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1 Introduction to Lagrange Equations of Motion for Conservative Forces

In Newtonian mechanics a system is made up of point masses and rigid bodies. These are subjected to known forces. To construct equations of motion you must determine the composition of the system an the forces which act on it. Then turn that understanding into a series of equations of motion. Once the system is described an initial condition is required in order to then calculate system's behavior.

The Lagrangian description of a mechanical system is different. For Lagrangian mechanics it is assumed that the position of the system at 2 instances of time (t₁ and t₂) are known (or knowable) and fixed. The system must move/behave between times t₁ and t₂ such that the system has the least "action". The "action" being defined by

$LaTeX: \int_{t_{1}}^{t_{2}} L\left(q_{i}, \dot q_{i}\right) dt$

"Action"

where

$LaTeX: L\left(q_{i}, \dot q_{i}\right)$ is called the Lagrange function or Lagrangian

$LaTeX: q_{i}$ is a generalized coordinate which can represent the x, y, z of Cartesian coordinates or the θ, R of Polar coordinates, etc.

2 Degrees of Freedom

A particle without constraints can be anywhere in space. Thus we need 3 coordinates - x, y, z in Cartesian space - to describe its position. In the case of m dimensions and p particles there are n degrees of freedom where

$LaTeX: n=mp$

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However, when c constraints are imposed on the particles, then the number of degrees of freedom is

$LaTeX: n=mp-c$

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2.1 Generalized Coordinates

Since the coordinates could be angular displacement or electric charge we need a generalized system of coordinates. For 3 degrees of freedom we have

$LaTeX: \begin{array}{lcl} <p>x & = & x\left(q_{1}, q_{2}, q_{3}\right) \\ y & = & y\left(q_{1}, q_{2}, q_{3}\right) \\ z & = & z\left(q_{1}, q_{2}, q_{3}\right) </p> \end{array}$

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then

$LaTeX: \dot x = \frac{\partial x}{\partial q_{1}}\dot q_{1}+\frac{\partial x}{\partial q_{2}}\dot q_{2}+\frac{\partial x}{\partial q_{3}}\dot q_{3}$

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If the particle is constrained to move along a surface then the problem becomes one of 2 degrees of freedom and the equations become

$LaTeX: \begin{array}{lcl} <p>x & = & x\left(q_{1}, q_{2}\right) \\ y & = & y\left(q_{1}, q_{2}\right) \\ z & = & z\left(q_{1}, q_{2}\right) </p> \end{array}$

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and

$LaTeX: \delta x=\frac{\partial x}{\partial q_{1}}\delta q_{1} + \frac{\partial x}{\partial q_{2}}\delta q_{2}$

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$LaTeX: \delta y=\frac{\partial y}{\partial q_{1}}\delta q_{1} + \frac{\partial y}{\partial q_{2}}\delta q_{2}$

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$LaTeX: \delta z=\frac{\partial z}{\partial q_{1}}\delta q_{1} + \frac{\partial z}{\partial q_{2}}\delta q_{2}$

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3 Lagrange Equations: Work

The Lagrange equation of motion for the q₁ coordinate is

$LaTeX: \frac{d}{dt}\ \left(\frac{\partial K}{\partial \dot q_{1}}\right)-\frac{\partial K}{\partial q_{1}}=F_{q_{1}}$.

Lagrange Equation of Motion for Conservative Forces only (no Friction)

This is where we want to end up.

3.1 Kinetic Energy minus Potential Energy

Work is force times displacement. For our purposes the scalar value for work is

$LaTeX: \delta W = F_{x}\delta x + F_{y}\delta y + F_{z}\delta z$

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This work is also equal to the change in kinetic energy which leads us to D'Alembert's equation

$LaTeX: m\left(\ddot x \delta x + \ddot y \delta y + \ddot z \delta z \right)=F_{x}\delta x + F_{y}\delta y + F_{z}\delta z$

1a, Zak

Formulation #2

$LaTeX: \delta W=\sum_{i=1}^{n} \left(\mathbf{F}_{i} - m_{i} \mathbf{a}_{i} \right) \cdot \ delta \mathbf{r}_{i}=0$

1b, Wikipedia

Back to Zak... The total work can be found by

$LaTeX: \delta W= \sum_{i=1}^{p} \left(F_{x_{i}}\delta x_{i}+F_{y_{i}}\delta y_{i} + F_{z_{i}}\delta z_{i} \right)$

2

where:

• $LaTeX: \mathbf{F}_{1}, \cdots, \mathbf{F}_{p}$ are the forces acting upon the particles and
• $LaTeX: \delta \mathbf{s}_{1}, \cdots, \mathbf{s}_{p}$ are displacements of those particles.

Substituting Eqn. (2) into Eqn. (1a) and rearranging we get

$LaTeX: \begin{alignat}{2} <p>\delta W & = & m \left(\ddot x \frac{\partial x}{\partial q_{1}} + \ddot y \frac{\partial y}{\partial q_{1}} + \ddot z \frac{\partial z}{\partial q_{1}} \right) \delta q_{1} + m \left(\ddot x \frac{\partial x}{\partial q_{2}} + \ddot y \frac{\partial y}{\partial q_{2}} + \ddot z \frac{\partial z}{\partial q_{2}} \right) \delta q_{2} \\ & = \left(F_{x} \frac{\partial x}{\partial q_{1}} + F_{y} \frac{\partial y}{\partial q_{1}} + F_{z} \frac{\partial z}{\partial q_{1}} \right) \delta q_{1} + \left(F_{x} \frac{\partial x}{\partial q_{2}} + F_{y} \frac{\partial y}{\partial q_{2}} + F_{z} \frac{\partial z}{\partial q_{2}} \right) \delta q_{2} </p> \end{alignat}$

3

Let

$LaTeX: F_{q_{1}}=F_{x} \frac{\partial x}{\partial q_{1}} + F_{y} \frac{\partial y}{\partial q_{1}} + F_{z} \frac{\partial z}{\partial q_{1}}$

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and

$LaTeX: F_{q_{2}}=F_{x} \frac{\partial x}{\partial q_{2}} + F_{y} \frac{\partial y}{\partial q_{2}} + F_{z} \frac{\partial z}{\partial q_{2}}$

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Then Eqn. (3) becomes

$LaTeX: \delta W=F_{q{1}}\delta q_{1}+F_{q_{2}}\delta q_{2}$

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3.2 Useful Equalities

Consider the $LaTeX: \ddot x \frac{\partial x}{\partial q_{1}}$ portion of Eqn. (3). What does $LaTeX: \ddot x \frac{\partial x}{\partial q_{1}}$ equal? Start with the differentiation of $LaTeX: \frac{d}{dt}\left(\dot x \frac{\partial x}{\partial q_{1}}\right)$ and using the product rule we get

$LaTeX: \frac{d}{dt}\left(\dot x \frac{\partial x}{\partial q_{1}}\right)=\ddot x \frac{\partial x}{\partial q_{1}}+\dot x \frac{d}{dt}\left(\frac{\partial x}{\partial q_{1}}\right)$

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Rearranging we come to the following useful equality

$LaTeX: \ddot x \frac{\partial x}{\partial q_{1}}=\frac{d}{dt}\left(\ddot x \frac{\partial x}{\partial q_{1}}\right)-\dot x \frac{d}{dt}\left(\frac{\partial x}{\partial q_{1}}\right)$

4

We also need to consider the time derivative of $LaTeX: x=x\left(q_{1}, q_{2}\right)$ which is

$LaTeX: \dot x=\frac{\partial x}{\partial q_{1}}\dot q_{1}+\frac{\partial x}{\partial q_{2}}\dot q_{2}$

5

Using partial differentiation with respect to $LaTeX: \dot q_{1}$ we acquire our second useful equality

$LaTeX: \frac{\partial x}{\partial \dot q_{1}}=\frac{\partial x}{\partial q_{1}}$

6

For our final useful equality we start with

$LaTeX: \frac{d}{dt}\left(\frac{\partial x}{\partial q_{1}}\right)=\frac{\partial}{\partial q_{1}}\left(\frac{\partial x}{\partial q_{1}}\right)\dot q_{1} + \frac{\partial}{\partial q_{2}}\left(\frac{\partial x}{\partial q_{1}}\right)\dot q_{2}$

7

Keeping in mind that $LaTeX: x=x\left(q_{1}, q_{2}\right)$ and the partial derivative $LaTeX: \frac{\partial x}{\partial q_{1}}$ is in general a function of q₁ and q₂. Taking the partial derivative of Eqn. (5) with respect to q₁ leads to

$LaTeX: \begin{alignat}{2} <p>\frac{\partial}{\partial q_{1}}\left(\dot x\right) & = &\frac{\partial}{\partial q_{1}}\left(\frac{\partial x}{\partial q_{1}}\dot q_{1} + \frac{\partial x}{\partial q_{2}}\dot q_{2}\right) \\ & =\frac{\partial}{\partial q_{1}}\left(\frac{\partial x}{\partial q_{1}}\right)\dot q_{1}+\frac{\partial}{\partial q_{1}}\left(\frac{\partial x}{\partial q_{2}}\right)\dot q_{2} </p> \end{alignat}$.

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Remember that we assumed $LaTeX: x,y,z \in \mathcal{C}^{2}$ and therefore

$LaTeX: \frac{\partial}{\partial q_{1}}\left(\frac{\pratial x}{\partial q_{2}}\right)=\frac{\partial}{\partial q_{2}}\left(\frac{\partial x}{\partial q_{1}}\right)$.

8

Using Eqn. (8) we can rearrange the $LaTeX: \dot q_{2}$ portion of the equation

$LaTeX: \begin{alignat}{2} <p>\frac{\partial}{\partial q_{1}}\left(\dot x\right) & = &\frac{\partial}{\partial q_{1}}\left(\frac{\partial x}{\partial q_{1}}\right)\dot q_{1}+\frac{\partial}{\partial q_{1}}\left(\frac{\partial x}{\partial q_{2}}\right)\dot q_{2} \\ & = \frac{\partial}{\partial q_{1}}\left(\frac{\partial x}{\partial q_{1}}\right)\dot q_{1}+\frac{\partial}{\partial q_{2}}\left(\frac{\partial x}{\partial q_{1}}\right)\dot q_{2} \\ & = \frac{\partial \dot x}{\partial q_{1}} </p> \end{alignat}$

9

Then we are left with

$LaTeX: \begin{alignat}{2} <p>\frac{d}{dt}\left(\frac{\partial x}{\partial q_{1}}\right) & = & \frac{\partial}{\partial q_{1}}\left(\frac{\partial x}{\partial q_{1}}\right)\dot q_{1}+\frac{\partial}{\partial q_{2}}\left(\frac{\partial x}{\partial q_{1}}\right)\dot q_{2} = \frac{\partial \dot x}{\partial q_{1}} \\ & = \frac{\partial \dot x}{\partial q_{1}} </p> \end{alignat}$

10

3.3 Back to the Derivation of the Lagrange Equations of Motion for a Single Particle

Comparing the right-hand sides of Eqns. (7) and (9), we get Eqn. (10). We now use Eqns. (4), (6), and (10) to arrive at the Lagrange equation of motion for a single particle. First substitute Eqns. (6) and (10) into (4) to get

$LaTeX: \ddot x\frac{\partial x}{\partial q_{1}}=\frac{d}{dt}\left(\dot x \frac{\partial \dot x}{\partial \dot q_{1}}\right)-\dot x\frac{\partial \dot x}{\partial q_{1}}$

11

For reference

$LaTeX: \frac{\partial x}{\partial \dot q_{1}}=\frac{\partial x}{\partial q_{1}}$

6

If you have a particle with a constant acceleration a and you double integrate to get position then you get

$LaTeX: x=a\frac{t^{2}}{2}$

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assuming zero initial conditions. If you then took the partial derivative with respect to a generalized coordinate q then

$LaTeX: \begin{alignat}{2} <p>\frac{\partial}{\partial q}\left(x\right) & = & a\frac{\partial}{\partial q}\frac{t^{2}}{2} \\ & = a\left(t\frac{\partial t}{\partial q}\right) </p> \end{alignat}$

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Consider the above in the context of the Lagrange equation of motion for a single particle and realize that

$LaTeX: \frac{\partial}{\partial \dot q_{1}}\left(\frac{\dot x^2}{2}\right)=\dot x\frac{\partial \dot x}{\partial \dot q_{1}}$

12

and

$LaTeX: \frac{\partial}{\partial q_{1}}\left(\frac{\dot x^2}{2}\right)=\dot x\frac{\partial \dot x}{\partial q_{1}}$

13

Substituting Eqns. (12) and (13) into Eqn. (11) results in

$LaTeX: \ddot x\frac{\partial x}{\partial q_{1}}=\frac{d}{dt}\left(\frac{\partial \frac{\dot x^{2}}{2}}{\partial \dot q_{1}}\right)-\frac{\partial \frac{\dot x^{2}}{2}}{\partial q_{1}}$

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Similarly expressions for y and z can be found. Realizing this, when $LaTeX: \delta q_{2}=0$, Eqn. (3) becomes

$LaTeX: \begin{alignat}{2} <p>\delta W & = & \left(\frac{d}{dt}\left(\frac{\partial}{\partial \dot q_{1}}\frac{m\left(\dot x^{2} + \dot y^{2} + \dot z^{2}\right)}{2}\right)-\frac{\partial}{\partial q_{1}}\frac{m\left(\dot x^{2} + \dot y^{2} + \dot z^{2}\right)}{2}\right)\delta q_{1}\\ & = \left(F_{x}\frac{\pratial x}{\partial q_{1}}+F_{y}\frac{\pratial y}{\partial q_{1}}+F_{z}\frac{\pratial z}{\partial q_{1}}\right)\delta q_{1}\\ & = F_{q_{1}}\delta q_{1} </p> \end{alignat}$

14

Let

$LaTeX: K=\frac{1}{2}m\left( \dot x^{2} + \dot y^{2} + \dot z^{2} \right)$

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denote the kinetic energy of the particle. Then, we can represent (14) as

$LaTeX: \frac{d}{dt}\ \left(\frac{\partial K}{\partial \dot q_{1}}\right)-\frac{\partial K}{\partial q_{1}}=F_{q_{1}}$.

15

Eqn. (15) is called the Lagrange equation of motion for the q₁ coordinate. Using the same arguments as above, we cna derive the Lagrange equation of motion for the q₂ coordinate. In general there are as many Lagrange equations of motion as there are degrees of freedom of the particle.

4 Notes

• Zak, Stanislaw H. Systems and Control. Oxford University Press, New York, 2003. ISBN 0195150112.
• Wikipedia: Lagrangian Mechanics
• Wikibooks: Classic Mechanics, Lagrangian