2

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
  • 201,751
372191
  • 27

3 Answers3

2

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

Jerrold Franklin
  • 1,748
  • He's not saying it's zero he's saying it's divergence is zero. – jensen paull Jul 08 '22 at 16:09
  • The divergence of a delta function is singular, and not zero. – Jerrold Franklin Jul 08 '22 at 16:14
  • This is a good answer but incomplete. You could complete it noting that \begin{equation} \boldsymbol{\nabla\cdot}\mathbf E=\dfrac{\rho}{\epsilon_0}=\dfrac{q\delta\left(\mathbf r-\mathbf r_q\right)}{\epsilon_0} \tag{a01}\label{a01} \end{equation} where $\mathbf r_q$ the position vector of the charge $q$. So \begin{equation} \dfrac{\mathbf u\left(\boldsymbol{\nabla\cdot}\mathbf E\right)}{c^2}=\dfrac{q,\mathbf u,\delta\left(\mathbf r-\mathbf r_q\right)}{c^2\epsilon_0}=\mu_0 \mathbf j \tag{a02}\label{a02} \end{equation} – Frobenius Jul 08 '22 at 22:30
  • ....where \begin{equation} \mathbf j=q,\mathbf u,\delta\left(\mathbf r-\mathbf r_q\right) \tag{a03}\label{a03} \end{equation} – Frobenius Jul 08 '22 at 22:30
  • It remains to prove that \begin{equation} -\left(\mathbf u\boldsymbol{\cdot\nabla}\right)\mathbf E=\dfrac{\partial \mathbf E}{\partial t} \tag{a04}\label{a04} \end{equation} – Frobenius Jul 08 '22 at 22:37
  • 1
    @Frobenius From: $$ \frac {dE}{dt}= \frac {\partial E}{\partial t}+\frac {\partial E}{\partial x}\frac {dx}{dt}+\frac {\partial E}{\partial y}\frac {dy}{dt}+\frac {\partial E}{\partial z}\frac {dz}{dt} $$ if taken in direction of $(x+u_x dt,y+u_y dt,z+u_z dt,t+dt)$ for r wont change since the source charge will also move the same amount and neither will v of source at new location (assumption of constant speed for formula). Hence equating the LHS to 0 as E is also only a function of those two, we get the above. – 372191 Jul 10 '22 at 20:43
  • @N1N74 : Well done !!! Very nice proof. – Frobenius Jul 10 '22 at 21:16
1

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
  • 6,624
  • Hey, thanks for the answer, which is a good guess but in the derivation of those equations the particle was merely at origin when its own clock was 0 and r corresponds to that instant. I didn't have to worry about it since time is kept constant by whatever we do with nabla. Turns out that additional term you suggested in the other comment gave the rest of the ampere's law and I had used a wrong formula. Thanks! – 372191 Jul 08 '22 at 17:34
  • The electric field at a point $\mathrm A\left(\mathbf x,t\right)$ is \begin{equation} \mathbf E\left(\mathbf x,t\right) =\dfrac{q}{4\pi \epsilon_{0}}\dfrac{(1-\beta^2)}{\left[1!\boldsymbol-\beta^2\sin^2\phi\left(t\right)\right]^{\frac32}}\dfrac{\mathbf r\left(t\right)}{::\Vert\mathbf r\left(t\right)\Vert^{3}} \tag{b01}\label{b01}
    \end{equation} where $\mathbf r\left(t\right)$ the position vector of point $\mathrm A$ with respect to the moving charge $q$, so a function of time $t$.     – Frobenius Jul 08 '22 at 23:23
  • Similarly, a function of time $t$ is the angle $\phi\left(t\right)$ between $\mathbf r\left(t\right)$ and the constant velocity $\mathbf u$ of the charge $q$. See Figure-01 in my answer here Electric field associated with moving charge. – Frobenius Jul 08 '22 at 23:23
  • So in this modified formula without the retarded time, am I correct in saying that $$\vec{r}(t) = \vec{x} - \vec{R}_{s}(t)$$ Where the last expression is the position vector of the point charge? I was very confused by the OP's non use of defining that it is E(x,t) and not E(r,t) as otherwise if your evaluating it with respect to distance from the charge as your origin, de/dt must always be zero – jensen paull Jul 09 '22 at 09:49
0

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
  • 19,536
  • I don't think this is correct. This formula, is somewhat broken, in the sense that it follows the charge as its center of coordinate system. As such, the curl of B is amperes law, as de/dt = 0 because E doesn't change with respect time with this formula. OP is correct with his analysis, im pretty sure. $$\nabla\times{\bf B=\nabla\times(u\times E)=u(\nabla\cdot E)-(u\cdot\nabla)E}$$ I suspect that the computation of $$-(u\cdot\nabla)E$$ is zero for this result. – jensen paull Jul 08 '22 at 17:29
  • OH WAIT, I got the Ampere's equation. Turns out, I recalled the wrong equation for curl of cross product. I knew Maxwell Eq, I was expecting to work them out from relativity and it works. Thanks! – 372191 Jul 08 '22 at 17:29
  • @N1N74 what did you compute -$$(u\cdot\nabla)E$$ as? Where theta is the angle between the position vector and the velocity vector remember. – jensen paull Jul 08 '22 at 17:30
  • Divergence of E is a relativistic invariant (Purcell). So divergence of E will always be charge density in whatever reference frame divided by epsilon. That term is multivariable expansion of dE/dt. – 372191 Jul 08 '22 at 17:37
  • @jensenpaull You wrote "I don't think this is correct." What don't you think is correct? Maxwell's equations? – hft Jul 08 '22 at 17:37
  • Ofcourse maxwells equations are correct. However the issue is that the formula doesn't really follow maxwells equations, since the charge density is always at the origin of coordinates , as such, the field SHOULD be the coulomb field. But it isn't, it is the field of a moving charge. This is an artificial changing of coordinates, without changing the field. Clearly de/dt in this formula is zero, nothing depends upon time. Hence it is amperes law. – jensen paull Jul 08 '22 at 17:52
  • @jensenpaull Maybe OP has some typos or something in his other equations. Or hiding some time dependence... But the question states he is looking at a moving charge. Otherwise J is zero as well. – hft Jul 08 '22 at 18:07
  • @jensenpaull His $r$ is probably supposed to be $\sqrt{x^2 + y^2 + \gamma^2(z-ut)^2}$ or something... – hft Jul 08 '22 at 18:08
  • I think it's supposed to be time invariant, I remember seeing it somewhere else before. but I guess that depends upon the evaluation of that term stated above! [In what the curl evaluates to] – jensen paull Jul 08 '22 at 18:10
  • I think that you must give a detailed proof. – Frobenius Jul 08 '22 at 19:18
  • @Frobenius Who must? Me? A proof of what? Maxwell's equations? – hft Jul 08 '22 at 20:24
  • This proof : \begin{equation} \texttt{why } \boldsymbol{\nabla\times}\left(\dfrac{\mathbf u\boldsymbol\times\mathbf E}{c^2}\right)=\mu_0 \mathbf j+\epsilon_0\mu_0\dfrac{\partial \mathbf E}{\partial t} ::\texttt{???} \tag{01}\label{01} \end{equation} and what is $\mathbf j$ in this case. – Frobenius Jul 08 '22 at 21:59
  • I have not provided any such proof in my answer, and I don't plan on doing so. Feel free to add your own answer. – hft Jul 08 '22 at 22:06
  • @jensenpaull Copy pasting here since I took the time to type it out elsewhere: From: $$ \frac {dE}{dt}= \frac {\partial E}{\partial t}+\frac {\partial E}{\partial x}\frac {dx}{dt}+\frac {\partial E}{\partial y}\frac {dy}{dt}+\frac {\partial E}{\partial z}\frac {dz}{dt} $$ if taken in direction of $(x+u_x dt,y+u_y dt,z+u_z dt,t+dt)$ for r wont change since the source charge will also move the same amount and neither will v of source at new location (assumption of constant speed for formula). Hence equating the LHS to 0 as E is also only a function of those two, we get the required formula. – 372191 Jul 10 '22 at 20:46