# Can a curvature in time (and not space) cause acceleration?

I realize that the curvature of space-time causes acceleration (gravity).

Is it possible to have a curvature only of space, or a curvature only of time?

If so, would a curvature only of space, or a curvature only of time, also cause acceleration?

• Isn't that basically the Newton-Cartan theory of gravity? Keep space flat, but curve the temporal dimension, and you get a geometric description of Newton's theory of gravity. See, e.g., Willie Wong's blogpost or wikipedia. Commented Jun 29, 2014 at 21:09
• See this quote; web.mit.edu/6.055/old/S2009/notes/bending-of-light.pdf '' Newton’s theory is the limit of general relativity that considers only time curvature; general relativity itself also calculates the space curvature. Since most objects move much slower than the speed of light, meaning that they travel much farther in time than in space, they feel mostly the time curvature. ''
– Riad
Commented Sep 28, 2018 at 4:41

## 2 Answers

In general it doesn't make sense to talk of curvature being only in space or only in time.

The geometry of a spacetime is described by the metric. Normally we start with some distribution of matter/energy and solve the Einstein equations to calculate the metric. Alternatively you can start with the desired metric and use the Einstein equations to work out what distribution of matter/energy is needed to create it, though more often than not you'll end up with an unphysical distribution of matter (e.g. requiring exotic matter).

Anyhow, the metric is a coordinate independant object - it is the same whatever coordinate system you use. However to write down a representation of the metric (usually as a 4 x 4 matrix) you need to choose a coordinate system (with one timelike and three spacelike coordinates) and it's only when you do this that you can start talking about curvature in coordinates.

The trouble is that there can be many different choices of coordinate system to describe the same spacetime. Even the humble static uncharged black hole can be described using Schwarzschild coordinates, Gullstrand-Painleve coordinates, Eddington-Finkelstein coordinates, Kruskal-Szekeres coordinates and probably many others that I don't know about. So the timelike coordinate you choose, and the curvature associated with it, won't be the same as the timelike coordinate that other general relativists might choose.

So you might well be able to come up with some choice of coordinates that is curved only in the time direction. But someone else using a different coordinate system might find the curvature is in the space coordinates or more likely both time and spatial coordinates. In all cases, any curvature will cause a freely moving abject to accelerate.

So if you ask Can a curvature in time (and not space) cause acceleration? then the answer is yes, but this is a somewhat empty answer because your condition curvature in time (and not space) is a statement about your coordinate system and not about the geometry of the spacetime.

• And, of course, there is the even more trivial problem that $R_{tttt}$ is always identically zero. Commented Jul 1, 2014 at 17:12
• I feel that Schwarzschild geometry isn't a good example, because it is static, and hence there's a distinguished timelike direction, in this case Schwarzschild time. But the point certainly holds for most spacetimes. Commented Jul 1, 2014 at 17:22
• @StanLiou: I only chose the Schwarzschild metric because I could think of several coordinate systems for it. The FLRW metric would be a better example but as far as I know we only using comoving coordinates for this, or at least I've never seen a different coordinate system used. If you can think of a better example please shout! Commented Jul 1, 2014 at 17:24

First off, I'm not entirely sure of what you are asking, or what you are thinking of as curvature. There are certainly coordinate systems which are non-Euclidean that are not considered to be "curved." For instance, standard cylindrical coordinates have zero Riemann Curvature, but they are "curvy looking."

My take though is that in GR you have a coordinate system which has 4 coordinates. You also have a metric tensor $$g$$ which describes / determines some properties of this coordinate system, such as arc length.

Without getting too mathematical, think of a sheet of paper with the vertical of the paper being the time coordinate, and the horizontal of the paper being a space coordinate. Now roll the paper up so that the space coordinate is like the round surface of the cylinder. There is a metric g to describe that. But if you wanted to do the same thing so that the time coordinate became the round part of the cylinder, there is a metric to describe that too. Incidentally, neither of these spaces are actually curved in the Reimann curvature sense, but some people would initially guess they were.

But imagine the cylinder being like a rubber tube and bend it like a donut:

Now, if you draw two "straight" lines around the donut in one direction, they would remain the same distance apart--parallel. But in the other direction they do not.

Does this mean curvature? Not necessarily. What determines curvature, I believe, isn't that the lines get closer or further apart per se, but more so the rate at which they get closer or further apart being non-constant. So to do that I think you would have to make the donut something like more oblong or squashed. This is related to 'geodesic deviation' and characterized by the Reimann tensor which is determined by the metric g.

Geodesic deviation basically constitutes an analogy to acceleration. So can you have acceleration (deviation) in the time direction and not the space direction or vice versa? hmmm . . .

. . . having trouble seeing it precisely from geometry, but I'm going to lean towards no. If something is accelerating in space relative to me, the way its time elapses compared to my time would also be non-constant. So if something is accelerating in time, at least one of the spatial components of velocity would have to be changing in Einstein theory, because its 4-speed overall is c in relativity.