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Whats the physical meaning of a covariant derivative?

by faeriewhisper
Tags: covariant, derivative, meaning, physical
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faeriewhisper
#1
Mar14-12, 06:02 PM
P: 4
Hi there!
I saw this exercise that we have to calculate the covariant derivative of a vector field (in polar coordinates). Most of them equals zero, but two of them are non-zero, sugesting that this vector field is not constant. What i want to understand is the physical meaning of this values.
The exercise itself is:

Consider in polar coordinates, the vector field with the following component:
A = (Aρ,A) = (0,1)
Is this vector field constant? Compute the covariant derivative.

The Christoffel symbols gave me (assuming ζ as the connection symbol):

ζσμε= 0
Except for:
ζρ∅∅=-ρ
ζρ∅∅ρ=1/ρ

The covariant derivate:

Aσ;ε = Aσ,ε in most cases
Except for:
A; = A, + ζρ∅A = 1/p
A;ρ = A,p + ζ∅ρA = 1/p

The question is, what is the physical meaning of this values? What do they really represent?

And be soft please guys, i'm only starting :)
Thanks for the help
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pervect
#2
Mar14-12, 06:23 PM
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Basically, the covariant derivative [itex]\nabla_\mu[/itex] is a directional derivative. (I hope the notation I'm using is familiar ).

If you specify a direction by a vector u with components [itex]u^\mu[/itex], the covariant derivative of some vector field with components [itex]v^\nu[/itex] is given by

[tex]
u^\mu \nabla_\mu v^\nu
[/tex]

In semicolon notation you'd write this something like:
[itex]
v^\nu{}_{; \, \mu}[/itex]

The physical meaning of this is you evaluate the vector field at two points, which you can think of as P and P + epsilon * u, parallel transport them to P, subtract them, and divide by epsilon, taking the limit as epsilon approaches zero. So, it's just like the directional derivative defined in such a manner that it works on vector fields as well as the more usual scalar functions.

You might not specify the direction beforehand, in which case the u becomes a placeholder for when you do specify it. The result is that you get a tensor of one higher rank than the tensor you apply it to - if you apply the covariant derivative to a rank 1 vector, you get a map from a vector specifying the direction to another vector specifying the result, which is a rank 2 tensor.

For a physical meaning of parallel transport, I'd suggest looking up "Schild's ladder", though you might not find a lot on it unless you look at MTW's "Gravitation", which I know has a section.

If you've ever seen a drafting machine for drawing parallel lines, though - and if you havean't, I think it's called a pantograph. try http://en.wikipedia.org/wiki/Pantograph, you might get the basic idea, which is that parallelograms with equal sides form a working definition of "parallel". If you replace the rigid bars of the pantograph with geodesics on your curved surface of constant length, you've basically defined parallel transport which is how you parallel transport a vector from one location to another so you can take the covariant derivative.

You can take other approaches mathematically - basically, I'm using the metric (some concept of distance) to define parallel transport, sometimes you'll see the reverse. If you have one, you can find the other.

MTW also has a rather long section with diagrams that goes into more detail and some diagrams.
Bill_K
#3
Mar15-12, 12:12 PM
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P: 4,160
Pervect, this explanation is really complicated. Not to mention backwards! What you have started with is not the covariant derivative but the "absolute derivative", usually written δvν/δs where s is a parameter along some curve. This you claim depends on the concept of parallel transport.

But in fact it goes the other way: parallel transport is defined by imposing the condition δvν/δs = 0, where δvν/δs is defined in terms of the covariant derivative, δvν/δs = uμvν.

The covariant derivative is defined by taking initially a coordinate system which is locally Minkowskian, writing the partial derivatives of the vector components in these coordinates, ∂vν/∂xμ, and then transforming them to any other coordinate system by assuming they transform as a tensor.

Chestermiller
#4
Mar15-12, 06:57 PM
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Whats the physical meaning of a covariant derivative?

In curvilinear coordinates, a vector field can be expressed in component form as the summation of components times coordinate basis vectors. In general, the coordinate basis vectors are functions of spatial position. For example, in cylindrical coordinates, the radial- and theta coordinate basis vectors vary with theta (and r). So if you want to evaluate the differential change in a vector field with spatial position, you need to include both the changes in the components, and the changes in the basis vectors. The Christoffel symbols take care of the changes in the basis vectors. The covariant derivative describes the gradient of a vector field (i.e., the effect of applying the gradient vector operator) to the vector, and properly includes the partial derivatives along the coordinate directions of both the vector components and the coordinate basis vectors.
pervect
#5
Mar15-12, 07:35 PM
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Quote Quote by Bill_K View Post
Pervect, this explanation is really complicated. Not to mention backwards! What you have started with is not the covariant derivative but the "absolute derivative", usually written δvν/δs where s is a parameter along some curve. This you claim depends on the concept of parallel transport.
My approach is (or was supposed to be) basically a quick rehash of MTW's ("Gravitation", the big black book) approach in section $10.3, around pg 248 or so.

Is the absolute derivative defined for a vetor field? The OP's post was about a taking the covariant derivative of a vector field, at least the example was and that's how I interpreted the question.

If you want to take the covariant derivative of a scalar, I'd explain it as being equal to the gradient, which is what I think you said.

I don't see how you're going to talk about the covariant derivative of a vector field without getting into parallel transport, though. Am I missing something? Or are we answering different questions?

Wald defines the covariant derivative as a derivative operator that satifies [itex]\nabla_a g_{bc} = 0[/itex], but this doesn't seem very "physical" to me, though it has some advantages for a purely mathematical approach - Wald also specifies the mathematical properties a "derivative" operator must have eariler, one of which is basically that it's equal to the gradient if you apply it to a scalar-valued function.

we could go into Wald's approach if the OP or anyone is especially interested, but while mathematically clear I didn't think it was as "physical" as MTW's approach.


But in fact it goes the other way: parallel transport is defined by imposing the condition δvν/δs = 0, where δvν/δs is defined in terms of the covariant derivative, δvν/δs = uμvν.
I'm not familiar with this as written, though the "setting to zero" part sounds like Wald's approach.

There's lots of ways of defining parallel transport, but I still think Schild's latter is among the most "physical", though MTW is really the only author I can recall who uses this particular approach.
faeriewhisper
#6
Mar20-12, 11:52 AM
P: 4
Thank you guys, for the replies. So, let me see if i understood.
Please look at the picture and tell me if i'm right.

http://www.freeimagehosting.net/nsqeh

This picture represents the vector field with the corresponding basis vectors, so, for what i've understood, the covariant derivative kind of assumes all the vectors possible and transport their basis vectors, subtract them and divide by dλ to see if they are equal in all possibilities. It's a derivative of the basis vectors. I'm i right?
Sorry, but i'm just starting with GR and since i'm from engineering this is like a big fury monster to me :) but i'm trying very hard. I just want to visualize it, so i can understand it completely. Thank's for the support :)
Chestermiller
#7
Mar20-12, 03:45 PM
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I'm not sure from reading what you've written whether you get it or not. What you wrote was a little confusing to me. So let me say it a little differently.

I'm an engineer also, so I know you must remember from freshman physics how to represent the velocity vector in cylindrical coordinates. It's the component in the radial direction times the unit vector in the radial direction plus the component in the theta direction times the unit vector in the theta direction. These unit vectors are both functions of theta. Now, if you want to get the acceleration, you have to take the derivative of the velocity vector with respect to time. That involves taking the derivatives both of the components and of the unit vectors with respect to time. If the velocity has a component in the theta direction, then the unit vectors must be changing direction along the trajectory, since theta is changing. You have to take this into account in calculating the acceleration. This is very closely related to what happens when you do covariant differentiation, although in the case of covariant differentiation, the operation is done with much more mathematical formalism.

Chet
pervect
#8
Mar20-12, 04:53 PM
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Quote Quote by faeriewhisper View Post
Thank you guys, for the replies. So, let me see if i understood.
Please look at the picture and tell me if i'm right.

http://www.freeimagehosting.net/nsqeh

This picture represents the vector field with the corresponding basis vectors, so, for what i've understood, the covariant derivative kind of assumes all the vectors possible and transport their basis vectors, subtract them and divide by dλ to see if they are equal in all possibilities. It's a derivative of the basis vectors. I'm i right?
Sorry, but i'm just starting with GR and since i'm from engineering this is like a big fury monster to me :) but i'm trying very hard. I just want to visualize it, so i can understand it completely. Thank's for the support :)
I can't tell if you've got it either, so I'm guessing you probably don't. So let me rehash a bit. If you have a scalar function, I assume you know how to find the gradient. One way of describingthe result (perhaps not the best way) will be a vector field that points in the direction that the scalar function changes most rapidly and the length is proportional to how fast it changes.

You can think of it another way - if you take the dot product of the gradient vector with a direction (represented by a unit vector), it gives you a scalar, which is the rate of change of that scalar function in that direction. This is rather like thinking of the gradient one-form, because one-forms are maps from vectors to scalars.

If you take [itex]\nabla_a[/itex] or [itex]_;a[/itex], because the a is a subscript and not a superscript, you are more precisely calculating the gradient one-form than the gradient as a vector.

But in your example, you have, not a scalar function, but a vector-valued function, a vector field.

The covariant derivative of this field is a rank-2 tensor. So, it may be difficult to find a physical interpretation at first.

It may be helpful to think about what the comma operator, the ordinary derivative, does to a vector field first.

You can write your vector function as <f(r,theta), g(r,theta)>, then the ordinary derivative of this vector function is a matrix:

[tex] \left( \begin{matrix} \partial f / \partial r & \partial f / \partial \theta \\ \partial g / \partial r & \partial g / \partial \theta \end{matrix} \right)
[/tex]

Now, if you multiply this matrix by a unit vector, you get a vector, you get the ordinary derivatives of your vector function in that direction. For instance, if your unit vector is [itex]\hat{r}[/itex], a unit vector in the r direction, you just get

[tex]\partial f / \partial r, \partial g / \partial r[/tex]

The ordinary derivative [itex]\partial f / \partial r[/itex] is just f(r+delta) - f(r) / delta, with no parallel transport before you do the subtraction.

The difference between the covariant derivative and the ordinary derivative is that you do the parallel transport operation , rather than just subtracting component values.


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