Why does one form need to be used over the other?

[Chris Hillman]

Recommend some reading

well you may have interested me in learning the physics of these matters, since i have realized it is also connected to hodge theory, so thank you!

Glad you are interested in learning how physics employs exterior calculus! One good books for mathematicians who wish to learn a bit about physics might be:

Frankel, Geometry of Physics, Cambridge University Press. (Just skim the mathematical exposition; there is plenty of exposition of physics to be found.)

Going in the other direction:

Isham, Modern Differential Geometry for Physicists, World Scientific, 2005. (Covers basic manifold theory and exterior calculus too.)

 

[bhale]

The two vector expressions for the curl (with contravariant and covariant basis vectors, respectively, inserted into the sums) are not equal. Only the second form (with covariant basis vectors and contravariant components) can be derived rigorously from the curl, a vector defined in the Cartesian frame. (Note, you must equate vectors, not components.)

You cannot transform the second (correct) form into the first form because elements of the metric tensor do not commute with the differential operator. If the generalized frame metric tensor components are all constants the two forms give the same result for the curl.

 

[timur]

I did not read the pages 2-4, but think about what happens when there is no metric: if you can define something without using a metric, it is more fundamental than the dual.

 

[mmcf]

Sorry about this, but none of these formulas seem to work out for me so
could someone tell me where I'm going wrong? I
guess starting with the simplest one, the proposed form for the gradient of
a scalar field in terms of the exterior derivative is:

$$(d\phi)^t$$

where the transpose of a covariant vector is obtained by applying the
contravariant metric tensor. I'm going to write a one-form as $$df$$ and a directional derivative along a coordinate x as $$\partial x$$. I'll only write $$\partial_x$$
when I'm taking a derivative of a function rather than talking about a vector field.
In a euclidean space the contravariant metric tensor can be expressed in cartesian
coordinates as:

$$g = \partial x \otimes \partial x + \partial y \otimes \partial y +<br /> \partial z \otimes \partial z$$

Let's pretend I'm interested in cylindrical coordinates
$$(r,\theta,z)$$ because they require the least typing:

$$x = rcos(\theta)$$
$$y = rsin(\theta)$$
$$z = z$$

I can take some derivatives to obtain:
$$\partial x = (x/r)\partial r + (-y/r^2)\partial \theta = cos(\theta)\partial<br /> r - (sin(\theta)/r) \partial \theta$$
$$\partial y = (y/r) \partial r + (x/r^2) \partial \theta =<br /> sin(\theta)\partial r + (cos(\theta)/r) \partial \theta$$

I then will get by substitution that:
$$\partial x \otimes \partial x = cos^2(\theta) \partial r \otimes \partial r<br /> + (sin^2(\theta)/r^2) \partial \theta \otimes \partial \theta +<br /> (-cos(\theta)sin(\theta)/r) \partial \theta \otimes \partial r +<br /> (-cos(\theta)sin(\theta)/r) \partial r \otimes \partial \theta$$
$$\partial y \otimes \partial y = sin^2(\theta) \partial r \otimes \partial r<br /> + (cos^2(\theta)/r^2) \partial \theta \otimes \partial \theta +<br /> (cos(\theta)sin(\theta)/r) \partial \theta \otimes \partial r +<br /> (cos(\theta)sin(\theta)/r) \partial r \otimes \partial \theta$$

Substituting those into the metric tensor in cartesian coordinates I get:
$$g = \partial r \otimes \partial r + (1/r^2) \partial \theta \otimes \partial \theta<br /> + \partial z \otimes \partial z$$

I'm pretty sure that's the standard result you should get because it's the inverse of the covariant metric guy.

So to obtain the cylindrical coordinates gradient, I start by taking the
0-form $$\phi$$ and applying an exterior derivative:
$$d\phi = (\partial_r \phi) dr + (\partial_\theta \phi) d\theta + (\partial_z \phi) dz$$

He's a one-form, and I'm interested in a vector, so I apply the metric tensor:
$$(d\phi)^t = \partial r <\partial r,(\partial_r \phi) dr + (\partial_\theta \phi) d\theta + (\partial_z \phi) dz> + (1/r^2) \partial \theta <\partial \theta, (\partial_r \phi) dr + (\partial_\theta \phi) d\theta + (\partial_z \phi) dz> + \partial z <\partial z, (\partial_r \phi) dr + (\partial_\theta \phi) d\theta + (\partial_z \phi) dz>$$
$$= (\partial_r \phi)\partial r + ((\partial_\theta \phi)/r^2) \partial \theta<br /> + (\partial_z \phi) \partial z$$
That's sort of messy but those $$<x,y>$$ are supposed to be a natural pairing as Bishop and Goldberg say it.

If I'm supposed to assume that the classical gradient is this where you substitute
$$\partial r \rightarrow \hat{r}$$
$$\partial \theta \rightarrow \hat{\theta}$$
$$\partial z \rightarrow \hat{z}$$

then this is off by a factor of $$1/r$$ along the $$\theta$$ direction from the normal definition:
$$(\partial_r \phi)\hat{r} + ((\partial_\theta \phi)/r) \hat{\theta}<br /> + (\partial_z \phi) \hat{z}$$

Any thoughts on where I went wrong?

 

[llarsen]

If I'm supposed to assume that the classical gradient is this where you substitute
$$\partial r \rightarrow \hat{r}$$
$$\partial \theta \rightarrow \hat{\theta}$$
$$\partial z \rightarrow \hat{z}$$

then this is off by a factor of $$1/r$$ along the $$\theta$$ direction from the normal definition:
$$(\partial_r \phi)\hat{r} + ((\partial_\theta \phi)/r) \hat{\theta}<br /> + (\partial_z \phi) \hat{z}$$

Any thoughts on where I went wrong?

In your derivation above, you used coordinate vectors. This is very natural when you are using the modern differential geometry approach. However, the equation for the gradient is often presented using unit vectors rather than coordinate vectors. I think this is the source of your problem. It looks like $$\hat{\theta}$$ represents a unit vector, whereas $$\partial \theta$$ is a coordinate vector. The correct substitution is:

$$\partial \theta = r \hat \theta$$

I believe that the overall derivation was correct. It seems to just be the last step substituting coordinate vectors with unit vectors that caused problem. The unit vector substitutions for r and z were OK since unit vectors and coordinate vectors are the same length for r and z in cylindrical coordinates.

 

[mmcf]

Alright, thanks. That seems to be my whole problem. It looks like I get the right expression for div A from
$$*d(*A^t)$$
and curl A from
$$(*dA^t)^t$$
just like it says in the thread as long as I consider my components to be in the form of the $$A^i$$ in
$$\Sigma_i \frac{A^i \partial_i}{g(\partial_i,\partial_i)^{1/2}}$$