Christoffel Symbol / Covariant derivative

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SUMMARY

The discussion centers on demonstrating that the covariant derivative \(\nabla_i V^i\) is a scalar. The equation provided is \(\nabla_i V^i = \frac{\partial V^{i}}{\partial q^{i}} + \Gamma^{i}_{ik} V^{k}\), where \(\Gamma^{i}_{ik}\) is derived as \(\Gamma^{i}_{ik} = \frac{1}{\sqrt{g}} \frac{\partial \sqrt{g}}{\partial q^{k}}\). The transformation of the expression \(\nabla_i V^i\) into \(\frac{1}{\sqrt{g}} \partial_i (\sqrt{g} V^{i})\) is highlighted, emphasizing the application of the product rule and chain rule in tensor calculus.

PREREQUISITES
  • Understanding of covariant derivatives in differential geometry
  • Familiarity with the properties of Christoffel symbols
  • Knowledge of tensor calculus and manipulation
  • Basic principles of differential equations and calculus
NEXT STEPS
  • Study the derivation and properties of Christoffel symbols in Riemannian geometry
  • Learn about the product rule and chain rule in the context of tensor calculus
  • Explore the implications of scalar quantities in covariant derivatives
  • Investigate the role of the determinant of the metric tensor in differential geometry
USEFUL FOR

Students of differential geometry, physicists working with general relativity, and mathematicians interested in tensor analysis will benefit from this discussion.

cristina89
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Homework Statement


My teacher solved this in class but I'm not understanding some parts of tis solution.

Show that \nabla_i V^i is scalar.


Homework Equations


\nabla_i V^i = \frac{\partial V^{i}}{\partial q^{i}} + \Gamma^{i}_{ik} V^{k}


The Attempt at a Solution


To start this, I'll solve this \Gamma^{i}_{ik} first.

\Gamma^{i}_{ik} = \frac{1}{2} g^{il} (\frac{\partial g_{lk}}{\partial q^{i}} + \frac{\partial g_{il}}{\partial q^{k}} - \frac{\partial g_{ki}}{\partial q^{l}}

\Gamma^{i}_{ik} = \frac{1}{2} g^{il} \frac{\partial g_{il}}{\partial q^{k}} = \frac{1}{2g} \frac{\partial g}{\partial q^{k}}

\Gamma^{i}_{ik} = \frac{1}{\sqrt g}\frac{\partial \sqrt{g}}{\partial q^{k}}

(THIS PART: how this \sqrt{g} appeared??)

Continuing...

\nabla_i V^i = \frac{\partial V^{i}}{\partial q^{i}} + \frac{V^{k}}{\sqrt{g}} \frac{\partial \sqrt{g}}{\partial q^{k}}

\nabla_i V^i = \frac{1}{\sqrt g} \partial_i (\sqrt g V^{i})

And this last part... What happened to \partial q^{i} and V^{k}?
 
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take this with a grain of salt as I haven't played with tensors in a while... anyway here I go... I think both these are tricks that can be understood by working back

I think the first part is just normal chain rule differentiation, what is the standard derivative of sqrt(g)?

then try expanding the very last expression using the standard product rule, what do you end up with?
 

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