Covariant derivative in spherical coordinate

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

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

I don't know why we need the factor g_{kk} 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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I think the 1/g_{kk} 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, \partial_k is the same as the operator \partial/\partial x^k.
 
bcrowell said:
I think the 1/g_{kk} 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, \partial_k is the same as the operator \partial/\partial x^k.

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

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

\frac{1}{r} \frac{\partial A_\theta}{\partial \theta} ?

I think the usual covariant derivative is defined as

D_\mu A_\nu = \partial_\mu A_\nu + \cdots

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

\frac{\partial A_\theta}{\partial \theta}

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

D_\mu A_\nu = \partial_\mu A_\nu + \cdots

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

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

hmm... so you mean mathworld writes

\partial_\mu A_\nu = \frac{\partial}{g^{\mu\alpha} \partial x_\alpha} A_\nu

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. D_\theta A_\theta has the partial derivative term like this

\frac{1}{r}\frac{\partial A_\theta}{\partial \theta}

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

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

\partial_\mu A_\nu = \frac{\partial}{g^{\mu\alpha} \partial x_\alpha} A_\nu

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. D_\theta A_\theta has the partial derivative term like this

\frac{1}{r}\frac{\partial A_\theta}{\partial \theta}

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

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

A_{r;r}

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

A_{r;r}=\frac{\partial}{\partial x^1} A_r

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

\frac{\partial}{\partial x^1}=e_1=e_r,

is a component of the basis vector e_a.

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

\omega^b e_a=\delta^b_a,

and that g_{ab}=e_ae_b

to obtain

g_{ab} \omega^b=e_a.

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

g_{bb} \omega^b=e_b,

and finally

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

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

AB
 
ismaili said:
hmm... so you mean mathworld writes

\partial_\mu A_\nu = \frac{\partial}{g^{\mu\alpha} \partial x_\alpha} A_\nu

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. D_\theta A_\theta has the partial derivative term like this

\frac{1}{r}\frac{\partial A_\theta}{\partial \theta}

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

Thanks for your discussion!

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

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

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

I don't know why we need the factor g_{kk} 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.
 
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