# Is the description of the gravitational field as a vector field and a tensor field compatible?

By electric or magnetic fields we mean the vector fields $\vec{E}(\vec{r},t)$ and $\vec{B}(\vec{r},t)$ respectively. But a gravitational field in Newtonian theory is a vector field that $\vec{g}(\vec{r})$ that obeys $\nabla\times\vec{g}=0$ and $\nabla\cdot\vec{g}=4\pi G\rho_{m}$. But in Einstein's theory we describe gravity as a tensor field $g_{\mu\nu}(x)$. How are these two descriptions of gravity, one in terms of the vector field $\vec{g}(\vec{r})$ and in other in terms of the tensor field $g_{\mu\nu}(x)$, compatible with each other in terms of the number of degrees of freedom? Thanks!

In Newtonian mechanics, the gravity does not consist of only one vector, there is another vector associated with the gravity.

The conventional gravity is actually a static theory of gravitation and the field associated with it is denoted by $\mathbf{g}$, or sometimes $\mathbf{E}_g$ in an analogy to electrostatics. The other vector is called gravitomagnetic field and denoted by $\mathbf{B}_g$, however, I will denote it by $\mathbf{\Gamma}$ to avoid confusion with electromagnetism. So, in static case $$\tag{1} \nabla \cdot \mathbf{g} = - 4\pi G \, \rho$$ $$\tag{2} \nabla \times \mathbf{g} = 0$$ where $G$ is the Newton's gravitational constant and $\rho$ is the mass denstiy.

Now, let's introduce a mass current density, $\mathbf{J_g}$, such that it satisfies the continuity equation: $$\frac{\partial \rho}{\partial t} = - \nabla \cdot \mathbf{J}_g$$ If we take the time derivative of Eq.(1), and use the continuity equation, we get $$\nabla \cdot \left( \frac{\partial \mathbf{g}}{\partial t} - 4 \pi G \mathbf{j} \right) = 0$$ Here, the expression in parenthesis is a vector with a vanishing divergence. For any vector $\mathbf{V}$, the divergence of its curl is zero, i.e., $\nabla \cdot \left( \nabla \times \mathbf{V} \right) = 0$. Therefore, $$\nabla \times \mathbf{\Gamma} = \frac{1}{4 \pi G} \frac{\partial \mathbf{g}}{\partial t} - \mathbf{j}$$ is the gravitational analogy of Ampere's Law in electromagnetism. Here, I have chosen the sign according to the attractive nature of gravity, in electromagnetism it is opposite so that different sign of charges attract each other while in gravity same sign of masses attract each other.

The vector, $\mathbf{\Gamma}$, represents the rotational degrees of freedom in gravity and it is useful if you are working with orbits. For instance, the solution for the equatorial gravitational field of a point-like satellite orbiting around a rotating planet would be the following: $$\mathbf{\Gamma} = \frac{4 \pi G \mathbf{L}}{c_g^2 r^3}$$ if we assume that the speed of the gravitational interactions is finite as $c_g < \infty$. However, in Newtonian gravity there is no such a constraint and indeed for slow-rotating bodies like Earth this contribution is very weak.

As a matter of fact, in order to obtain the full analogous equations to electromagnetism which are relativistic in their nature, we need to introduce a finite speed for the gravitational interaction, as in electromagnetic waves. However, in Newtonian mechanics, there is no such speed. So, $$\tag{3} \nabla \times \mathbf{g} + \frac{4 \pi G}{c_g^2} \frac{\partial \mathbf{\Gamma}}{\partial t}$$ can not be proven to be zero, where $c_g$ is the speed of the gravitational interaction. If we were to assume this expression to vanish, then one would show that $c_g = c$, which is strictly relativistic as in electromagnetism. Nevertheless, it is evident that the second term in the expression (3) is negligible if $c_g \rightarrow \infty$ and it would reduce to the following: $$\nabla \times \mathbf{g} = 0$$ which is just like the usual case. This why there are no gravitational wave in Newtonian gravity. However once you have the general theory of relativity, you can express everything in this notation, just like electromagnetism, however you need to have more degrees of freedom for gravitational fields because of the metric tensor. Above, we have 6 degrees of freedom, three from $\mathbf{g}$ and three from $\mathbf{\Gamma}$. However in Einsteinian gravity the metric has 10 degrees of freedom.

In terms of the potential, $g_{\mu\nu}$, the considerations above match completely if we take the torsion tensor, $T_{\mu\nu}^a$, together with the curvature tensor, $R_{\mu\nu\beta}^\beta$, to be analogous to $\mathbf{g}$ and $\mathbf{\Gamma}$. The time-time component of the metric is analogous to the gravitational scalar potential, $\phi$, and the 3-column components of the metric, $g_{i0}$, are analogous to the vector potential, $\mathbf{A}_g$, where the curl of the latter is $\mathbf{\Gamma}=\nabla \times \mathbf{A}_g$.

Actually the rest of the metric is other degrees of freedom that are not mentioned in the Newtonian gravity but play an important role in the polarization of the gravitational field which consist of tidal behaviors (it is the reason that hypothetical graviton is supposed to be spin-2 in quantum gravity). The most natural way to express the gravity in terms of its potentials and their geometric and gauge-theoretical aspects is to use the tetrad (or Cartan) notation where intead of the metric, it is used the so-called "square-root of the metric", $e_\mu^a$, such that $g_{\mu \nu} = e_\mu^a e_{a\nu}$. The you would have 4 copies of the equations similar to electromagnetism but 6 of them would be gauged away because of the translation invariance of the theory.

• Is the scalar potential you mention $\mathbf{g}=-\nabla\phi$ (if one takes the limit where the curl does vanishes)? – Emil Jun 18 '18 at 18:24
• Assuming eq. (1) and (2), i.e., static case, the gravitational acceleration would be $\mathbf{g} = - \nabla \phi$. So, yes, that $\phi$. – Oktay Doğangün Jun 20 '18 at 15:43

The metric is actually analogous to the Newtonian gravitational potential, not the gravitational field. The number of degrees of freedom doesn't match, because in Newtonian gravity there are no gravitational waves.

Firstly, the difference between Newton and Einstein's description of gravity goes beyond simply representing one as a vector and one as a tensor. General relativity introduces a radically different concept, that a particle will move along geodesics of a manifold, whose curvature is determined by the stress-energy of the matter present.

Contrast this to Newton's description, wherein the field $\vec g$ directly acts on the particle. The two approaches are different, and valid in different regimes. What is important is not that the maths of each are equivalent in the Newtonian limit, but more so that the physical predictions are.

In general relativity, the Newtonian limit would correspond to velocities much smaller than the speed of light, which is to say that,

$$\frac{\mathrm dx^i}{\mathrm d\tau} \ll \frac{\mathrm dx^0}{\mathrm d\tau}$$

and that $g_{\mu\nu}$ is static. In such a limit, it can be shown that the geodesic equation is $\ddot x^i = \frac12 \partial_i h_{00}$ where $h_{\mu\nu}$ is the first order perturbation to a flat background. This corresponds to,

$$\ddot x^i = -\partial_i \phi$$

for $h_{00} = \phi$, where we can interpret $\phi$ as the Newtonian potential.

The gravity metric tensor $g_{\mu\nu} (x)$ is actually the square of vector $e_\mu^I(x)$: $$g_{\mu\nu} = e_\mu^I e_\nu^J\eta_{IJ}.$$ The frame/tetrad/vierbein/vielbein vector (one-form) $e_\mu^I$ has $4*4 = 16$ degree of freedom (one $4$ comes from Greek index $\mu$ and the other $4$ comes from Roman index $I$). But you have to subtract the Lorentz gauge freedom $6$ ($3$ rotations + $3$ boosts) to arrive at the correct physical degree of freedom $16-6 = 10$, which matches with the $10$ degree of freedom of symmetric tensor $g_{\mu\nu} (x)$.

The Newtonian vector field $\vec{g}(x)$ you mentioned is the gradient of zero-zero component of $e_\mu^I$ vector, $$\vec{g}(x) = \frac{1}{2}\nabla{g_{00}} \simeq \nabla{e_0^0}.$$ In other words, in the non-relativistic Newtonian limit we are only concerned with one degree of freedom $e_0^0$.

And FYI, the electromagnetic fields $\vec{E}(x)$ and $\vec{B}(x)$ are parts of curvature/field strength bivector (two-form) $F_{\mu\nu}(x)$, under the disguise of vectors. The gauge connection field $A_{\mu}(x)$ is the true vector (one-form).

• And FYI, the electromagnetic fields E⃗ (x) and B⃗ (x) are parts of curvature/field strength bivector (two-form) Fμν(x), Not sure what you mean by the reference to curvature here. The F tensor has nothing to do with curvature. – user4552 Jun 18 '18 at 17:50
• @BenCrowell The field strength two-form $F(x)$ is the curvature of gauge connection field one-form$A(x)$: $F = dA + A\bigwedge A$. For Abelian Electromagnetic gauge field $A_{EM} \bigwedge A_{EM} = 0$, thus $F = dA _{EM}$. – MadMax Jun 18 '18 at 18:01