Calculating divergence using covariant derivative

In summary, the conversation discussed the use of the definition of divergence in relation to the Levi-Civita connection. The attempt at a solution involved calculating the contraction of a vector field and a volume element, and using the fact that the Riemannian metric can be written as a determinant and transformed into an orthonormal basis. The final solution was found by using the fact that the basis 1-forms are not necessarily closed when using an orthonormal basis.
  • #1
v2536
7
0

Homework Statement



Using the definition of divergence [itex]d(i_{X}dV) = (div X)dV[/itex] where [itex]X:M\rightarrow TM[/itex] is a vector field, [itex]dV[/itex] is a volume element and [itex]i_X[/itex] is a contraction operator e.g. [itex]i_{X}T = X^{k}T^{i_{1}...i_{r}}_{kj_{2}...j_{s}}[/itex], prove that if we use Levi-Civita connection then the divergence can also be written as
[itex]div X = X^{i}_{;i}[/itex]

2. The attempt at a solution

This is what i tried:
since [itex]dV = dx^{1} \wedge ... \wedge dx^{n}[/itex]
after some calculation i conclude that [itex]i_{X}dV = \sum_{i=1}^{n}(-1)^{i}X_{i}dx^{1} \wedge ... \wedge dx^{i-1} \wedge dx^{i+1} \wedge ... \wedge dx^{n}[/itex]
so [itex]d(i_{X} dV) = (\partial _{i}X^{i})dV[/itex]
Then i attempt the use the fact that [itex]\Gamma^{i}_{jk} = \Gamma^{i}_{kj}[/itex] to get a lot of cancellation and show that [itex]\partial _{i}X^{i} = X^{i}_{;i}[/itex]
but i couldn't.
So can anyone please help? Thx in advanced :)
 
Last edited:
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  • #2
Hey v2536 and welcome to the forums.

I don't know much about differential geometry, but I do understand in essence what you are trying to do.

Maybe you should look at this website (http://en.wikipedia.org/wiki/Divergence#Generalizations). A lot of the results you need are found in the coordinate free form you are trying to prove.
 
  • #3
Your first step is wrong. You should have

[tex]dV = \sqrt{g} \, dx^1 \wedge \ldots \wedge dx^n[/tex]
where [itex]\sqrt{g}[/itex] is the square root of the determinant of the Riemannian metric.
 
  • #4
Thank you chiro and Ben for your reply :)
I see that i forgot about [itex]\sqrt{g}[/itex] in front but can't i always find an orthonormal basis and work in that?
and by doing that g should become 1 right?
 
  • #5
ok i think i understand now, if i use orthonormal basis then the connection may no be Levis-Civita.
 
  • #6
If you use an orthonormal basis, then your basis 1-forms are not necessarily closed (so it would be incorrect to write them as dx).
 
  • #7
yes. that make sense.
Now i got the solution, thanks for your help.
 

Related to Calculating divergence using covariant derivative

1. What is the purpose of calculating divergence using covariant derivative?

The purpose of calculating divergence using covariant derivative is to determine the rate at which a vector field diverges or converges at a given point in space. This is important in understanding the behavior of fluids, electric and magnetic fields, and other physical phenomena.

2. How do you calculate divergence using covariant derivative?

To calculate divergence using covariant derivative, you first need to determine the covariant derivative of the vector field in question. Then, you can use the appropriate formula to calculate the divergence, which involves taking the trace of the covariant derivative matrix. This will give you a scalar value representing the divergence at a specific point.

3. What is the difference between covariant derivative and partial derivative?

Covariant derivative and partial derivative are two different types of derivatives used in mathematics and physics. Partial derivative is a measure of how a function changes with respect to one of its variables, while covariant derivative takes into account the effect of changes in the coordinate system. In other words, covariant derivative takes into consideration the curvature of the space in which the function is defined.

4. What is the physical significance of the divergence of a vector field?

The divergence of a vector field has physical significance in that it represents the rate of change of the density of a quantity flowing through a surface. In other words, it tells us how much of a quantity is entering or leaving a given point in space. This can be applied to various physical phenomena such as fluid flow, electric and magnetic fields, and more.

5. Can divergence be negative?

Yes, divergence can be negative. This indicates that the vector field is converging at a given point in space. Conversely, a positive divergence indicates that the vector field is diverging at that point. A divergence of zero means that the vector field is neither converging nor diverging, and is often referred to as a source or a sink.

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