3
$\begingroup$

I was thinking about the following scenario:

Consider a particle which causes a metric $g_{\mu\nu}$ on an otherwise Minkowski spacetime (or any manifold). Now, consider another particle, somewhere in the vicinity of the first particle, which causes a metric $h_{\mu\nu}$ on a spacetime which would have been Minkowski if not for these two particles.

Then, what what would the metric in the vicinity of these two points be? I am guessing that it is: $$(g_{\mu\nu}-\eta_{\mu\nu})+(h_{\mu\nu}-\eta_{\mu\nu}) + \eta_{\mu\nu} = g_{\mu\nu}+h_{\mu\nu} - \eta_{\mu\nu}$$

Also, does the Riemann Curvature Tensor $R_{\mu\nu\rho}^\sigma$add up directly? I don't think it should because the Einstein tensor $G_{\mu\nu}$ does (I think) and it is dependent on the Ricci Curvature AND the spacetime metric tensor.

Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ You can do this addition at weak field, so long as your coordinates make the metric nonsingular (rectangular coordinates as the unperturbed metric), and add the usually negligible corrections perturbatively. $\endgroup$ – Ron Maimon Aug 22 '13 at 23:57
3
$\begingroup$

Unlike classical electromagnetism, General Relativity is highly nonlinear--this means that the gravitational field can serve as its own source. A consequence of this fact is that fields decidedly do not superpose, and you can get all sorts of effects even from vacuum relativity. The most notable of these effects are things such as Brill waves and Geons, where gravitational waves collide or collapse to form black holes. You can work out solutions where this happens even when the spaces initially are empty outside of the two regions before overlap occurs.

$\endgroup$
4
$\begingroup$

The metrics don't simply add together as you suggest. In fact there is no known solution for the metric when you have two point masses (thought there are approximate solutions). If there was it would make calculating the motion of binary black holes a great deal simpler than it currently is. The curvature has to be calculated numerically.

$\endgroup$
2
$\begingroup$

One way of phrasing this is that the nonlinearity of the equations means that you can't say that the sum (or difference) of two solutions to the Einstein field equations is also a solution. So even taking the delta with respect to the Minkowski metric, as you would like to do, isn't allowed (unless you work perturbatively as John alluded to).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.