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- #2

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$$

\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

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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

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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

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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

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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.

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$$

\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

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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

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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

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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

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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

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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.

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