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

1. Jan 31, 2010

### Michael_1812

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?

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

Last edited: Jan 31, 2010
2. Jan 31, 2010

### Michael_1812

a relativistic counterpart to the angular-momentum-conservation law??

Here comes an even more wicked question.

Varying the Hilbert action with respect to gauge-like ripples of the metric, i.e., with respect to small shifts of the coordinate chart, we arrive at G^{\mu\nu}_{ ; \nu} = 0 , which seems to be a generalisation of the momentum-conservation law.

Now, what if we vary the Hilbert action with respect to gauge-like ripples of the metric generated by infinitesimal rotations? Evidently, that should yield a relativistic generalisation of conservation of the angular-momentum tensor. How does it look? Has anyone tried to do this?

Many thanks,
Michael

3. Jan 31, 2010

### DrGreg

[noparse]$$G^{\mu\nu}_{ ; \nu} = 0$$[/noparse]​

which gives

$$G^{\mu\nu}_{ ; \nu} = 0$$​

Use "itex" instead of "tex" within a paragraph. If you read this within 24 hours of your post, you can go back and edit it.

4. Feb 1, 2010

### Altabeh

$$R^{\mu\nu}_{ ; \nu} = 0$$ results in $$T^{\mu\nu}_{ ; \nu} = 0$$ which then you can easily get the momentum conservation law. But remember that this is the differential version of conservation law and we have to make an integral version out of it to get MC in GR which I think Landau and Lifgarbagez themselves do that in their book.

The angular momentum conservation can be derived by making use of the Landau-Lifgarbagez pseudotensor $${\mathfrak{t}}^{\mu}_{\nu}$$ and I've never seen something beyond this when wanting to provide the AMC.

AB

5. Feb 1, 2010

### George Jones

Staff Emeritus
Do you mean for vacuum spacetimes, like Schwarzschild and Kerr?

While $\nabla_\nu R^{\mu \nu} = 0 \Rightarrow \nabla_\nu G^{\mu \nu} = 0$ is always true, $\nabla_\nu G^{\mu \nu} = 0 \Rightarrow \nabla_\nu R^{\mu \nu} = 0$ is not, in general, true. The Einstein tensor is

$$G^{\mu \nu} = R^{\mu \nu} - \frac{1}{2} g^{\mu \nu} R,$$

so

$$\nabla_\nu G^{\mu \nu} = 0 \Rightarrow \nabla_\nu R^{\mu \nu} = \frac{1}{2} g^{\mu \nu} \frac{\partial R}{\partial x^\nu},$$

the (twice) contracted Bianchi identity. This is why Einstein used $G^{\mu \nu}$, not $R^{\mu \nu}$, in his field equation.

6. Feb 1, 2010

### Altabeh

Since I get uncomfortable by not giving a reasoning to show why $$\nabla_\nu R^{\mu \nu}=0 \Rightarrow \nabla_\nu G^{\mu \nu} = 0$$ is neccessarily true, I think the following would be interesting and logical:

$$\nabla_\nu G^{\mu \nu} = \nabla_\nu (R^{\mu \nu}-\frac{1}{2}g^{\mu \nu}R)$$
$$=\nabla_\nu (R^{\mu \nu}-\frac{1}{2}g^{\mu \nu}g^{\alpha\beta}R_{\alpha\beta})$$
$$= \nabla_\nu R^{\mu \nu} - \frac{1}{2}g^{\mu \nu}g^{\alpha\beta}\nabla_\nu R_{\alpha\beta}.$$

So if $$\nabla_\nu R^{\mu \nu}=0$$, then from the above calculation we must put

$$\nabla_\nu R_{\alpha\beta}=0.$$ (1)

to get

$$\nabla_\nu G^{\mu\nu}=0.$$ (2)

Eq. (1) means that each component of the Ricci tensor must vanish when being covariantly differentiated wrt any coordinates. Let's write down the second Bianchi Identity for $$\nabla_\nu R_{\alpha\beta}$$ or similarly $$\nabla_\nu R^{\kappa}_{\alpha\beta\kappa}$$:

$$R^{\kappa}_{\alpha[\beta\kappa ;\nu]}=0\rightarrow$$
$$R^{\kappa}_{\alpha\beta\kappa ;\nu}+ R^{\kappa}_{\alpha\nu\beta ;\kappa}+R^{\kappa}_{\alpha\kappa\nu ;\beta}=0,$$ (3)

where we used the symbol ";" in place of nabla. Here the last and first terms vanishes because of (1), thus leaving a vanishing second term on the left hand side of (3):

$$R^{\kappa}_{\alpha\nu\beta ;\kappa}=0.$$

This is itself equivalent to

$$R^{\kappa}_{\alpha\nu\beta ;\kappa}=R_{\alpha\nu ;\beta }-R_{\alpha\beta;\nu}=0.$$

So that from this last equation and (1) we finally get that our first assumption*, i.e. $$\nabla_\nu R^{\mu \nu}=0$$ neccessarily leads to (2).

* This is the logical part: we led to something true from something true.

AB

Last edited: Feb 1, 2010