# Lagrangian classical action for particle with constant force

1. Feb 23, 2014

### micole

1. The problem statement, all variables and given/known data
for particle with lagrangian L = m/2 dx/dt^2 + fx where x is constant force, what is ScL (classical action)

2. Relevant equations
d/dt (∂L/∂(dx/dt)) = ∂L/∂x
ScL = ∫m/2 dx/dt^2 + fx dt from ti to tf

3. 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 im 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 thats 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 im assuming im 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

2. Feb 24, 2014

### micole

3. Feb 24, 2014

### hilbert2

Well, the action is just the time integral of the Lagrangian over the time interval that is being considered:

$S=\int_{t_{1}}^{t_{2}}L(t)dt=\int_{t_{1}}^{t_{2}}\left[\frac{m}{2}\left(\frac{dx(t)}{dt}\right)^{2}+Fx(t)\right]dt$, where $F$ is the constant force.

As it's basic physics that the trajectory of the particle is $x(t)=x_{0}+v_{0}t+\frac{F}{2m}t^{2}$, it's easy to calculate the integral provided that the force $F$, the initial position and velocity $x_{0},v_{0}$ and the time interval $[t_{1},t_{2}]$ are known.

4. Feb 25, 2014

### micole

Thanks hilbert. None of those are known. If you take a look at my work you see that what we are being asked is to simplify this integral from a long function of t (as you and I both indicate, it's basic physics to write x as a function of t) into a succinct statement. The point is not to substitute numbers (see final paragraph of my question)
Thanks for any help

5. Feb 25, 2014

### micole

And I see an error in my statement. The coefficient of integral of Fx is (2-m)/2, not m/2

6. Feb 25, 2014

### hilbert2

How is that possible? You are subtracting a dimensional quantity $m$ from a bare number $2$.

I'm not sure how the integral $\int_{t_{1}}^{t_{2}}\left[\frac{m}{2}\left(\frac{dx(t)}{dt}\right)^{2}+Fx(t)\right]dt$ could be simplified without solving for $x(t)$ with the initial conditions $x_{0}$, $v_{0}$ known. Integration by parts doesn't seem to help.

7. Feb 25, 2014

### micole

Me neither, exactly my question

Last edited: Feb 25, 2014
8. Feb 25, 2014

### Goddar

Hi.
Is this the exact problem statement? It doesn't suggest that you should get a particularly succinct (or enlightening) result... You are given a specific potential, so you're indeed supposed to solve for x(t) in terms of a (= the constant acceleration f/m), v0 and x0, then integrate.
You can always make abbreviations like L[x(t0)] in your final result, but you'll always get something with at least four terms and no immediate connection to something familiar.

The thing is, this kind of Lagrangian is defined as L = T – V, where T and V are determined by your choice of coordinates; then it's the principle of least action that gives you the familiar equations of motion upon varying S and you would be going in circles by trying to get something edifying from S. It's just what it's supposed to be: its specific form may look uneventful but that's just how things are sometimes... :)