Simple Nature - Light and Matter

af / Breaking trail, by WalterE. Bohl. The pack horse is notdoing any work on the pack,because the pack is moving in ahorizontal line at constant speed,and therefore there is no kineticor gravitational energy beingtransferred into or out of it.which is a geometric interpretation for the dot product.The result of example 76 is very useful. It gives us a way tofind the angle between two vectors if we know their components. Itcan be used to show that the dot product of any two perpendicularvectors is zero. It also leads to a nifty proof that the dot productis rotationally invariant — up until now I’ve only proved that if arotationally invariant product exists, the dot product is it — becauseangles and lengths aren’t affected by a rotation, so the right side ofthe equation is rotationally invariant, and therefore so is the leftside.I introduced the whole discussion of the dot product by way ofgeneralizing the equation dE = F dx to three dimensions. In termsof a dot product, we havedE = F · dr .If F is a constant, integrating both sides gives∆E = F · ∆r .(If that step seemed like black magic, try writing it out in termsof components.) If the force is perpendicular to the motion, as infigure af, then the work done is zero. The pack horse is doing workwithin its own body, but is not doing work on the pack.Pushing a lawnmower example 77⊲ I push a lawnmower with a force F=(110 N)ˆx − (40 N)ŷ, and thetotal distance I travel is (100 m)ˆx. How much work do I do?⊲ The dot product is 11000 N·m = 11000 J.A good application of the dot product is to allow us to write asimple, streamlined proof of separate conservation of the momentumcomponents. (You can skip the proof without losing the continuityof the text.) The argument is a generalization of the one-dimensionalproof on page 130, and makes the same assumption about the typeof system of particles we’re dealing with. The kinetic energy ofone of the particles is (1/2)mv · v, and when we transform into adifferent frame of reference moving with velocity u relative to theoriginal frame, the one-dimensional rule v → v + u turns into vectoraddition, v → v + u. In the new frame of reference, the kineticenergy is (1/2)m(v + u) · (v + u). For a system of n particles, wehaven∑ 1K =2 m j(v j + u) · (v j + u)j=1⎡= 1 n∑n∑⎣ m j v j · v j + 2 m j v j · u +2j=1j=1⎤n∑m j u · u⎦ .j=1214 Chapter 3 Conservation of Momentum