- #1

- 290

- 36

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 dsaun777
- Start date

The Christoffel symbols are used to represent a covariant derivative in a specified coordinate system. They are a feature of the interaction between the coordinate system and the tensor fields involved. The covariant derivative itself is a coordinate-independent object and does not rely on Christoffel symbols. Taking a covariant derivative is a tensor operation, and hence is coordinate-independent. It takes in a tensor field and a vector and returns a tensor.f

- #1

- 290

- 36

- #2

- 4,089

- 1,677

$$

\Gamma^a_{bc} = \frac12 g^{bd}\left(g_{db,c} + g_{dc,b} - g_{bc,d} \right)

$$

That formula is simply a coordinate-dependent, algebraic expression for calculating the Christoffel symbols for a given metric and coordinate system. It is purely

For example, given a coordinate system and a metric tensor, ##g_{12,2}## is ##\frac{\partial }{\partial x^2} g_{12}(x^1,x^2,x^3,x^4)## which is a partial derivative of the scalar field whose value is the component in the first row and

second column of the 4-by-4 matrix that expresses the metric tensor ##\mathbf g## in that coordinate system, with respect to the second input to the function ##g_{12}:\mathbb R^4\to \mathbb R## that represents that scalar field in the given coordinate system.

- #3

- 290

- 36

The christoffels depend on the coordinate system which is why they are not tensors. So g has to be a function of a specific coordinate system before calculating the chistroffels? But you still can take a tensor covariant derivative of the the metric correct? I can calculate the christoffel symbol without this formula?

$$

\Gamma^a_{bc} = \frac12 g^{bd}\left(g_{db,c} + g_{dc,b} - g_{bc,d} \right)

$$

That formula is simply a coordinate-dependent, algebraic expression for calculating the Christoffel symbols for a given metric and coordinate system. It is purelyformaland is not a tensor derivative.

For example, given a coordinate system and a metric tensor, ##g_{12,2}## is ##\frac{\partial }{\partial x^2} g_{12}(x^1,x^2,x^3,x^4)## which is a partial derivative of the scalar field whose value is the component in the first row and

second column of the 4-by-4 matrix that expresses the metric tensor ##\mathbf g## in that coordinate system, with respect to the second input to the function ##g_{12}:\mathbb R^4\to \mathbb R## that represents that scalar field in the given coordinate system.

Last edited:

- #4

- 4,089

- 1,677

Yes that's right.So g has to be a function of a specific coordinate system before calculating the chistroffels?

Taking a tensor covariant derivative is a tensor operation, and hence is coordinate-independent. It takes in a tensor field and a vector and returns a tensor. It cannot be used to return a Christoffel symbol as Christoffel symbols are coordinate-dependent. Any formula for a Christoffel symbol will be coordinate-dependent and hence not a tensor formula.But you still can take a tensor covariant derivative of the the metric correct? I can calculate the christoffel symbol without this formula?

- #5

- 290

- 36

But taking a covariant derivative involves christoffels does it not?Yes that's right.

Taking a tensor covariant derivative is a tensor operation, and hence is coordinate-independent. It takes in a tensor field and a vector and returns a tensor. It cannot be used to return a Christoffel symbol as Christoffel symbols are coordinate-dependent. Any formula for a Christoffel symbol will be coordinate-dependent and hence not a tensor formula.

- #6

- 4,089

- 1,677

The Christoffel symbols are used toBut taking a covariant derivative involves christoffels does it not?

Usually in practice, tensors and vectors are represented in coordinate systems, so that for a covariant derivative, we want to know its representation in a particular coordinate system. But coordinate systems are not needed in all cases, and in coordinate-free cases covariant derivatives can be specified perfectly well without using Christoffel symbols.

- #7

- 20,004

- 10,663

$$

\nabla_a \partial_b = \Gamma_{ab}^c \partial_c.

$$

In the case of the Levi-Civita connection, the Christoffel symbols take the form mentioned in #2 in any coordinate system. There is nothing strange about this form, it contains a number of sums of derivatives of functions, that is it. Each metric component is a function of the coordinates and the derivatives are your regular partial derivatives with respect to the coordinates. However, that expression is notoriously cumbersome from a bookkeeping perspective to actually use to compute the Christoffel symbols.

A less cumbersome way is noticing that the Euler-Lagrange equations that make the integral

$$

\frac{1}{2}\int g_{ab} \dot x^a \dot x^b ds

$$

stationary are the geodesic equations. From writing down the EL equations, you therefore obtain the geodesic equations and you can just identify the Christoffel symbols of the Levi-Civita connection from there (noticing that the Levi-Civita connection by definition is torsion-free, i.e., the Christoffel symbols are symmetric in the lower two indices).

- #8

- 290

- 36

What is used in place of christoffels in coordinate free covariant derivative?The Christoffel symbols are used torepresenta covariant derivative in a specified coordinate system. They are a feature of the interaction between the coordinate system and the tensor fields involved. The covariant derivative itself is a coordinate-independent object and does not rely on Christoffel symbols.

Usually in practice, tensors and vectors are represented in coordinate systems, so that for a covariant derivative, we want to know its representation in a particular coordinate system. But coordinate systems are not needed in all cases, and in coordinate-free cases covariant derivatives can be specified perfectly well without using Christoffel symbols.

- #9

- 290

- 36

Should that integral be with respect to dt not ds?

$$

\nabla_a \partial_b = \Gamma_{ab}^c \partial_c.

$$

In the case of the Levi-Civita connection, the Christoffel symbols take the form mentioned in #2 in any coordinate system. There is nothing strange about this form, it contains a number of sums of derivatives of functions, that is it. Each metric component is a function of the coordinates and the derivatives are your regular partial derivatives with respect to the coordinates. However, that expression is notoriously cumbersome from a bookkeeping perspective to actually use to compute the Christoffel symbols.

A less cumbersome way is noticing that the Euler-Lagrange equations that make the integral

$$

\frac{1}{2}\int g_{ab} \dot x^a \dot x^b ds

$$

stationary are the geodesic equations. From writing down the EL equations, you therefore obtain the geodesic equations and you can just identify the Christoffel symbols of the Levi-Civita connection from there (noticing that the Levi-Civita connection by definition is torsion-free, i.e., the Christoffel symbols are symmetric in the lower two indices).

- #10

- 20,004

- 10,663

What you call the curve parameter is of course conpletely irrelevant. I would strongly recommendShould that integral be with respect to dt not ds?

- #11

- 290

- 36

It's just that ds is normally used to calculate an interval.What you call the curve parameter is of course conpletely irrelevant. I would strongly recommendagainstcalling it t as that would be easily confused with the time coordinate of many coordinate systems.

- #12

- 20,004

- 10,663

So what? That does not mean it is not useful for anything else. In this case it is precisely the same s.It's just that ds is normally used to calculate an interval.

Share:

- Replies
- 2

- Views
- 125

- Replies
- 4

- Views
- 327

- Replies
- 34

- Views
- 1K

- Replies
- 3

- Views
- 319

- Replies
- 12

- Views
- 4K

- Replies
- 7

- Views
- 432

- Replies
- 37

- Views
- 7K

- Replies
- 0

- Views
- 400

- Replies
- 4

- Views
- 775

- Replies
- 3

- Views
- 513