Covariant derivative of a vector field

In summary, we need to use the equations \nabla_a V^b = \partial_a V^b+\Gamma^b_{ac}V^c and \Gamma^a_{ab} = \frac{1}{\sqrt{det\;h}}\partial_b\sqrt{det\;h} in order to show that \nabla_a(\sqrt{-det\;h}S^a)=\partial_a(\sqrt{-det\;h}S^a), where h is the metric and S^a is a vector. This involves correcting a mistake in the second relevant equation and using the fact that one of the square roots should be outside the derivative on one side.
  • #1
physicus
55
3

Homework Statement


Show that [itex]\nabla_a(\sqrt{-det\;h}S^a)=\partial_a(\sqrt{-det\;h}S^a)[/itex]
where [itex]h[/itex] is the metric and [itex]S^a[/itex] a vector.


Homework Equations


[itex]\nabla_a V^b = \partial_a V^b+\Gamma^b_{ac}V^c[/itex]
[itex]\Gamma^a_{ab} = \frac{1}{2det\;h}\partial_b\sqrt{det\;h}[/itex]
[itex]\nabla_a\sqrt{-det\;h}[/itex] (is that right??) since [itex]\nabla_a h[/itex]


The Attempt at a Solution


I can't quite figure out how to get the result:
[itex]\nabla_a(\sqrt{-det\;h}S^a)[/itex]
[itex]=\nabla_a(\sqrt{-det\;h})S^a+\sqrt{-det\;h}\nabla_a(S^a)[/itex]
[itex]=\sqrt{-det\;h}\;\partial_a S^a+\sqrt{-det\;h}\Gamma^a_{ab}S^b[/itex]
[itex]=\sqrt{-det\;h}\;\partial_a S^a+\sqrt{-det\;h}\frac{1}{2det\;h}\partial_b\sqrt{det\;h}S^b[/itex]
[itex]=\sqrt{-det\;h}\;\partial_a S^a+\frac{1}{2\sqrt{-det\;h}}\partial_b\sqrt{det\;h}S^b[/itex]
[itex]=\ldots[/itex]
 
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  • #2
I'd be interested to find out if someone figures out what's going on here, because to me, it does seem unusual that the connection term must drop out for this to work.
 
  • #3
Are you sure you have the right equations? I'm not entirely sure, but I don't think there should be square root in the second relevant equation. Also I think the problem would make more sense if one of the square roots was outside the derivative on one of the sides of the original equation.
 
  • #4
Thank you very much for you help! You are right. There is a mistake in my second relevant equation. It should be: [itex]\Gamma^a_{ab} = \frac{1}{\sqrt{det\;h}}\partial_b\sqrt{det\;h}[/itex]

Then I get:
[itex]\nabla_a(\sqrt{-det\;h}S^a)[/itex]
[itex]=\nabla_a(\sqrt{-det\;h})S^a+\sqrt{-det\;h}\;\nabla_a(S^a)[/itex]
[itex]=\sqrt{-det\;h}\;\partial_a S^a+\sqrt{-det\;h}\;\Gamma^a_{ab}S^b[/itex]
[itex]= \sqrt{-det\;h}\;\partial_a S^a+\sqrt{-det\;h}\frac{1}{\sqrt{det\;h}}\partial_b \sqrt{det\;h}\;S^b[/itex]
[itex]=\sqrt{-det\;h}\;\partial_a S^a+\partial_b\sqrt{-det\;h}\;S^b[/itex]
[itex]=\partial_a (\sqrt{-det\;h}\;S^a)[/itex]
 
  • #5


The covariant derivative of a vector field is used to describe how the vector field changes as one moves along a curved space. It takes into account the curvature of the space and is defined using the Christoffel symbols, which are related to the metric of the space. In this problem, we are asked to show that the covariant derivative of the vector field S^a is equal to its partial derivative, which is the derivative taken in flat space.

To solve this problem, we start by using the definition of the covariant derivative, which states that \nabla_a V^b = \partial_a V^b+\Gamma^b_{ac}V^c, where \Gamma^a_{ab} is the Christoffel symbol. Next, we expand the expression for the covariant derivative of the vector field S^a:

\nabla_a(\sqrt{-det\;h}S^a) = \partial_a(\sqrt{-det\;h}S^a)+\Gamma^a_{ac}(\sqrt{-det\;h}S^c)

We then use the definition of the Christoffel symbol, \Gamma^a_{ab} = \frac{1}{2det\;h}\partial_b\sqrt{det\;h}, to simplify the expression:

\nabla_a(\sqrt{-det\;h}S^a) = \partial_a(\sqrt{-det\;h}S^a)+\frac{1}{2\sqrt{-det\;h}}\partial_b\sqrt{det\;h}S^b

At this point, we can see that the first term on the right-hand side is just the partial derivative of \sqrt{-det\;h}S^a, which is equal to \partial_a(\sqrt{-det\;h}S^a). The second term can be simplified using the fact that \partial_a(\sqrt{-det\;h}) = \frac{1}{2\sqrt{-det\;h}}\partial_b\sqrt{det\;h}, which can be easily verified by taking the derivative of \sqrt{-det\;h} with respect to a particular coordinate. Thus, we have:

\nabla_a(\sqrt{-det\;h}S^a) = \partial_a(\sqrt{-det\;h}S^a)+\partial_a(\sqrt{-det\
 

1. What is a covariant derivative of a vector field?

The covariant derivative of a vector field is a mathematical operation that describes how the vector field changes as one moves along a curved surface or manifold. It takes into account the intrinsic curvature of the surface and is used in differential geometry and tensor calculus.

2. How is the covariant derivative different from the ordinary derivative?

The ordinary derivative is defined for functions on a flat surface, while the covariant derivative is defined for functions on a curved surface. The covariant derivative also takes into account the curvature of the surface, while the ordinary derivative does not.

3. How is the covariant derivative calculated?

The covariant derivative is calculated by taking the ordinary derivative of the vector field along a specified direction and then correcting for the effects of the curvature of the surface. This is done using a connection, which is a mathematical tool that describes the relationship between points on a curved surface.

4. What is the significance of the covariant derivative in physics?

The covariant derivative is used in many areas of physics, including general relativity and fluid dynamics. It allows for the description of physical quantities on curved surfaces, which is important in understanding the behavior of objects in the presence of gravity or other forces.

5. Can the covariant derivative be extended to higher dimensions?

Yes, the concept of the covariant derivative can be extended to higher dimensions through the use of differential forms and exterior calculus. This allows for a more general and powerful formulation of the covariant derivative in mathematical and physical theories.

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