Today, someone asked me why "the warped space-time warps itself" (he read it in Kip Thorne's: The Science of Interstellar). I guess this is related to the gravitational self-interaction. But I don't really understand the gravitational self-interaction. Why the curvature in general relativity influences itself?

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    $\begingroup$ Technically, because the field equations are non-linear. $\endgroup$ – Sebastian Riese Feb 22 '16 at 17:21
  • $\begingroup$ So implies the non-linearity of the field equation the non-linearity of the lagrange density, which is neeeded for the self-interaction term? Ok I accept this answer in the mathematically sense. But what is the physically interpration that the geometry influences itself? But the mass is still the cause of the curvature? $\endgroup$ – Core2016 Feb 22 '16 at 17:35
  • $\begingroup$ Physically, the interpretation is, that the gravitational field carries energy and all energy is a source of gravitation, although that is subtle as the stress-energy of the gravitational field is coordinate dependent (and not a proper tensor). $\endgroup$ – Sebastian Riese Feb 22 '16 at 17:37
  • $\begingroup$ Why is there energy associated with the pure geometric phenomenon of GR? $\endgroup$ – Peter R Feb 22 '16 at 20:29
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    $\begingroup$ @Timaeus I totally disagree with your analogy to electromagnetism. For me self-interaction means self-interaction. In a way, interaction means non-linearity by definition. Interacting solutions cannot be superposed, non-interacting ones can. In this sense the electromagnetic field does not influence itself: Given two vacuum solutions you can always add them, they will again be a solution. This is not true for general relativity, in this sense on could say the non-linearity is the self-interaction. $\endgroup$ – Sebastian Riese Feb 23 '16 at 22:32

There are a million ways to answer the "why" here, but here is the simplest way to see that there has to be a gravity-gravity interaction in GR:

we have two things baked into the theory:

1) locally, we can only move at the speed of light, which means that we can only travel at the speed of light, as measured by the metric tensor

2) we can transmit signals with gravitational waves

So, let's set up a gravitational field, somehow. This will bake in some energy into the metric tensor, and create a non-trivial spacetime geometry.

Now, send a localized gravitational wave with a small energy relative to the curvature through this geometry. It will travel, to first approximation, along a null geodesic in the background geometry. This is a different path than it would travel in the absence of the spacetime curvature, obviously. Well, here you go -- the gravitational field is interacting with itself.

  • $\begingroup$ I like how this shows there is self-interaction of the field even in a weak-field limit. $\endgroup$ – Sebastian Riese Feb 23 '16 at 22:48
  • $\begingroup$ Where is the gravitational wave come from? $\endgroup$ – Core2016 Feb 23 '16 at 23:28
  • $\begingroup$ @Core2016: it doesn't ultimately matter -- some localized source $\endgroup$ – Jerry Schirmer Feb 24 '16 at 0:03
  • $\begingroup$ Ok but why is it a self-interaction of the gravitational field. More precisely why does in this case the field influence itself? Isn't this only possible if the gravitational wave cames from the same gravitational source? $\endgroup$ – Core2016 Feb 24 '16 at 22:04
  • $\begingroup$ @Core2016: No, this is explicitly independent of that. The static background field actually scatters the wave that passes through it. Gravity is influencing gravity. $\endgroup$ – Jerry Schirmer Feb 24 '16 at 22:15

The same thing happens in electromagnetism. And the first step in electromagnetism is to admit that the electromagnetic field exists and has particular values at every event. Then you note that an initial slice can be extended forwards or backwards by the Maxwell Equation. You can think of $$\frac{\partial \vec B}{\partial t}=-\vec \nabla\times \vec E$$ as evolving the magnetic part. And think of $$\frac{\partial \vec E}{\partial t}=\frac{1}{\epsilon_0}\left(-\vec J+\frac{1}{\mu_0}\vec \nabla\times \vec B\right)$$ as evolving the electric part.

So electromagnetic fields don't need something to force then to be nonzero, they have a particular value and they evolve in particular ways. That's exactly what happens with the metric. It has particular values. And just like Maxwell tells the electromagnetic field how to evolve, Einstein tells the metric how to evolve.

You literally can't object to Einstein and think Maxwell is fine. It's the exact same thing going on. Yes, the evolution equation is nonlinear and there are more of them (second order and for a symmetric rank two tensor). It just makes it harder to right them down, but it doesn't change what is going on.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Feb 25 '16 at 5:42

A good example (assume e.g. black hole is accelerating and creates gravitational waves):

Step 1. Linearize general relativity (assume that space does not warp itself)

Step 2. Curvature creates gravitational waves

Step 3. Gravitational waves create curvature

Step 4. Re-do the calculation: Curvature creates gravitational waves (gravitational waves technically create gravitational waves, although the effect is miniscule)

  • $\begingroup$ Does a static black hole in no double star system creates gravitational waves? $\endgroup$ – Core2016 Feb 24 '16 at 18:31
  • $\begingroup$ @Core2016 No. And a purely radial electromagnetic field also creates no waves. $\endgroup$ – Timaeus Feb 24 '16 at 22:05
  • $\begingroup$ So why create curvature gravitational waves (according to 2.)? $\endgroup$ – Core2016 Feb 24 '16 at 22:11
  • $\begingroup$ I should be more specific. A black hole must be accelerating to create gravitational waves. $\endgroup$ – Otto Feb 25 '16 at 2:41

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