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What the mechanics arises if to take limit of general relativity with massless particles interacting with strong fields? Suppose there a system of attracting particles that have zero rest mass. What would be their interaction in GR? Can they form stable orbits around each other?

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Related: physics.stackexchange.com/q/10759/2451 –  Qmechanic Dec 6 '12 at 14:02
    
Either this question makes no sense, or the answer is the post-Newtonian formalism. If it's the latter, then the answer is yes, because the zeroth order term in PPN is Newtonian gravity. –  Jerry Schirmer Dec 6 '12 at 14:22
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What the mechanics arises if to take limit of general relativity with massless particles interacting with strong fields? Suppose there a system of attracting particles that have zero rest mass. What would be their interaction in GR?

This is standard GTR with the restriction that the energy-momentum tensor is of the "pure radiation" (aka "null dust") form $$\begin{eqnarray*}T^{\mu\nu} = \Phi^2 k^\mu k^\nu, &\;\text{where}\;& k_\mu k^\mu = 0\end{eqnarray*}.$$ This could, but need not, arise from a Maxwell field, and represents radiation in lightlike direction $k$. Since the trace is $T = 0$, the Ricci curvature is also $R_{\mu\nu} = \Phi^2 k_\mu k_\nu$ in units of $8\pi G = c = 1$. Thus, any spacetime whatsoever with Ricci curvature in this form satisfies your condition, the most trivial case being a vacuum (having zero radiation intensity), e.g., Schwarzschild spacetime

If $k$ is a geodesic field and a Killing field, then it is automatically twist-free and the field equations are reducible to a system of two-dimensional Poisson equations. Then either the solution is a $pp$-wave ($k_{\mu;\nu} = 0$) and can be put in the form $$\mathrm{d}s^2 = -H(u,x,y)\mathrm{d}u^2 - 2\mathrm{d}u\mathrm{d}v + \mathrm{d}x^2 + \mathrm{d}y^2$$ for any smooth function $H$, or else ($k_{\mu;\nu} \neq 0$) it can be put in the form $$\mathrm{d}s^2 = x^{-1/2}(\mathrm{d}x^2+\mathrm{d}y^2) - 2x\mathrm{d}u\left[\mathrm{d}v + M(x,y,u)\mathrm{d}u\right]$$ where $(xM_{,x})_{,x}+xM_{,yy} > 0$. The simplest nontrivial example of the latter case is the van Stockum dust.

For details, additional constraints for Maxwell fields, and other conditions, see Exact Solutions of Einstein's Field Equations by Stephani, et al.

Can they form stable orbits around each other?

If that means a stable closed orbit, then I don't know. I suspect not, but can't prove it in general.

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What happens if other fields also affect such massless particle besides gravity? My question was motivated by a preon model where preons were taken to be massless but actively interacting. –  Anixx Dec 7 '12 at 10:51
    
Having only massless particles constrains the $T^{\mu\nu}$ to have a certain form, but is otherwise agnostic about how they interact, and so can include other non-gravitational interactions. I don't think anything more can be said in the general case, but having forces other than gravity has may make stable (non-gravitational) lightlike orbits. But I'm not familiar with how preons are supposed to work, sorry. –  Stan Liou Dec 7 '12 at 11:15
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