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We know from classical mechanics that the gravitational field equation for the scalar potential takes the form $$\nabla^2\phi=4\pi \rho,$$ where $\rho$ is mass density (which, can depend on time and space). Also, the associated equation of motion for point particle takes $$\ddot{x}+\nabla\phi=0.$$ One of the basic requirement for a classical theory is that it should not depend on the inertial reference frame we are choosing. In particular, for a non-relativist theory such as the one described above, I would expect the theory to keep its form under Galilean transformations. I am, however, not sure how to do this rigorously with a general $\phi$. Any ideas?

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  • $\begingroup$ The equation $\nabla^2\phi=4\pi\rho$ is not a dynamical equation, it's more like a constraint. Cf. physics.stackexchange.com/a/20072/4552 . In your two equations, $\rho$ only appears in one, so we can just take it as a definition of $\rho$. Although $\rho$ transforms trivially, even if it didn't, we wouldn't care; it wouldn't affect the truth-value of the equations. To make this a predictive theory, you need to couple your two equations somehow, probably by adding in an equation of continuity or something that relates motion of particles ($\ddot{x}$) to changes in $\rho$. $\endgroup$ – Ben Crowell Feb 6 at 6:55
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The Galilean group consists of three different types of coordinate transformations between two different inertial reference frames: translations, rotations, and boosts.

A translation looks like

$$x'=x-X\\y'=y-Y\\z'=z-Z$$

where $X$, $Y$, and $Z$ are constants.

A rotation looks like

$$x_i'=R_{ij}x_j$$

where $R$ is a constant rotation matrix.

A boost looks like

$$x'=x-V_xt\\y'=y-V_yt\\z'=z-V_zt$$

where $V_x$, $V_y$, and $V_z$ are constants.

Under any Galilean transformations, the potential $\phi$ is assumed to be scalar satisying $\phi’(\mathbf{r}’, t)=\phi(\mathbf{r}, t)$. Here $\mathbf{r}$ and $\mathbf{r}’$ represent the same point in two different reference frames. The potential is just a single value at each point, and all observers agree on what that value is.

The same applies to the mass density $\rho$.

The Laplacian operator can be shown to be a scalar with transformation $\nabla’^2=\nabla^2$. The easy argument is that it is the scalar product of the gradient vector operator with itself. For a more careful argument, work out what happens to $\partial^2/\partial x^2+\partial^2/\partial y^2+\partial^2/\partial z^2$ under translations, rotations, and Galilean boosts, using the transformation equations above.

Therefore your first equation

$$\nabla^2\phi=4\pi\rho$$

has the covariant form scalar=scalar under translations, rotations, and boosts. Put differently

$$\nabla^2\phi(\mathbf{r},t)=4\pi\rho(\mathbf{r},t)$$

implies

$$\nabla’^2\phi’(\mathbf{r’},t)=4\pi\rho’(\mathbf{r’},t),$$

which shows that it is form-invariant.

The second equation,

$$\ddot{\mathbf{r}}=-\nabla\phi,$$

is covariant because both acceleration and the gradient operator are vectors under rotations and scalars under translations and boosts; and the potential is a scalar under all three.

So under rotations, this equation has the covariant form vector=vector, and under translations and boosts it has the covariant form scalar=scalar.

Put another way, this equation implies

$$\ddot{\mathbf{r’}}=-\nabla’\phi’,$$

so it is form-invariant.

Note: In the case of rotations, you get these same-form equations after “cancelling” the rotation matrix that the rotation introduces on both sides. Just multiply both sides by the inverse matrix to get rid of it and restore the original form.

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