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In several posts on Quora, here Howard Landman writes that the Schwarzschild metric shows/proves that around a smaller body, like the Earth or the Sun, 99.9999% of the warping of spacetime is time, and only 0.0001% is the warping of space. And that space warping only reaches 50% at a black hole or neutron star. First, is this interpretation of the SM correct?

I created a spreadsheet based on the SM, and the graphs agree with Landman that the time component crosses into negative numbers at the Schwarzschild radius. And the space component curves in the opposite direction, increasing.

How do I interpret this result to understand what Landman is saying? And why is this never discussed, anywhere as far as I can see. Or have I just missed something obvious?

Urb
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foolishmuse
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  • Please see this recent closely related question: https://physics.stackexchange.com/q/609217/123208 – PM 2Ring Jan 25 '21 at 17:09
  • In particular, see JEB's remark about being on the Earth's surface sitting at rest. Of course, you aren't at rest (relative to the centre of the Earth) because the Earth is rotating, and moving observers don't agree on which direction in spacetime is the purely time direction. – PM 2Ring Jan 25 '21 at 17:13

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What Landman writes is just nonsense. The components of the metric are not measures of curvature. Measures of curvature involve second derivatives of the metric's components. Spacetime is locally flat, so we can always choose local coordinates such that the components of the metric are the same as in flat spacetime.

There is no such thing as purely temporal curvature. The components of the Riemann curvature tensor vanish if you use only a timelike axis for all four indices.

user286987
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    In a metric $ds^2=f(r)^2dt^2-g(r)^2dr^2-r^2dΩ^2$, the proper acceleration of a stationary worldline is $f'(r)/f(r)g(r)$. If $f$ and $g$ are close to $1$ then $f'$ dominates. So it's more or less true that the acceleration is just the gradient of the "time dilation factor" $f$. He shouldn't have said "all of the curvature is in the time dimension" but aside from that I think it's okay. "Nonsense" is way too harsh. – benrg Jan 26 '21 at 03:32
  • @benrg At 99.999...%, his claim that "all of the curvature is in the time dimension" seems reasonable. Or have I missed something? Is there still curvature in the space dimension, but just not contributing to gravity at that point? This is exactly what my OP is about, how to interpret the Schwarzchild Metric to arrive at Landman's statement that it's all time dilation and virtually no space warping around the earth. Not even considering whether time dilation is the cause or the effect of gravity. – foolishmuse Jan 26 '21 at 17:30
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The question is: how GR explains that we (and all objects) are under an acceleration of $g$ at the surface of the earth?

The argument of the link is that for small velocities compared to the light speeds and $r_s \lll r$, the time dilation is basically a function of $r$. But it doesn't answer the question above. It only notes that now there is a time dilation. As there is not such a thing in Minkowski spacetime for objects at rest in its frame, it concludes that time dilation is the "cause" of gravity.

In order to get an answer, we can write the geodesic equation for the coordinate $r$. After eliminating the null terms, using the Schwarzshild equation:

$$\frac{\partial^2 r}{\partial \tau^2} + \Gamma^1_{00} \frac{\partial t}{\partial \tau}\frac{\partial t}{\partial \tau} + \Gamma^1_{11} \frac{\partial r}{\partial \tau}\frac{\partial r}{\partial \tau} + \Gamma^1_{22} \frac{\partial \theta}{\partial \tau}\frac{\partial \theta}{\partial \tau} + \Gamma^1_{33} \frac{\partial \phi}{\partial \tau}\frac{\partial \phi}{\partial \tau} = 0$$

Now we use the argument of small velocities: $dr, d\theta, d\phi \lll dt$, so only 2 terms matter:

$$\frac{\partial^2 r}{\partial \tau^2} + \Gamma^1_{00} \frac{\partial t}{\partial \tau}\frac{\partial t}{\partial \tau} \approx 0$$

Using Schwarzshild metric: $$\Gamma^1_{00} = \frac{2GM}{r^2}\frac{1-\frac{2GM}{r}}{2}$$

From the Schwarzshild equation, where $dr, d\theta, d\phi \lll dt$:

$$d\tau^2 \approx \left(1-\frac{2GM}{r}\right)dt^2$$

Leading to the desired $g$ acceleration: $$\frac{\partial^2 r}{\partial \tau^2} \approx -\frac{GM}{r^2}$$

Claudio Saspinski
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  • What you have shown is that the time component of the Schwarzschild Metric is equal to Newtonian gravity, correct? But you have also raised the question of the chicken and the egg; does time dilation cause gravity, or does gravity cause time dilation - but that is another question not for today. – foolishmuse Jan 26 '21 at 17:24
  • I used the time component of SE and the geodesic equation for $r$. For small velocities it leads to Newtonian gravity. – Claudio Saspinski Jan 26 '21 at 19:57