Covariant derivative and geometry of tensors

In summary, the conversation discusses the difficulty in understanding general relativity and differential geometry concepts, specifically the notion of smooth vector fields and the use of tensor products and contractions. The conversation also touches on the issue of parallel transport and the limitations of using the standard trick of passing to R^n. Suggestions for understanding these concepts are also requested.
  • #1
avorobey
14
0
I'm trying to teach myself GR from Wald's General Relativity, and it's very tough going. I do have basic knowledge of differential geometry, but I think my geometric intuition is next to nonexistent. I'd very much appreciate some help in understanding several basic questions, or pointers to texts with good explanations.

1. The text stresses that in the absence of something like covariant derivative, we have no way of connecting tangent spaces of different points with each other. What confuses me is how can we have a notion of a smooth vector field in that case? I understand the definition of a smooth vector field (tangent vectors acting as derivations turn smooth functions to smooth functions); since we have it, doesn't it mean that we do have a notion of what it means for tangent vectors from two nearby tangent spaces to be very close?

Also, why doesn't the standard trick of passing to R^n work? Given two nearby points on the manifold, why can't we translate them into R^n by a chart and identify their tangent spaces in R^n? If this leads to different results in different charts, why wouldn't the smoothness of chart transformations make everything alright, and is there an instructive example to show this failure?

2. Is there a good way of understanding tensor product and contraction geometrically? I can follow all the indices around, I just completely fail to understand what it *means*, and it's very frustrating. For example, Wald defines parallel transport of vector [tex]v^{b}[/tex] along a curve with tangent [tex]t^{a}[/tex] by the equation [tex]t^{a}\nabla_{a}v^{b} = 0[/tex]. But I don't understand why this should capture the notion of parallel transport. [tex]\nabla_{a}v^{b}[/tex] is some (1,1)-tensor, and while I can write out the definition of what it means to multiply it by [tex]t^{a}[/tex] and contract over index a, I don't really understand what it means. Even worse, for a general tensor T with both covariant and contravariant indices, I have no clue how to imagine [tex]t^{a}\nabla_{a}T = 0[/tex]. Is there a helpful way to visualize/understand this?

Many thanks in advance!
 
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  • #2
avorobey said:
1. The text stresses that in the absence of something like covariant derivative, we have no way of connecting tangent spaces of different points with each other. What confuses me is how can we have a notion of a smooth vector field in that case? I understand the definition of a smooth vector field (tangent vectors acting as derivations turn smooth functions to smooth functions); since we have it, doesn't it mean that we do have a notion of what it means for tangent vectors from two nearby tangent spaces to be very close?

No, because how close they are depends on a coordinate system. When you have a coordinate derivative, you can compare tangent vectors at different points depending on a path that connects these two points but is independent of a a local coordinate system.

Also, why doesn't the standard trick of passing to R^n work? Given two nearby points on the manifold, why can't we translate them into R^n by a chart and identify their tangent spaces in R^n? If this leads to different results in different charts, why wouldn't the smoothness of chart transformations make everything alright, and is there an instructive example to show this failure?

Smoothness does not depend on a chart, but other things do depend.

2. Is there a good way of understanding tensor product and contraction geometrically? I can follow all the indices around, I just completely fail to understand what it *means*, and it's very frustrating.

Tesnor products of vector space is a part of multilinear algebra. Some of that has a geometrical interpretation in terms of Grassmann (or Clifford) algebra. But otherwise algebra is just algebra - a very useful tool.

For example, Wald defines parallel transport of vector [tex]v^{b}[/tex] along a curve with tangent [tex]t^{a}[/tex] by the equation [tex]t^{a}\nabla_{a}v^{b} = 0[/tex]. But I don't understand why this should capture the notion of parallel transport. [tex]\nabla_{a}v^{b}[/tex] is some (1,1)-tensor, and while I can write out the definition of what it means to multiply it by [tex]t^{a}[/tex] and contract over index a, I don't really understand what it means. Even worse, for a general tensor T with both covariant and contravariant indices, I have no clue how to imagine [tex]t^{a}\nabla_{a}T = 0[/tex]. Is there a helpful way to visualize/understand this?

It is good to check some other book where these concepts are explained in detail, especially parallel transport along along a path. I like Bishop, Crittenden "Geometry of manifolds", but you will certainly find other good books.
 
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  • #3
avorobey said:
Also, why doesn't the standard trick of passing to R^n work?

It does, at least formally. See, for example, Dirac, General Theory of Relativity, section 6, where he uses an embedding to define parallel transport. Try searching in the forum for some interesting discussions of how big an n you need. (Oops, sorry, you're right, OP is not asking about that.)

[tex]t^{a}\nabla_{a}v^{b} = 0[/tex]

[tex]t^a\nabla_a[/tex] is how one writes out the directional derivative along the curve in Wald's s notation. It's the same as

[tex]t^a \nabla_{\frac{\partial}{\partial x^a}} = \nabla_{t^a\frac{\partial}{\partial x^a}}[/tex]

in Koszul notation, where it's easier to see that it's the covariant derivative in the direction of the tangent vector field..
 
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  • #4
Daverz said:
It does, at least formally. See, for example, Dirac, General Theory of Relativity, section 6, where he uses an embedding to define parallel transport.

I don't think this is what the OP was asking about when he asked his question in the context of smoothness. Moreover the embedding trick is not really "geometrical", because it works on an infinitesimal level as regards the law of the parallel transport.
 
  • #5
arkajad said:
Tesnor products of vector space is a part of multilinear algebra. Some of that has a geometrical interpretation in terms of Grassmann (or Clifford) algebra. But otherwise algebra is just algebra - a very useful tool.

Right, I understand this. What I meant is, about the most typical thing that I see done with two tensors in Wald's book so far, is multiplying them and contracting the product on one of the indices. For example, this is done all the time to raise/lower indices using the metric; but while I understand the formalism and see that it e.g. changes an (n,m) tensor into (n-1,m+1), I don't understand the *meaning* of the operation.

I worked out that if I have a dual vector and a vector, and if I treat them as tensors and multiply then contract those tensors, what I get is the same as simply acting on the vector with the dual vector. That's about the simplest example imaginable, and that's now clear to me. But I completely lack intuition into what it means e.g. to tackle on the metric to an (n,m) tensor, contracting on one index. Sure, formally it's a (n-1,m+1)-tensor, a multilinear function on n-1 V*'s and m+1 V's, and given a sample set of n-1 dual vectors and m+1 vectors, I can write out the formula of the value of the function as a sum over one basis of a product of the metric values and the original tensor's values. But that formal sum is all I have; I can't help thinking I should have some kind of (vague? geometric? algebraic?) intuitive understanding of what was actually *done*.

Anyway, I wanted to thank you, and Daverz, for your kind answers and some pointers. I've also started reading Schutz's GR book, and it seems to go into more details setting up tensors and covariant derivatives (only just getting to that part); perhaps it'll clear things up for me. I've looked at Bishop & Crittenden and it's also very nice, though perhaps a bit more mathematically-minded than I need for getting through Wald (e.g. they discuss everything in terms of arbitrary fiber bundles, while Wald only looks at the tensor bundle w/o calling it so); but I'll try looking things up there when they're too vague for me elsewhere.
 
  • #6
avorobey said:
For example, this is done all the time to raise/lower indices using the metric; but while I understand the formalism and see that it e.g. changes an (n,m) tensor into (n-1,m+1), I don't understand the *meaning* of the operation. .

I think it all boils down to the question what is the geometrical (or whatever) *meaning* of the trace of a matrix. All these contractions are generalized partial traces. These are *invariants*. In physics we are trying to associate some geometrical and geometrical content to the invariants and to invariant operations. For some it is easy, for some it is not so easy. In quantum theory we are even taking traces over a continuous index, like x or p - we call then "integrals".
 
  • #7
avorobey said:
I'm trying to teach myself GR from Wald's General Relativity, and it's very tough going. I do have basic knowledge of differential geometry, but I think my geometric intuition is next to nonexistent. I'd very much appreciate some help in understanding several basic questions, or pointers to texts with good explanations.

You might try the Stanford Leonard Susskind lectures on General Relativity (GR starts at lecture 28):

http://academicearth.org/courses/foundations-of-modern-physics

or you can just google: "Stanford Susskind Lectures" and more places to get or view the video will show up. They are also available via iTunes University.

I found the discussion of parallel transport to be easy to understand. His treatment of covariant derivatives is also good.
 

What is a covariant derivative?

A covariant derivative is a mathematical operation that describes how a tensor field changes as one moves along a curve or manifold. It takes into account the curvature and geometry of the space in which the tensor field is defined.

How is a covariant derivative different from an ordinary derivative?

An ordinary derivative measures the rate of change of a function with respect to a single variable. A covariant derivative, on the other hand, takes into account the influence of the underlying space on the function, making it more suitable for curved spaces or manifolds.

What is the relationship between tensors and the geometry of a space?

Tensors are mathematical objects that describe the geometric properties of a space. They provide a way to measure distances, angles, and other geometric quantities in a way that is independent of the coordinate system used to describe the space.

How is a covariant derivative used in general relativity?

In general relativity, the curvature of spacetime is described by the Einstein field equations. The covariant derivative is used to calculate the curvature of spacetime and how it changes in the presence of matter and energy.

What is meant by the "tensorial nature" of a covariant derivative?

A covariant derivative is said to have a tensorial nature because it transforms like a tensor when the coordinates of the space are changed. This means that the covariant derivative is independent of the coordinate system used to describe the space, making it a useful tool in tensor calculus and differential geometry.

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