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In every electrodynamics book there is one chapter on special relativity which includes one section about" covariant formulation of electrodynamics" which uses tensor to describe the two fields and the four potential and lorentz force etc. and unifies maxwell's equation in a simple way. The problem encountered is the relation between E field and A field:some book gives $\vec{E}=\frac{\partial \vec A}{∂t} - \nabla{V}$, while some $\vec{E}=-\frac{∂ \vec A}{∂t} - \nabla{V}$. (c.f. Jackson or Landau for the first choice, Goldstein's CM or Dubrovin, Fomenko, Novikov's modern geometry for the second choice) I do understand the second one, for Faraday's law(here particularly Lenz's law) needs to be satisfied, and I do understand the first one as well, for the four-d Lorentz force becomes clear in such way. E field, however, is not like A field which can be given a gauge; E field can be measured directly (in an inertial reference frame), so the choice mustn't be arbitrary, and one of them must be wrong. This is my question.

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  • $\begingroup$ Precisely which equations in Jackson and Landau are you referring to? $\endgroup$ – Qmechanic Oct 6 '13 at 11:45
  • $\begingroup$ L&Lvol2 eq17.3 Jackson eq11.134 $\endgroup$ – Wilson of Gordon Oct 7 '13 at 3:58
  • $\begingroup$ (i) Landau & Lifshitz vol. 2 eq. (17.3), (ii) Jackson eq. (11.134), and (iii) Goldstein ed. 3 eq. (1.61a) all have the second choice $\vec{E}=-\frac{\partial\vec{A}}{\partial t}-\vec{\nabla}\phi$. All the books that I've checked have the second choice. Where do you see the first choice? $\endgroup$ – Qmechanic Oct 7 '13 at 13:06
  • $\begingroup$ o please i said c.f. goldstein and gtm93 for the first choice.. $\endgroup$ – Wilson of Gordon Oct 7 '13 at 22:13
  • $\begingroup$ What you are saying now seems to be the opposite of what is written in your post (v4). What is gtm93? Please provide equation numbers for all books to facilitate comparison. $\endgroup$ – Qmechanic Oct 7 '13 at 22:21
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Let us work in units where speed of light in vacuum is equal to one $c=1$. In all the Refs 1-3 that OP mentions, they have

$$ \vec{E}~=~-\vec{\nabla}\phi-\frac{\partial \vec{A}}{\partial t}, \tag{A} $$

except for Ref. 4, which has the wrong equation

$$ \vec{E}~=~-\vec{\nabla}\phi+\frac{\partial \vec{A}}{\partial t} \qquad(\leftarrow \text{Wrong!}).\tag{B} $$

So let us investigate the conventions of Ref. 4 to check if it is a conventional issue or a typo. From e.g. Section 6.1 one deduces that Ref. 4 uses signature $(+,-,-,-)$ for the Minkowski metric $\eta_{\mu\nu}$. Thus if one uses the Minkowski metric $\eta_{\mu\nu}$ to lowers the magnetic vector potential $\vec{A}=(A^1, A^2,A^3)$ to a magnetic co-vector/one-form potential $(A_1, A_2,A_3)=-(A^1, A^2,A^3)$, that would cost a minus! Perhaps that could explain the wrong sign in eq. (B)? Not really. It is a typo, because Ref. 4 writes just below on p. 389 that

$$\vec{B}~=~\vec{\nabla}\times\vec{A},$$

and if one e.g. compares with their Faraday induction law

$$ \vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} ~=~ \vec{0}\tag{20}$$

on p. 392, it becomes clear that eq. (B) is wrong.

Eq. (B) is not the only sign mistake of Ref. 4. In the very same section 37.3, there is a wrong sign in front of the Maxwell's correction term in their Ampere's law (18) on p. 392.

References:

  1. J.D. Jackson, Classical Electrodynamics, eq. (11.134).

  2. L.D. Landau and E.M. Lifshitz, The Classical Theory of Fields, Vol. 2., eq. (17.3).

  3. H. Goldstein, Classical Mechanics, 3rd edition, eq. (1.61a).

  4. B.A. Dubrovin, A.T. Fomenko and S.P. Novikov, Modern Geometry: Methods and Applications, Vol. I, section 37.3, second-last eq. on p. 389.

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  • $\begingroup$ good answer thank you and btw the 3 russian book,i think, may be differentiating w.r.t. co-variant coordinates. $\endgroup$ – Wilson of Gordon Oct 15 '13 at 1:31
  • $\begingroup$ The space time coordinates $(x^0,x^1,x^2,x^3)$ in Ref. 4 carry upper (as opposed to lower) indices, see e.g. the bottom half of p. 391. $\endgroup$ – Qmechanic Oct 15 '13 at 16:38
  • $\begingroup$ so that must have been a typo? or the may be using co-variant 4-potential?"The three spatial components A(upper)1, A(upper)2 , A(upper)3 of the 4-vector (A(upper)i) obtained by raising the index of the tensor (A(lower)i) (for this purpose resorting, of course, to the Minkowski metric), define a 3-vector A called the vector-potential of the field. " this is the original sentence, so i think the author may be differentiating co-variant 4-potential w.r.t. contra-variant coordinates so that we have anti-symmetry. $\endgroup$ – Wilson of Gordon Oct 15 '13 at 22:30

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