Does spacetime have symmetric curvature around an object? If yes, then why do planets revolve around the Sun in elliptical (as opposed to circular) orbits?

  • $\begingroup$ General relativity is highly non-linear, and for a generic object it's likely that it does not admit a symmetry. Also, keep in mind that surrounding objects also contribute to space time curvature in a non-linear fashion. $\endgroup$ – JamalS Mar 19 '15 at 9:20
  • $\begingroup$ Also keep in mind that the relevant curvature is the one of space-time. The orbits you are thinking about are orbits in space. And space can be flat, and there is no unique way to choose space and time. $\endgroup$ – MBN Mar 19 '15 at 10:33
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    $\begingroup$ Comment to the question (v3): Before going to GR, do you understand the resolution to the corresponding Newtonian question: Is the magnitude of the Newtonian gravitational force symmetric around an object? If yes, then why do planets revolve around the Sun in elliptical (as opposed to circular) orbits? $\endgroup$ – Qmechanic Mar 19 '15 at 10:58

To a first approximation the spacetime curvature around the Sun is indeed spherically symmetric. I say to a first approximation because the masses of the planets (particularly Jupiter) also produce curvature and this breaks the spherical symmetry. However let's ignore this for now because I don't think it's relevant to your question.

If I understand you correctly you're asking how a spherically symmetric curvature can produce orbits that are not circularly symmetric. The answer is that spherically symmetric curvature only means that angular momentum is conserved. The explanation for this is given by Noether's theorem, though I suspect the maths involved in this will be too hard going for non-nerds.

Anyhow, the angular momentum is given by:

$$ L = mr^2\omega $$

where $m$ is the mass of the orbiting object, $r$ is the orbital radius and $\omega$ is the angular velocity. We know that $L$ must be constant, and one way to achieve this is to have a circular orbit so $r$ and $\omega$ are constant. But we can also have constant $L$ if any change in $r$ is balanced out by a change in $\omega$ so that the product $r^2\omega$ stays constant. This is exactly what happens in an elliptical orbit, and in fact it is just Kepler's second law:

A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

In a small time $dt$ the area swept out is:

$$ A = \tfrac{1}{2}r^2 \omega dt $$

If $A/dt$ is constant, as Kepler's law requires, this means $r^2\omega$ must be constant and this is exactly what we concluded above.

So spherical symmetry does not require circular orbits, only that the orbit must conserve angular momentum.


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