From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Lagrangian formalism in general relativity Date: 20 May 2000 13:19:05 -0700 Newsgroups: sci.physics.research Summary: [missing] In article <9V4q4VAyEkI5EwbP@upthorpe.demon.co.uk>, Oz wrote: >In article <8fs3i5$din$1@pravda.ucr.edu>, John Baez > writes >>"For a classical point particle, the LAGRANGIAN is the kinetic energy >>minus the potential energy. We integrate the Lagrangian over time to >>get the ACTION. Thus to any little stretch of time we associate a little >>bit of action; we total these up to get the whole action. >Would a little simple math be easier for me to follow? >Nothing too difficult you understand. Yeah, I think you gotta see how this trick works to really believe that it DOES work. >Lets see if I can get what you are on about. We have some particle mass >m at point x in a uniform grav field. In time dt it has a variety of >choices of places to go. It can get to x+vdt. I assume v is >unconstrained and for small dt is constant. Dimensions 1. > >Now, being hugely simplistic lets take the Lagrangian as L=1/2mv^2 - mgh >so the action is > >dS = Ldt = 1/2m(v^2)dt -mg(vdt)dt Okay, the first problem here is that we're really gonna need a formula for the TOTAL action - folks call it S so I've changed your notation to reflect that - not just a little bit of action dS. The second problem is that "(vdt)dt" gives me the chills - it gets pretty confusing when you start squaring something like dt, and besides, you seem to have got this term by setting h = vdt, which is an equation coming out of nowhere. >Now I want dS to be a minimum. Er with respect to v I guess. What you really want is for S to be a minimum, with respect to all paths x(t) that start and end at fixed locations at fixed times. This stuff is called "variational calculus" - we're gonna take a derivative of S, not with respect to a number, but to the whole path x(t). Let me show you how it goes. Your calculation was actually sorta close to the one I'm gonna do. In what follows, "D" will stand for the variational derivative with respect to x(t). I'm not gonna explain the steps, I'll just do 'em and you see if you can figure it out. By the way, I'm not gonna use the variable "h" - I'm using x(t) to stand for the height of the rock at time t, and I'm using v(t) to stand for its velocity at time t, which is just x'(t). Here goes: L(t) = (m/2) v(t)^2 - mg x(t) S = integral L(t) dt = integral [(m/2) v(t)^2 - mg x(t)] dt DS = D integral [(m/2) v(t)^2 - mg x(t)] dt = integral D [(m/2) v(t)^2 - mg x(t)] dt = integral [(m/2) D v(t)^2 - mg Dx(t)] dt = integral [(m/2) 2 v(t) Dv(t) - mg Dx(t)] dt = integral [m x'(t) D dx/dt - mg Dx(t)] dt = integral [m x'(t) (d/dt) Dx(t) - mg Dx(t)] dt (*) = integral [-m x''(t) Dx(t) - mg Dx(t)] dt = integral -m[x''(t) + mg] Dx(t) dt Now we want DS = 0 no matter what the little change in our path, Dx(t), happens to be. So we must have x''(t) + mg = 0 or in other words x''(t) = -mg !!!! The crucial tricky step in the calculation was the step right after the line I marked (*). Do you see how that step works? All the magic happens right there. ============================================================================== From: toby@ugcs.caltech.edu (Toby Bartels) Subject: Re: Lagrangian formalism in general relativity Date: 21 May 2000 07:51:10 GMT Newsgroups: sci.physics.research John Baez wrote at last: > = integral [-m x''(t) Dx(t) - mg Dx(t)] dt > = integral -m[x''(t) + mg] Dx(t) dt >Now we want DS = 0 no matter what the little change in our path, >Dx(t), happens to be. So we must have >x''(t) + mg = 0 >or in other words >x''(t) = -mg !!!! Wow, John! you've just rederived Aristoteles's result that the motion of a falling body depends on its mass. (Oz, check that factorisation at the beginning of the quoted text.) -- Toby toby@ugcs.caltech.edu