Covariant derivative summation convention help

[Mathematicsresear]

Homework Statement

Assume that you want to the derivative of a vector V with respect to a component Z^k, the derivative is then ∂VⁱZ_i/∂Z^k=Z_i∂Vⁱ/∂Z^k+Vⁱ∂Z_i/∂Z^k = Z_i∂Vⁱ/∂Z^k+VⁱΓ^m_ikZ_m Now why is it that I can change m to i and i to j in VⁱΓ^m_ikZ_m?

 

[nrqed]

Homework Statement

Assume that you want to the derivative of a vector V with respect to a component Z^k, the derivative is then ∂VⁱZ_i/∂Z^k=Z_i∂Vⁱ/∂Z^k+Vⁱ∂Z_i/∂Z^k = Z_i∂Vⁱ/∂Z^k+VⁱΓ^m_ikZ_m Now why is it that I can change m to i and i to j in VⁱΓ^m_ikZ_m?

What do you mean by "I can change m to i and i to j"?

 

[Chestermiller]

What are the boldface $Z_m$s. Are they supposed to be coordinate basis vectors? If so, and you are really expressing the covariant derivative in this way, then there should be $\mathbf{Z_k}$ also, so that there are dyadic products of coordinate basis vectors. The gradient of a vector is a 2nd order tensor.

 

[vela]

The Insight posted the other day might be of interest to you.

https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/

 

[facenian]

1. Now why is it that I can change m to i and i to j in VⁱΓ^m_ikZ_m?

Because in that expression "m" is a dummy index and so is "i". The name of a dummy index can be changed without changing the meaning of the equation