Is the Contravariant Derivative Defined by Contracting with the Metric?

Click For Summary

Discussion Overview

The discussion centers around the definition and properties of the contravariant derivative in the context of general relativity (GR). Participants explore whether the contravariant derivative can be defined similarly to the covariant derivative and the role of the metric in this definition.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions whether the notation v^{a;b} represents the contravariant derivative and seeks clarification on its definition, particularly in relation to the Christoffel symbols and the metric.
  • Another participant suggests that the contravariant derivative can be expressed as v^{a;b} = g^{bc} v^a{}_{;c}, implying a relationship between raising indices and the covariant derivative.
  • A third participant discusses the meaning of "covariant" in the context of tensor transformation and the necessity of the covariant derivative to ensure tensorial behavior under coordinate transformations.

Areas of Agreement / Disagreement

Participants express differing views on the definition and calculation of the contravariant derivative, indicating that multiple competing perspectives remain unresolved.

Contextual Notes

There are limitations regarding the assumptions made about the definitions of the contravariant and covariant derivatives, as well as the dependence on the metric for raising and lowering indices. The discussion does not resolve these aspects.

Pacopag
Messages
193
Reaction score
4
Hi everyone;
I'm new to both PF and GR, so please bear with me if I'm not being very clear, or using standard syntax and such. Here is my question.

Given a vector v^a, the covariant derivative is defined as v^a _;b = v^a _,b + v^c GAMMA^a _bc.
(here I'm using ^ before upper indices and _ before lower indices).
The object I'm interested in now is v^a;b (where the whole thing a;b is upper). Is this called the contravariant derivative? Is there a similar definition in terms of the Christoffel symbol? or can we only obtain it from contracting v^a _;b with the metric? I can't seem to find the definition of this object in any books, and when I try to do my calculation via contraction with the metric, I'm getting the wrong answer.

Thanks.
 
Physics news on Phys.org
You might want to look at the Latex capabilities of PF. Quote this post for examples.

The short answer is that you should already know how to or be able to find out how to raise and lower indices with tensors. Then

v^{a;b} = g^{bc} v^a{}_{;c}
 
Ok. Thanks pervect.
 
The term "covariant" is used in the sense of how the components of a tensor transform. But covariance also means that the laws of physics can be written down tensorially, and so have a certain behaviour under coordinate transformations.

The covariant derivative is the demand that we use a derivative in our physical formulation that transforms tensorially. The partial derivative doesn't, so we need an extra term. This is the connection, but to specify it you have to demand certain properties on it. The same happens in the standard model: you impose a local symmetry, and demand that the derivative of your fields transforms exactly as the field itself under the symmetry transformation.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
2
Views
2K
  • · Replies 50 ·
2
Replies
50
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K