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Bear with me while I try to explain exactly what the question is. The question Can a curvature in time (and not space) cause acceleration? is imagining a coordinate system in which the curvature is only in the time coordinate. I want to be as precise as possible about what we mean by curvature in the time coordinate.

It seems to me that a good starting point is the geodesic equation:

$$ {d^2 x^\mu \over d\tau^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} = 0 $$

because if we stick to Cartesian coordinates then in flat space all the Christoffel symbols vanish and we're left with:

$$ {d^2 x^\mu \over d\tau^2} = 0 $$

So a coordinate system in which spacetime is only curved in the time coordinate, $x^0$, would be one in which:

$$\begin{align} {d^2 x^0 \over d\tau^2} &\ne 0 \\ {d^2 x^{\mu\ne 0} \over d\tau^2} &= 0 \end{align}$$

So my question is whether this is a sensible perspective.

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    $\begingroup$ I've never heard of a coordinate being curved, only a manifold. If it were me, I'd be inclined to reply with that statement to the other question, although now that you bring it up it is an interesting question whether one can meaningfully say a coordinate can be curved. $\endgroup$ – David Z Jul 1 '14 at 18:06
  • $\begingroup$ Pardon my ignorance, but wouldn't it just be that the measurement of the distance traveled in a particular coordinate is different for a local observer vs. an observer in a different inertial frame? So if only a single coordinate is curved then only the distance measured for that coordinate would be different? sigh I wish I understood the math... $\endgroup$ – Freudian Slip Jul 1 '14 at 19:44
  • $\begingroup$ Thanks to everyone who responded. I think I understand things better now. I wish I could accept all the answers - Jerry gets the tick by virtue of being first to respond. $\endgroup$ – John Rennie Jul 3 '14 at 10:05
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Well, by the presentation you give, you're going to have $\frac{d^{2}x^{i}}{d\tau^{2}}\neq 0$, because you have those $\Gamma_{0i}{}^{j}$ terms. For instance ${\ddot y} + 2\frac{\dot a}{a}{\dot y}{\dot t} = 0$ (I abuse notation and mean the obvious things with dots, but obviously, $a = a(t(s))$ and $y=y(s)$)

The condition you want is $\Gamma_{\mu\nu}{}^{i} = 0$, with at least one $\Gamma_{\mu\nu}{}^{0}\neq 0$. I'm sure there are metrics that satisfy this condition, but I don't know any (non-trivial ones${}^{1}$) off of the top of my head.

EDIT: note that even the minimally coupled "perturbative spherical potential" metric $ds^{2} = -(1-2\Phi(r))dt^{2} + dr^{2} + r^{2}d\theta^{2} + r^{2}\sin^{2}\theta d\phi^{2}$ will have a nonzero component for $\Gamma_{tt}{}^{r}$, so it might be a bit tricky to find a nontrivial example.



${}^{1}$For instance, you could define $g_{ab} = -f(t)dt^{2} + \delta_{ij}dx^{i}dx^{j}$. This will have a nonzero value for $\Gamma_{tt}{}^{t}$, but the space is really just Minkowski space because it can be changed to it by the substitution $T = \int \sqrt{f(t)}dt$

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In order for a manifold to be curved (intrinsic curvature), it must be of dimension $\geq 2$, which means that at least two principal curvatures must be non-zero, since Gauss's curvature is the product of them. This cannot be done with only one curved basis vector or dimension, as you put it; actually, there's no way to define intrinsic curvature in dimension 1, a circle has intrinsic curvature 0.

Another way to see this, more rigorous, follows. For a globally hyperbolic spacetime $(M,g_{ab})$ of dimension 4 with a time vector (physically relevant), one can always perform ADM decomposition, which implies that $M = \mathbb{R} \times \Sigma_t$ for $\{t\} \subseteq \mathbb{R}$ and $\Sigma_t$ is the spatial hypersurface of the foliation. If $\Sigma_t$ is flat, then the Gauss-Codazzi equation $$^{(3)}R_{abc}{}^d = h_a{}^f h_b{}^g h_c{}^k h^d{}_j R_{fgk}{}^j - K_{ac} K_b{}^d + K_{bc} K_a{}^d,$$ with $K_{ab}$ the extrinsic curvature of $\Sigma_t$ and $h_{ab}$ its metric, which also acts as a projection operator in the form $h_a{}^b$, implies that if $\Sigma_t$ is flat ($^{(3)}R_{abc}{}^d =0$, $K_{ab} \equiv h_a{}^c \nabla_c n_b =0$, with $n_a$ the unit normal to $\Sigma_t$), then so is $M$.

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    $\begingroup$ But if you decompose $\mathbb{R}^{3}$ into cylindrical coordinates, the leaves are flat, but have nonzero extrinsic curvature, so ${}^{3}R_{abcd} = 0$ does not imply that $K_{ab}$ equals zero, just that that sum $K_{ac}K_{b}{}^{d} - K_{bc}K_{a}{}^{d}$ does. $\endgroup$ – Jerry Schirmer Jul 1 '14 at 18:35
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I think it is better to argue from the curvature tensor $R_{ab}{}^\mu{}_\nu$. It is defined by $$R_{ab}{}^\mu{}_\nu x^\nu = (\nabla_a \nabla_b - \nabla_b \nabla_a)x^\mu$$ so it tells you the degree to which covariant derivatives along the $a$ and $b$ axes do not commute. You can see formally from this that curvature requires two dimensions, so it does not make much sense to speak of a curved coordinate. It does not make sense to say "the curvature in the $t$-direction", it is always "the curvature in the $tx$-plane". You should speak of a curved slice: if you can find two coordinates such that $$R_{12}{}^\mu{}_\nu = 0$$ in some region, the surfaces defined by $x^0 = t_0, x^3 = z_0$ are flat, otherwise they are curved.

The previous is in the general case. The FLR case is a little special since you can find three-dimensional flat slices. Since spacetime is four-dimensional, any plane in which the curvature does not vanish must contain the remaining orthogonal direction, which is the $t$-direction.

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It seems to me that a good starting point is the geodesic equation: [...]

This apparently refers to some particular (image of) curve $\gamma$; indeed to some particular time-like curve $\gamma$ for which $$\int_{\gamma} d \tau = \Delta \tau \mid_{\gamma} ~ \gt 0.$$

Given two (not necessarily distinct) (images of) time-like curves $\gamma$ and $\psi$ the corresponding real number value of the ratio

$$ \int_{\gamma} d \tau ~ / ~ \int_{\psi} d \tau $$

is of course a geometric quantity, and independent of any particular assignment (if any) of coordinate tuples to these two (images of) curves, or to the given set of events as a whole.

So a metric in which only the time coordinate, $x^0$, is curved would be one in which:

$\frac{d^2}{d\tau^2}x^0 \neq 0$ [...]

If the time coordinate, $x^0$, is assigned to a given (image of a) time-like curve $\gamma$ such that

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

then the assignment is called "good" (cmp. MTW Fig. 1.9) or "affine" (MTW § 10.1); especially if referring to geodesic curves.

Conversely, if the time coordinate, $x^0$, is assigned such that

$\frac{d^2}{d\tau^2}x^0 \neq 0$, or such that the derivative $\frac{d^2}{d\tau^2}x^0$ doesn't exist at all
then the assignment would consequently be called "not good", or "not affine".

(The assignments of the other "space related" coordinates can be discussed separately.)

In contrast, curvature is a geometric characteristic of a given set of events (or of curves as subsets of given events); and thus independent of any particular assignment (if any) of coordinate tuples.

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  • $\begingroup$ This actually does not answer the question. $\endgroup$ – Danu Oct 10 '15 at 11:31

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