Covariant derivative in spherical coordinate

[ismaili]

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

 

[bcrowell]

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 $\partial/\partial x^k$.

 

[ismaili]

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?

 

[bcrowell]

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.

 

[ismaili]

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!

 

[Altabeh]

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

 

[bcrowell]

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.

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.

 

[Altabeh]

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

 

[AEM]

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.

 

[bcrowell]

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.