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From electric field of a point charge: $$ \vec{E} = \frac{k Q \vec{r}}{\gamma^{2}r^3(1-\beta^2sin^2\theta)^{\frac{3}{2}}}, \vec{B} = \frac{\vec{u} \times \vec{E}}{c^2} $$ taking curl of B gives $$ \nabla \times B = {\mu_0J}$$ Taking divergence of above gives $$ \nabla . J = \frac{\nabla .(\nabla \times B)}{\mu_0} =0 = -\frac{\partial q}{\partial t} $$ which implies stationary charge distribution despite the point charge being in motion. Expressions for E and B derived using special relativity where the charge Q moves across space with constant velocity to inertial observer. So why exactly does the expressions for E and B for moving field fail?

I have found answers on this post claim it is due to not considering speed of light limit but assuming we let enough time for the charge in its rest frame to produce electric field in all of its space, it doesn't seem to be a factor since we only need to lorentz transform four force experienced in charge's rest frame to get above same equation for E and B, where the charge will be in motion. Other answer on the same post attribute the error to lack of consideration of charge conservation which if we let q in its rest frame be constant (avoids need to re-consider speed of light limit) would also be constant in inertial frame where the charge is moving, by invariance of charge (already assumed for the derivation of above equations).

Qmechanic
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3 Answers3

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$\nabla\times{\bf B=\nabla\times(u\times E)=u(\nabla\cdot E)-(u\cdot\nabla)E}$. $\quad{\bf\nabla\cdot E}$ leads to a delta function, not zero.

Jerrold Franklin
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Taking the curl of the B field for the field of a moving charge yields ampere maxwell law.

Your curl equation apparently yields amperes law without the addition of maxwell hence the error. [Haven't tried this can you show me how you got there?]

I believe that this may be due to the fact that your formula isn't complete, I think.

Off the top of my head. The correct formula is slightly different, in that the coordinate system is stationary, and the charge density physically moves with respect to that coordinate system.

I think the formula you have, is defining a new set of coordinates Where r is radial distance from the charge, that moves WITH the charge, such that the charge density is always at the center, hence there is no change of charge density.

I could be completely wrong, however nothing about your formula seems to have any time variable present which the formula from the lienard wichert potentials does.

jensen paull
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taking curl of B gives $$ \nabla \times B = {\mu_0J}$$

No, it gives: $$ \nabla \times \vec B = {\mu_0\vec J} +\epsilon_0\mu_0\frac{\partial \vec E}{\partial t} $$

This is literally one of Maxwell's equations.

hft
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