- #36
PeterDonis
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mieral said:
This goes beyond the scope of a "B" level thread; the math required is at least "I" level. However, a starting point for a heuristic solution is the Schwarzschild-de Sitter solution, which is described here:
https://en.wikipedia.org/wiki/De_Sitter–Schwarzschild_metric
The metric for this solution is:
$$
ds^2 = - \left( 1 - \frac{2M}{r} - 3 \Lambda r^2 \right) dt^2 + \frac{1}{1 - 2M / r - 3 \Lambda r^2} dr^2 + r^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)
$$
This metric describes a vacuum region around a central mass ##M## which is spherically symmetric, in which there is a cosmological constant (dark energy) with density ##\Lambda##. The proper acceleration of an object at rest in these coordinates (i.e., at constant ##r##) is given by
$$
a = \frac{1}{\sqrt{1 - 2M / r - 3 \Lambda r^2}} \left( \frac{M}{r^2} - 3 \Lambda r \right)
$$
where positive ##a## points radially outward (i.e., normal weight) and negative ##a## points radially inward (i.e., negative "weight"). As you can see, the terms in ##M## and ##\Lambda## have opposite signs--this is why dark energy is often said to cause "antigravity".
The above metric isn't quite what we want, though, because it assumes that the cosmological constant ##\Lambda## is present throughout the region out to radius ##r##. So if we take ##r## to be just above the surface of the Earth, this would require the entire Earth to be permeated with dark energy of density ##\Lambda##, where that density is some normal density rather than the actual dark energy density in our universe (which is so tiny that it has no measurable effect, by many orders of magnitude, on length scales like the size of the Earth--the product ##\Lambda r^2## for ##r## the radius of the Earth is many orders of magnitude smaller than ##2M / r## for the Earth). What we really want is a metric where, heuristically, the ##2M / r## term is as above, but the ##3 \Lambda r^2## term uses an ##r## that is much, much smaller--the size of the small bundle of dark energy that gets placed under the automobile. Such a metric, as I have stated it, is not, on its face, a solution of the Einstein Field Equation (since we can't just arbitrarily change ##r## in one term). But it shows the general kind of thing that I based my intuitive guess on in my previous post: basically, the relative magnitudes of ##M / r^2## for the Earth and ##3 \Lambda R##, where ##R## is the radius of the dark energy bundle and ##\Lambda## is its density, will, heuristically, determine whether the automobile "falls upward" or just has less weight.
If you want to develop your understanding of GR, I suggest that you try working out a solution yourself.