I Metric tensor derivatives

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Calculating the christoffel symbols requires taking the derivatives of the metric tensor. What are you taking derivatives of exactly? Are you taking the derivatives of the inner product of the basis vectors with respect to coordinates? In curvilinear coordinates, for instance curved spacetime in GR, the christoffel levi civita connection allows you to take derivatives across a manifold of two different vectors that are separated by some interval. What is being differentiated here? A little confused ...please help.
 

andrewkirk

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Are you referring to the following 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 purely formal and 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.
 
78
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Are you referring to the following 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 purely formal and 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.
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?
 
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andrewkirk

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So g has to be a function of a specific coordinate system before calculating the chistroffels?
Yes that's right.
But you still can take a tensor covariant derivative of the the metric correct? I can calculate the christoffel symbol without this formula?
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.
 
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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.
But taking a covariant derivative involves christoffels does it not?
 

andrewkirk

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But taking a covariant derivative involves christoffels does it not?
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.

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.
 

Orodruin

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More to the point, the Christoffel symbols (or, more generally, connection coefficients) are defined as the components of the covariant derivatives of your basis vectors. Sticking to a holonomic basis, you have
$$
\nabla_a \partial_b = \Gamma_{ab}^c \partial_c.
$$
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).
 
78
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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.

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.
What is used in place of christoffels in coordinate free covariant derivative?
 
78
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More to the point, the Christoffel symbols (or, more generally, connection coefficients) are defined as the components of the covariant derivatives of your basis vectors. Sticking to a holonomic basis, you have
$$
\nabla_a \partial_b = \Gamma_{ab}^c \partial_c.
$$
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).
Should that integral be with respect to dt not ds?
 

Orodruin

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Should that integral be with respect to dt not ds?
What you call the curve parameter is of course conpletely irrelevant. I would strongly recommend against calling it t as that would be easily confused with the time coordinate of many coordinate systems.
 
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What you call the curve parameter is of course conpletely irrelevant. I would strongly recommend against calling it t as that would be easily confused with the time coordinate of many coordinate systems.
It's just that ds is normally used to calculate an interval.
 

Orodruin

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

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