Talk:PlanetPhysics/Conservation of Linear Momentum
Original TeX Content from PlanetPhysics Archive[edit source]
%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: conservation of linear momentum %%% Primary Category Code: 45.50.-j %%% Filename: ConservationOfLinearMomentum.tex %%% Version: 1 %%% Owner: mdo %%% Author(s): mdo %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}
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\textit{If the net \htmladdnormallink{force}{http://planetphysics.us/encyclopedia/Thrust.html} acting on a \htmladdnormallink{particle}{http://planetphysics.us/encyclopedia/Particle.html} is zero, then to total linear \htmladdnormallink{momentum}{http://planetphysics.us/encyclopedia/Momentum.html} of that particle is conserved. }
Recall that the momentum of a particle of \htmladdnormallink{mass}{http://planetphysics.us/encyclopedia/Mass.html} $m$ is given by $\mathbf{p} = md\mathbf{r}/dt$, where $\mathbf{r}$ is the \htmladdnormallink{position vector}{http://planetphysics.us/encyclopedia/PositionVector.html} of that particle. Now, by Newton's First Law, if no force is acting on a particle, then it will remain in constant (or zero) \htmladdnormallink{velocity}{http://planetphysics.us/encyclopedia/Velocity.html}. This can be phrased mathematically as $\mathbf{F} = d\mathbf{p}/dt = 0$, which leads directly to the desired result; that if the net force acting upon a particle is zero, then the total linear momentum of that particle is constant in time, and hence, conserved.
The above result can be taken further, by considering what happens when the net force in a given direction is zero. Let $\mathbf{s}$ be a \htmladdnormallink{vector}{http://planetphysics.us/encyclopedia/Vectors.html} such that $\mathbf{F}\cdot\mathbf{s}=0$. This means that in the direction of the vector $\mathbf{s}$, the force vanishes. Substituting the \htmladdnormallink{relation}{http://planetphysics.us/encyclopedia/Bijective.html} between force and momentum into the equation, it is seen that $d\mathbf{p}/dt\cdot\mathbf{s} = 0$. Assume that $\mathbf{s}$ is independent of time, and integrate this equation. Trivially, the result is that $\mathbf{p}\cdot\mathbf{s} = c$, where $c$ is a constant. This result means that in the direction of $\mathbf{s}$, the component of total linear momentum is conserved. Since the force vanishes in this direction, it means that the component of momentum in the direction of vanishing force is conserved.
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