Classical physics in curved space

I'm trying to learn General Relativity. As a stepping-stone between classic (Newtonian) physics and complete relativity, imagine a universe where space is curved, but time is perfectly flat. Also, the curvature of space is just an intrinsic property of that universe and is not changing with time. What would the laws of physics look like in such a universe? For instance, shouldn't gravity follow Poisson's equation $$\nabla^2 \phi = 4 \pi G \rho$$ Where $\nabla^2$ is the Laplace-Beltrami Operator of the manifold describing space. A free particle moving in this space should be subject to 'inertial' forces caused by the curvature of space, similar to the constraining forces in Lagrangian mechanics. Is it possible (or even meaningful) in General Relativity to have curvature in space but not in time?

• As far as, I know you cannot really separate time from space in general or special relativity – physicopath Dec 10 '15 at 20:59

In relativity we are free to choose any coordinate system we want, and in some circumstances it is possible to choose a time coordinate that is not curved. The obvious example of this is the FLRW metric expressed in comoving coordinates:

$$ds^2 = -dt^2 + a^2(t)(dx^2 + dy^2 + dz^2)$$

For any comoving observer, i.e. any observer for whom $dx = dy = dz = 0$, the metric reduces to:

$$ds^2 = -dt^2$$

which is exactly the same as in flat spacetime. However this is a result of our choice of coordinates. If we chose to use conformal time as the time coordinate then the metric would not reduce to the flat space limit.

To address what I think is the main point of your question, studying GR breaks down into two general areas. The first is how the metric is obtained from the stress-energy tensor using Einstein's equations, and the second is taking some metric as a given then investigating its properties. If I understand your question you're not too concerned with the first part. You're effectively saying let's assume the metric is xxx and then study the properties such as geodesics and the motion of test particles. There's nothing wrong with this, and it can be an entertaining pastime, However it is (usually) devoid of physical significance. I also don't see why it is any easier than starting with a real physically relevant metric, as the mathematical treatment is the same.

Is it possible (or even meaningful) in General Relativity to have curvature in space but not in time?

The principle of general relativity itself precludes the notion of curved space without curved time because these notions are not coordinate independent; that is, one observer can make measurements which determine that space is curved but not time, but other observers will in general disagree.

A free particle moving in this space should be subject to 'inertial' forces caused by the curvature of space

Genuine spacetime curvature cannot be discerned from the presence of fictitious forces, as accelerated frames in flat spacetime also experience such forces. One intuitive way of thinking about spacetime curvature is in terms of geodesic deviation. Imagine two initially parallel geodesics separated by a small distance $\epsilon$. If the geodesics remain parallel along their entire length, then $\epsilon$ will be constant (i.e. the separation between the two geodesics does not change). This will always be true in flat space, no matter what coordinates you choose. In curved space, however, $\epsilon$ is not constant. Instead, the separation between the two geodesics varies as you move along them. In other words, initially parallel geodesics do not remain parallel. This can be taken as a defining property of curvature.

Formalizing this notion leads one to define the Riemann curvature tensor, $R_{\mu\nu\alpha\beta}$, which encodes information about precisely how the separation (now promoted to a vector) changes infinitesimally at each point in spacetime.

If one considers a radial symmetric potential $\phi(r)$ in a curved (Schwarzschild) backround by solving the (outer) partial differential equation $$\Delta\phi=0,$$ while $\Delta$ is the associated Laplace-Beltrami operator corresponing to the Schwarzschild metric. This equation can be rewritten by using the fact, that the metric is not explicitly time-dependent, i.e. $\partial_0 g^{\mu\nu}=0$. We find $$\Delta\phi=\frac{1}{r^2}\frac{\partial}{\partial r}\left(g^{rr}r^2\frac{\partial\phi}{\partial r}\right),$$ while $g^{rr}=1-r_s/r$, and $r_s$ is the Schwarzschild radius of the central mass $M$. Solving this equation with respect to the associated boundary conditions, we obtain $$\phi(r)= \frac{c^2}{2}\log\left(g^{rr}\right)=-\frac{GM}{r}+O\left(\frac{1}{r^2}\right).$$ At this point it looks like that there are additional terms with subject to certain 'inertial' forces caused by the curvature of space. But if we write down the Newton law of gravity, we have by definition $$F^r=-m(\text{grad}\,\phi)^r=m g^{rr}\frac{\partial \phi}{\partial r}=-\frac{\text{GMm}}{r^2},$$ which is corresponding to the ordinary Newton law of gravitation, whithout any additional terms.