- #1

micole

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## Homework Statement

for particle with lagrangian L = m/2 dx/dt^2 + fx where x is constant force, what is ScL (classical action)

## Homework Equations

d/dt (∂L/∂(dx/dt)) = ∂L/∂x

ScL = ∫m/2 dx/dt^2 + fx dt from ti to tf

## The Attempt at a Solution

d/dt (∂L/∂(dx/dt)) = ∂L/∂x implies f = md2x/dt2 (f = ma)

on aside, from this you can calculate the familiar equations

dx/dt (t) = f/m(t) + dx/dt(0) or v(t) = at + v0

x(t) = 1/2 f/m t^2 + dx/dt(0) t + x(0)

Question: is it helpful to define ti as zero so Xi = x(ti) = X(0) = 0? And if I do this, is my

solution generalizable? I think yes

Scl = ∫m/2 dx/dt^2 + fx dt

for ∫m/2 dx/dt^2 I integrate by parts with

u =dx/dt --> du = (d2x/dt2) dt

and

dv = dx/dt dt --> v = x

Scl = m/2 (dx/dt)(x)

*from ti to tf*-m/2∫fx dt + ∫fx dt

Scl = m/2 {(xf )(f/m tf +dx/dt(0)) - xi(f/m ti + dx/dt(0))} + m/2 ∫fx dt

where xi(f/m ti + dx/dt(0)) goes away if its okay to set xi = x(0) = 0

so potentially I am left with

Scl = m/2 (xf )(f/m tf +dx/dt(0)) + fm/2 ∫x dt

I have X as a function of t so i can integrate the right hand term ( ∫x dt) but I end up with a big equation that's a function with terms t^3/3 and many other terms. Most of the solutions to these problems have ended up simple and elegant so I am assuming I am either making an error or there is a slick math trick that simplifies by function of ti, tf and xf, xi into something lovely.

Thats the part i would like help on please

Thanks very much in advance