# Factor of 4 (or 2) in the gravitoelectromagnetic (GEM) Lorentz-force law. Which is correct? Why is it there?

I realize that the Gravitoelectromagnetic equations (GEM) are derived from the Einstein field equation (EFE) in the degenerate case of reasonably flat spacetime, which is the case for the propagation of gravitational waves in free space reasonably far away from nasties such as black holes or neutron stars or the like.

Now, I understand Maxwell's equations pretty well, and how to derive EM radiation (at the wave speed of $c$) from them. I also understand the electromagnetic force as a manifestation of the sole electrostatic force, but with the consequences of special relativity taken into consideration.

So, in both cases, EM and GEM, we have in the limiting case, an inverse-square static interaction and a dynamic interaction that propagates at the same speed of $c$. The inverse-square interaction leads to Gauss's law and the differential counterpart (divergence) in Maxwell's equations (or in the GEM equations).

Okay, just for comparison, the static EM and graviation inverse-square laws are:

$$F_\mathrm{e} = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$$ and $$F_\mathrm{g} = -G \frac{m_1 m_2}{r^2}$$

In both cases, a positive force $F_\mathrm{e}$ or $F_\mathrm{g}$ is repulsive. This is why the minus sign needs to be attached to $G$ in the static gravitation law.

Then Maxwell's equations for EM are:

\begin{align} \nabla \cdot \mathbf{E} &= \frac {1}{\epsilon_0} \rho \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{E} &= -\frac{1}{c}\frac{\partial \mathbf{B}} {\partial t} \\ \nabla \times \mathbf{B} &= \frac{1}{c} \left( \frac{1}{\epsilon_0}\rho\mathbf{v}_\rho + \frac{\partial \mathbf{E}} {\partial t} \right) \end{align}

and the GEM counterparts:

\begin{align} \nabla \cdot \mathbf{E}_\mathrm{g} &= -4\pi G \ \rho \\ \nabla \cdot \mathbf{B}_\mathrm{g} &= 0 \\ \nabla \times \mathbf{E}_\mathrm{g} &= -\frac{1}{c}\frac{\partial \mathbf{B}_\mathrm{g}} {\partial t} \\ \nabla \times \mathbf{B}_\mathrm{g} &= \frac{1}{c} \left( -4\pi G \ \rho\mathbf{v}_\rho + \frac{\partial \mathbf{E}_\mathrm{g}} {\partial t} \right) \end{align}

In the EM case, I have eliminated $\mu_0$ and expressed it in terms of $\epsilon_0$ and $c$. In both cases, I have expressed current density $\mathbf{J}$ as charge density (or mass density) times the velocity of the differential charge or mass. And, in both cases, I expressed in terms of Lorentz-Heaviside units which makes the $\mathbf{B}$ field have the same dimensions (and units) as the $\mathbf{E}$ field. This is consistent with most of the papers dealing with GEM.

Both the inverse-square and EM/GEM expressions are totally consistent with replacing charge with mass, charge density with mass density, and $\frac{1}{4 \pi \epsilon_0}$ with $-G$. Both EM/GEM sets of equations degenerate to the inverse-square laws and to an interaction that propagates at a speed of $c$.

So far, this agrees with the expressions for EM or GEM in the Wikipedia articles on either. The difference comes with the Lorentz force equations acting on a small test charge $q$ or small test mass $m$ (moving at a velocity independent of the charge current density or mass current density above):

For EM it's: $$\mathbf{F}_\mathrm{e} = q\mathbf{E} + \frac{q}{c}\mathbf{v}_q \times \mathbf{B}$$

For GEM, in the Wikipedia article it's: $$\mathbf{F}_\mathrm{g} = m\mathbf{E}_\mathrm{g} + \frac{m}{c}\mathbf{v}_m \times (\color{red}{4}\mathbf{B}_\mathrm{g})$$

The first term (right of the "$=$" sign) is the electrostatic or static gravitational force and the the latter term is the electromagnetic or gravitomagnetic force.

Now, for the GEM Lorentz force, where does that factor of $\color{red}{4}$ come from? And there are other papers that show the GEM equations as above, but have a factor of $2$ instead:

$$\mathbf{F}_\mathrm{g} = m\mathbf{E}_\mathrm{g} + \frac{m}{c}\mathbf{v}_m \times (2\mathbf{B}_\mathrm{g})$$

or no fudge factor in the Lorentz force, but a $\tfrac12\mathbf{B}_\mathrm{g}$ in the GEM, which is equivalent.

I don't know why either the $4$ or the $2$ would come into this, but I would like to know who is correct; the $4\mathbf{B}_\mathrm{g}$ advocates or the $2\mathbf{B}_\mathrm{g}$ advocates?

TL;DR: The factor of $$\color{red}{4}$$ in the Lorentz force comes morally speaking from trying to mimic a spin-2 field as a spin-1 field. There is no unique/canonical/"correct" normalization convention: It is still possible to normalize/scale the fields $$\phi$$, $${\bf A}$$, $${\bf E}$$ & $${\bf B}$$, as we please, but that only moves the factor of $$\color{red}{4}$$ around: It doesn't disappear everywhere!

In detail:

1. Consider the linearized EFE$$^1$$ $$G^{\mu\nu}~=~-\frac{1}{2}\Box \bar{h}^{\mu\nu}~=~\kappa T^{\mu\nu},\qquad \kappa~\equiv~\frac{8\pi G}{c^4}, \tag{1}$$ in the Lorenz gauge $$\partial_{\mu} \bar{h}^{\mu\nu} ~=~0. \tag{2}$$ Here \begin{align} g_{\mu\nu}~=~&\eta_{\mu\nu}+h_{\mu\nu}, \cr \bar{h}_{\mu\nu}~:=&~h_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}h \qquad\Leftrightarrow\qquad h_{\mu\nu}~:=~\bar{h}_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}\bar{h}. \end{align}\tag{3} The matter is assumed to be dust: $$T^{\mu 0}~=~cj^{\mu}, \qquad j^{\mu}~=~\begin{bmatrix} c\rho \cr {\bf J} \end{bmatrix}, \qquad T^{ij}~=~{\cal O}(c^0). \tag{4}$$

2. In our convention, the GEM ansatz reads \begin{align} A^{\mu}~=~&\begin{bmatrix} \phi/c \cr {\bf A} \end{bmatrix}, \qquad\bar{h}^{ij}~=~{\cal O}(c^{-4}),\cr -\frac{1}{4}\bar{h}^{\mu\nu} ~=~&\begin{bmatrix} \phi/c^2 & {\bf A}^T /c\cr {\bf A}/c & {\cal O}(c^{-4})\end{bmatrix}_{4\times 4}\cr ~\Updownarrow~& \cr -h^{\mu\nu} ~=~&\begin{bmatrix} 2\phi/c^2 & 4{\bf A}^T/c \cr 4{\bf A}/c & (2\phi/c^2){\bf 1}_{3\times 3}\end{bmatrix}_{4\times 4} \cr ~\Updownarrow~& \cr g_{\mu\nu} ~=~&\begin{bmatrix} -1-2\phi/c^2 & 4{\bf A}^T/c \cr 4{\bf A}/c & (1-2\phi/c^2){\bf 1}_{3\times 3}\end{bmatrix}_{4\times 4}. \end{align}\tag{5} The gravitational Lorenz gauge (2) corresponds to the Lorenz gauge condition $$c^{-2}\partial_t\phi + \nabla\cdot {\bf A}~\equiv~ \partial_{\mu}A^{\mu}~=~0 \tag{6}$$ and the "electrostatic limit" $$\partial_t {\bf A}~=~{\cal O}(c^{-2}).\tag{7}$$

3. Next define the field strength \begin{align} F_{\mu\nu}~:=~&\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}, \cr -{\bf E}~:=~&{\bf \nabla} \phi+\partial_t{\bf A}, \cr {\bf B}~:=~&{\bf \nabla}\times {\bf A}.\end{align} \tag{8} Then the tempotemporal & the spatiotemporal sectors of the linearized EFE (1) become the gravitational Maxwell equations with sources $$\partial_{\mu} F^{\mu\nu}~=~\frac{4\pi G}{c}j^{\mu}. \tag{9}$$ Note that the gravitational (electric) field $${\bf E}$$ should be inwards (outwards) for a positive mass (charge), respectively. For this reason, in this answer/OP/Wikipedia, the GEM equations (9) and the Maxwell equations have opposite$$^2$$ signs. See also this related Phys.SE post.

4. The Lagrangian for a massive point particle in curved space in the static gauge $$x^0=ct$$ is \begin{align}L~=~&-m_0c\sqrt{ -g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\mu}}\cr ~\stackrel{(5)}{=}~&-m_0c\sqrt{ c^2+2\phi -8{\bf A}\cdot{\bf v}-(1-2\phi/c^2){\bf v}^2 }\cr ~=~&-\frac{m_0c^2}{\gamma}\sqrt{ 1+\frac{2\gamma^2}{c^2}\left( (1+{\bf v}^2/c^2)\phi -\color{red}{4} {\bf v}\cdot{\bf A}\right)}\cr ~\stackrel{(12)}{=}~& -\frac{m_0c^2}{\gamma}~-~U ~+~{\cal O}(A^2), \end{align}\tag{10} $$\gamma~:=~\frac{1}{\sqrt{ 1-{\bf v}^2/c^2}} .\tag{11}$$ Here the velocity-dependent potential for the gravitational Lorentz force is \begin{align} U~=~&m_0\gamma\left((1+{\bf v}^2/c^2)\phi -\color{red}{4} {\bf v}\cdot{\bf A}\right)\cr ~\stackrel{NR}{=}~& m_0\left(\phi -\color{red}{4} {\bf v}\cdot{\bf A}\right)~+~{\cal O}({\bf v}^2).\end{align} \tag{12} The gravitational Lorentz force becomes in the non-relativistic (NR) limit \begin{align} {\bf F}~=~&\frac{d}{dt}\frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}\cr ~\stackrel{NR}{\approx}~& m_0\left(-{\bf \nabla} \phi + \color{red}{4}({\bf v}\times {\bf B}-\partial_t{\bf A})\right) \cr ~\stackrel{(7)}{=}~& m_0({\bf E}+ \color{red}{4}{\bf v}\times {\bf B}) .\end{align}\tag{13}

References:

1. B. Mashhoon, Gravitoelectromagnetism: A Brief Review, arXiv:gr-qc/0311030.

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$$^1$$ In this answer we use Minkowski sign convention $$(-,+,+,+)$$ and work in the SI-system. Space-indices $$i,j,\ldots \in\{1,2,3\}$$ are Roman letters, while spacetime indices $$\mu,\nu,\ldots \in\{0,1,2,3\}$$ are Greek letters.

$$^2$$ Warning: In Mashhoon (Ref. 1) the GEM equations (9) and the Maxwell equations have the same sign. For comparison, in this Phys.SE answer $$\phi~=~-\phi^{\text{Mashhoon}}, \qquad {\bf E}~=~-{\bf E}^{\text{Mashhoon}},$$ $${\bf A}~=~-\frac{1}{2c}{\bf A}^{\text{Mashhoon}}, \qquad {\bf B}~=~-\frac{1}{2c}{\bf B}^{\text{Mashhoon}}.\tag{14}$$

• i added +1 and check-marked the answer, even though i dunno enough GR to decode the answer. so it appears that "$\color{red}{4}$" it is. even so, in this answer i demonstrate how little this electrical engineer knows of physics by setting up a thought experiment with two lines of charge, and the acceleration of their repulsion is reduced by either the magnetic force or by time dilation. Now if you were to.... Jun 25, 2018 at 20:45
• ... replace those two lines of charge with lines of mass and consider their attractive acceleration in the case of the lines of mass stationary to you, the observer, and then again with them moving relative to you the observer, from a POV of special relativity, the acceleration would be reduced the same as with magnetism. but from the POV of the GEM equations with that factor of $4$, wouldn't that mean the attractive acceleration would be reduced by 4 times as much? could that attractive acceleration be reduced to zero? or even to a negative acceleration and gravity repels the two lines? Jun 25, 2018 at 20:49
• Q, should i ask that question in a separate question? Jun 25, 2018 at 20:50
• Yes, that seems to be a new separate question. Jun 26, 2018 at 8:37
• okay, Q, i will try to post a separate question in an attempt to resolve this disparity in my understanding. it just seems to me that the gravito-magnetic force is 4 times too strong and, thinking of this simple pair of infinite parallel lines of mass, it seems to me that the same thought experiment should work for GEM as it does for EM. Jun 27, 2018 at 1:58