Lagrangian of a driven pendulum (Landau problem)

In summary, Landau's problem is that he doesn't understand what did we omit exactly and why. He did the calculation himself and got the following result: L= \frac{1}{2}ml^2\dot{\phi^2} + mla\gamma sin(\phi - \gamma t)\dot{\phi} +mgl cos(\phi) + \frac{1}{2}ma^2\gamma^2. He neglected the last term for sure, but that's a constant, it has no time-dependence. He also left the first term here. The 'total derivative' (as he mentions) should mean both terms together, what makes the first one negligable? Landau
  • #1
Robin04
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Homework Statement


There's the following problem (the task is to construct the Lagrangian) in the first Landau (part a):
K-pkiv-g-s.png

My problem is that I don't understand what did we omit exactly and why.

Homework Equations

The Attempt at a Solution


I did the calculation myself (even checked with Mathematica) and got the following result:
##L= \frac{1}{2}ml^2\dot{\phi^2} + mla\gamma sin(\phi - \gamma t)\dot{\phi} +mgl cos(\phi) + \frac{1}{2}ma^2\gamma^2##
The solution says that we omit the terms that depend explicitly only on time, but there's no such term here. They omitted the last term for sure, but that's a constant, it has no time-dependence.

Also, the time derivative of that cosine term:
##\frac{d}{dt}[mla\gamma cos(\phi - \gamma t)]=...=-mla\gamma sin(\phi - \gamma t)\dot{\phi}+mla\gamma^2 sin(\phi-\gamma t)##
Landau seems to only leave the first term here, but why? The 'total derivative' (as he mentions) should mean both terms together, what makes the first one negligable?
 

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  • #2
Robin04 said:
##L= \frac{1}{2}ml^2\dot{\phi^2} + mla\gamma sin(\phi - \gamma t)\dot{\phi} +mgl cos(\phi) + \frac{1}{2}ma^2\gamma^2##
The solution says that we omit the terms that depend explicitly only on time, but there's no such term here. They omitted the last term for sure, but that's a constant, it has no time-dependence.
There's a term associated with the gravitational potential energy that depends only on time.

Also, the time derivative of that cosine term:
##\frac{d}{dt}[mla\gamma cos(\phi - \gamma t)]=...=-mla\gamma sin(\phi - \gamma t)\dot{\phi}+mla\gamma^2 sin(\phi-\gamma t)##
Landau seems to only leave the first term here, but why? The 'total derivative' (as he mentions) should mean both terms together, what makes the first one negligable?

Note that the middle term of the right-hand side of your expression for ##L## is not quite the same as Landau's middle term for ##L##. You can use your expression for ##\frac{d}{dt}[mla\gamma cos(\phi - \gamma t)]## to express your middle term of ##L## in terms of Landau's middle term.
 
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  • #4
TSny said:
There's a term associated with the gravitational potential energy that depends only on time.
I realized that I didn't really think about the potential energy just accepted it as it is.
So let's call ##d## the vertical displacement of the point of support relative to its lowest possible position, and let's put the zero-potential at the lowest possible position of the pendulum. Then the height of the mass above this level is ##h=d+l(1-cos(\phi))##, and ##d=a(1-sin(\gamma t))##
If I leave the sine term as it only depends explicitly on time I still have ##U=mga+mgl(1-cos\phi)##
I assume I have to leave all the constants too, and this seems so trivial that Landau doesn't even mention it, but I don't really understand why can we do this. I suppose this was also the case with the ##\frac{1}{2}ma^2\gamma^2## term.
TSny said:
Note that the middle term of the right-hand side of your expression for LLL is not quite the same as Landau's middle term for LLL. You can use your expression for ddt[mlaγcos(ϕ−γt)]ddt[mlaγcos(ϕ−γt)]\frac{d}{dt}[mla\gamma cos(\phi - \gamma t)] to express your middle term of LLL in terms of Landau's middle term.
Oh, so that's what he means by leaving the total derivatives. But I don't understand this either. Why can we leave the total derivatives? I had classical mechanics this semester and my teacher neglected lots of things too, but I don't really see why are the neglected terms less important than the others. What's so special about only time dependent, total derivative or constant terms?
kuruman said:
I haven't noticed it, thank you! However, they don't discuss why are those terms neglected.
 
  • #5
Robin04 said:
What's so special about only time dependent, total derivative or constant terms?
Such terms do not affect the equations of motion (i.e., the Euler-Lagrange equations given by equation (2.6) in Landau on page 3). Any two Lagrangians that differ only by such terms, will produce the same equations of motion. So, you can neglect such terms in the Lagrangian when deriving the equations of motion.

It should be fairly clear by inspection of the Euler-Lagrange equations that terms in the Lagrangian that are either constant or depend only on time will not affect the equations of motion.

As far as terms that are total derivatives, see Landau's explanation on page 4 in the last paragraph before section 3.
 
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  • #6
TSny said:
Such terms do not affect the equations of motion (i.e., the Euler-Lagrange equations given by equation (2.6) in Landau on page 3). Any two Lagrangians that differ only by such terms, will produce the same equations of motion. So, you can neglect such terms in the Lagrangian when deriving the equations of motion.

It should be fairly clear by inspection of the Euler-Lagrange equations that terms in the Lagrangian that are either constant or depend only on time will not affect the equations of motion.

As far as terms that are total derivatives, see Landau's explanation on page 4 in the last paragraph before section 3.
It's all clear now, thank you very much!
 

Related to Lagrangian of a driven pendulum (Landau problem)

What is the Lagrangian of a driven pendulum?

The Lagrangian of a driven pendulum, also known as the Landau problem, is a mathematical function that describes the dynamics of a pendulum that is driven by an external force. It takes into account the kinetic and potential energies of the pendulum and the external force acting on it.

How is the Lagrangian of a driven pendulum derived?

The Lagrangian of a driven pendulum is derived using the Lagrangian formalism, which is a mathematical framework for describing the dynamics of a system. It involves defining the kinetic and potential energies of the system and then using the Euler-Lagrange equations to obtain the equations of motion.

What is the significance of the Lagrangian of a driven pendulum?

The Lagrangian of a driven pendulum is significant because it allows us to understand the behavior of a pendulum that is driven by an external force. It provides a mathematical description of the system's dynamics and can be used to predict the motion of the pendulum under different conditions.

How does the Lagrangian of a driven pendulum differ from that of a simple pendulum?

The Lagrangian of a driven pendulum differs from that of a simple pendulum in that it takes into account the external force acting on the pendulum. This force can be periodic, which leads to interesting behaviors such as chaos and resonance.

What are some real-life applications of the Lagrangian of a driven pendulum?

The Lagrangian of a driven pendulum has applications in various fields such as physics, engineering, and mathematics. It is used to study the dynamics of pendulums in mechanical systems, to model the behavior of molecules in chemistry, and to understand the behavior of waves in optics and acoustics. It is also used in the study of chaos and nonlinear dynamics.

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