^3}$$
but when I calculate it by transforming the electrostatic field (taking the motion along x) I get,
$$\frac{\gamma \mu_0 q \vec v \times \vec r}{4\pi ((\gamma x)^2+y^2+z^2)^{3/2}}$$

I can't find anything like this version anywhere. I just want to confirm that the version I keep seeing online is just the low speed limit where the gamma factor is approximately one? Or am I wrong?
 

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  • #3
Hiero
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Thank you @Dale, though I’m still not sure if I’m correct.

I can manipulate mine to look exactly like their version (which is just the first term) except where they have ##(1-\vec \beta \cdot \hat n)^3## in the denominator, I instead have ##(1-\beta ^2 - (\vec \beta \cdot \hat n)^2)^{3/2}##

I thought this meant I was wrong until I noticed the little sub t_r on their parenthesis indicating that they’re evaluating it at the retarded time whereas I’m using the instantaneous displacement, so my {n, |r-rs|} is not the same as theirs.

I suspect mine is still right because they show the same retarded denominator for the electric field, and my instantaneous denominator is the same as I saw for the electric field of a uniformly moving charge in Purcell’s book, but I’ll have to wait until tomorrow to think about it and be sure.
 
  • #4
vanhees71
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See Sect. 3.6

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

There the derivation is given in terms of the manifestly covariant formulation of the Lienard-Wiechert retarded potentials.

You can of course save this complication by simply first looking at the problem in the rest frame of the charge. There a solution for the potential is of course the Coulomb potential,
$$A^0=\frac{q}{4 \pi |\vec{x}|}, \vec{A}=0.$$
Now simply write this in terms of covariant quantities. For that purpose you only need to introduce the four-velocity of the point charge. In the here used rest frame it's ##(u^{\mu})=(1,0,0,0)##. It's clear that
$$A^{\mu}=\frac{q}{4 \pi |\vec{x}|} u^{\mu}.$$
Now you only need to express the scalar (sic!) factor in a covariant way:
$$|\vec{x}|=\sqrt{(u_{\mu} x^{\mu})-x_{\mu} x^{\mu}}=\sqrt{(u \cdot x)^2-x^2},$$
which gives immediately the same result as in my manuscript, Eq. (3.6.8).

This is a manifestly covariant expression and thus valid in any frame. For a particle moving with three-velocity ##\vec{v}## you simply have
$$u^{\mu}=\gamma \begin{pmatrix}1\\ \vec{v}/c \end{pmatrix}, \quad \gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}.$$
 

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