# How do spatial curvature and temporal curvature differ?

You need to be cautious about treating a *time curvature* and *spatial curvature* separately because this split is not observer independent. In some cases a metric can be written in coordinates where the $dt^2$ term is $c^2$ (or unity in geometric units) but this is just a choice of coordinates.

If you take, for example, the FLRW metric then we usually write it as:

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

where $t$, $x$, $y$ and $z$ are the comoving coordinates. However it can also be written using conformal coordinates as:

$$ ds^2=a(\eta)^2(-d\eta^2+dx^2+dy^2+dz^2) $$

It's the same metric, describing the same spacetime geometry, but in one case the time coordinate looks as though it is curved while in the other case it looks as if it is flat. Both metrics are perfectly good descriptions of the geometry and we choose whichever version happens to be most convenient for our purposes.

But back to your question: the trajectory of a freely falling particle, i.e. its geodesic, is given by the geodesic equation:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = -\Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu \tag{1} $$

In this equation $\mathbf x$ is the position $(t,x,y,z)$ of the particle in spacetime, $\mathbf U$ is the four velocity and the symbols $\Gamma^\alpha_{\,\,\mu\nu}$ are the Christoffel symbols that describe the curvature of the spacetime. You can think of this as a kind of equivalent to Newton's second law in that it relates the second derivative of position to the curvature.

Suppose we consider a stationary particle (stationary in our coordinates that is). Since the particle is stationary in space the components of the four velocity $U^x = U^y = U^z = 0$ and only $U^t$ is non-zero. In that case the geodesic equation (1) simplifies to:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = -\Gamma^\alpha_{\,\,tt}U^t U^t \tag{2} $$

Calculating the Christoffel symbols is a huge pain unless you have a copy of Mathematica to hand, but you can usually find them by Googling as indeed is the case for the Ellis wormhole (NB that link is a PDF) and the only non-zero Christoffel symbols are (I'll list them all in case the link above breaks):

$$\begin{align} \Gamma^\rho_{\theta\theta} &= -\rho \\ \Gamma^\rho_{\phi\phi} &= -\rho\sin^2\theta \\ \Gamma^\theta_{\theta \rho} = \Gamma^\theta_{\rho\theta} &= \frac{\rho}{n^2+\rho^2} \\ \Gamma^\theta_{\phi\phi} &= -\sin\theta\cos\theta \\ \Gamma^\phi_{\phi \rho} = \Gamma^\phi_{\rho\phi} &= \frac{\rho}{n^2+\rho^2} \\ \Gamma^\phi_{\phi\theta} &= \Gamma^\phi_{\theta\phi} = cot \theta \end{align}$$

Note that all the symbols $\Gamma^\alpha_{tt}$ are zero, so our geodesic equation (2) becomes:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = 0 $$

Or in other words in the Ellis wormhole a stationary particle remains stationary.

But even this result needs to be treated with some care because you have to understand your coordinates to interpret it. To show this consider the FLRW metric that I referred to above. I won't go through the details but you can do exactly the same calculation for the FLRW metric and reach the same conclusion:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = 0 $$

But remember that in the FLRW metric the coordinates are comoving coordinates, not the coordinates that you or I use when e.g. measuring the distances to distant galaxies, and the comoving coordinates are moving relative to *everyday* coordinates (which is why distant galaxies are moving and indeed accelerating relative to us). Even when we find that in a particular coordinate system a stationary particle remains stationary, this doesn't mean we would actually observe a stationary object to remain stationary.

(Though as it happens in the Ellis wormhole spacetime you and I would observe that a stationary object remains stationary.)

I think this addresses your questions 1 to 4. As for your questions 5 and 6, as it happens I asked exactly the same question in What makes a coordinate curved? and the answer is that at least two principal curvatures must be non-zero. So you cannot find a geometry/coordinate system where the curvature is only in the time coordinate.

Consider a local Lorentz frame, $g_{ij} = \mathrm{diag}(1,-1,-1,-1)$. An observer (really a congruence of observers) at rest with respect to this frame has velocity vector $u^i = \delta^i_0$. It experiences no force (geodesic deviation) if it obeys the geodesic equation, which in this case becomes simply $\gamma^i{}_{00} \equiv 0$. From the compatibility of the connection with the metric we know that $\gamma^i{}_{00} \equiv 0$ if and only if $\gamma^0{}_{i0} \equiv 0$, and from the first Cartan equation $$ d\omega^i = \omega^j \wedge \gamma^i{}_j = \gamma^i{}_{jk}\omega^j\wedge\omega^k, $$ we know that this is certainly the case if $d\omega^0 \equiv 0$. Here $\gamma^i{}_j$ are the connection forms and $\gamma^i{}_{jk}$ are the components (Ricci rotation coefficients). Given a set of coordinates it is natural to consider an observer to be static if the velocity vector is given by $$ u^\mu = \frac{1}{\sqrt{g_{00}}}\delta^\mu{}_0 $$ (here I let $\mu,\nu,\ldots$ signify coordinate indices). Considering then the static case ($g_{0\mu} \equiv 0$ for all $\mu = 1,2,3$) we find that upon setting $e^\mu_0 = u^\mu$ we have $d\omega^0 \equiv 0$ whenever $g_{00}$ is constant. Such is the reasoning behind the statement that gravitational attraction arises from non-constant $g_{00}$. As you can see it is most definitely a simplification.

As is apparent from the above exposure, any observer not at rest with respect to our local Lorentz frame may experience a force (though the nature of its correspondence to high speeds depends on the exact form of the metric and/or connection forms).

As to your questions regarding the effects on light, it is important to recall that light follows null geodesics. Therefore they will always be affected by the nature of the rotation coefficients $\gamma^i{}_{00}$ but also by at least some other coefficients. It would require a speed greater than that of light (spacelike observer) to escape to effects of $\gamma^i{}_{00}$, but this is clearly unphysical.

Although, as John Rennie links to in his answer, it is meaningless to talk about curvature in one direction, in light of the above considerations we might ponder the case where $\gamma^i{}_{00} = \gamma^0{}_{i0}$ are the *only* non-zero rotation coefficients. This corresponds concretely to the simplest case of the greater the speed with respect to our frame, the smaller the "curvature effects" on the motion (though as noted above, one would require a speed greater than that of light to escape them altogether). Then $d\omega^i \equiv 0$ for all $i = 1,2,3$. By the second Cartan equation
$$
d\gamma^i{}_j = \gamma^k{}_j\wedge\gamma^i{}_k + \frac{1}{2}R^i{}_{jk\ell}\omega^k\wedge\omega^\ell,
$$
we immediately find
$$
d\gamma^0{}_i = -\gamma^0{}_{i0|j} \omega^0 \wedge \omega^j = R^0{}_{i0j}
\omega^0 \wedge \omega^j
$$
to give the only (potentially) non-zero curvature components, up to symmetries. Note that we take $i,j \neq 0$, whence in particular it follows that the Ricci tensor is zero if and only if the Riemann tensor is. Therefore we can at least conclude that such solutions cannot be vacuum, and therefore cannot describe the exterior of any object.

**EDIT:** In fact, I was a bit lazy in concluding the above. Making the contractions, and ignoring any cosmological constant, we find that the Einstein field equations yield $T_{0i} = 0$ for all $i$, whence any (non-flat) solution must violate the dominant energy condition. Thus we can further conclude that such a solution is unphysical, since there are timelike observers that observe energy to flow faster than the speed of light, i.e. timelike vectors $v^i$ such that $T_i{}^jv^i$ is spacelike (namely all observers not at rest with respect to our frame).