Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Physics
Classical Physics
Mechanics
The Lagrange equations from mechanics
Reply to thread
Message
[QUOTE="etotheipi, post: 6481089"] 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 [B]specified force [/B](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 [B]any[/B] 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? [/QUOTE]
Insert quotes…
Post reply
Forums
Physics
Classical Physics
Mechanics
The Lagrange equations from mechanics
Back
Top