# Reversing Indices in Contractions: Can it be Done?

• unscientific
In summary: You should write out this covariant derivative first in terms of F, and then in that expression write out F in terms of A. If your connection is symmetric, F can be expressed in terms of partial derivatives only.
unscientific
Suppose I have something like
$$\left( \nabla_\mu \nabla_\beta - \nabla_\beta \nabla_\mu \right) V^\mu = R_{\nu \beta} V^\nu$$
Can since all the terms involving ##\mu## on the left and ##\nu## on the right are contractions, can I simply do:
$$\left( \nabla^\mu \nabla_\beta - \nabla_\beta \nabla^\mu \right) V_{\mu} = R^\nu_{\beta} V_{\nu}$$?

As long as the connection is metric compatible.

Orodruin said:
As long as the connection is metric compatible.

But my concern is that in the first term on the LHS, the gradient operator is acting on the gradient operator via ##\nabla_\mu (\nabla_\beta) V^\mu##, and not on the vector ##V^\mu##. I thought for contractions, these have to be acting on one another?

unscientific said:
But my concern is that in the first term on the LHS, the gradient operator is acting on the gradient operator via ##\nabla_\mu (\nabla_\beta) V^\mu##, and not on the vector ##V^\mu##. I thought for contractions, these have to be acting on one another?
No, this is not how to interpret the LHS. The operators are each acting on ##V^\mu##, one after the other.

Orodruin said:
No, this is not how to interpret the LHS. The operators are each acting on ##V^\mu##, one after the other.
Ok, now I'm confused. I thought the chain rule applies:
$$\nabla^\mu \left( \nabla_\mu A_\nu \right) = \nabla^\mu \left( \nabla_\mu\right) A_\nu + \nabla_\mu \nabla^\mu A_\nu$$

Orodruin said:
No, this is not how to interpret the LHS. The operators are each acting on ##V^\mu##, one after the other.
If I have something like ##F_{\mu \nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu##, what will ##\nabla^\mu F_{\mu \nu}## look like?

What do you mean by ##\nabla_\mu(\nabla^\mu)##? This is an operator applied to an operator. The only way to make sense of this is to apply the right operator to something first and then apply the left to the result.

unscientific said:
If I have something like ##F_{\mu \nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu##, what will ##\nabla^\mu F_{\mu \nu}## look like?
You should write out this covariant derivative first in terms of F, and then in that expression write out F in terms of A. If your connection is symmetric, F can be expressed in terms of partial derivatives only.

## 1. Can contractions be reversed in a sentence?

Yes, contractions can be reversed in a sentence. This process is known as "expanding" the contraction.

## 2. What is the purpose of reversing indices in contractions?

The purpose of reversing indices in contractions is to make the sentence clearer and easier to read by expanding the contracted words.

## 3. How is the process of reversing indices in contractions done?

The process involves identifying the contracted words in a sentence, determining the expanded form of each word, and rearranging the words in the sentence accordingly.

## 4. Are there any exceptions to reversing indices in contractions?

Yes, there are a few exceptions to reversing indices in contractions. For example, the contraction "won't" cannot be reversed as "n'two" because it is not a valid word.

## 5. Is reversing indices in contractions necessary in all cases?

No, reversing indices in contractions is not necessary in all cases. It is a matter of style and personal preference. Some writers may choose to use contractions, while others may prefer to expand them for clarity.

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