Covariant derivative in spherical coordinate

In summary, the conversation discusses the confusion regarding the use of spherical coordinates in a 2D polar coordinate system. The mathworld website presents an equation for the covariant derivative that includes a factor of 1/g_{kk} in the first term, leading to confusion about its necessity. The conversation includes discussions about the role of the metric and transformations in understanding this equation. Ultimately, the conversation ends with a clarification that the use of 1/g_{kk} is a typo and should not be included in the first term of the equation.
  • #1
ismaili
160
0
I am confused with the spherical coordinate.
Say, in 2D, the polar coordinate [tex] (r, \theta) [/tex]
The mathworld website says that
http://mathworld.wolfram.com/SphericalCoordinates.html

[tex] D_k A_j = \frac{1}{g_{kk}} \frac{\partial A_j}{\partial x_k} - \Gamma^i_{ij}A_i [/tex]

I don't know why we need the factor [tex] g_{kk} [/tex] in the first term.
I think there should be no such a scale factor, but just the partial derivative for the first term, am I right?

Thanks
 
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  • #2
I think the [itex]1/g_{kk}[/itex] is there because they're defining the lower-index D, but taking the partial with respect to the lower-index x would give you the upper-index partial derivative. That is, [itex]\partial_k[/itex] is the same as the operator [itex]\partial/\partial x^k[/itex].
 
  • #3
bcrowell said:
I think the [itex]1/g_{kk}[/itex] is there because they're defining the lower-index D, but taking the partial with respect to the lower-index x would give you the upper-index partial derivative. That is, [itex]\partial_k[/itex] is the same as the operator [itex]\partial/\partial x^k[/itex].

Sorry, I don't get it.
For example, their eq(61) should correspond to [tex] D_\theta A_\theta [/tex] in 2D polar coordinate.

However, for the partial derivative term of the covariant derivative,
why is it

[tex] \frac{1}{r} \frac{\partial A_\theta}{\partial \theta} [/tex] ?

I think the usual covariant derivative is defined as

[tex] D_\mu A_\nu = \partial_\mu A_\nu + \cdots [/tex]

where there in NO scale factors as [tex]1/r [/tex] in the partial derivative term.
That is, if [tex] \mu, \nu[/tex] both are [tex]\theta[/tex], the partial derivative term would just be

[tex] \frac{\partial A_\theta}{\partial \theta} [/tex]

right?
 
  • #4
ismaili said:
I think the usual covariant derivative is defined as

[tex] D_\mu A_\nu = \partial_\mu A_\nu + \cdots [/tex]

where there in NO scale factors as [tex]1/r [/tex] in the partial derivative term.

But [itex]\partial_\mu=\partial/\partial x^\mu[/itex], and [itex]x^\mu[/itex] differs from [itex]x_\mu[/itex]. To lower the index on [itex]x^\mu[/itex], you use the metric.
 
  • #5
bcrowell said:
But [itex]\partial_\mu=\partial/\partial x^\mu[/itex], and [itex]x^\mu[/itex] differs from [itex]x_\mu[/itex]. To lower the index on [itex]x^\mu[/itex], you use the metric.

hmm... so you mean mathworld writes

[tex] \partial_\mu A_\nu = \frac{\partial}{g^{\mu\alpha} \partial x_\alpha} A_\nu [/tex]

But mathworld's metric is lower-indexed, which is NOT upper-indexed, this is contradiction.

Moreoever, I still don't get why their eq(61), i.e. [tex] D_\theta A_\theta [/tex] has the partial derivative term like this

[tex] \frac{1}{r}\frac{\partial A_\theta}{\partial \theta} [/tex]

If the metric is involved, since [tex] g_{\theta\theta} = 1/r^2 [/tex],
I don't understand how they could get the scale factor [tex]1/r[/tex].

Thanks for your discussion!
 
  • #6
ismaili said:
hmm... so you mean mathworld writes

[tex] \partial_\mu A_\nu = \frac{\partial}{g^{\mu\alpha} \partial x_\alpha} A_\nu [/tex]

But mathworld's metric is lower-indexed, which is NOT upper-indexed, this is contradiction.

Moreoever, I still don't get why their eq(61), i.e. [tex] D_\theta A_\theta [/tex] has the partial derivative term like this

[tex] \frac{1}{r}\frac{\partial A_\theta}{\partial \theta} [/tex]

If the metric is involved, since [tex] g_{\theta\theta} = 1/r^2 [/tex],
I don't understand how they could get the scale factor [tex]1/r[/tex].

Thanks for your discussion!

You seem to have gotten it all confused!

They use an standard metric for a space associated with spherical coordinates-based metric. For example, to compute

[tex] A_{r;r}[/tex]

one must know that if there are one-forms and basis vectors involved in the equation, then it is also mandatory to know how to deal with their transformations. Here obviously the second term in the covariant derivative vanishes and the first gives

[tex] A_{r;r}=\frac{\partial}{\partial x^1} A_r[/tex]

where the coordinates are taken to be [tex](x^1,x^2,x^3).[/tex] Of course the reason why I have written [tex]\frac{\partial}{\partial x^1}[/tex] independently of the r-component of the tensor [tex]A_a[/tex] is that it is the r-component of a basis vector, to wit,

[tex]\frac{\partial}{\partial x^1}=e_1=e_r,[/tex]

is a component of the basis vector [tex]e_a.[/tex]

Therefore to transform it into its dual vector (or one-form) [tex]\omega^a[/tex] we can use the fact that

[tex]\omega^b e_a=\delta^b_a,[/tex]

and that [tex]g_{ab}=e_ae_b[/tex]

to obtain

[tex]g_{ab} \omega^b=e_a.[/tex]

Since the metric chosen for our coordinates is diagonal, this turns out to be

[tex]g_{bb} \omega^b=e_b,[/tex]

and finally

[tex]g_{rr} \omega^r=e_r \Rightarrow A_{r;r}= g_{rr} \frac{\partial}{\partial x_1}. [/tex]

But don't be tricked into laying your finger at this because Mathworld uses the matrix multiplication and so, in MW's sense, [tex]g_{rr}[/tex] is equivalent to [tex]g^{-1}_{rr}.[/tex] Now if you got this right, everything would be well set!

AB
 
  • #7
ismaili said:
hmm... so you mean mathworld writes

[tex] \partial_\mu A_\nu = \frac{\partial}{g^{\mu\alpha} \partial x_\alpha} A_\nu [/tex]

But mathworld's metric is lower-indexed, which is NOT upper-indexed, this is contradiction.
I could be wrong, but it looks to me like whoever wrote the mathworld article simply doesn't use the tensor-gymnastics conventions of upper and lower indices.

ismaili said:
Moreoever, I still don't get why their eq(61), i.e. [tex] D_\theta A_\theta [/tex] has the partial derivative term like this

[tex] \frac{1}{r}\frac{\partial A_\theta}{\partial \theta} [/tex]

If the metric is involved, since [tex] g_{\theta\theta} = 1/r^2 [/tex],
I don't understand how they could get the scale factor [tex]1/r[/tex].
I haven't worked it out myself, but if you have two versions that differ in their exponent of r, you can tell which is right and which is wrong simply by checking the units.
 
  • #8
ismaili said:
If the metric is involved, since [tex] g_{\theta\theta} = 1/r^2 [/tex],
I don't understand how they could get the scale factor [tex]1/r[/tex].

Thanks for your discussion!

Well don't take it hard on you. It is just a typo!

AB
 
  • #9
ismaili said:
I am confused with the spherical coordinate.
Say, in 2D, the polar coordinate [tex] (r, \theta) [/tex]
The mathworld website says that
http://mathworld.wolfram.com/SphericalCoordinates.html

[tex] D_k A_j = \frac{1}{g_{kk}} \frac{\partial A_j}{\partial x_k} - \Gamma^i_{ij}A_i [/tex]

I don't know why we need the factor [tex] g_{kk} [/tex] in the first term.
I think there should be no such a scale factor, but just the partial derivative for the first term, am I right?

Thanks

Isn't there a problem with the indices here, or am I missing something? Three k's and 3 i's in each term doesn't ring true to me.
 
  • #10
AEM said:
Isn't there a problem with the indices here, or am I missing something? Three k's and 3 i's in each term doesn't ring true to me.

I think these non-summed repeated indices arise because the metric is diagonal.
 

Related to Covariant derivative in spherical coordinate

What is a covariant derivative in spherical coordinate?

A covariant derivative in spherical coordinate is a mathematical concept used in differential geometry to describe the change of a vector or tensor field on a curved surface. It takes into account the curvature of the surface and how it affects the direction and magnitude of the vector or tensor field.

How is a covariant derivative calculated in spherical coordinate?

To calculate a covariant derivative in spherical coordinate, you first need to define a set of basis vectors that are tangent to the surface at a specific point. Then, you use these basis vectors to calculate the partial derivatives of the vector or tensor field. Finally, you use the Christoffel symbols, which represent the curvature of the surface, to calculate the covariant derivative.

What is the significance of the covariant derivative in spherical coordinate?

The covariant derivative in spherical coordinate is significant because it allows us to define a consistent way to differentiate vector and tensor fields on curved surfaces. This is important in many areas of physics and engineering, such as general relativity and fluid mechanics.

Can the covariant derivative in spherical coordinate be extended to other coordinate systems?

Yes, the concept of covariant derivative can be extended to any coordinate system, not just spherical coordinates. This allows us to describe the change of vector and tensor fields on various types of curved surfaces, such as cylinders, cones, and more.

What are some practical applications of the covariant derivative in spherical coordinate?

The covariant derivative in spherical coordinate has many practical applications in physics and engineering. It is used in general relativity to describe the curvature of spacetime, in fluid mechanics to calculate the acceleration of a fluid particle on a curved surface, and in geodesy to measure the shape and size of the Earth.

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