Could someone please explain to me in simple words (i.e., without referring to forms on the frame bundle, etc) why the Bianchi identity is the relativistic generalisation of the momentum-conservation law?(adsbygoogle = window.adsbygoogle || []).push({});

Here comes my hypothesis, but I am not 100% convinced that it is correct. In Newtonian mechanics, we used to get momentum (and energy) conservation from the action's invariance under infinitesimal spatial and time displacements. In the GR, the Hilbert action should stay stationary under gauge-like changes of the metric, i.e., under infinitesimal displacements of the coordinates. Variation of the Hilbert action with respect to these entails:

G^{\mu\nu}_{ ; \nu} = 0 .

To get this, repeat verbatim eqns (94.5 - 94.7) in Landau & Lifgarbagez, with the Lagrangian changed to R, and T_{\mu\nu} changed to G_{\mu\nu}.

[Had we varied the Hilbert action with respect to NONgauge variations of the metric, we would have arrived to the equations of motion G^{\mu\nu} = 0. The difference stems from the fact that the NONgauge variations of the metric are all independent, up to symmetry. The gauge variations are dependent and, thus, must be expressed via the coordinate shifts, like in section 94 of Landau & Lifgarbagez.]

Clearly, G^{\mu\nu}_{ ; \nu} = 0 is equivalent to

R^{\mu\nu}_{ ; \nu} = 0 ,

which, in its turn, is identical to Bianchi. (Well, it follows from Bianchi, but I guess they are just identical.)

This line of reasoning seems to show that Bianchi (or, to be more exact, its corollary

R^{\mu\nu}_{ ; \nu} = 0) is the relativistic analogue to momentum conservation.

Is that true?

Can this conclusion be achieved by less cumbersome arguments?

Many thanks,

Michael

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# The Bianchi identity as a new incarnation of the momentum-conservation law

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