I Generating Divergence Equation In Cylindrical Coordinates

[Hardy]

TL;DR Summary

Question about how to apply Taylor's Formula to a vector field to derive equation for divergence in cylindrical coordinates.

This is from an old E&M exam question where we were asked to derive the formula for the divergence of a vector field in cylindrical coordinates using Taylor's Approximation and the fundamental definition of the divergence:

∇⋅A = Lim V→0 { ( ∫_S A⋅da ) / V }​

The vector field, A, is defined in cylindrical components such that:

A = A_ρ(ρ,θ,z)ρ + A_θ(ρ,θ,z)θ + A_z(ρ,θ,z)z​

Where ρ, θ and z are the unit vectors in cylindrical coordinates, and A_ρ, A_θ and A_z are everywhere continuous and well defined functions.

The formula we are looking for (divergence in cylindrical coordinates):

∇⋅A = ( 1/ρ ) (δ/δρ) (ρA_ρ) + ( 1/ρ ) (δ/δθ)A_θ + (δ/δz)A_z​

I have found several similar analysis on-line, but most simply use the definition of the ∇ function in cylindrical coordinates and not Taylor's Formula. I am having a problem completing the analysis and it seems to stem from how Taylor's Formula is applied to the vector field. Hopefully the following is enough to convey the issue, I have lots more so that I could add if needed.

The vector field is represented by a Taylor's Series approximation about a point: P_o(ρ_o,θ_o,z_o)

At this point I would apply Taylor's Formula such that:

A = AAA_ρ(P_o)ρ + (ρ - ρ_o) (δ/δρ)(A_ρρ) + (θ - θ_o) (δ/δθ)(A_ρρ) + (z - z_o) (δ/δz)(A_ρρ)
A = + A_θ(P_o)θ + (ρ - ρ_o) (δ/δρ)(A_θθ) + (θ - θ_o) (δ/δθ)(A_θθ) + (z - z_o) (δ/δz)(A_θθ)
A = + A_z(P_o)z + (ρ - ρ_o) (δ/δρ)(A_zz) + (θ - θ_o) (δ/δθ)(A_zz) + (z - z_o) (δ/δz)(A_zz)

Which almost provides the correct answer, but not quite:
( 1/ρ ) (δ/δρ) (ρA_ρ) + ( 1/ρ ) { (δ/δθ)A_θ + A_θ } + (δ/δz)A_z

Notice the only difference is the A_θ term.

But if I apply Taylor's Formula like:
A = AAA_ρ(P_o)ρ + (ρ - ρ_o) (δ/δρ)(A_ρρ)
A = + A_θ(P_o)θ + (θ - θ_o) (δ/δθ)(A_θθ)
A = + A_z(P_o)z + (z - z_o) (δ/δz)(A_zz)

It provides the correct answer.

At this point, the rest of the analysis goes the same with both representations:
- The equation for A is developed, carrying out the derivatives, taking care since the unit vectors are not constant
- Collecting terms for each unit vector in the equation for A,
- The six elements for differential area, da is from :

da ≡ (+/-)ρδθδzρ ; (+/-)δρδzθ ; (+/-)ρδρδθz ​

where plus is applied for the sides in the direction from P_o of the unit vector, minus if in the other direction
- and six surface integrals are defined by taking the dot products and then evaluated

I could include more gory details in the double integrals. But, once the integrals are set up, I have gotten the same answer solving them manually and with a MAPLE tool (symbolic math calculations). Just hoping the error is in how this Taylor's Formula is applied. It seems that adding the extra terms is correct since the vector components are functions of all three cylindrical variables and the unit vectors are not constant, so not including them seems wrong when applying a directional derivative, but it isn't providing the correct answer. Thanks for any insight.

 

[vanhees71]

Just use an infinitesimal volume with boundary surfaces defined by the coordinate lines. That's an infinitesimal cuboid.

Start from
$$\vec{x}=\begin{pmatrix} \rho \cos \theta \\ \rho \sin \theta \\ z \end{pmatrix}.$$
The tangent vectors are
$$\partial_\rho \vec{x}=\vec{e}_{\theta}, \quad \partial_{\theta} \vec{x}=\rho \vec{e}_{\theta}, \quad \partial_z \vec{x}=\vec{e}_z,$$
where the $\vec{e}_{\rho}$, $\vec{e}_{\theta}$ and $\vec{e}_z$ are the tangent unit vectors. The order of the cylinder coordinates $(\rho,\theta,z)$ is chosen such that $(\vec{e}_{\rho} \times \vec{e}_{\theta}) \cdot \vec{e}_z=+1$, i.e., in this order the tangent unit vectors everywhere build a right-handed triad.

The said cuboid-volume element is then given by
$$\mathrm{d}^3 x =\mathrm{d} \rho \mathrm{d} \theta \mathrm{d} z (\partial_{\rho} \vec{x} \times \partial_{\theta} \vec{x}) \cdot \partial_z \vec{x}=\mathrm{d} \rho \mathrm{d} \theta \mathrm{d} z \rho.$$
The surface-elements are given by
$$\mathrm{d}^2 \vec{f}_{\rho} = \pm \mathrm{d} \theta \mathrm{d} z \partial_{theta} \vec{x} \times \partial_z \vec{x}=\pm \mathrm{d} \theta \mathrm{d} z \partial_{theta} \rho \vec{e}_{\rho},$$
$$\mathrm{d}^2 \vec{f}_{\theta}=\pm \mathrm{d} \rho \mathrm{d} z \partial_{\rho} \vec{x} \times \partial_z \vec{x}=\pm \mathrm{d} \rho \mathrm{d} z \vec{e}_{\theta},$$
$$\mathrm{d}^2 \vec{f}_z = \pm \mathrm{d} \rho \mathrm{d} \theta (\partial_{\rho} \vec{x} \times \partial_{\theta} \vec{x}) = \pm \mathrm{d} \rho \mathrm{d} \theta \rho \vec{e}_z.$$
Now evaluate the surface integral
$$\begin{split}
\int_{\partial \mathrm{d} V} \mathrm{d}^2 \vec{f} = & \mathrm{d} \theta \mathrm{d} z [(\rho+\mathrm{d} \rho) A_{\rho}(\rho+\mathrm{d} \rho,\theta,z)-\rho A_{\rho}(\rho,\theta,z)] \\ & + \mathrm{d} \rho \mathrm{d} z [A_{\theta}(\rho,\theta+\mathrm{d} \theta,z) - A_{\theta}(\rho,\theta,z)] \\
&+ \mathrm{d} \rho \mathrm{d} \theta \rho [A_z(\rho,\theta,z+\mathrm{d} z)-A_z(\rho,\theta,z)].
\end{split}
$$
Expand the square brackets:
$$\int_{\partial \mathrm{d} V} \mathrm{d}^2 \vec{f} = \mathrm{d} \rho \mathrm{d} \theta \mathrm{d} z [\partial_{\rho} (\rho A_{\rho}) + \partial_{\theta} A_{\theta} + \rho \partial_z A_z].$$
Dividing by $\mathrm{d}^3 x$ finally gives
$$\vec{\nabla} \cdot \vec{A}=\frac{1}{\rho} \partial_{\rho}(\rho A_{\rho}) + \frac{1}{\rho} \partial_{\theta} A_{\theta} + \partial_z A_z.$$

 

[Hardy]

vanhees71 wrote:
"Just use an infinitesimal volume with boundary surfaces defined by the coordinate lines. That's an infinitesimal cuboid."

Thank you for providing the analysis.

It has several interesting aspects that I have not seen before. I have been trying to understanding how you seemed to evaluate an integral without defining integral limits or even applying any anti-derivatives. If you have any references that might explain the math technique in a little bit more depth, I would appreciate it.

As for the EM exam question, I would wonder how the prof would react. The use of Taylor's formula nor the fact that the derivative of the cylindrical unit vectors not being constant is not evident. But, what is evident, from my technique, is that your derivation appears more like the application of Taylor's formula in the expansion of the vector field that provides the correct formula.

The only explanation I can think of why the representation of the vector field that includes more terms does not provide the correct answer is that the extra terms are not incorrect, but simply higher order terms in the expansion that diminish and eventually have no impact when taken to the infinitesimal level. For example, the term (θ - θo) (δ/δθ)(Aρρ) says there is a change in two variables, while (ρ - ρo) (δ/δρ)(Aρρ) is looking at a change in only one variable.

One thing that helped me getting to this thought was looking at EM analysis of charge distributions. A monopole expansion is sufficient at some point, and is the correct answer. But, when getting relatively closer to the charges themselves, you need to account for dipole terms to get the correct answer, and then again having to going to multi-poles components at some point. Each is a correct answer at some point - relatively anyway. This is kinda of the opposite though, the higher order terms go away when 'zooming in' to a point.

Now, maybe I'll spend some time looking at the formula for the curl, and also, how these techniques look using spherical formulas. If it’s found that the same Taylor’s formula expansion does not work, it will be a head scratcher! Thanks again.

 

[vanhees71]

vanhees71 wrote:
"Just use an infinitesimal volume with boundary surfaces defined by the coordinate lines. That's an infinitesimal cuboid."

Thank you for providing the analysis.

It has several interesting aspects that I have not seen before. I have been trying to understanding how you seemed to evaluate an integral without defining integral limits or even applying any anti-derivatives. If you have any references that might explain the math technique in a little bit more depth, I would appreciate it.

A very nice treatment of vector calculus is in A. Sommerfeld, Lectures on Theoretical Physics vol. 2.

As for the EM exam question, I would wonder how the prof would react. The use of Taylor's formula nor the fact that the derivative of the cylindrical unit vectors not being constant is not evident. But, what is evident, from my technique, is that your derivation appears more like the application of Taylor's formula in the expansion of the vector field that provides the correct formula.

Of course, I don't know, what your professor expects as an answer. You should use the techniques taught in the lecture or provided in the textbook used.

The only explanation I can think of why the representation of the vector field that includes more terms does not provide the correct answer is that the extra terms are not incorrect, but simply higher order terms in the expansion that diminish and eventually have no impact when taken to the infinitesimal level. For example, the term (θ - θo) (δ/δθ)(Aρρ) says there is a change in two variables, while (ρ - ρo) (δ/δρ)(Aρρ) is looking at a change in only one variable.

One thing that helped me getting to this thought was looking at EM analysis of charge distributions. A monopole expansion is sufficient at some point, and is the correct answer. But, when getting relatively closer to the charges themselves, you need to account for dipole terms to get the correct answer, and then again having to going to multi-poles components at some point. Each is a correct answer at some point - relatively anyway. This is kinda of the opposite though, the higher order terms go away when 'zooming in' to a point.

Now, maybe I'll spend some time looking at the formula for the curl, and also, how these techniques look using spherical formulas. If it’s found that the same Taylor’s formula expansion does not work, it will be a head scratcher! Thanks again.

The coordinate independent definitions of the derivative operators grad, div, curl always work. It's a good idea to check it for spherical coordinates.