- #1

GR191511

- 76

- 6

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter GR191511
- Start date

In summary, the covariant derivative differs from the partial derivative with respect to the coordinates only because the basis vectors change.

- #1

GR191511

- 76

- 6

Physics news on Phys.org

- #2

malawi_glenn

Science Advisor

Homework Helper

- 6,735

- 2,458

The book does the example of polar coordinates just a few pages earlier.

- #3

- 24,488

- 15,026

$$\overleftrightarrow{T} = \vec{\nabla} \otimes \vec{V} = \vec{b}^i \otimes \partial_i (V^j \vec{b}_j) = \partial_i V^j (\vec{b}^i \otimes \vec{b}_j),$$

where ##\vec{b}^i## denotes the dual basis of ##\vec{b}_j##.

Now if you introduce an arbitrary basis ##\vec{B}_k##, which depends on the position, you get

$$\overleftrightarrow{T} = \vec{B}^i \otimes (\vec{B}_j\partial_i V^j + V^j \partial_i \vec{B}_j).$$

Now you define the connection symbols by

$$\partial_i \vec{B}_j = \vec{B}_k {\Gamma^k}_{ij},$$

and you see that these simply come from the position dependence of your basis (and dual basis) vectors.

- #4

Ibix

Science Advisor

- 12,383

- 14,402

Partial derivatives tell you how the vector components change as you move. Covariant derivatives also account for the fact that the basis vectors might be different, so tell you how the vector has actually changed (subject to fine print about what "actually changed" means, which is what the connection coefficients encode).

For example, consider flying a great circle path around the Earth, one not passing through the poles or along the equator. Clearly you are in some sense traveling straight forward (so ##\nabla_av^b=0##), yet your compass bearing varies between due East and whatever the inclination of the circle is, so in longitude/latitude basis vectors the components change (so ##\partial_av^b\neq 0##).

- #5

- 22,106

- 13,604

At this point of the book the reader is not yet familiar with curved spaces. The chapter in question is called ”Preface to curvature”. I don’t have a copy readily at hand, but if I do not misremember what is being done is exclusively working in Euclidean space similar to what I do in chapter 2 of my book. With a generally position dependent vector basis ##\vec E_a##, any vector can be written as ##\vec v = v^a \vec E_a##. Taking the derivative of this in the ##b## direction would lead toIbix said:For example, consider flying a great circle path around the Earth, one not passing through the poles or along the equator.

$$

\partial_b \vec v = \vec E_a \partial_b v^a + v^a \partial_b \vec E_a

$$

by the product rule for derivatives. Since we can write ##\partial_b \vec E_a = \Gamma_{ba}^c \vec E_c## this results in

$$

\partial_b \vec v = (\partial_b v^a + \Gamma_{bc}^a v^c) \vec E_a \equiv (\nabla_b v^a) \vec E_a.

$$

The point being that the partial derivatives ##\partial_b v^a## of the vector field components are generally not the components of ##\partial_b\vec v## unless the basis is constant.

An example would be the vector field ##\vec v = 2\vec E_r## in polar coordinates. It’s components are constant but the direction of ##\vec E_r## depends so the position so generally the vector field ##\vec v## is not constant. The discrepancy is described by the change of the basis vector ##\vec E_r##, which is encoded in the ##\Gamma_{br}^a##.

- #6

Ibix

Science Advisor

- 12,383

- 14,402

Sure, but OP has recently been asking questions based on much later in the book, so I suspect is currently on a re-read.Orodruin said:At this point of the book the reader is not yet familiar with curved spaces.

Covariant and partial derivatives are mathematical operations used in general relativity to describe how quantities change in curved spacetime. They are used to calculate the rate of change of a quantity with respect to a coordinate system or along a specific direction.

Covariant derivatives take into account the curvature of spacetime, while partial derivatives do not. This means that covariant derivatives are better suited for describing changes in curved spacetime, while partial derivatives are better for describing changes in flat spacetime.

In general relativity, spacetime is described as being curved due to the presence of mass and energy. This means that traditional methods of calculating changes in quantities, such as using partial derivatives, do not accurately describe the behavior of objects in this curved spacetime. Covariant derivatives take into account this curvature and provide a more accurate description of how quantities change in general relativity.

Covariant derivatives are used to define the geodesic equation, which describes the path that a free-falling object will take in curved spacetime. They are also used in the Einstein field equations, which describe the relationship between the curvature of spacetime and the distribution of matter and energy.

Yes, understanding covariant and partial derivatives is crucial for accurately predicting and describing the behavior of objects in the presence of strong gravitational fields, such as around black holes. This knowledge is also important for various applications in astrophysics, such as predicting the behavior of gravitational waves and understanding the evolution of the universe.

- Replies
- 7

- Views
- 629

- Replies
- 22

- Views
- 2K

- Replies
- 11

- Views
- 663

- Replies
- 16

- Views
- 3K

- Replies
- 124

- Views
- 7K

- Replies
- 5

- Views
- 2K

- Replies
- 2

- Views
- 1K

- Replies
- 17

- Views
- 2K

- Replies
- 7

- Views
- 2K

- Replies
- 15

- Views
- 5K

Share: