A Problem in Lagrangian Mechanics

Today, as well as saying goodbye to Sally Church, I managed to finish my lecture course on Theoretical Physics. There’s still another week of teaching to go, but I have covered all the syllabus now and can use the remaining sessions for revision. The last bit of the course module concerned the calculus of variations and a brief introduction to Lagrangian mechanics so for a bit of fun I included this example.

Professor Percy Poindexter of the University of Neasden has invented a new theory of mechanics in which the one-dimensional motion of a particle in a potential $V(x)$ is governed by a Lagrangian of the form

$L=mx\ddot{x} +2V(x)$ .

Use Hamilton’s Principle and an appropriate form of the Euler equation to derive the equation of motion for such a particle and comment on your answer.

UPDATE: Since nobody has commented I’ll just reveal the point of this question, which is that if you follow the instructions the equation of motion you should obtain is

$m\ddot{x}= -\frac{\partial V}{\partial x},$

which is exactly the same as you would have got using the usual Lagrangian

$L= \frac{1}{2}m\dot{x}^{2} - V(x).$

Anyone care to comment on that?

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This entry was posted on April 28, 2016 at 4:47 pm and is filed under Cute Problems with tags Euler's Equation, Hamilton's Principle, Lagrangian Mechanics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

11 Responses to “A Problem in Lagrangian Mechanics”

ejlflop Says:
April 29, 2016 at 11:30 am

$S[x + \delta x] = \ldots = S[x] + \int dt \left( m (\delta x \ddot{x} + \delta \ddot{x} x ) + 2 V'(x) \delta x \right)$

Integrate by parts twice to turn $\delta x \ddot{x}$ into $\delta x \ddot{x}$ , then minimise the variation of the action by setting $\delta x = 0$ for all $x$ , which means that $m \ddot{x} + V'(x) = 0$ when the equations of motion are satisfied

Maybe the moral is that actions which involve third (and higher) derivatives are more physically plausible than we might have expected? So we might have to worry about more than just the tangent space to our manifold.

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Anton Garrett Says:
April 29, 2016 at 3:38 pm

It’s not great that x itself appears in the Lagrangian.

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- telescoper Says:
  April 29, 2016 at 3:42 pm
  
  This Lagrangian has a number of problems…
  
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  - Jim Fry Says:
    April 29, 2016 at 6:32 pm
    
    with the right signs and factors of 2, this Lagrangian differs from the usual by the time derivative d/dt (m x xdot).
Cesar Uliana Says:
April 29, 2016 at 7:44 pm

Hi Peter, very interesting, as someone commented above this is just a case of throwing out a total derivative, but nevertheless interesting.

If you indulge me, I would suggest another good one, the hamiltonian $H=\omega q\cosh p$ . Through Hamilton’s equation you find it is a well known system in a good disguise, for different reasons from your example if I am correct.

All the best

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R. Gerl Says:
April 29, 2016 at 11:49 pm

For Percy Poindexter’s Lagrangian to have ma = -V’ for equations of motion, the boundary term n'(t2)*E(t2)-n'(t1)*E(t1) must equal zero, where E is Q minus dQ/dt with Q being the partial derivative of L with respect to x” and n(x) is any perturbing function. That constraint is not required for the Lagrangian m*v^2/2 minus V. I typed this in quickly, so sorry for any errors.

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moons R us Says:
April 30, 2016 at 1:11 am

It has been decades, but from what I recall
a position dependent potential means that
any test particle will experience a force.
That is assuming said particle can respond to such a force.

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moons R us Says:
April 30, 2016 at 1:19 am

Let me amend my previous comment by removing the last sentence.

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Mickey Says:
April 30, 2016 at 9:38 pm

Getting to the intended answer requires assuming an extra boundary condition than is usually used in these problems. Not only are variations of x assumed to vanish at the boundary, but the variations of its derivative must also vanish. Picking a mixture of boundary conditions seems a bit odd and contrived considering the usual route to Newton’s second law with the other Lagrangian. Not sure if that is the comment you were looking for…

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GG Says:
May 2, 2016 at 9:09 am

Recent graduate speaking: the Lagrangian you presented differs from the usual one by a total derivative over time, hence it gives the same E-L equations.

😉

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domenico Says:
May 3, 2016 at 7:46 pm

I wrote, many years ago, a dynamic for a Lagrangian with greater derivative; some time after, I found the same equation in a russian book of Elsgolts about the Calculus of variations.
If the indipendent variables are coordinate and velocity, then the accelerations are a function x”(x,x’), and you need to derivate the acceleration (there is another term); if the acceleration are an indipendent variable, then the Euler-Lagrange equations obtained from the Hamilton’s principle have and additional term (I tried an infinite number of terms to obtain a dynamic of each possible real system, like a mathematical game, many years ago).

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