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?
Tell me more
×
Physics Stack Exchange is a question and answer site for
active researchers, academics and students of physics. It's 100% free, no registration required.
|
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.
If that means a stable closed orbit, then I don't know. I suspect not, but can't prove it in general. |
|||||
|
