Is the gravity we feel in our everyday life mainly curvature of the time coordinate?

The question is pretty clear I think.

I only started learning GR but the results I've seen so far and from some lectures I saw from Leonard Susskind on YouTube I understood that for low energy densities the curvature of space-time is mainly in the time coordinate.

Is this true, is the gravity we feel here on earth mainly the result of curvature of the time coordinate?

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There is no "curvature" in just one coordinate. –  Siyuan Ren May 22 '12 at 2:01

3 Answers

The quantity that determines day-to-day gravity we feel is the difference is clock-rate at different places, but yes, it's all in the time coordinate, in the sense that the space we see is flat, and only the time coordinate is mismatched from point to point.

Gravity describes spaces that curve too, distances that change from place to place, but this effect does not matter since it is very small.

The reason we see time-component of metric and nothing else is simply because we are so much longer in time than in any other direction--- when you sit still for a few seconds, you extend over a meter or so, but your time extent is one light-second, or 300,000 meters. So on our scales, thing are very very long in time, and short in space, and we can see a slight curvature between different places in the time coordinate, because we go by so much time coordinate.

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There is a clear explanation of the point in Rindler's Essential Relativity (a gorgeous book for conceptual understanding of Relativity) Chapter 9. In short and very roughly, for small speeds (compared to c) the world-line is nearly time-like, therefore the geodesics can be approximately obtained from the time-time component of the metric. As Karsus says, however, "curvature of the time coordinate" is probably not a great term to use!

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Generally speaking we write the Schwarzschild metric in units where $c$ = $G$ = 1. If we don't use these units the metric is:

$$ds^2 = -\left(1 - \frac{r_s}{r}\right)c^2dt^2 + \left(1 - \frac{r_s}{r}\right)^{-1} dr^2 + r^2 \left( d\theta^2 + sin^2\theta d\phi^2 \right)$$

Suppose you take 1 second to move 1 metre (radially so $d\theta = d\phi = 0$). The Schwarzshild radius of the Earth, $r_s$, is $8.87 \times 10^{-3}m$ and the radius of the Earth, $r$, is $6.37 \times 10^6$m, then:

$$ds^2 \approx -8.9820089875 \times 10^{16} + 1.00000000139$$

where the first term on the right comes from $dt = 1$ second, and the second term comes from $dr = 1$m, so the line element is dominated by the time as expected. Suppose we do the same calculation for the flat space (Minkowski) metric, then we get (using the same number of decimal places):

$$ds_{flat}^2 = -8.9820090000 \times 10^{16} + 1.00000000000$$

so, again keeping the time and space parts separate:

$$ds^2 - ds_{flat}^2 \approx 1.25 \times 10^{8} + 1.39 \times 10^{-9}$$

The curvature in the time contributes about 11 kilometres while the curvature in space contributes about 40 microns.

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This is not a great answer--- there is no coordinate independent meaning to $ds^2 - ds^2_{flat}$. –  Ron Maimon May 22 '12 at 15:56
Agreed absolutely, this is just meant to give a feel for the difference. Given the tone of the question I didn't think it was appropriate to attempt anything more sophisticated. –  John Rennie May 22 '12 at 17:06