# Help with covariant differentiation

• A
• redtree
In summary, the conversation discusses the difficulty of evaluating an expression involving covariant differentiation. The next steps to complete the differentiation involve using basis and formulas, as well as understanding how the covariant derivative acts on a general type (1,1) tensor. The conversation also mentions considering the tensor to be a function of ##z_k## and using contraction to simplify the expression.

#### redtree

I'm having trouble evaluating the following expression (LATEX):

##\nabla_{i}\nabla_{j}T^{k}= \nabla_{i} \frac{\delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{i m} \frac{\delta T^{m}}{\delta z^{i}} + \Gamma^{k}_{i m} \Gamma^{m}_{i l} T^{l}##

What are the next steps to complete the covariant differentiation? (This is not a homework assignment)

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redtree said:
What are the next steps to complete the covariant differentiation?
Getting the first step correct. Your expression even has different free indices on both sides.

If you will
redtree said:
I'm having trouble evaluating the following expression (LATEX):

##\nabla_{i}\nabla_{j}T^{k}= \nabla_{i} \frac{\delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{i m} \frac{\delta T^{m}}{\delta z^{i}} + \Gamma^{k}_{i m} \Gamma^{m}_{i l} T^{l}##

What are the next steps to complete the covariant differentiation? (This is not a homework assignment)

Just use basis and these formulas will be easy to write. For example let ##A=a_i^je^i\otimes e_j## then $$(\nabla_k A)e^k=(\frac{\partial a_i^j}{\partial x^k}e^i\otimes e_j+a_i^j(\nabla_ke^i)\otimes e_j+a_i^je^i\otimes(\nabla_k e_j))\otimes e^k$$ and remember that $$\nabla_ke_i=\Gamma_{ik}^se_s,\quad \nabla_ke^i=-\Gamma_{sk}^ie^s$$

Sorry. Better?

##\nabla_{i}\nabla_{j}T^{k}= \nabla_{i} \frac{\delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} \frac{\delta T^{l}}{\delta z^{i}} + \Gamma^{l}_{i m} \Gamma^{k}_{j l} T^{m}##

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The first covariant derivative I get. I'm having trouble with the second covariant differentiation ##\nabla_{i}## of ##\nabla_{j} T^{k}##, where ##\nabla_{j} T^{k} = \frac{ \delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} T^{l}##.

redtree said:
The first covariant derivative I get. I'm having trouble with the second covariant differentiation ##\nabla_{i}## of ##\nabla_{j} T^{k}##, where ##\nabla_{j} T^{k} = \frac{ \delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} T^{l}##.

How does the covariant derivative act on a general type (1,1) tensor? How can you apply that to the type (1,1) tensor ##\nabla_j T^k##?

Orodruin said:
How does the covariant derivative act on a general type (1,1) tensor? How can you apply that to the type (1,1) tensor ##\nabla_j T^k##?

You mean the invariant ##T##? Where ##T=T^{k} z_{k}##?##\nabla_{j} T = \nabla_{j} T^{k} z_{k} = (\frac{ \delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} T^{l}) z_{k}##?

redtree said:
You mean the invariant ##T##? Where ##T=T^{k} z_{k}##?##\nabla_{j} T = \nabla_{j} T^{k} z_{k} = (\frac{ \delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} T^{l}) z_{k}##?
No, I mean a general type (1,1) tensor ##T_i^j##.

##\nabla_{k} T^{i}_{j} = \frac{ \delta T^{i}_{j}}{\delta z^{k}} + \Gamma^{i}_{k m} T^{m}_{j} - \Gamma^{m}_{j k} T^{i}_{m} ##

The tensor ##T^{i}_{j}## can be contracted ## T^{i}_{j} z^{j} = T^{i} ##?

##\nabla_{i}\nabla_{j}T^{k}= \nabla_{i} \frac{\delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} \frac{\delta T^{l}}{\delta z^{i}} + \Gamma^{l}_{i m} \Gamma^{k}_{j l} T^{m}##

##\frac{\delta T }{\delta z^{j}} = \frac{\delta (T^{k} z_{k})}{\delta z^{j}} ## and then solve for ##\frac{\delta T^{k}}{\delta z^{j}}##?

One thought I had was to consider ##T^{k}## a function of ##z_{k}##, i.e., ##T^{k} = T^{k}(z_{k})##, such that ##\frac{\delta T^{k}}{\delta z^{j}} = \frac{\delta T^{k}}{\delta z^{k}} \frac{\delta z_{k}}{\delta z^{j}}## and ##\frac{\delta z_{k}}{\delta z^{j}} = \Gamma^{m}_{k j} z_{m}##

redtree said:
##\nabla_{k} T^{i}_{j} = \frac{ \delta T^{i}_{j}}{\delta z^{k}} + \Gamma^{i}_{k m} T^{m}_{j} - \Gamma^{m}_{j k} T^{i}_{m} ##

The tensor ##T^{i}_{j}## can be contracted ## T^{i}_{j} z^{j} = T^{i} ##?
No, put ##T^i_j = \nabla_j T^i##.

Ahhh. Thanks!

## What is covariant differentiation?

Covariant differentiation is a mathematical concept used in differential geometry and tensor calculus. It is a way of differentiating vector and tensor fields on curved spaces in a way that respects the properties of those fields.

## Why is covariant differentiation important?

Covariant differentiation is important because it allows us to study the behavior of vector and tensor fields on curved spaces, which are often more realistic and relevant to real-world situations than flat, Euclidean spaces. It is also a key tool in general relativity, where curved spacetime is the foundation of the theory.

## How is covariant differentiation different from regular differentiation?

Regular differentiation, also known as partial differentiation, is used to find the rate of change of a function with respect to one of its variables. Covariant differentiation, on the other hand, is used to find the rate of change of a vector or tensor field with respect to a curve or surface on which it is defined. It takes into account the curvature and other geometric properties of the space in which the field is defined.

## What are some common applications of covariant differentiation?

Covariant differentiation has many applications in physics and engineering, particularly in the fields of general relativity, fluid dynamics, and electromagnetism. It is also used in computer graphics and computer vision, where it is used to analyze and manipulate geometric objects and shapes.

## What are some common mistakes when using covariant differentiation?

One common mistake when using covariant differentiation is confusing it with regular differentiation and applying the rules of regular differentiation to curved spaces. This can lead to incorrect results. Another mistake is forgetting to take into account the properties of the space, such as its curvature, when performing covariant differentiation. It is important to have a good understanding of the underlying geometric concepts in order to use covariant differentiation correctly.