Gravitational self-interaction

  • Thread starter Thread starter Neutrinos02
  • Start date Start date
  • Tags Tags
    Gravitational
Neutrinos02
Messages
43
Reaction score
0
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?

What term in the Lagrangian will cause this effect? Is there something like an energy tensor of the gravitation field (I read something in the textbook of Landau and Lifschitz).
 
Physics news on Phys.org
Neutrinos02 said:
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?
The Einstein field equation is a system of differential equations. Given a specification of the stress-energy tensor throughout a contiguous region of spacetime, the spacetime curvature at the boundaries of the region, and a value for the cosmological constant, the solution is a specification of the curvature of spacetime inside that region. It interacts with itself in the same way that all solutions of differential equations interact with themselves.

Consider a hanging chain, suspended from two points. The shape it makes is the solution of a differential equation. The shape 'interacts with itself' because the position of every part of the chain depends on the position of every other part of the chain. Ditto for the angle and curvature. The position of each link is determined by the force of gravity on it, and the forces applied to it by the links on either side, so the latter means that the chain 'bends itself'.

Unless I'm missing something, the quote means no more than that.
 
Neutrinos02 said:
What term in the Lagrangian will cause this effect? Is there something like an energy tensor of the gravitation field (I read something in the textbook of Landau and Lifschitz).
The only term in the action relevant for gravity and itself interactions is that containing the Ricci scalar. This term (together with the ##\sqrt{-g}## from the volume element) is not quadratic in the metric. This leads to non-linear terms in the field equations.
 
andrewkirk said:
Consider a hanging chain, suspended from two points. The shape it makes is the solution of a differential equation. The shape 'interacts with itself' because the position of every part of the chain depends on the position of every other part of the chain. Ditto for the angle and curvature. The position of each link is determined by the force of gravity on it, and the forces applied to it by the links on either side, so the latter means that the chain 'bends itself'.

Unless I'm missing something, the quote means no more than that.

Should it be interpreted in that way? I thougth about a self-interaction like the gluon field interacts with it self.
So that self-interaction is not satisfied for every differential equation. The electromagnetic field does not interact with itself (there is no photon-photon vertex in QED).

Orodruin said:
The only term in the action relevant for gravity and itself interactions is that containing the Ricci scalar. This term (together with the ##\sqrt{-g}## from the volume element) is not quadratic in the metric. This leads to non-linear terms in the field equations.

And the non-linearity gives rise to the self interaction?
 
Neutrinos02 said:
Should it be interpreted in that way? I thougth about a self-interaction like the gluon field interacts with it self.
So that self-interaction is not satisfied for every differential equation. The electromagnetic field does not interact with itself (there is no photon-photon vertex in QED).
And the non-linearity gives rise to the self interaction?
Yes. Non-linearities result in the field acting as its own source. By definition, this is self-interaction. It is how the self-interactions in QCD work as well.
 
Orodruin said:
Yes. Non-linearities result in the field acting as its own source. By definition, this is self-interaction. It is how the self-interactions in QCD work as well.

Thank you. This effect is given for all solutions of the einstein equation or only for the gravitational waves?

Now I read something that there is no need for the pseudotensor (which i mentioned in my question) and so I'm a bit confussed. Where can I find the energy of the gravitational field if I don't insert this pseudotensor. Landau worked in his textbook the second order corrections to this pseudotensor out and interpreted this as the self-interaction of the waves.
So do we need this pseudotensor in our theory?
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...

Similar threads

Back
Top