Lagrangian mech.: Action for a particle under constant force

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Homework Help Overview

The discussion revolves around finding the action for a particle under a constant force using Lagrangian mechanics. The original poster presents a Lagrangian and attempts to derive the action, expressing confusion about the dependencies involved in the integration process.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the Euler-Lagrange equation and integrate the Lagrangian but struggles with the dependencies of variables. Some participants question the algebraic manipulations made by the original poster and suggest solving the differential equation for position and velocity as a next step.

Discussion Status

Participants are actively engaging with the original poster's attempts, providing corrections and guidance on the proper approach to derive the equations of motion. There is a recognition of algebraic mistakes, and some participants offer suggestions for how to proceed with the integration once the correct expressions for position and velocity are established.

Contextual Notes

The original poster expresses uncertainty about the implicit and explicit dependencies in the Lagrangian, indicating a potential gap in understanding the foundational concepts of Lagrangian mechanics.

devaluator
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Homework Statement


Find Scl for a particle under constant force f, that is:

L = (m/2)v2 + fx


Homework Equations



S = ∫Ldt

d(∂L/∂[itex]q^{.}[/itex])/dt = ∂L/∂q


The Attempt at a Solution



Apologies if this belongs in the Introductory Physics section. Apologies for terrible formatting.

∂L/∂v = mv
d(mv)/dt = ma

∂L/∂x = f

From the Euler eqn it follows that f = ma... surprise surprise!

This makes the Lagrangian:

L = (m/2)v2 + m.a.x

So when we integrate Ldt over ta and tb the dependencies confuse me... the m.a.x part, which is also f.x., should have the m.a (or f) just taken out of the integral because it's constant over time. On the left hand side m/2 is constant so it can be taken out. This leaves me with

S = m^2 * a/2 ∫ (v^2)xdt

I know this is wrong, but I can't tell you why... even if it's not wrong I have no idea on how to continue because Lagrangian mechanics confuses me on implicit / explicit dependencies. I can't tell what is a function of what and which one can be taken as a constant over dt. I'd appreciate any help since I'm lost on the subject.
 
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It's certainly not true that
$$\frac{1}{2}mv^2 + max = \frac{m^2a}{2}v^2x.$$ You can't just ignore the rules of algebra.

The Euler-Lagrange equation told you ##m\ddot{x} = f##. From it, you can solve for x(t) and v(t). Then you can find L as a function of t, which you can then integrate.
 
vela said:
It's certainly not true that
$$\frac{1}{2}mv^2 + max = \frac{m^2a}{2}v^2x.$$ You can't just ignore the rules of algebra.

The Euler-Lagrange equation told you ##m\ddot{x} = f##. From it, you can solve for x(t) and v(t). Then you can find L as a function of t, which you can then integrate.

Whoops, that obviously wasn't correct... i made an algebraic mistake while writing the equations down here.

What I meant was simply taking m/2 out of the first term and ma out of the second term since they should be constant:

S = (m/2) ∫v^2dt + (ma) ∫ xdt.

And can you please provide an example of which equation to use to solve for x(t) and v(t)... the dependencies totally throw me off, and I can't come up with what to use.
 
Solve the differential equation ##m\ddot{x} = f##.
 
Last edited:
Okay... So that produces:

x(t) = t2f/2m + t
v(t) = f.t/m

That makes the Lagrangian

L = (m/2)f2t2/m2 + f (t2f/2m + t)

So this is L as nothing but a function of t (and f is constant)

If we integrate this wrt time over ta-tb that should be the answer?
 
Your expressions for x(t) and v(t) aren't quite correct. You forgot the constant of integration when finding v(t). It should be v(t) = v0 + (f/m)t. Hopefully, you recognize that equation from intro physics. You made the same mistake with x(t). Also, I'm not sure where that lone t came from.
 

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