Levi-Civita Connection & Riemannian Geometry for GR

• binbagsss
In summary: Anyways, there is a structure that IS anti-commutative that sometimes we use for manifolds, and that is called a symplectic structure. The basic quantity of interest there; however, is not called a metric, but the symplectic form. We use this for, e.g. geometric Hamiltonian mechanics.

binbagsss

Conventional GR is based on the Levi-Civita connection.

From the fundamental theorem of Riemann geometry - that the metric tensor is covariantly constant, subject to the metric being symmetric, non-degenerate, and differential, and the connection associated is unique and torsion-free - the connection can be determined by the metric from a relativity simple formula.

BUT , doesn't whether the metric is symmetric or not, depend upon the choice of coordinates when specifying the metric. E.g- Schwarzschild metric given in spherical polar coordinates is diagonal, and so symmetric.
However, the extended Schwarzschild metric in Eddington-Finkelstein coordinates is not diagonal, contains a dudr term, but not a drdu term. Does this mean that for the metric in Eddington-Finkelstein coordinates , the connection can no longer be simply computed from the metric?

The metric is always symmetric. This is in the definition of a metric.

When people write out the metric in the familiar ##ds^2=-dt^2+...## form, they are actually writing out the contraction ##ds^2=g_{ij}dx^i dx^j##. But since ##dx^idx^j=dx^jdx^i## the ##g_{ij}## and ##g_{ji}## terms just get added together so you don't see both terms, just the combination ##g_{ij}+g_{ji}=2g_{ij}##.

Matterwave said:
The metric is always symmetric. This is in the definition of a metric.
Is this true for any geometry and not just Riemannian?

binbagsss said:
Is this true for any geometry and not just Riemannian?

If the metric were not symmetric, then the dot product of two vectors would not be commutative.

Mathematically, inner products are symmetric by definition. In a coordinate system you are merely expressing the inner product ( in fact any tensor) in terms of the coordinates. You are not changing it in any way.

Nugatory said:
If the metric were not symmetric, then the dot product of two vectors would not be commutative.

It would be anti-commutative , like Grassmann variables...
But wouldn't that make a very weird universe?? It blows my mind to think that $r \theta = - \theta r$

ChrisVer said:
It would be anti-commutative , like Grassmann variables...
But wouldn't that make a very weird universe?? It blows my mind to think that $r \theta = - \theta r$

It wouldn't have to be "anti-commutative" unless it were totally anti-symmetric. A general tensor will have both symmetric and anti-symmetric parts. Usually the term "anti-commutative" means that variables anti-commute ##\{a,b\}=0##.

Anyways, there is a structure that IS anti-commutative that sometimes we use for manifolds, and that is called a symplectic structure. The basic quantity of interest there; however, is not called a metric, but the symplectic form. We use this for, e.g. geometric Hamiltonian mechanics.

1. What is the Levi-Civita connection?

The Levi-Civita connection is a mathematical tool used in Riemannian geometry to define a unique way of differentiating vector fields on a curved manifold. It allows for the calculation of covariant derivatives, which are essential in understanding the curvature and geodesic paths on a manifold.

2. How does the Levi-Civita connection relate to General Relativity?

In General Relativity, the fundamental equations describe how matter and energy affect the curvature of spacetime. The Levi-Civita connection is used to calculate the curvature of spacetime, which is crucial in understanding the behavior of matter and energy in the presence of gravity.

3. What is Riemannian geometry?

Riemannian geometry is a branch of mathematics that studies curved spaces, also known as manifolds. It is the mathematical framework used to describe the curvature of spacetime in General Relativity. It involves concepts such as distances, angles, and geodesic paths, which are essential in understanding the behavior of objects in curved spaces.

4. Why is Riemannian geometry important in General Relativity?

Riemannian geometry is crucial in General Relativity because it provides the mathematical tools needed to describe the curvature of spacetime. This curvature is what determines how matter and energy move and interact in the presence of gravity. Without Riemannian geometry, it would be impossible to understand the fundamental principles of General Relativity.

5. How are the Levi-Civita connection and Riemannian geometry used in practical applications?

The Levi-Civita connection and Riemannian geometry have many practical applications, including in physics, engineering, and computer science. In physics, they are used to understand the behavior of matter and energy in the presence of gravity. In engineering, they are used in fields such as robotics and navigation systems. In computer science, they are used in the development of algorithms and data analysis techniques.