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In SR in any inertial frame electric charge is invariant, i.e. is independent of the frame. I have seen a claim that it was confirmed experimentally with some accuracy.

Is this law true in GR, i.e. in an arbitrary non-inertial frame?

Moreover the proof in SR of the fact that $(c\rho,\vec j)$ is a Lorentz vector is based on the charge invariance law. Is there analogous property in GR?

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  • $\begingroup$ "In SR in any inertial frame electric charge is invariant, i.e. is independent of the frame." I don't understand this: e.g., take a body with total charge $Q$.. its charge is $Q$ whether we re orbiting around it or it's spinning on an axis or any other strange motion (in both SR as well as GR). $\endgroup$
    – Quillo
    Commented Feb 21, 2022 at 14:46
  • $\begingroup$ @Quillo: (1) Do you want to say that the answer to my first question is positive, in your opinion? (2) The charge invariance law is not self evident: it requires experimental confirmation and, as far as I heard, was confirmed in SR relativity to certain degree (this is mentioned in Matveev's course on general physics). $\endgroup$
    – MKO
    Commented Feb 21, 2022 at 14:55
  • $\begingroup$ 1 - I would say yes, but I suggest you to wait for more detailed answers from someone that is "GR expert". 2 - you' re right, physics is experimental, but "net" charge conservation is a super well established fact and a building block of every theory (valid in QFT as well as in GR+Maxwell). In SR we test net charge conservation every day at particle colliders. $\endgroup$
    – Quillo
    Commented Feb 21, 2022 at 15:16
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    $\begingroup$ You can take a look here: en.wikipedia.org/wiki/Maxwell%27s_equations_in_curved_spacetime , see the section "Electric current". Moreover, in both SR and GR, there is a geometrical interpretation of Maxwell equations in terms of exterior calculus, which makes it evident that charge conservation is the same in both GR and SR: $d*F=J$, implying $dJ=0$. See "Differential forms approach" here en.wikipedia.org/wiki/… $\endgroup$
    – Quillo
    Commented Feb 21, 2022 at 15:32
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    $\begingroup$ In SR, the charge is the invariant $\int j^0d^3x$; in GR, the charge is the invariant $\int j^0\sqrt{|g|}d^3x$. $\endgroup$
    – J.G.
    Commented Feb 21, 2022 at 15:35

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Given a stress energy tensor $T_{ab}$, one can define a conserved 4-current $J^a$ associated to a timelike Killing-Vector field $k$ as: $$J^a=T^{ab}k_b$$which is conserved , i.e. $\nabla^aJ_a=0$. This conservation law follows from (1) the continuity equation $\nabla^aT_{ab}=0$ and (2) the Killing equation $\nabla^{(a}k^{b)}=0$. The conserved charge associated to this current density is simply the integral:$$Q[k]=\int_{\Sigma}J_ad\sigma^a=\int_{\Sigma}T_{ab}k^bd\sigma^a$$In Minkowski space, we can choose the timelike Killing vector field $k=\partial_t$ as the observer's velocity field, then the conserved charge is simply three dimensional volume integral $$Q=\int J^0d^3x$$ Let's say there is another observer with a time like velocity field $u$. Then there exists a three dimensional hypersurface $\Sigma '$ which is orthogonal to this velocity field $u$. The above definition of charge only depends on topology of hypersurface $\Sigma$ , so as long as $\Sigma'\cong_{top} \Sigma $, the charge corresponding to a particular $k$ will be invariant.

This definition is not the most general definition for conserved charge, since it relies on existence of Killing vector field. A general curved space-time may not admit any Killing vector field. The above definition agrees with the "charge" we see in Maxwell's theory. However, we can re-express the above definition for $Q$ in terms of the intrinsic curvature ($R_{abcd}$) content, called "Penrose's Quasi-Local Momentum and Angular Momentum" and it is independent of existence of Killing vector field. Such a definition encodes information about the 4-momentum and 6-angular momentum of the source. The brief idea of Quasi-Local momentum and angular-momentum is as follows:

Consider a linearized metric $g_{ab}=\eta_{ab}+uh_{ab}+\mathcal{O}(u^2)$, where $|u|<<1$, the linearized Riemann tensor $K_{abcd}=\lim_{u\to 0}(u^{-1}R_{abcd}(u))$ has all the symmetries of a linearized Weyl tensor $u^{-1}C_{abcd}(u)$ upto first order in $u$. Now, if $Q^{ab}$ is a bivector which satisfies $$\nabla^{(a}Q^{b)c}-\nabla^{(a}Q^{c)b}+g^{a[b}\nabla_dQ^{c]d}=0$$then the contraction $K_{abcd}Q^{cd}$ can be written as dual of a bi-vector $F_{ab}$: $$K_{abcd}Q^{cd}=-2\star F_{ab}$$. If the curvature tensor satisfies the vacuum Einstein equation $\nabla^aK_{abcd}=0\leftrightarrow \nabla^aC_{abcd}=0$ , then $F_{ab}$ satisfies the free Maxwell's equation $\nabla^aF_{ab}=0$. We can write the conserved charge $$Q=\int_{\Sigma}J_ad\sigma^a=-\frac{1}{8\pi}\int_{\partial\Sigma}\star F=\frac{1}{16\pi}\int_{\partial\Sigma}K_{abcd}Q^{cd}d\sigma^{ab}$$This definition is then extended for finite distance as $$Q=\frac{i}{16\pi G}\int_{S}R_{abcd}Q^{cd}d\sigma^{ab}$$In this definition, the role of Killing vector $k$ is replaced by "2-surface twistors" defined on the compact and finite space-like 2-surface $S$ enclosing the source. Actually, the vector $\zeta^a=\frac{1}{3}\nabla_bQ^{ab}$ is a Killing vector, but a solution for $Q^{ab}$ for general space-time may not exist. So, instead one can project the differential equation for $Q^{ab}$ on 2-surface $S$, and if $S$ is topologically $S^2$ we can represent the projected differential equation as four linearly independent spinor equations called "2-surface twistor equations". As stated earlier, this definition carries information about the momentum and angular momentum of the source. For detailed calculations and mathematical arguments, refer to chapter 13 of "Introduction to Twistor theory" by S.A.Huggett and K.P.Tod and Section 9.9 of "Spinors and Space-time volume II" by Penrose and Rindler.

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