# Three questions related to the principle of general covariance in GR

1. Oct 30, 2009

### Michael_1812

1. A (presumably) simple question:

We are used to think that the affine connections emerge whenever one wants to
differentiate a vector (tensor, spinor) on a curved manifold in general relativity. Now suppose that we are still on a flat background of special relativity, though in a curvilinear grid. Will the connections show up there?

2. Now comes a tougher one:

According to the (weak) equivalence principle, at each point of the general-relativity manifold there exists a reference frame (the freely falling one) wherein the motions are the same as in an inertial frame of the special relativity.

According to the principle of general covariance, all the fundamental laws of physics stay form-invariant under a transition between any two coordinate charts (provided the derivatives in these laws are covariant and are taken with respect to a true scalar - the interval). Stated alternatively, the fundamental laws can be expressed in a coordinate-independent way.

How do these two fundamental principles square?

I think it was Weinberg's book where I saw a sentence saying that the principle of general
covariance is a mathematical implementation of the equivalence principle. Why is that??

3. ... and the toughest one:

Above, when formulating the equivalence principle, I used the wording "at each point of the general-relativity manifold". Is this wording at all legitimate? I mean: is it right to assume that each point of the manifold is a physical event? I am asking, because we have gauge invariance, so it may happen that a whole orbit corresponds to one and the same event.
Please correct me if I am wrong.

Many thanks for your time and help,
Michael

Last edited: Oct 30, 2009
2. Oct 30, 2009

### George Jones

Staff Emeritus
Yes. And they even show up in Galilean/Newtonian mechanics in curvilinear coordinates.

Sorry for not getting back to you. I kept putting this off in the hope of writing a more detailed answer than I give above.

3. Oct 30, 2009

### atyy

I believe Weinberg does say GC=EP BUT he does not define GC as you do, and says that GC as normally defined is physically empty. (But check his book to be sure.)

I think it's something like:
GC (as normally defined, not Weinberg's def) is physically empty.
EP implies gravity can be geometrically formulated, but does not lead uniquely to GR - see eg. Newton-Cartan theory or Nordstrom's second theory.
GR requires one more principle: "no prior geometry" (MTW).
And GR in full form does not have particles travelling on geodesics - it's a field with "metric" symmetry, whose equations of motion are derived from a Lagrangian, and matter is just other fields, which are also derived from a Lagrangian.

Edit: In addition to "no prior geometry", GR also requires an assumption about the highest order derivative that enters the field equations.

Last edited: Oct 30, 2009
4. Oct 30, 2009

### atyy

It's legitimate. Strictly speaking, you should say isometry equivalence class of manifolds and metrics or something like that, but it's taken for granted you can just make true statements about one member of the equivalence class. See eg. Ellis & Hawking p56.

Last edited: Oct 30, 2009
5. Oct 30, 2009

### atyy

Sean Carroll's http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll4.html gives more detail. A quick summary of the essential points. Suppose we already decided from the EP that the "metric" should become a field that models gravity. Then we should presumably seek a Lagrangian whose terms contain the metric. Because the Lagrangian is a scalar, a further restriction on these terms is that they should be scalars. This allows Lagrangians such as Eq 4.76. To obtain the Hilbert action 4.55, a further restriction is made to include only terms containing up to 2 derivatives of the metric.

Also useful:
Luca Bombelli, http://www.phy.olemiss.edu/~luca/Topics/grav/higher_order.html [Broken].
Donoghue, http://arxiv.org/abs/gr-qc/9512024, especially Eq 21.

Last edited by a moderator: May 4, 2017
6. Oct 30, 2009

### atyy

Hmmm, I didn't realise that there are many subtleties about the relationship between metric theories of gravity and the equivalence principle:

http://arxiv.org/abs/0707.2748
Theory of gravitation theories: a no-progress report
Thomas P Sotiriou, Valerio Faraoni, Stefano Liberati

Last edited: Oct 30, 2009
7. Jan 31, 2010

### Michael_1812

particles (not??) following geodesics

Dear atyy,
Thank you for answering my questions.
You say: "And GR in full form does not have particles travelling on geodesics..."
What do you mean by this? How can a free particle deviate from its geodesic?
(Or did you imply that particles were interacting with fields other than gravity?)
Many thanks,
Michael

8. Jan 31, 2010

### Michael_1812

The Bianchi identity as a reincarnation of the momentum-conservation law

Guys,

Can anyone 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 stays 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 with 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.]

Now, 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 means?

Learning the math sometimes is easier than grasping the physics beneath it...

Great many thanks,
Michael

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

### bcrowell

Staff Emeritus
This thread is three months old and on a different topic. Your new question is one I'd be interested in hearing people's thoughts on, but I think you should really start a new thread for it rather than posting it in this old thread.

10. Jan 31, 2010

### Michael_1812

which of my two questions do you have in mind - geodesics or Bianchi?

11. Jan 31, 2010

### bcrowell

Staff Emeritus
Bianchi was the one I was interested in.

12. Jan 31, 2010

### Michael_1812

I have re-posted the latter two questions as new threads