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Flow Curves

Given a vector field, F(r)\boldsymbol{F}(\boldsymbol{r}) parameterized by r(t)\boldsymbol{r}(t), the tangent vector is equal to the vector field:

dr(t)dt=F(r(t))\frac{d\boldsymbol{r}(t)}{dt} = \boldsymbol{F}(\boldsymbol{r}(t))

For 2D coordinates, this would imply:

dxdtx^+dydty^=Fx(x(t),y(t))x^+Fy(x(t),y(t))y^,dxdt=Fx(x(t),y(t)),dydt=Fy(x(t),y(t)). \begin{align*} \frac{dx}{dt} \boldsymbol{\hat{x}} + \frac{dy}{dt} \boldsymbol{\hat{y}} &= F_x(x(t), y(t)) \boldsymbol{\hat{x}} + F_y(x(t), y(t)) \boldsymbol{\hat{y}}, \\[1.5ex] \frac{dx}{dt} &= F_x(x(t), y(t)), \\ \frac{dy}{dt} &= F_y(x(t), y(t)). \end{align*}

Consider the following vector field:

F(x,y)=−yx^+xy^\boldsymbol{F}(x, y) = -y \boldsymbol{\hat{x}} + x \boldsymbol{\hat{y}}

Constructing a system of two differential equations:

dxdt=−y,dydt=x. \begin{align*} \frac{dx}{dt} &= -y, \\ \frac{dy}{dt} &= x. \end{align*}

The flow curves seem to plot circles, the parameterized coordinates are equal to:

x(t)=acos⁡t,y(t)=asin⁡t. \begin{align*} x(t) &= a \cos t, \\ y(t) &= a \sin t. \end{align*}

The derivatives are equal to:

dxdt=−asin⁡t,dydt=acos⁡t. \begin{align*} \frac{dx}{dt} &= - a \sin t, \\ \frac{dy}{dt} &= a \cos t. \end{align*}

This satisfies the original differential equations.

The inverse square law in vector form is equal to:

F(rs)=krs2r^s=krsrs3, \begin{align*} \boldsymbol{F}(\boldsymbol{r}_s) &= \frac{k}{r_s^2} \boldsymbol{\hat{r}_s} \\ &= k \frac{\boldsymbol{r}_s}{r_s^3}, \end{align*}

where kk is a constant.

The vector rs\boldsymbol{r}_s is the separation vector:

rs=r−rp,\boldsymbol{r}_s = \boldsymbol{r} - \boldsymbol{r}_p,

where r\boldsymbol{r} is the current position that we are evaluating at and rp\boldsymbol{r}_p is a constant position of the point creating the field.

Converting to 3D cartesian coordinates:

rs=r−rp=(x−xp)x^+(y−yp)y^+(z−zp)z^,rs=(x−xp)2+(y−yp)2+(z−zp)2,F(x,y,z)=k(x−xp)x^+(y−yp)y^+(z−zp)z^rs3=kx−xprs3x^+ky−yprs3y^+kz−zprs3z^. \begin{align*} \boldsymbol{r}_s &= \boldsymbol{r} - \boldsymbol{r}_p \\ &= (x - x_p) \boldsymbol{\hat{x}} + (y - y_p) \boldsymbol{\hat{y}} + (z - z_p) \boldsymbol{\hat{z}}, \\ r_s &= \sqrt{(x - x_p)^2 + (y - y_p)^2 + (z - z_p)^2}, \\ \boldsymbol{F}(x, y, z) &= k \frac{(x - x_p) \boldsymbol{\hat{x}} + (y - y_p) \boldsymbol{\hat{y}} + (z - z_p) \boldsymbol{\hat{z}}}{r_s^3} \\ &= k \frac{x - x_p}{r_s^3} \boldsymbol{\hat{x}} + k \frac{y - y_p}{r_s^3} \boldsymbol{\hat{y}} + k \frac{z - z_p}{r_s^3} \boldsymbol{\hat{z}}. \end{align*}

Separating into a system of differential equations:

dxdt=kx−xprs3,dydt=ky−yprs3,dzdt=kz−zprs3. \begin{align*} \frac{dx}{dt} &= k \frac{x - x_p}{r_s^3}, \\ \frac{dy}{dt} &= k \frac{y - y_p}{r_s^3}, \\ \frac{dz}{dt} &= k \frac{z - z_p}{r_s^3}. \end{align*}

In 2D, the zz coordinate is always zero for every object.

Dividing the third equation by the second:

dzdy=z−zpy−yp,∫z0zdzz−zp=∫y0ydyy−yp,[ln⁡z−zp∣]z0z=[ln⁡∣y−yp∣]y0y,ln⁡∣z−zpz0−zp∣=ln⁡∣y−ypy0−yp∣,∣z−zpz0−zp∣=∣y−ypy0−yp∣,∣z−zp∣=∣(y−yp)(z0−zp)y0−yp∣,z=zp±(y−yp)(z0−zp)y0−yp. \begin{align*} \frac{dz}{dy} &= \frac{z - z_p}{y - y_p}, \\ \int_{z_0}^z \frac{dz}{z - z_p} &= \int_{y_0}^y \frac{dy}{y - y_p}, \\ \left[\ln z - z_p|\right]_{z_0}^z &= \left[\ln |y - y_p|\right]_{y_0}^y, \\ \ln \left|\frac{z - z_p}{z_0 - z_p}\right| &= \ln \left|\frac{y - y_p}{y_0 - y_p}\right|, \\ \left|\frac{z - z_p}{z_0 - z_p}\right| &= \left|\frac{y - y_p}{y_0 - y_p}\right|, \\ |z - z_p| &= \left|\frac{(y - y_p) (z_0 - z_p)}{y_0 - y_p}\right|, \\ z &= z_p \pm \frac{(y - y_p) (z_0 - z_p)}{y_0 - y_p}. \end{align*}

Dividing the second equation by the first:

dydx=y−ypx−xp,∫y0ydyy−yp=∫x0xdxx−xp,[ln⁡∣y−yp∣]y0y=[ln⁡∣x−xp∣]x0x,ln⁡∣y−ypy0−yp∣=ln⁡∣x−xpx0−xp∣,∣y−ypy0−yp∣=∣x−xpx0−xp∣,∣y−yp∣=∣(x−xp)(y0−yp)x0−xp∣,y=yp±(x−xp)(y0−yp)x0−xp. \begin{align*} \frac{dy}{dx} &= \frac{y - y_p}{x - x_p}, \\ \int_{y_0}^y \frac{dy}{y - y_p} &= \int_{x_0}^x \frac{dx}{x - x_p}, \\ \left[\ln |y - y_p|\right]_{y_0}^y &= \left[\ln |x - x_p|\right]_{x_0}^x, \\ \ln \left|\frac{y - y_p}{y_0 - y_p}\right| &= \ln \left|\frac{x - x_p}{x_0 - x_p}\right|, \\ \left|\frac{y - y_p}{y_0 - y_p}\right| &= \left|\frac{x - x_p}{x_0 - x_p}\right|, \\ |y - y_p| &= \left|\frac{(x - x_p) (y_0 - y_p)}{x_0 - x_p}\right|, \\ y &= y_p \pm \frac{(x - x_p) (y_0 - y_p)}{x_0 - x_p}. \end{align*}

Substituting into the equation for zz:

z=zp±(yp±(x−xp)(y0−yp)x0−xp−yp)(z0−zp)y0−yp=zp±±(x−xp)(y0−yp)x0−xp(z0−zp)y0−yp=zp+(x−xp)(y0−yp)(z0−zp)(x0−xp)(y0−yp)=zp+(x−xp)(z0−zp)x0−xp. \begin{align*} z &= z_p \pm \frac{(y_p \pm \frac{(x - x_p) (y_0 - y_p)}{x_0 - x_p} - y_p) (z_0 - z_p)}{y_0 - y_p} \\ &= z_p \pm \frac{\pm \frac{(x - x_p) (y_0 - y_p)}{x_0 - x_p} (z_0 - z_p)}{y_0 - y_p} \\ &= z_p + \frac{(x - x_p) (y_0 - y_p) (z_0 - z_p)}{(x_0 - x_p) (y_0 - y_p)} \\ &= z_p + \frac{(x - x_p) (z_0 - z_p)}{x_0 - x_p}. \end{align*}

Substituting into the equation for rr:

r=(x−xp)2+(yp±(x−xp)(y0−yp)x0−xp−yp)2+(zp+(x−xp)(z0−zp)x0−xp−zp)2=(x−xp)2+(±(x−xp)(y0−yp)x0−xp)2+((x−xp)(z0−zp)x0−xp)2=(x−xp)1+(y0−yp)2(x0−xp)2+(z0−zp)2(x0−xp)2=(x−xp)(x0−xp)2+(y0−yp)2+(z0−zp)2(x0−xp)2=(x−xp)rs,0x0−xp. \begin{align*} r &= \sqrt{(x - x_p)^2 + \left(y_p \pm \frac{(x - x_p) (y_0 - y_p)}{x_0 - x_p} - y_p\right)^2 + \left(z_p + \frac{(x - x_p) (z_0 - z_p)}{x_0 - x_p} - z_p\right)^2} \\ &= \sqrt{(x - x_p)^2 + \left(\pm \frac{(x - x_p) (y_0 - y_p)}{x_0 - x_p}\right)^2 + \left(\frac{(x - x_p) (z_0 - z_p)}{x_0 - x_p}\right)^2} \\ &= (x - x_p) \sqrt{1 + \frac{(y_0 - y_p)^2}{(x_0 - x_p)^2} + \frac{(z_0 - z_p)^2}{(x_0 - x_p)^2}} \\ &= (x - x_p) \sqrt{\frac{(x_0 - x_p)^2 + (y_0 - y_p)^2 + (z_0 - z_p)^2}{(x_0 - x_p)^2}} \\ &= (x - x_p) \frac{r_{s,0}}{x_0 - x_p}. \end{align*}

Substituting into the equation for dxdt\frac{dx}{dt} and solving:

dxdt=kx−xp((x−xp)rs,0x0−xp)3=kx−xp(x−xp)3rs,03(x0−xp)3=k(x0−xp)3(x−xp)2rs,03,dtdx=rs,03k(x0−xp)3(x−xp)2,∫0tdt=rs,03k(x0−xp)3∫x0x(x−xp)2dx,t=rs,03k(x0−xp)3[13(x−xp)3]x0x=rs,033k(x0−xp)3((x−xp)3−(x0−xp)3)=rs,033k(x0−xp)3(x−xp)3−rs,033k,rs,033k(x0−xp)3(x−xp)3=t+rs,033k,(x−xp)3=3k(x0−xp)3rs,03t+(x0−xp)3,x−xp=x0−xprs,03kt+rs,033,x=xp+3kt+rs,033x0−xprs,0. \begin{align*} \frac{dx}{dt} &= k \frac{x - x_p}{\left((x - x_p) \frac{r_{s,0}}{x_0 - x_p}\right)^3} \\ &= k \frac{x - x_p}{(x - x_p)^3 \frac{r_{s,0}^3}{(x_0 - x_p)^3}} \\ &= k \frac{(x_0 - x_p)^3}{(x - x_p)^2 r_{s,0}^3}, \\ \frac{dt}{dx} &= \frac{r_{s,0}^3}{k(x_0 - x_p)^3} (x - x_p)^2, \\ \int_0^t dt &= \frac{r_{s,0}^3}{k(x_0 - x_p)^3} \int_{x_0}^x (x - x_p)^2 dx, \\ t &= \frac{r_{s,0}^3}{k(x_0 - x_p)^3} \left[\frac{1}{3} (x - x_p)^3\right]_{x_0}^x \\ &= \frac{r_{s,0}^3}{3k(x_0 - x_p)^3} \left((x - x_p)^3 - (x_0 - x_p)^3\right) \\ &= \frac{r_{s,0}^3}{3k(x_0 - x_p)^3} (x - x_p)^3 - \frac{r_{s,0}^3}{3k}, \\ \frac{r_{s,0}^3}{3k(x_0 - x_p)^3} (x - x_p)^3 &= t + \frac{r_{s,0}^3}{3k}, \\ (x - x_p)^3 &= \frac{3k(x_0 - x_p)^3}{r_{s,0}^3}t + (x_0 - x_p)^3, \\ x - x_p &= \frac{x_0 - x_p}{r_{s,0}} \sqrt[3]{3kt + r_{s,0}^3}, \\ x &= x_p + \sqrt[3]{3kt + r_{s,0}^3} \frac{x_0 - x_p}{r_{s,0}}. \end{align*}

Substituting into the equation for yy:

y=yp±(x0−xprs,03kt+rs,033+xp−xp)(y0−yp)x0−xp=yp±x0−xprs,03kt+rs,033(y0−yp)x0−xp=yp±3kt+rs,033y0−yprs,0. \begin{align*} y &= y_p \pm \frac{(\frac{x_0 - x_p}{r_{s,0}} \sqrt[3]{3kt + r_{s,0}^3} + x_p - x_p) (y_0 - y_p)}{x_0 - x_p} \\ &= y_p \pm \frac{\frac{x_0 - x_p}{r_{s,0}} \sqrt[3]{3kt + r_{s,0}^3} (y_0 - y_p)}{x_0 - x_p} \\ &= y_p \pm \sqrt[3]{3kt + r_{s,0}^3} \frac{y_0 - y_p}{r_{s,0}}. \end{align*}

Finally, substituting into the equation for zz:

z=zp+(xp+x0−xprs,03kt+rs,033−xp)(z0−zp)x0−xp=zp+x0−xprs,03kt+rs,033(z0−zp)x0−xp=zp+3kt+rs,033z0−zprs,0. \begin{align*} z &= z_p + \frac{(x_p + \frac{x_0 - x_p}{r_{s,0}} \sqrt[3]{3kt + r_{s,0}^3} - x_p) (z_0 - z_p)}{x_0 - x_p} \\ &= z_p + \frac{\frac{x_0 - x_p}{r_{s,0}} \sqrt[3]{3kt + r_{s,0}^3} (z_0 - z_p)}{x_0 - x_p} \\ &= z_p + \sqrt[3]{3kt + r_{s,0}^3} \frac{z_0 - z_p}{r_{s,0}}. \end{align*}

The parameterized coordinates are equal to (choosing the plus sign):

x(t)=xp+3kt+rs,033x0−xprs,0,y(t)=yp+3kt+rs,033y0−yprs,0,z(t)=zp+3kt+rs,033z0−zprs,0. \begin{align*} x(t) &= x_p + \sqrt[3]{3kt + r_{s,0}^3} \frac{x_0 - x_p}{r_{s,0}}, \\ y(t) &= y_p + \sqrt[3]{3kt + r_{s,0}^3} \frac{y_0 - y_p}{r_{s,0}}, \\ z(t) &= z_p + \sqrt[3]{3kt + r_{s,0}^3} \frac{z_0 - z_p}{r_{s,0}}. \end{align*}

Or in vector form:

r(t)=rp+3kt+rs,033r^s,0,r^s,0=rs,0rs,0,rs,0=r0−rp, \begin{align*} \boldsymbol{r}(t) &= \boldsymbol{r}_p + \sqrt[3]{3kt + r_{s,0}^3} \boldsymbol{\hat{r}_{s,0}}, \\ \boldsymbol{\hat{r}_{s,0}} &= \frac{\boldsymbol{r}_{s,0}}{r_{s,0}}, \\ \boldsymbol{r}_{s,0} &= \boldsymbol{r}_0 - \boldsymbol{r}_{p}, \end{align*}

where rp\boldsymbol{r}_p is the position vector of the source and r0\boldsymbol{r}_0 is the initial position vector.