Why should the covariant derivative of the metric tensor be 0 ?

In summary, the GR assumption that the length and angle between vectors would not change under parallel transport can be relaxed, giving you more freedom in your theory. This is the simplest possibility and is based on the assumption that raising and lowering indices are the same thing.
  • #1
lalbatros
1,256
2
That's a crucial point of GR !
And I have always problems with that.

Back to the basics, with your help.

Thanks

Michel
 
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  • #2
The vanishing of the nonmetricity tensor in GR has to with the natural assumption that under parallel transport of vectors, neither the length of one vector, nor the angle between 2 such vectors would change.

It's instructive to perform one of the 2 calculations referred to in the phrase above.

Daniel.
 
  • #3
This assumption can probably be relaxed. But this gives you more freedom. I would say that this is merely the simplest (and geometrically the most natural) possibility.
 
  • #4
A quick algebraic reason is that raising (lowering) indices then differentiating should be the same thing as differentiating then raising (lowering) indices.
 
  • #5
That condition is also needed for the "Fundamental Theorem of (Pseudo-) Riemannian Geometry".
 
  • #6
As far as I know, it's a defintional thing. If one adopts the defintion that two sides of a quadralateral are parallel when opposite sides have the same length, one can use a geometric construction based on this defintion of the parallelness of two different vectors to define parallel transport (also, the more elegant gometric construction known as "Schild's ladder") to perform covariant differentiation.

(Note that given a metric, one can measure the length of the sides of a parallelogram.)

Schild's ladder is discussed in MTW (but not many other textbooks include it).

I've never quite understood torsion, though, or torsion-based theories of gravity.
 
  • #7
Here's another way to look at it.
Given the metric tensor, if you don't choose the unique torsion-free metric-compatible derivative operator, you have to specify some more structure (say, the nonmetricity tensor or the torsion tensor or something equivalent) to pick out the derivative operator you prefer. From a physics point of view, you'd probably want to physically justify the choice.
 
  • #8
pervect said:
I've never quite understood torsion, though, or torsion-based theories of gravity.

I also don't know much know about torsion-based theories of gravity, but I think one story goes something like the following.

Some theories relate torsion to particle spin. This is basically a coupling of spin to spacetime geometry, so that what constitutes a geodesic depends on a particle's spin.
 
  • #9
My understanding is that, if there is no force acting on your drawing compass, physically you'd expect the angle it subtends to remain constant. Mathematically: [itex]0 = \nabla_a (\boldsymbol{guv}) = \nabla_a \boldsymbol g[/itex] QED.

To define the connection you also assume the covariant derivative is torsion free on scalar fields, and I'm not sure what that means physically.
 

1. What is the significance of the covariant derivative of the metric tensor being 0?

The covariant derivative of the metric tensor being 0 implies that the metric tensor is constant throughout the space or spacetime. This means that the distances and angles between any two points in the space or spacetime remain the same, regardless of the coordinates used to measure them.

2. How does the covariant derivative of the metric tensor relate to the curvature of space or spacetime?

The covariant derivative of the metric tensor being 0 is a necessary condition for a space or spacetime to be flat or have zero curvature. This means that the space or spacetime is Euclidean, and the rules of Euclidean geometry apply.

3. Can the covariant derivative of the metric tensor be non-zero in certain cases?

Yes, the covariant derivative of the metric tensor can be non-zero in cases where the space or spacetime is curved. This means that the distances and angles between two points will vary depending on the coordinates used to measure them, and the rules of non-Euclidean geometry will apply.

4. What is the mathematical definition of the covariant derivative of the metric tensor?

The covariant derivative of the metric tensor is defined as the partial derivative of the metric tensor with respect to one of its indices, plus the Christoffel symbols multiplied by the metric tensor. This can be written as Dgαβ/dxγ = ∂gαβ/∂xγ + Γαγρgρβ, where Γαγρ are the Christoffel symbols.

5. How is the covariant derivative of the metric tensor used in general relativity?

In general relativity, the covariant derivative of the metric tensor is used to define the curvature tensor, which is a measure of the curvature of space or spacetime. It is also used in the Einstein field equations, which describe how matter and energy affect the curvature of space or spacetime.

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