# Two Covariant Derivatives (Chain Rule)?

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• berlinspeed
In summary, the conversation was about trying to find information on the internet about the notation ∇ϒ∇δ𝒆β and its relation to 𝒆μΓμβδ,ϒ+(𝒆νΓνμϒ)Γμβδ. The conversation included a discussion of the use of the comma notation for partial derivatives and the difference between using a basis vector in notation and not using one. The participants also clarified that the "chain rule" in this context is actually the product rule for covariant derivatives.
berlinspeed
TL;DR Summary
Failed find information on the internet, really appreciate any help.
Summary: Failed find information on the internet, really appreciate any help.

Can someone tell me what is ∇ϒδ𝒆β? It seems to be equal to 𝒆μΓμβδ,ϒ+(𝒆νΓνμϒμβδ. Is this some sort of chain rule or is it by any means called anything?

What do you get when you try to compute it yourself?

Orodruin said:
What do you get when you try to compute it yourself?
At first look I just get the second term, so still trying to figure out where went wrong..

Orodruin said:
Okay I think where went wrong is I need to expand ∇ϒ(𝒆μΓμβδ), which now looks like something to apply chain rule to, and if I do that I get the answer except shouldn't the first term be 𝒆μΓμβδ;ϒ? The "," stands for partial derivative?

Yes, the , stands for a regular partial derivative. You are here including the basis vector ##e_\mu## in your notation. When you do that you are using a notation where anything multiplying it is just a scalar function. It holds that ##\nabla_a e_b = \Gamma_{ab}^c e_c## by definition. The covariant derivative is linear and satisfies the product rule (this is not chain rule)
$$\nabla_a (fV) = V \nabla_a f + f \nabla_a V,$$
where ##f## is a scalar field and ##V## is a vector. Note that ##\nabla_a f = \partial_a f## for any scalar field. In your case, therefore
$$\nabla_a(\Gamma^c_{bd}e_c) = \Gamma^c_{bd} \nabla_a e_c + e_c \nabla_a \Gamma^c_{bd} = \Gamma^c_{bd} \Gamma^f_{ac} e_f + e_c \Gamma^c_{bd,a}.$$

Do not confuse this notation with the usual notation without writing out the vector basis, where we typically use the convention
$$\nabla_a V^b \equiv (\nabla_a V)^b,$$
i.e., the LHS is defined as the ##b##-component of ##\nabla_a V##. Actually writing out the basis in the RHS gives you
$$(\nabla_a V)^b = [\nabla_a (V^c e_c)]^b = [V^c \nabla_a e_c]^b + [e_c \nabla_a V^c]^b = V^c [\Gamma^d_{ac} e_d]^b + [V^c_{,a} e_c]^b = V^c \Gamma^b_{ac} + V^b_{,a},$$
since ##[e_c]^b = \delta^b_c##.

Orodruin said:
Yes, the , stands for a regular partial derivative. You are here including the basis vector ##e_\mu## in your notation. When you do that you are using a notation where anything multiplying it is just a scalar function. It holds that ##\nabla_a e_b = \Gamma_{ab}^c e_c## by definition. The covariant derivative is linear and satisfies the product rule (this is not chain rule)
$$\nabla_a (fV) = V \nabla_a f + f \nabla_a V,$$
where ##f## is a scalar field and ##V## is a vector. Note that ##\nabla_a f = \partial_a f## for any scalar field. In your case, therefore
$$\nabla_a(\Gamma^c_{bd}e_c) = \Gamma^c_{bd} \nabla_a e_c + e_c \nabla_a \Gamma^c_{bd} = \Gamma^c_{bd} \Gamma^f_{ac} e_f + e_c \Gamma^c_{bd,a}.$$

Do not confuse this notation with the usual notation without writing out the vector basis, where we typically use the convention
$$\nabla_a V^b \equiv (\nabla_a V)^b,$$
i.e., the LHS is defined as the ##b##-component of ##\nabla_a V##. Actually writing out the basis in the RHS gives you
$$(\nabla_a V)^b = [\nabla_a (V^c e_c)]^b = [V^c \nabla_a e_c]^b + [e_c \nabla_a V^c]^b = V^c [\Gamma^d_{ac} e_d]^b + [V^c_{,a} e_c]^b = V^c \Gamma^b_{ac} + V^b_{,a},$$
since ##[e_c]^b = \delta^b_c##.
Okay cleared now, thanks much.

Last edited:
Orodruin said:
Yes, the , stands for a regular partial derivative. You are here including the basis vector ##e_\mu## in your notation. When you do that you are using a notation where anything multiplying it is just a scalar function. It holds that ##\nabla_a e_b = \Gamma_{ab}^c e_c## by definition. The covariant derivative is linear and satisfies the product rule (this is not chain rule)
$$\nabla_a (fV) = V \nabla_a f + f \nabla_a V,$$
where ##f## is a scalar field and ##V## is a vector. Note that ##\nabla_a f = \partial_a f## for any scalar field. In your case, therefore
$$\nabla_a(\Gamma^c_{bd}e_c) = \Gamma^c_{bd} \nabla_a e_c + e_c \nabla_a \Gamma^c_{bd} = \Gamma^c_{bd} \Gamma^f_{ac} e_f + e_c \Gamma^c_{bd,a}.$$

Do not confuse this notation with the usual notation without writing out the vector basis, where we typically use the convention
$$\nabla_a V^b \equiv (\nabla_a V)^b,$$
i.e., the LHS is defined as the ##b##-component of ##\nabla_a V##. Actually writing out the basis in the RHS gives you
$$(\nabla_a V)^b = [\nabla_a (V^c e_c)]^b = [V^c \nabla_a e_c]^b + [e_c \nabla_a V^c]^b = V^c [\Gamma^d_{ac} e_d]^b + [V^c_{,a} e_c]^b = V^c \Gamma^b_{ac} + V^b_{,a},$$
since ##[e_c]^b = \delta^b_c##.
Btw oddly enough it is named "chain rule" in the Charles & Wheeler book..

## What is the concept of Two Covariant Derivatives (Chain Rule)?

The concept of Two Covariant Derivatives (Chain Rule) is a mathematical tool used to calculate the derivative of a function with respect to another function. It is used to find the rate of change of a function with respect to two different variables, and takes into account how those variables are related to each other.

## How is the Two Covariant Derivatives (Chain Rule) different from the regular Chain Rule?

The Two Covariant Derivatives (Chain Rule) is an extension of the regular Chain Rule, which only takes into account one variable. The Two Covariant Derivatives (Chain Rule) takes into account two variables and their relationship, allowing for a more accurate calculation of the derivative.

## What are some real-world applications of the Two Covariant Derivatives (Chain Rule)?

The Two Covariant Derivatives (Chain Rule) has various applications in fields such as physics, engineering, and economics. It is used to calculate the rate of change of a system with respect to two variables, such as velocity and time, or price and demand.

## What are the limitations of using the Two Covariant Derivatives (Chain Rule)?

The Two Covariant Derivatives (Chain Rule) can only be used for functions that are continuously differentiable. It also assumes that the two variables are independent of each other, which may not always be the case in real-world scenarios.

## How can I improve my understanding of the Two Covariant Derivatives (Chain Rule)?

To improve your understanding of the Two Covariant Derivatives (Chain Rule), it is important to have a strong foundation in calculus and the regular Chain Rule. Practicing with various examples and problems can also help in gaining a deeper understanding of this concept.

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