Lagrangian mech.: Action for a particle under constant force

In summary: It should be another constant of integration, let's call it x0. So the correct expressions should be:x(t) = x0 + v0t + (f/2m)t^2v(t) = v0 + (f/m)tPlugging these into your Lagrangian, we get:L = (m/2)(v0 + (f/m)t)^2 + f(x0 + v0t + (f/2m)t^2)Now, we can integrate this with respect to time to get our action S:S = ∫Ldt = (m/2)∫(v0 + (f/m)t)^2dt + f∫(x0 + v
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
Solve the differential equation ##m\ddot{x} = f##.
 
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  • #5
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?
 
  • #6
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.
 

FAQ: Lagrangian mech.: Action for a particle under constant force

What is Lagrangian mechanics?

Lagrangian mechanics is a mathematical framework used to describe the motion of particles and systems in classical mechanics. It is based on the principle of least action, which states that the motion of a system is determined by minimizing the action, a quantity that combines the kinetic and potential energies of the system.

What is the action for a particle under constant force?

The action for a particle under constant force is given by the equation S = ∫(L - U)dt, where L is the Lagrangian of the system and U is the potential energy. This means that the action is the difference between the kinetic and potential energies integrated over time. In simpler terms, it represents the total energy of the system over a certain period of time.

How is the Lagrangian of a system determined?

The Lagrangian of a system is determined by summing the kinetic energy of all particles in the system and subtracting the potential energy due to external forces. In the case of a particle under constant force, the Lagrangian would be L = 1/2mv^2 - Fx, where m is the mass of the particle, v is its velocity, and x is its position.

What is the significance of the principle of least action in Lagrangian mechanics?

The principle of least action is significant because it provides a more elegant and efficient way of describing the motion of systems in classical mechanics. It allows us to derive the equations of motion for a system using only the Lagrangian, rather than having to consider all the individual forces acting on each particle.

How is Lagrangian mechanics used in real-world applications?

Lagrangian mechanics is used in a wide range of applications, including physics, engineering, and even economics. It is particularly useful in situations where the system is complex and involves multiple interacting particles, such as in celestial mechanics, fluid dynamics, and quantum mechanics. It also provides a powerful tool for analyzing and predicting the behavior of physical systems.

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