How do you derive relativistic tensors in an orthonormal basis?

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SUMMARY

The discussion centers on deriving the Einstein tensor and stress-energy momentum tensor using an orthonormal basis rather than a coordinate basis for a traversable wormhole metric. The correct approach involves deriving basis vectors from the metric tensor and ensuring orthogonality, which is crucial for accurate tensor calculations. The steps include calculating the Christoffel symbols and Ricci tensor, but the focus shifts to changing bases to resolve unit discrepancies in the stress-energy tensor components. Understanding linear algebra, multilinear algebra, and differential geometry is essential for progressing in this topic.

PREREQUISITES
  • Linear algebra concepts, specifically understanding bases
  • Multilinear algebra, particularly the abstract definition of tensors
  • Differential geometry, including vector spaces associated with spacetime
  • Metric tensor properties and their implications in tensor calculus
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  • Study the derivation of basis vectors from the metric tensor in differential geometry
  • Learn about the properties and calculations of the Christoffel symbols in orthonormal bases
  • Research the process of changing bases for tensors and its implications in physics
  • Explore the relationship between curvature tensors and nonholonomic bases in curved manifolds
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Physicists, mathematicians, and students specializing in general relativity, differential geometry, and tensor calculus who are looking to deepen their understanding of tensor derivation in curved spacetime.

space-time
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I have been recently trying to derive the Einstein tensor and stress energy momentum tensor for a certain traversable wormhole metric. In my multiple attempts at doing so, I used a coordinate basis. My calculations were correct, but the units of some of the elements of the stress energy momentum tensor were wrong because of the fact that angles don't have units of length like the radial coordinate and the temporal coordinate (ct). The metric is a spherical basis by the way. Anyway, I was told that I should use an orthonormal basis to derive the tensors as opposed to a coordinate basis.

Now, in a coordinate basis, the steps to derive your relativistic tensors (assuming you already have your metric tensor and inverse metric tensor) are as follows:

1. Plug your metric and inverse metric tensors into the Christoffel symbol formula to get your Christoffel symbols: Γmij = ½ gmk [ ∂gki/∂xj + ∂gjk/∂xi - ∂gij/∂xk ]

2. Plug your Christoffel symbol into the Ricci tensor formula:
Rbv= ∂Γavb/∂xa - ∂Γaab/∂xv + ΓaacΓcvb - ΓavcΓcab

3. Contract your Ricci tensor using an inverse metric tensor to get your curvature scalar:
R= gbvRbv

Then all you have to do is plug in all of these tensors into the Einstein field equations and you get the Einstein tensor. This was all in a coordinate basis.Apparently in an orthonormal basis however, the steps are as follows:

1. Get your metric tensor.

2. Derive some basis vectors.

How to derive those basis vectors:
Do gijdxidxj where dxi and dxj are treated as vectors instead of just coordinates. You must then make sure that this expression equals 0 whenever i ≠ j and this expression equals 1 whenever i = j .

Example: If my metric tensor component g11 = sin(θ) then my basis vector e1 must be < 0, 1/squrt(sin(θ)) , 0, 0 > . This is because if you dot product this basis vector with itself and then multiply that dot product by g11, then the result will be 1. Of course this is assuming that your basis vector is orthogonal to the other ones.

In short, that is how you derive your basis vectors (so I am told).

Now my question to you is:

What do I do next after I have derived these basis vectors if I am trying to get to stuff like the Christoffel symbol or any of the curvature tensors?
 
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If by an orthonormal coordinate system (I would not say "basis"), you mean your coordinate axes are straight lines that are orthogonal at every point, then the Christoffel symbols are all 0 and so the curvature is 0- such a coordinate system can exist ONLY in a "flat" space.
 
HallsofIvy said:
If by an orthonormal coordinate system (I would not say "basis"), you mean your coordinate axes are straight lines that are orthogonal at every point, then the Christoffel symbols are all 0 and so the curvature is 0- such a coordinate system can exist ONLY in a "flat" space.

His question comes from an earlier thread. There, we were talking about 'nonholonomic', i.e., noncoordinate, orthonormal bases on the tangent spaces. These can exist in curved manifolds since they don't have an associated coordinate system. The thing that fails is that they don't commute and in fact the curvature can be expressed in terms of the non-zero commutators. If all the commutators vanish, then the curvature is zero, and this is consistent with what you say, since in this case there exists a coordinate system associated to the orthonormal basis and the usual formulas for the curvature of course imply it's zero.

As I told you, @space-time, you don't need to calculate the whole curvature again with this new method if your goal is only to solve the units issue you had. This is because you already calculated the components of the SET in the coordinate basis; so, you only need to make a change of basis in order to obtain the components of the SET in the orthonormal basis. I already did the explicit calculation in the other thread.

The basic ingredients you need are some concepts from linear algebra (what a basis is), some others from multilinear algebra (the abstract algebraic definition of tensors and how to make change of bases there), and finally a basic fact from differential geometry (that at each point of spacetime you associate a vector space and that these vectors and orthonormal basis we are talking about actually live in these spaces). If you don't understand these notions first, we can't make more progress here. Google all this, ask in the geometry forum, read about it, and then come back.
 

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