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The non-relativistic version of Lorentz equation has the form $$m\frac{d\vec{v}}{dt}=q(\vec{E}+\vec{v}\times\vec{B}) $$ Where $\vec{v}, \vec{E}, \vec{B}$ refers to the velocity of charged particle, electric field and magnetic field measured with respect to a given observer respectively. The usual extension of this equation in flat spacetime has the form $$\frac{dp^a}{d\tau}=q{F^a}_bv^b$$ where $p^a$, $v^a$ has the interpretation of 4-momentum, 4-velocity of the charged particle. What I'm unsure of is whether these two expressions of the Lorentz equation are really equivalent. There are two issues with the relativistic form:

  1. $F_{ab}$ can be written as $2E_{[a}u_{b]}+e_{abc}B^c$ where E, B are electric and magnetic fields measured wrt a time like observer travelling with velocity $u$ (unit time-like). On plugging this expression in the relativistic Lorentz equation we see that it has the form of ($E+v\times B$) only for the special case where $u=(1, 0,0,0) $

  2. Unlike in non-relativistic case, here we interpret $v^a$ as the 4-velocity of the charged particle and not the relative velocity measured wrt the observer $u$

It seems to me that if $\beta^a$ refers to relative velocity between $v^a$ and $u^a$ then the following expression can also be the candidate for relativistic Lorentz equation: $$m\frac{d\beta^a}{d\tau}=q(E^a+{e^a}_{bc}\beta^bB^c) $$ I don't know if there is any flaw in my interpretation of standard Lorentz equation

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  • $\begingroup$ "$F_{ab}$ can be written as $2E_{[a}u_{b]}+e_{abc}B^c$" I do not see how this can work. Do you have a reference, preferably a good textbook? Your notation $v$ is ambiguous. In your 2nd and 3d equation $a,b,c$ mean different things. $\endgroup$
    – my2cts
    Commented Apr 4 at 10:46
  • $\begingroup$ What is the distinction between "the four-velocity of the charged particle" and "the relative velocity measured w.r.t. the observer"? An object's four-velocity in a given reference frame is its (four-)velocity relative to that frame. $\endgroup$
    – Michael Seifert
    Commented Apr 4 at 11:35
  • $\begingroup$ @my2cts The decomposition of F_ab into its electric and magnetic parts are given, for eg, in "Relativistic Cosmology" by G. F. R. Ellis, Roy Maartens and A. H. McCallum. The indices a, b, c are spacetime indices everywhere. The $E^a$ field for eg has zero time component. Same for the magnetic field. This essentially follows from projection of F_ab parallel and orthogonal to $u^a$ . The spatial components are orthogonal to $u^a$.The same applies for 3d levi-civita which is $e_{abc}=e_{abcd}u^d$ (the time component is naturally zero due to antisymmetry of levi-civita) $\endgroup$
    – paul230_x
    Commented Apr 4 at 16:52
  • $\begingroup$ @MichaelSeifert I think this is more of a semantics thing : when it comes to "relative velocity" in the context of SR or GR, I have seen the following definition : $v^a=\Gamma (u^a+\beta^a) $ such that $u^a\beta_a=0$ , which gives $\Gamma = 1/\sqrt{1-\beta^2}$ . The $\beta^a$ is the relative velocity (see arxiv:gr-qc/0506032v5). If I take your definition and the decomposition of F_ab which I mentioned above, it would seem that the standard Lorentz force is defined wrt some co-moving observer $u^a$. But is this really the case? $\endgroup$
    – paul230_x
    Commented Apr 4 at 17:19
  • $\begingroup$ If this is true then I think the final expression expression I wrote at the bottom is for arbitrary reference frame where the observer is no longer static anymore $\endgroup$
    – paul230_x
    Commented Apr 4 at 17:24

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The first two expression are clearly not equivalent. The first is the low velocity limit of the second.

The second equation is correct if $p^a$ denotes $mu^a = m \gamma dx^a/dt$ and $u^a = dx^a/dt$, setting $c=1$. $a,b,c$ should run over $t,x,y,z$.

The third equation is incorrect. Also it is not Lorentz covariant as $a,b,c$ run over $x,y,z$, since $\beta^t$, $E^t$ and $B^t$ have no meaning.

Not the answer you wanted to hear perhaps :-).

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