# The Lagrange equations from mechanics

• aclaret
In summary, the conversation discusses the use of Lagrangian mechanics to derive Lagrange's equations of motion. It is possible to do so by using d'Alembert's principle, which states that if constraint forces do no work, then the specified forces satisfy a certain equation. This equation can be manipulated to obtain Lagrange's equations, and in this way, the Lagrangian can be derived from the mechanics of the system. This process may involve some guessing, similar to the process of determining the correct forces in Newton's second law.
aclaret
I was having a doubt about the Lagrangian mechanics. Possible we can derive the lagranges equations of by extremisation principle of action, that is assume we already guess what is the lagrangian of the systeme. I say that minimisation procedure rely on assume a lagrangian, and then show it derive correct motions. that okay, but backward. can derive the lagranges equation from mechanics?

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?

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wrobel
What etotheipi says is basically correct (though it has to be modified for situations with variable masses https://aapt.scitation.org/doi/10.1119/1.4885349), but I want to address the conceptual issue the OP arises.

"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.

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

andresB said:
"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.
thank, yes that what i ask. i am satisfied

aclaret said:
how does result follow? you make error here
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'll prescribe a certain virtual motion with ##\dot{q}^{\xi} = 1## and ##\dot{q}^i = 0## for ##i \neq {\xi}##. Then, by the chain rule, ##d \boldsymbol{x}_a^* / dt = (\partial \boldsymbol{x}_a / \partial q^i) \dot{q}^i = \partial \boldsymbol{x}_a / \partial q^{\xi}## and$$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^{\xi}} = \sum_a \mathbf{F}^{(s)}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^{\xi}}$$Just replace the free index ##{\xi} \mapsto i## and you've got the next line.

vanhees71
aclaret said:
you make error here
there is no error
aclaret said:
I say that minimisation procedure
the motion is not obliged to provide a minimum for the Action.

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vanhees71 and etotheipi

## 1. What are the Lagrange equations from mechanics?

The Lagrange equations from mechanics, also known as the Euler-Lagrange equations, are a set of equations that describe the motion of a system by taking into account the kinetic and potential energies of the system. They are based on the principle of least action, which states that the path taken by a system between two points is the one that minimizes the action, a quantity defined as the integral of the difference between the kinetic and potential energies along the path.

## 2. What is the significance of the Lagrange equations in mechanics?

The Lagrange equations are significant because they provide a powerful and elegant method for determining the equations of motion of a system. They can be used to solve a wide range of problems in classical mechanics, including those involving multiple bodies and complex geometries. They also allow for the incorporation of constraints and can be extended to include non-conservative forces.

## 3. How are the Lagrange equations derived?

The Lagrange equations are derived from the principle of least action, which is based on the concept of virtual work. This principle states that the variation of the action, or the difference between the kinetic and potential energies, must be equal to zero for the true path of the system. By applying this principle to a system, the Lagrange equations can be obtained by taking the partial derivatives of the action with respect to the generalized coordinates and their time derivatives.

## 4. What is the difference between the Lagrange equations and Newton's laws of motion?

The Lagrange equations and Newton's laws of motion both describe the motion of a system, but they approach the problem from different perspectives. Newton's laws focus on the forces acting on a system, while the Lagrange equations focus on the energies of the system. Additionally, Newton's laws are limited to systems with a small number of particles, while the Lagrange equations can be applied to systems with a large number of particles and complex geometries.

## 5. Can the Lagrange equations be applied to systems with non-conservative forces?

Yes, the Lagrange equations can be extended to include non-conservative forces, such as friction or air resistance. This is done by including these forces in the expression for the potential energy and taking them into account when deriving the equations of motion. However, this may result in more complex equations and may require numerical methods for solving the equations.

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