User:Theanphibian/ring equations

Problem Statement

A cylindrical coordinate system with origin O, polar axis A, and longitudinal axis L. The dot is the point with radial distance ρ = 4, angular coordinate φ = 130°, and height z = 4.

Integral of ${\frac {1}{r^{3}}}{\vec {r}}$ for a line source laid out in a ring in the x-z plane with a linear density of 1 unit mass/unit length and radius of 1.

${\vec {F}}\left({\vec {r}}\right)=\oint \limits _{C}{{\frac {{\vec {r}}-{\vec {s}}}{\left|{\vec {r}}-{\vec {s}}\right|^{3}}}\,\mathrm {d} {\vec {s}}}=\int \limits _{0}^{2\pi }{{\frac {{\vec {r}}-{\vec {s}}\left(1,\varphi ,0\right)}{\left|{\vec {r}}-{\vec {s}}\left(1,\varphi ,0\right)\right|^{3}}}\,\mathrm {d} \varphi }$

A Cylindrical coordinate system, (ρ, φ, z) is used to specify the ring and give the answer.

$\rho =1$

$z=0$

Cartesian Formulation

The above equation yields a fully mathematical problem by substituting for $\varphi$ according to the coordinate system transformation.

$x=\rho \cos \varphi =\cos \varphi$

$y=\rho \sin \varphi =\sin \varphi$

${\vec {s}}\left(1,\varphi ,0\right)_{cylindrical}=(\cos \varphi ,\sin \varphi ,0)_{Cartesian}$

This is used at the while $r=(x,0,z)=(\rho ,0,z)$, (given x>0) is substituted in. This means that the equations will calculate the field only for single xz plane. Then, any other given plane about the z-axis will have the same form so the full problem will be addressed by symmetry after this step is completed.

${\vec {F}}\left({\vec {r}}\right)=\int \limits _{0}^{2\pi }{{\frac {(\rho ,0,z)-(\cos \varphi ,\sin \varphi ,0)}{\left|(\rho ,0,z)-(\cos \varphi ,\sin \varphi ,0)\right|^{3}}}\,\mathrm {d} \varphi }=\int \limits _{0}^{2\pi }{{\frac {(\rho -\cos \varphi ,\sin \varphi ,z)}{\left|(\rho -\cos \varphi ,-\sin \varphi ,z)\right|^{3}}}\,\mathrm {d} \varphi }$

Notation is changed.

${\vec {F}}\left({\vec {r}}\right)=\int \limits _{0}^{2\pi }{{\frac {(\rho -\cos \varphi ,y-\sin \varphi ,z)}{\left((\rho -\cos \varphi )^{2}+\sin ^{2}\varphi +z^{2}\right)^{(3/2)}}}\,\mathrm {d} \varphi }=\int \limits _{0}^{2\pi }{{\frac {(\rho -\cos \varphi ,y-\sin \varphi ,z)}{\left(\rho ^{2}-2\rho \cos \varphi +z^{2}\right)^{(3/2)}}}\,\mathrm {d} \varphi }$

Solution

An intermediate quantity is introduced here.

$\Phi \left({\rho ,z}\right)={\frac {1}{\sqrt {\left({\rho +1}\right)^{2}+z^{2}}}}$

The equations giving the integral are given here, these are the result of the integral specified above. The integral is in the form of a vector defined to be $(g'_{\rho },g'_{\varphi },g'_{z})$.

$g'_{\rho }={\frac {2\Phi }{\rho }}\left({E\left(\Phi 2{\sqrt {\rho }}\right){\frac {\rho ^{2}-1-z^{2}}{\left({\rho -1}\right)^{2}+z^{2}}}+K\left(\Phi 2{\sqrt {\rho }}\right)}\right)$

$g'_{\varphi }=0$

$g'_{z}={\frac {4\Phi E\left(\Phi 2{\sqrt {\rho }}\right)z}{\left({\rho -1}\right)^{2}+z^{2}}}$

This uses the Complete elliptic integral of the first kind, $K\left(\Phi \right)$ and the Complete elliptic integral of the second kind, $E\left(\Phi \right)$. The above gives a field as a function of the cylindrical coordinates, therefore the field can be computed given a set of Cartesian coordinates given that a function to find the cylindrical coordinates as a function of the Cartesian coordinates exists. Such a transformation is trivial regarding axial position, or z, and the radial distance, or $\rho$, is the computed distance from origin in the xy-plane.

$\rho ={\sqrt {x^{2}+y^{2}}}$

The azimuth, $\varphi$, is erroneous for the set of equations here. In pretty much every case discussed there is full rotational invariance about the z-axis.

Simplifications

A new value is introduced here, denoted just $r$, which represents the distance of a point from the center of the torus tube measured in its plane about the z-axis.

$r={\sqrt {(\rho -1)^{2}+z^{2}}}$

Objectively, it would be nice to get the prior equation for field in terms of this quantity. In particular, the argument for the elliptic functions, $\Phi 2{\sqrt {\rho }}$ must eliminate dependencies on $\rho$ and $z$ using the definition of $r$ above. This is done by taking a 3rd Taylor expansion of the quantity about the point $(\rho ,z)=(1,0)$. There are no 2nd order dependencies, and the expression resolves neatly to a value of 1 at that point. It is important to recognize that the Elliptic functions are undefined at 1.

$\Phi 2{\sqrt {\rho }}=1-{\frac {(\rho -1)^{2}}{8}}-{\frac {z}{8}}+O((\rho -1)^{3})+O(z^{3})\approx 1-{\frac {z^{2}}{8}}$

This is a very convenient approximation to use in order to do the next integral, in which we define the average net field at r from centerline by integrating over a new angle, denoted $\psi$. The integral still yields a very messy answer, but an approximation is applied to a simplified version.

$\rho =1+r\cos \psi$

$z=r\sin \psi$

${\overrightarrow {G}}(r)={\frac {1}{2\pi }}\int _{0}^{2\pi }\!(g'_{\rho },g'_{\varphi },g'_{z})\,d\psi \,=(-3/2+3*ln(2)-ln(r))+O(r^{2})$

Second solution

In this case the problem will be addressed with variables including $\lambda$, the charge per unit length, $R$, the radius of the circle, $k$ Coulomb's constant. In addition to these, two other variables that will be referenced are:

$r={\sqrt {(\rho -R)^{2}+z^{2}}}$

$l={\sqrt {(\rho +R)^{2}+z^{2}}}$

And these represent the closest distance to a point on the ring and furthest distance to a point on the ring, given some coordinates. Next, the electric potential is found using the $1/r$ integral because it comes out to be the most simple and is the easiest to derive a meaningful answer from.

$E(\rho ,z)={\frac {4Rk\lambda }{l}}K\left({\frac {2{\sqrt {R\rho }}}{l}}\right)$

The main challenge is simplifying this to a new form that can have a different approach applied to it. Given the assumption that $\rho <<R$ the following can be obtained.

$E(\rho ,z)=2k\lambda \left(ln({\frac {8R}{r}})+{\frac {\rho -R}{R}}\left(1-ln({\frac {8R}{r}})\right)\right)$

This allows the potential to be broken up into two forms. Keeping in mind that the offset is irrelevant as potential represents a relative value, the first term exactly represents the potential that would exist due to an infinite line charge, which is expected. The 2nd term then represents an effect from the larger ring.

$E_{line}(\rho ,z)=2k\lambda \left(ln({\frac {8R}{r}})\right)$

$E_{loop}(\rho ,z)=2k\lambda \left({\frac {\rho -R}{R}}\left(1-ln({\frac {8R}{r}})\right)\right)$

From here, the effective field acting on a differential length of a part of the charged wire of radius r can be assessed. It is known that the field from the local wire will not have an effect on this from the infinite line of charge example. So the calculations only need to be done with $E_{ring}$ and they only need to be done in the $\rho$ direction because we know that the loop is putting outward tension and not any in the up or down direction. Doing this requires taking the gradient of the electric potential in order to get the electric field, then integrating over a loop r away from the centerline.

$\rho (r,\psi )=R+r\cos {\psi }$

$z(r,\psi )=r\sin {\psi }$

$F_{effective}={\frac {1}{2\pi }}\int _{0}^{2\pi }{\frac {d}{d\rho }}E_{loop}(\rho (r,\psi ),z(r,\psi ))d\psi ={\frac {\lambda k}{R}}(ln({\frac {8R}{r}})-{\frac {3}{2}})$

Application

The conventional form of gravity can be stated in the following ways.

$\mathbf {F} _{g}=m\mathbf {g} (\mathbf {r} )$

where

m is the mass of a particle,

r is the position vector of the particle.

For a point source

$\mathbf {g} (\mathbf {r} )=-GM{\frac {\mathbf {e_{r}} }{r^{2}}}$

where

e_r is the radial unit vector,

r is the radius, |r|.

M is the mass of the particle, which is assumed to be a point mass located at the origin.

The distributed case is done for a volume as follows, and then narrowed to the specific case of a thin ring thus using a linear density.

$\mathbf {g} (\mathbf {r} )=-G\int _{V}{\frac {\rho (\mathbf {s} )(\mathbf {r} -\mathbf {s} )}{|\mathbf {r} -\mathbf {s} |^{3}}}dV(\mathbf {s} )\approx -G\int _{S}{\frac {\alpha (\mathbf {s} )(\mathbf {r} -\mathbf {s} )}{|\mathbf {r} -\mathbf {s} |^{3}}}dS(\mathbf {s} )$

where

$\rho$ is the volumetric density

$\alpha$ is the linear density

Radius Scaling

The radius was assumed to be 1 before, this needs to be scaled while preserving the linear mass density.

${\vec {g'}}=f\left(x,y\right)$