- #1

- 140

- 0

I've heard of something called a covariant derivative. what motivates it and what is it?

You should upgrade or use an alternative browser.

- Thread starter Terilien
- Start date

- #1

- 140

- 0

I've heard of something called a covariant derivative. what motivates it and what is it?

- #2

cristo

Staff Emeritus

Science Advisor

- 8,122

- 74

- #3

- 29

- 6

I'm somewhat in your shoes myself, but here's what I know:

Say you have a vector field defined on your mainfold. You would like to know what the derivative of this vector field is. The problem is that there are *two* ways that the field could appear to change:

1) The field really does change magnitude or direction over a small distance

2) The way you *measure* your magnitude or direction changes over a small distance.

In a "flat" manifold (one without curvature) then the derivative of the field is simply given by #1. But if there is curvature (say, on a sphere), then the vector field will appear to change, even though some (or all) of the apparent change is due to your coordinates changing.

Here's another way of looking at the same thing:

Say you want to compare two vectors at different locations to see their rate of change. In order to do so, you must "transport" one of the vectors to the other one and do the comparison. But how do you do this transport so that you don't "rotate" the vector en-route? This is the problem of parallel transport.

One of the definitions of a covariant derivative is that the covariant derivative of the velocity vector of a geodesic path (the "shortest" path between two points) should be zero. In other words, a geodesic path does this "parallel transport" of its velocity vector from one point on the path to any other point on the path.

So, with all that said, if you can find a copy of Wald's "General Relativity" book, he does an algebraic derivation of the covariant derivative in chapter 3.1 (page 30 in my copy)

If, on the other hand, you're more differential-geometry motivated (i.e. masochistic) you can follow the path I've been following which involves fibre bundles and Lie groups and algebras. I'm not sure I recommend it, but if I can find my way back down from a connection form on a principal fibre bundle (Bishop and Goldberg do an ok job of this) to getting the Christoffel symbols of GR, I'll try and post what I've learned.

ZM

- #4

- 406

- 8

[tex]\nabla_{\mathbf{w}}\mathbf{v}[/tex]

That really all there is to it. But, as zenmaster99 mentioned, if you are in a curvilinear coordinate system, then you have some additional bookkeeping to do because not only are your vector coordinates changing from point to point, but your basis vectors themeselves are also changing.

The bookkeeping here involves the dreaded Christoffel symbols and their partner in crime "index notation". The result is a haze of mathematical symbols and definitions amid which the meaning of the operation is all but lost completely. But that's all there is to it. Rate of change of one vector in the direction of another.

- #5

- 140

- 0

Yah I know what it is now but computations with it seems nigh impossible. Ok this question is stupid, but can't we just use the chain rule to calculate the directional derivative of a tensor field in an arbitrary direction(byt that I mean can the directional derivative be written as a linear combination of the covariant derivative along corrdinate axis)? I heard that you can't but don't know why you wouldn't be able to. If so how do we calculate it in a rbitrary direction. please don't tear me apart. There's something weird about the covariant derivative. the christoffel symbols make computation seem impossbile.

Last edited:

- #6

- 29

- 6

Ok, so let's look at what happens when we try and take a partial derivative of a vector field:

The partial derivative of a vector field [tex]V^\nu[/tex] is given by

[tex]\partial_\mu V^\nu[/tex]

But we would like this "derivative" to be covariant---it shouldn't change based on our choice of coordinate system. So let's transform to a new coordinate system and see what we get:

[tex]

\partial_\mu V^\nu=\left(\frac{\partial y^i}{\partial x^\mu}\partial_i\right)

\left(\frac{\partial x^\nu}{\partial y^j}V^j\right)\\

=\frac{\partial y^i}{\partial x^\mu}\frac{\partial x^\nu}{\partial y^j}\partial_iV^j

+V^j\frac{\partial y^i}{\partial x^\mu}\frac{\partial^2 x^\nu}{\partial y^i\partial y^j}

[/tex]

If it were not for the second term, then [tex]\partial_iV^j[/tex] would transform as a (1,1) tensor. However, in general, this object is*not* covariant and any "derivative" of a vector field defined this way will yield different results in different coordinate systems.

So you can't just blindly start taking derivatives of everything in sight. Indeed, the vector [tex]\partial_i[/tex] is only defined on real-valued functions anyway and*not* on vector fields.

In my wanderings, I've found three different "derivatives" besides the standard partial derivative:

1) Exterior Derivatives:

For some tensors (p-forms), no additional structure is necessary and there exists a derivative of a p-form which is a (p+1)-form. The down-side is that one is restricted to only taking derivatives of p-forms.

2) Lie Derivatives

If one constructs a diffeomorphism [tex]\phi\colon M\rightarrow M[/tex] , then we know that the push-forward of a vector [tex]\phi_*\vec v[/tex] will be a unique vector because the push-forward of a diffeomorphism is a vector-space isomorphism. Therefore, we can construct such a diffeomorphism and push vectors along them such that distant vectors may be compared and a derivative taken. The downside is that every different diffeomorphism will yield a different push-forward for all vectors and there is no preferred choice of diffeomorphism.

3) Covariant derivatives

Now one defines*a priori* a fixed "connection" between tangent spaces so that we can push vectors around in a consistent way. The downside is that a new level of structure is required: the connection. However, given a metric on the manifold, a particular connection which preserves the inner product is preferred over all others.

Making any more sense?

ZM

The partial derivative of a vector field [tex]V^\nu[/tex] is given by

[tex]\partial_\mu V^\nu[/tex]

But we would like this "derivative" to be covariant---it shouldn't change based on our choice of coordinate system. So let's transform to a new coordinate system and see what we get:

[tex]

\partial_\mu V^\nu=\left(\frac{\partial y^i}{\partial x^\mu}\partial_i\right)

\left(\frac{\partial x^\nu}{\partial y^j}V^j\right)\\

=\frac{\partial y^i}{\partial x^\mu}\frac{\partial x^\nu}{\partial y^j}\partial_iV^j

+V^j\frac{\partial y^i}{\partial x^\mu}\frac{\partial^2 x^\nu}{\partial y^i\partial y^j}

[/tex]

If it were not for the second term, then [tex]\partial_iV^j[/tex] would transform as a (1,1) tensor. However, in general, this object is

So you can't just blindly start taking derivatives of everything in sight. Indeed, the vector [tex]\partial_i[/tex] is only defined on real-valued functions anyway and

In my wanderings, I've found three different "derivatives" besides the standard partial derivative:

1) Exterior Derivatives:

For some tensors (p-forms), no additional structure is necessary and there exists a derivative of a p-form which is a (p+1)-form. The down-side is that one is restricted to only taking derivatives of p-forms.

2) Lie Derivatives

If one constructs a diffeomorphism [tex]\phi\colon M\rightarrow M[/tex] , then we know that the push-forward of a vector [tex]\phi_*\vec v[/tex] will be a unique vector because the push-forward of a diffeomorphism is a vector-space isomorphism. Therefore, we can construct such a diffeomorphism and push vectors along them such that distant vectors may be compared and a derivative taken. The downside is that every different diffeomorphism will yield a different push-forward for all vectors and there is no preferred choice of diffeomorphism.

3) Covariant derivatives

Now one defines

Making any more sense?

ZM

Last edited:

- #7

- 140

- 0

I think there is something lacking in physics books.

- #8

- 29

- 6

He also has a curvature text. I haven't gotten a chance to see it yet, though.

For a good algebraic run up of covariant derivatives, Wald does a pretty good job. But if you want to run through fibre bundles.... oooooh.... I have yet to find any single book that approaches clarity. If you can at all avoid this path, I think you'll sleep better at night. But if you can't be dissuaded, here's my bibliography on such things:

Norman Steenrod, "The Topology of Fibre Bundles," Princeton U. Press (1951)

Steenrod is the classic text, but gets impenetrable once you get past the definition of the transition functions and some related constructions.

Shigeyuki Morita, "Geometry of Differential Forms," Iwanami Shoten (1998)

Morita has a good book (orignally in japanese, but translated by the AMS) and actualy shows you why principal fibre bundles are required, but, again, there seem to be (to a physicist, anyway) some unecessary mathematical constructions which I haven't been able to penetrate.

Nash & Sen, "Topology and Geometry for Physicsts" Academic Press (1983)

Great overview... and then they jump straight to a connection formed by the structure constants of a Lie algebra---frustrating as hell.

You will also need to know things about Lie groups and their algebras. Lee does an ok job, but leaves out things like the adjoint representation. Again, there is no single book I'd recommend for this but

Anthony Knapp, "Lie Groups, Lie Algebras, and Cohomology," Princeton U. Press (1988)

was fairly helpful.

And, so far, I have *yet* to find the text that brings you from defining the connection in the principal fiber bundle back down to Earth so you can compute Christoffel symbols.

If you find anything helpful on that front, please, let me know.

The journey continues.....

ZM

- #9

- 2,952

- 1

That's not quite right. When one calculates the directional derivative of a vector you need two things. The vector field and a vector which determines the direction you're interested.Ok this question is stupid, but can't we just use the chain rule to calculate the directional derivative of a tensor field in an arbitrary direction(byt that I mean can the directional derivative be written as a linear combination of the covariant derivative along corrdinate axis)? I heard that you can't but don't know why you wouldn't be able to.

Also, to use the chain rule you must know how to do it in any system of coordinates (since that's what tensor analysis is all about). A

I worked out an example where the covariant derivative is used. I believe I showed it to you before. Its at my web site at

http://www.geocities.com/physics_world/ma/geodesics.htm

Look under the last section wich is entitled

I have a few examples where I determine the Christoffel symbols but I'm in the process of reworking my web site for greater surfacing ease. In the mean time those pages are unavailable. But I'll get to them soon. Meanwhile I can try to fix one of them tonight.

Calculating the Christoffel symbols can be laborious at times but once you've done it a dozen or so times it will become second nature to you.If so how do we calculate it in a rbitrary direction. please don't tear me apart. There's something weird about the covariant derivative. the christoffel symbols make computation seem impossbile.

More later

Best regards

Pete

- #10

- 2,952

- 1

Pete

- #11

- 140

- 0

how does one derive the general formula for the covariant derivative of a tensor field? To be more precise I took out sean carolls book at the library but did not understand equation 3.17 on page 97. Could someone derive it or prove it, or at the very least give me a better hint?

Last edited:

- #12

- 367

- 0

how does one derive the general formula for the covariant derivative of a tensor field? To be more precise I took out sean carolls book at the library but did not understand equation 3.17 on page 97. Could someone derive it or prove it, or at the very least give me a better hint?

http://eom.springer.de/c/c026870.htm

- #13

- 140

- 0

Care to explain Sean caroll's reasoning? That resource is too formal.

Share: