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Euler-Lagrange equation on Lagrangian in generalized coordinates

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Homework Statement


I need some help understanding a derivation in a textbook. It involves the Lagrangian in generalized coordinates.



Homework Equations



The text states that generalized coordinates {q_1, ..., q_3N} are related to original Cartesian coordinates [tex]q_\alpha = f_\alpha(\mathbf r_1, ..., \mathbf r_N)[/tex]
Makes sense to me...
Velocities in general coordinates are, by the chain rule,
[tex]\dot{\mathbf r_i} = \sum_{\alpha=1}^{3N} \frac{\partial \mathbf r_i}{\partial q_\alpha} \dot{q_\alpha}[/tex]
Ok so then kinetic energy in cartesian is
[tex] K = \frac{1}{2} \sum_{i=1}^N m_i \dot{r_i}^2 [/tex]
So KE in new velocities is then
[tex] \tilde{K}(q, \dot{q}) = \frac{1}{2} \sum_{\alpha=1}^{3N} \sum_{\beta=1}^{3N}[\sum_{i=1}^{N} m_i \frac{\partial \mathbf r_i}{\partial q_\alpha}\cdot \frac{\partial \mathbf r_i}{\partial q_\beta}] \dot{q_\alpha} \dot{q_\beta} [/tex]
[tex] = \frac{1}{2} \sum_{\alpha=1}^{3N} \sum_{\beta=1}^{3N} G_{\alpha\beta}(q_1, ..., q_3N) \dot{q_\alpha} \dot{q_\beta}[/tex]
where the text calls the expression in brackets the mass metric tensor.

Ok, so then given this new KE, the Lagrangian (KE - PE) is
[tex]L = \frac{1}{2} \sum_{\alpha=1}^{3N} \sum_{\beta=1}^{3N} G_{\alpha}{\beta}(q_1,...,q_{3N})\dot{q_\alpha} \dot{q_\beta} - U(\mathbf r_1(q_1,...,q_{3N}), ..., \mathbf r_N(q_1, ... , q_{3N})) [/tex]

All of the above makes sense to me, but then the text substitutes the lagrangian into the Euler-lagrange equation and gets the equation of motion for each [itex]q_\gamma, \gamma = 1, ... 3N[/itex]
[tex]\sum_{\beta=1}^{3N} G_{\gamma\beta}(q_1,...,q_{3N})\ddot{q_\beta} + \sum_{\alpha=1}^{3N} \sum_{\beta=1}^{3N} [\frac{\partial G_{\gamma\beta}}{\partial q_\alpha} - \frac{\partial G_{\alpha\beta}}{\partial q_\gamma} ]\dot{q_\alpha}\dot{q_\beta} = -\frac{\partial U}{\partial q_\gamma} [/tex]

(Euler-lagrange: [itex]\frac{d}{dt}(\frac{\partial L}{\partial \dot{q_\gamma}}) - \frac{\partial L}{\partial q_\gamma} = 0[/itex])

The Attempt at a Solution


Ok, so I can see where the first and third term in the equation of motion are coming from. The first is a result of [itex]\frac{d}{dt}(\frac{\partial L}{\partial \dot{q_\gamma}})[/itex] acting on the kinetic energy part of the Lagrangian. The third term is the result of [itex]\frac{\partial L}{\partial q_\gamma}[/itex] acting on the potential energy part of the Lagrangian. The mass metric tensor is a function of the q's, so I feel like the second term is coming from [itex]\frac{\partial L}{\partial q_\gamma}[/itex] acting on the kinetic energy term, but that's where I'm stuck.

Where does [itex]\frac{\partial G_{\gamma\alpha}}{\partial q_\alpha}[/itex] come from?
 
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Answers and Replies

  • #2
ehild
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It comes from the total derivative with respect to t. [tex]\frac{d}{dt}G(q_1,...,q_{3N})=\sum{\frac{\partial G}{\partial q_k}\dot q_k} [/tex]

ehild
 
  • #3
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That was tricky to see because of the way the terms were grouped.
I worked it through, and now I get it. Thank you.
 
  • #4
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Ok, so it goes on though. This text develops the Hamiltonian in generalized coordinates:
[tex]H(q_1, ... , q_{3N}, p_1, ..., p_{3N}) = \frac{1}{2} \sum_{\alpha=1}^{3N} \sum_{\beta=1}^{3N}p_\alpha G^{-1}_{\alpha\beta}(q_1,...,q_{3N}) p_\beta + U(\mathbf r_1(q1,...,q_{3N}),...,.\mathbf r_N(q1,...,q_{3N}))[/tex]

And says Hamilton's equations of motion are
[tex]\dot{q}_\alpha = \frac{\partial H}{\partial p_\alpha},\ \ \ \dot{p}_\alpha = -\frac{\partial H}{\partial q_\alpha}[/tex]

And how did they do that?!? This text is going to take me a while to get through if unless my math improves all of a sudden.

I tried [itex]\frac{\partial H}{\partial p_\alpha}[/itex], and I feel like it yields [itex]\frac{1}{2} \sum_{\beta=1}^{3N} G^{-1}_{\alpha\beta}(q_1,...,q_{3N}) p_\beta = \frac{1}{2} \dot{q}_\alpha[/itex]. This equivalency was defined a bit before in the text, and I guess it makes sense. Partial derivative of H with respect to p_alpha should zero out everything except for the terms with p_alpha. But the factor of 1/2 remains.

I don't even know how to begin to obtain the second equation of motion.


I was wondering if a book like this one, https://www.amazon.com/dp/0471198269/?tag=pfamazon01-20, would benefit me?
 
  • #5
ehild
Homework Helper
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In my book, he derivation of the Hamilton function starts with defining the generalized momenta, as

[tex] p_i = \frac{\partial L}{\partial \dot q_i}[/tex].

With that definition, the Euler-Lagrange equations become

[tex]\dot p_i = \frac{\partial L}{\partial q_i}[/tex]

Now you write up dL if t, qi and dot qi change by dt, dqi and d(dot qi). Using
[tex]p_i d \dot q_i = d ( p_i \dot q_i )-\dot q_i d p_i [/tex]
And defining H as

[tex]H=\sum{p_i \dot q_i}-L[/tex].

The total derivative of H is

[tex]dH=-\frac{\partial L}{\partial t}dt - \sum{ \left (\dot p_id q_i - \dot q_i dp_i \right )}[/tex].

H is function of time t, and of the generalized coordinates qi and momenta pi. Compared with

[tex]dH=\frac{\partial H}{\partial t}dt+ \sum{ \left (\frac{\partial H}{\partial q_i} dq_i+\frac{\partial H}{\partial p_i}dp_i \right )}[/tex]

you get the equations

[tex]\frac{\partial H}{\partial t}= -\frac{\partial L}{\partial t}[/tex]

[tex]\dot q_i= \frac{\partial H}{\partial p_i}[/tex] and [tex]\dot p_i=-\frac{\partial H}{\partial q_i}[/tex]

For conservative and scleronomic systems, H=T+V.
 
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  • #6
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Hi there,
thanks for you reply. I was able to understand your derivation after some studying of it. It does seem that my book offered no indication of this sort of derivation though, which disappoints me.

Note. I wrote above that the Euler-Lagrange in generalized coordinates is:
[tex]\sum_{\beta=1}^{3N} G_{\gamma\beta}(q_1,...,q_{3N})\ddot{q_\beta} + \sum_{\alpha=1}^{3N} \sum_{\beta=1}^{3N} [\frac{\partial G_{\gamma\beta}}{\partial q_\alpha} - \frac{\partial G_{\alpha\beta}}{\partial q_\gamma} ]\dot{q_\alpha}\dot{q_\beta} = -\frac{\partial U}{\partial q_\gamma} [/tex]
Which is verbatim from the text.
But it is actually,
[tex]\sum_{\beta=1}^{3N} G_{\gamma\beta}(q_1,...,q_{3N})\ddot{q_\beta} + \sum_{\alpha=1}^{3N} \sum_{\beta=1}^{3N} [\frac{\partial G_{\gamma\beta}}{\partial q_\alpha} - \frac{1}{2}\frac{\partial G_{\alpha\beta}}{\partial q_\gamma} ]\dot{q_\alpha}\dot{q_\beta} = -\frac{\partial U}{\partial q_\gamma} [/tex]
Correction as seen here: http://www.nyu.edu/classes/tuckerman/stat.mechII/StatMech_errata_2nd.pdf

I was too busy reasoning out how the 1/2 term is not present for the first and second terms of the equation that I didn't realize that it was missing in the third term.
 
  • #7
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(Euler-lagrange: ##\frac{d}{dt}(\frac{\partial L}{\partial \dot{q_\gamma}}) - \frac{\partial L}{\partial q_\gamma} = 0##)
How did you come to this equation? I derived Euler-Lagrange with Cartesian coordinates (i.e. ##\vec{r_i}##, ##\vec{v_i}##, as showed in that textbook) but there is no clue it should be valid for generalized coordinates ##q_{\gamma}, \dot{q_{\gamma}}##.
 

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