Understanding Covariant and Partial Derivatives in General Relativity

[GR191511]

In the 128 pages of 《A First Course in General Relativity - 2nd Edition》:"The covariant derivative differs from the partial derivative with respect to the coordinates only because the basis vectors change."Could someone give me some examples？I don't quite understand it.Tanks！

 

[malawi_glenn]

What would you get if all $\Gamma^\alpha{}_{\mu \beta} = 0$?
The book does the example of polar coordinates just a few pages earlier.

 

[vanhees71]

Take the case of an affine space as a special case of a Riemann space. Then there are basis vectors that are constant in the entire space. Let's call them $\vec{b}_k$. Then for a given vector field $\vec{V}=V^j \vec{b}_j$ you can define the 2nd-rank tensor
$$\overleftrightarrow{T} = \vec{\nabla} \otimes \vec{V} = \vec{b}^i \otimes \partial_i (V^j \vec{b}_j) = \partial_i V^j (\vec{b}^i \otimes \vec{b}_j),$$
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.

 

[Ibix]

If you have two vectors at two different points you might want to know if they are the same. A vector is expressed as a set of components that multiply the basis vectors, so you could compare the components. But how do you know the basis vectors are the same?

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 $\partial_av^b\neq 0$).

 

[Orodruin]

For example, consider flying a great circle path around the Earth, one not passing through the poles or along the equator.

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 to
$$
\partial_b \vec v = \vec E_a \partial_b v^a + v^a \partial_b \vec E_a
$$
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 $\Gamma_{br}^a$.

 

[Ibix]

At this point of the book the reader is not yet familiar with curved spaces.

Sure, but OP has recently been asking questions based on much later in the book, so I suspect is currently on a re-read.