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aclaret

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- Thread starter aclaret
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- #1

aclaret

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- #2

Not sure if I understood the question completely, but I think the answer is yes! You can derive those equations just using d'Alembert's principle. You have a system ##\mathcal{S}## with ##k## degrees of freedom, described by generalised co-ordinates ##\mathbf{q} = (q^1, \dots, q^k)## and generalised velocities ##\dot{\mathbf{q}} = (\dot{q}^1, \dots, \dot{q}^k)##. Acting on any particle ##\mathcal{P}_a \in \mathcal{S}## is a total force ##\mathbf{F}_a## which may be decomposed into the sum of an specified force ##\mathbf{F}^{(s)}_{a}##, which includes known external and internal forces, as well as an unknown constraint force ##\mathbf{F}^{(c)}_{a}##.

d'Alembert's principle states that if the constraint forces do zero work, then the**specified force **(alone) satisfies$$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \mathrm{d}\boldsymbol{x}_a^* = \sum_a \mathbf{F}^{(s)}_a \cdot \mathrm{d}\boldsymbol{x}_a^*$$where the ##\mathrm{d}\boldsymbol{x}_a^*## is **any** virtual infinitesimal displacement of ##\mathcal{P}_a##. It follows that$$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^i} = \sum_a \mathbf{F}^{(s)}_a \cdot \frac{ \partial \boldsymbol{x}_a}{\partial q^i}$$You can define the generalised force ##Q_i## corresponding to ##q^i##$$Q_i := \sum_a \mathbf{F}^{(s)}_a \cdot \frac{ \partial \boldsymbol{x}_a}{\partial q^i}$$Defining ##T := \sum \frac{1}{2} m_a \dot{\boldsymbol{x}}_a^2##, the left-hand side equals ##\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial T}{\partial \dot{q}^i} - \frac{\partial T}{\partial q^i}##; to show this notice that by the chain rule ##\dot{\boldsymbol{x}}_a = \left( \partial \boldsymbol{x}_a / \partial q^i \right) \dot{q}^i## from which it follows that$$\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial T}{\partial \dot{q}^i} = \frac{d}{dt} \sum_a m_a \dot{\boldsymbol{x}}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^i} = \sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^i} + \sum_a m_a \dot{\boldsymbol{x}}_a \cdot \frac{\partial^2 \boldsymbol{x}_a}{\partial q^i \partial q^j} \dot{q}^j$$and similarly$$\frac{\partial T}{\partial q_i} = \sum_a m_a \dot{\boldsymbol{x}}_a \cdot \frac{\partial \dot{\boldsymbol{x}}_a}{\partial q^i} = \sum_a m_a \dot{\boldsymbol{x}}_a \cdot \frac{\partial^2 \boldsymbol{x}_a}{\partial q^i q^j} \dot{q}^j$$which proves the result; you end up with Lagrange's equation$$\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial T}{\partial \dot{q}^i} - \frac{\partial T}{\partial q^i} = Q_i$$If the applied forces ##\mathbf{F}^{(s)}_a## are all conservative, it follows that the generalised forces themselves can be written as ##Q_i = - \partial \varphi / \partial q^i ## for some function ##\varphi = \varphi(\mathbf{q})##. Notice also that since ##\varphi## does not depend on the velocities ##\dot{\mathbf{q}}##, we have ##\partial \varphi / \partial \dot{q}^i = 0## and thus$$Q_i = - \frac{\partial \varphi}{\partial q^i} + 0 = - \frac{\partial \varphi}{\partial q^i} + \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial \varphi}{\partial \dot{q}_i}$$Inserting this into Lagrange's equations and defining ##\mathscr{L} := T - \varphi## gives$$\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial \mathscr{L}}{\partial \dot{q}^i} - \frac{\partial \mathscr{L}}{\partial q^i} = 0$$Is that sort of what you were after?

d'Alembert's principle states that if the constraint forces do zero work, then the

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- #3

andresB

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"guessing" the Lagrangian is not that different than "guessing" the correct forces to put into Newton 2nd law. They both requires a talk between the mathematics and the experience/experiments. So, I see nothing backwards with the Lagrangian approach.

- #4

aclaret

- 24

- 9

how does result follow? you make error here$$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \mathrm{d}\boldsymbol{x}_a^* = \sum_a \mathbf{F}^{(s)}_a \cdot \mathrm{d}\boldsymbol{x}_a^*$$where the ##\mathrm{d}\boldsymbol{x}_a^*## isanyvirtual infinitesimal displacement of ##\mathcal{P}_a##. It follows that$$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^i} = \sum_a \mathbf{F}^{(s)}_a \cdot \frac{ \partial \boldsymbol{x}_a}{\partial q^i}$$

thank, yes that what i ask. i am satisfied"guessing" the Lagrangian is not that different than "guessing" the correct forces to put into Newton 2nd law. They both requires a talk between the mathematics and the experience/experiments. So, I see nothing backwards with the Lagrangian approach.

- #5

You start from $$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \frac{\mathrm{d}\boldsymbol{x}_a^*}{dt} = \sum_a \mathbf{F}^{(s)}_a \cdot \frac{\mathrm{d}\boldsymbol{x}_a^*}{dt}$$For each generalised coordinate in turn, we'llhow does result follow? you make error here

- #6

wrobel

Science Advisor

- 997

- 859

there is no erroryou make error here

the motion is not obliged to provide a minimum for the Action.I say that minimisation procedure

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