Lagrangian mechanics

Lagrangian mechanics is a re-formulation of classical mechanics that combines conservation of momentum with conservation of energy. It was introduced by Joseph Louis Lagrange in 1788. In Lagrangian mechanics, the trajectory of a system of particles is derived by solving Lagrange's equation, given herein, for each of the system's generalized coordinates. The fundamental lemma of calculus of variations shows that solving Lagrange's equation is equivalent to finding the path that minimizes the action functional, a quantity that is the integral of the Lagrangian over time.

The use of generalized coordinates may considerably simplify a system's analysis. For example, consider a small frictionless bead traveling in a groove. If one is tracking the bead as a particle, calculation of the motion of the bead using Newtonian mechanics would require solving for the time-varying constraint force required to keep the bead in the groove. For the same problem using Lagrangian mechanics, one looks at the path of the groove and chooses a set of independent generalized coordinates that completely characterize the possible motion of the bead. This choice eliminates the need for the constraint force to enter into the resultant system of equations. There are fewer equations since one is not directly calculating the influence of the groove on the bead at a given moment.

Lagrange's equations

The equations of motion in Lagrangian mechanics are Lagrange's equations, also known as Euler-Lagrange equations. Below, we sketch out the derivation of Lagrange's equation. Please note that in this context, V is used rather than U for potential energy and T replaces K for kinetic energy. See the references for more detailed and more general derivations.

Start with D'Alembert's principle for the virtual work of applied forces, Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{F}_i , and inertial forces on a three dimensional accelerating system of n particles, i, whose motion is consistent with its constraints:^[1]:269

Failed to parse (Missing texvc executable; please see math/README to configure.): \delta W = \sum_{i=1}^n ( \mathbf {F}_{i} - m_i \mathbf{a}_i )\cdot \delta \mathbf r_i = 0

.

Failed to parse (Missing texvc executable; please see math/README to configure.): \delta W

is the virtual work

Failed to parse (Missing texvc executable; please see math/README to configure.): \delta \mathbf r_i

is the virtual displacement of the system, consistent with the constraints

Failed to parse (Missing texvc executable; please see math/README to configure.): m_i

are the masses of the particles in the system

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf a_i

are the accelerations of the particles in the system

Failed to parse (Missing texvc executable; please see math/README to configure.): m_i \mathbf a_i

together as products represent the time derivatives of the system momenta, aka. inertial forces

Failed to parse (Missing texvc executable; please see math/README to configure.): i

is an integer used to indicate (via subscript) a variable corresponding to a particular particle

Failed to parse (Missing texvc executable; please see math/README to configure.): n

is the number of particles under consideration

Break out the two terms:

Failed to parse (Missing texvc executable; please see math/README to configure.): \delta W = \sum_{i=1}^n \mathbf {F}_{i} \cdot \delta \mathbf r_i - \sum_{i=1}^n m_i \mathbf{a}_i \cdot \delta \mathbf r_i = 0

.

Assume that the following transformation equations from m independent generalized coordinates, Failed to parse (Missing texvc executable; please see math/README to configure.): q_j , hold:^[1]:260

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{r}_1=\mathbf{r}_1(q_1, q_2, ..., q_m, t)

,

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{r}_2=\mathbf{r}_2(q_1, q_2, ..., q_m, t)

, ...

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{r}_n=\mathbf{r}_n(q_1, q_2, ..., q_m, t)

.

Failed to parse (Missing texvc executable; please see math/README to configure.): m

(without a subscript) indicates the total number generalized coordinates

An expression for the virtual displacement (differential), Failed to parse (Missing texvc executable; please see math/README to configure.): \delta \mathbf{r}_i , of the system is^[1]:264

Failed to parse (Missing texvc executable; please see math/README to configure.): \delta \mathbf{r}_i = \sum_{j=1}^m \frac {\partial \mathbf {r}_i} {\partial q_j} \delta q_j

.

Failed to parse (Missing texvc executable; please see math/README to configure.): j

is an integer used to indicate (via subscript) a variable corresponding to a generalized coordinate

The applied forces may be expressed in the generalized coordinates as generalized forces, Failed to parse (Missing texvc executable; please see math/README to configure.): Q_j ,^[1]:265

Failed to parse (Missing texvc executable; please see math/README to configure.): Q_j = \sum_{i=1}^n \mathbf {F}_{i} \cdot \frac {\partial \mathbf {r}_i} {\partial q_j}

.

Combining the equations for Failed to parse (Missing texvc executable; please see math/README to configure.): \delta W , Failed to parse (Missing texvc executable; please see math/README to configure.): \delta \mathbf{r}_i , and Failed to parse (Missing texvc executable; please see math/README to configure.): Q_j

yields the following result after pulling the sum out of the dot product in the second term:[1]:269 

Failed to parse (Missing texvc executable; please see math/README to configure.): \delta W = \sum_{j=1}^m Q_j \delta q_j - \sum_{j=1}^m \sum_{i=1}^n m_i \mathbf{a}_i \cdot \frac {\partial \mathbf {r}_i} {\partial q_j} \delta q_j = 0

.


Substituting in the result from the kinetic energy relations to change the inertial forces into a function of the kinetic energy leaves[1]:270 

Failed to parse (Missing texvc executable; please see math/README to configure.): \delta W = \sum_{j=1}^m Q_j \delta q_j - \sum_{j=1}^m \left ( \frac {d}{d t} \left ( \frac {\partial T}{\partial \dot{q}_j} \right ) - \frac {\partial T}{\partial q_j} \right ) \delta q_j = 0

.


In the above equation, Failed to parse (Missing texvc executable; please see math/README to configure.): \delta q_j

is arbitrary, though it is—by definition—consistent with the constraints. So the relation must hold term-wise:[1]:270 

Failed to parse (Missing texvc executable; please see math/README to configure.): Q_j = \frac {d}{d t} \left ( \frac {\partial T}{\partial \dot{q}_j} \right ) - \frac {\partial T}{\partial q_j}

.


If the Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf F_i

are conservative, they may be represented by a scalar potential field, Failed to parse (Missing texvc executable; please see math/README to configure.): V

^{[1]:266 & 270}

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf F_i = - \nabla V \Rightarrow Q_j = - \sum_{i=1}^n \nabla V \cdot \frac {\partial \mathbf {r}_i} {\partial q_j} = - \frac {\partial V}{\partial q_j}

.

The previous result may be easier to see by recognizing that Failed to parse (Missing texvc executable; please see math/README to configure.): V

is a function of the Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf {r}_i

, which are in turn functions of Failed to parse (Missing texvc executable; please see math/README to configure.): q_j , and then applying the chain rule to the derivative of Failed to parse (Missing texvc executable; please see math/README to configure.): V

with respect to Failed to parse (Missing texvc executable; please see math/README to configure.): q_j

.

The definition of the Lagrangian is^[1]:270

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathcal{L} = T - V

.

Since the potential field is only a function of position, not velocity, Lagrange's equations are as follows:^[1]:270

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 = \frac {d}{d t} \left ( \frac {\partial \mathcal L}{\partial \dot{q}_j} \right ) - \frac {\partial \mathcal L}{\partial q_j}

.

This is consistent with the results derived above and may be seen by differentiating the right side of the Lagrangian with respect to Failed to parse (Missing texvc executable; please see math/README to configure.): \dot{q}_j

and time, and solely with respect to Failed to parse (Missing texvc executable; please see math/README to configure.): q_j

, adding the results and associating terms with the equations for Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf F_i

and Failed to parse (Missing texvc executable; please see math/README to configure.):  Q_j 

.

If there are forces Failed to parse (Missing texvc executable; please see math/README to configure.): \tilde{\mathbf F}_j

acting on the system that are not derived from potentials, then by a development similar to that presented above, a more general form of Lagrange's equations are

Failed to parse (Missing texvc executable; please see math/README to configure.): \tilde{\mathbf F}_j = \frac {d}{d t} \left ( \frac {\partial \mathcal L}{\partial \dot{q}_j} \right ) - \frac {\partial \mathcal L}{\partial q_j}

.

Another generalization can involve forces attributable to potentials and viscosity. If an appropriate transformation can be found from the Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf F_i , Rayleigh suggests using a dissipation function, Failed to parse (Missing texvc executable; please see math/README to configure.): D , of the following form:^[1]:271

Failed to parse (Missing texvc executable; please see math/README to configure.): D = \frac {1}{2} \sum_{j=1}^m \sum_{k=1}^m C_{j k} \dot{q}_j \dot{q}_k

.

Failed to parse (Missing texvc executable; please see math/README to configure.): C_{j k}

are constants that are related to the damping coefficients in the physical system, though not necessarily equal to them

If Failed to parse (Missing texvc executable; please see math/README to configure.): D

is defined this way, then[1]:271 

Failed to parse (Missing texvc executable; please see math/README to configure.): Q_j = - \frac {\partial V}{\partial q_j} - \frac {\partial D}{\partial \dot{q}_j}

and

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 = \frac {d}{d t} \left ( \frac {\partial \mathcal L}{\partial \dot{q}_j} \right ) - \frac {\partial \mathcal L}{\partial q_j} + \frac {\partial D}{\partial \dot{q}_j}

.

Kinetic energy relations

The kinetic energy, Failed to parse (Missing texvc executable; please see math/README to configure.): T
, for the system of particles is defined by[1]:269 

Failed to parse (Missing texvc executable; please see math/README to configure.): T = \frac {1}{2} \sum_{i=1}^n m_i \mathbf {v}_i \cdot \mathbf {v}_i

.


The partial derivative of Failed to parse (Missing texvc executable; please see math/README to configure.): T

with respect to the time derivatives of the generalized coordinates, Failed to parse (Missing texvc executable; please see math/README to configure.): \dot{q}_j

, is[1]:269 

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac {\partial T}{\partial \dot{q}_j} = \sum_{i=1}^n m_i \mathbf {v}_i \cdot \frac {\partial \mathbf {v}_i} {\partial \dot{q}_j}

.


The previous result may be difficult to visualize. As a result of the product rule, the derivative of a general dot product d(f(x)•g(x))/dx is f(x)•dg(x)/dx + g(x)•df(x)/dx. This general result may be seen by briefly stepping into a Cartesian coordinate system, recognizing that the dot product is (there) a term-by-term product sum, and also recognizing that the derivative of a sum is the sum of its derivatives. In our case, f and g are equal to v, which is why the factor of one half disappears.


According to the chain rule and the coordinate transformation equations given above for Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{r}
, it's time derivative, Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{v}
, is:[1]:264 

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf{v}_i = \sum_{j=1}^m \frac {\partial \mathbf{r}_i}{\partial q_j} \dot{q}_j + \frac {\partial \mathbf{r}_i}{\partial t}

.


Together, the definition of Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf v_i

and the total differential, Failed to parse (Missing texvc executable; please see math/README to configure.): d \mathbf {r}_i

, suggest that[1]:269 

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac {\partial \mathbf {v}_i}{\partial \dot{q}_j} = \frac {\partial \mathbf {r}_i}{\partial q_j}

.[clarify]


[ Remember that :Failed to parse (Missing texvc executable; please see math/README to configure.):  \frac {\partial } {\partial {\dot{q}_k}}  {A}{\dot{q}_k} = A   
, and it is easier to visualise the result if you replace the subscript Failed to parse (Missing texvc executable; please see math/README to configure.): j

with some other subscript Failed to parse (Missing texvc executable; please see math/README to configure.): k

. Also remember that in the sum, there is only one Failed to parse (Missing texvc executable; please see math/README to configure.):  {\dot{q}_k} 
. ]


Substituting this relation back into the expression for the partial derivative of Failed to parse (Missing texvc executable; please see math/README to configure.): T

gives[1]:269 

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac {\partial T}{\partial \dot{q}_j} = \sum_{i=1}^n m_i \mathbf v_i \cdot \frac {\partial \mathbf {r}_i}{\partial q_j}

.


Taking the time derivative gives[1]:270 

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac {d}{d t} \left ( \frac {\partial T}{\partial \dot{q}_j} \right ) = \sum_{i=1}^n \left [ m_i \mathbf a_i \cdot \frac {\partial \mathbf {r}_i}{\partial q_j} + m_i \mathbf {v}_i \cdot \frac {d}{d t} \left ( \frac {\partial \mathbf {r}_i}{\partial q_j} \right ) \right ]

.


Using the chain rule on the last term gives[1]:270 

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac {d}{d t} \left ( \frac {\partial \mathbf {r}_i}{\partial q_j} \right ) = \sum_{k=1}^m \frac {\partial^2 \mathbf r_i}{\partial q_j \partial q_k} \dot{q_k} + \frac {\partial^2 \mathbf r_i}{\partial q_j \partial t}

.


From the expression for Failed to parse (Missing texvc executable; please see math/README to configure.): \mathbf v_i
, one sees that[1]:270 

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac {d}{d t} \left ( \frac {\partial \mathbf {r}_i}{\partial q_j} \right ) = \frac {\partial \mathbf {v}_i}{\partial q_j}

.


This allows simplification of the last term,[1]:270 

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac {d}{d t} \left ( \frac {\partial T}{\partial \dot{q}_j} \right ) = \sum_{i=1}^n \left [ m_i \mathbf a_i \cdot \frac {\partial \mathbf {r}_i}{\partial q_j} + m_i \mathbf {v}_i \cdot \frac {\partial \mathbf {v}_i}{\partial q_j} \right ]

.


The partial derivative of Failed to parse (Missing texvc executable; please see math/README to configure.): T

with respect to the generalized coordinates, Failed to parse (Missing texvc executable; please see math/README to configure.): q_j

, is[1]:270 

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac {\partial T}{\partial q_j} = \sum_{i=1}^n m_i \mathbf {v}_i \cdot \frac {\partial \mathbf {v}_i} {\partial q_j}

.[clarify]


[This last result may be obtained by doing a partial differentiation directly on the kinetic energy definition represented by the first equation.] The last two equations may be combined to give an expression for the inertial forces in terms of the kinetic energy:[1]:270 

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac {d}{d t} \left ( \frac {\partial T}{\partial \dot{q}_j} \right ) - \frac {\partial T}{\partial q_j} = \sum_{i=1}^n m_i \mathbf a_i \cdot \frac {\partial \mathbf {r}_i}{\partial q_j}

Old Lagrange's equations

Consider a single particle with mass m and position vector Failed to parse (Missing texvc executable; please see math/README to configure.): \bold{r}
, moving under an applied force, Failed to parse (Missing texvc executable; please see math/README to configure.): \bold{F}
, which can be expressed as the gradient of a scalar potential energy function Failed to parse (Missing texvc executable; please see math/README to configure.): V (\bold{r},t)

Failed to parse (Missing texvc executable; please see math/README to configure.): \bold{F} = - \bold{\nabla} V.

Such a force is independent of third- or higher-order derivatives of Failed to parse (Missing texvc executable; please see math/README to configure.): \bold{r}
, so Newton's second law forms a set of 3 second-order ordinary differential equations. Therefore, the motion of the particle can be completely described by 6 independent variables, or degrees of freedom. An obvious set of variables is Failed to parse (Missing texvc executable; please see math/README to configure.): \{ \bold{r}_j, \dot{\bold{r}}_j | j = 1, 2, 3\}
, the Cartesian components of Failed to parse (Missing texvc executable; please see math/README to configure.): \bold{r}

and their time derivatives, at a given instant of time (i.e. position (x,y,z) and velocity Failed to parse (Missing texvc executable; please see math/README to configure.): (v_x,v_y,v_z)

). 


More generally, we can work with a set of generalized coordinates, Failed to parse (Missing texvc executable; please see math/README to configure.): q_j
, and their time derivatives, the generalized velocities, Failed to parse (Missing texvc executable; please see math/README to configure.): \dot{q_j}
. The position vector, Failed to parse (Missing texvc executable; please see math/README to configure.): \bold{r}
, is related to the generalized coordinates by some transformation equation:

Failed to parse (Missing texvc executable; please see math/README to configure.): \bold{r} = \bold{r}(q_i , q_j , q_k, t).

For example, for a simple pendulum of length l, a logical choice for a generalized coordinate is the angle of the pendulum from vertical, θ, for which the transformation equation would be

Failed to parse (Missing texvc executable; please see math/README to configure.): \bold{r}(\theta, \dot{\theta} , t) = (l \sin \theta, l \cos \theta)

.


The term "generalized coordinates" is really a holdover from the period when Cartesian coordinates were the default coordinate system.


Consider an arbitrary displacement Failed to parse (Missing texvc executable; please see math/README to configure.): \delta \bold{r}

of the particle. The work done by the applied force Failed to parse (Missing texvc executable; please see math/README to configure.): \bold{F}
is Failed to parse (Missing texvc executable; please see math/README to configure.): W = \bold{F} \cdot \delta \bold{r}

. Using Newton's second law, we write:

Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{matrix} \bold{F} \cdot \delta \bold{r} = m\ddot{\bold{r}} \cdot \delta \bold{r}. \end{matrix}

Since work is a physical scalar quantity, we should be able to rewrite this equation in terms of the generalized coordinates and velocities. On the left hand side,

Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{matrix} \bold{F} \cdot \bold{\delta} \bold{r} & = & - \bold{\nabla} V \cdot \displaystyle\sum_i {\partial \bold{r} \over \partial q_i} \delta q_i \\ \\ & = & - \displaystyle\sum_{i,j} {\partial V \over \partial r_j} {\partial r_j \over \partial q_i} \delta q_i \\ \\ & = & - \displaystyle\sum_i {\partial V \over \partial q_i} \delta q_i. \\ \end{matrix}

On the right hand side, carrying out a change of coordinates[clarify], we obtain:


Failed to parse (Missing texvc executable; please see math/README to configure.): m \ddot{\bold{r}} \cdot \delta \bold{r} = m \sum_{i,j} \ddot{r_i} {\partial r_i \over \partial q_j} \delta q_j 




Rearranging Slightly:


Failed to parse (Missing texvc executable; please see math/README to configure.): m \ddot{\bold{r}} \cdot \delta \bold{r} =  m \sum_j \left[ \sum_i \ddot{r_i} {\partial r_i \over \partial q_j} \right] \delta q_j 




Now, by performing an "integration by parts" transformation, with respect to t:


Failed to parse (Missing texvc executable; please see math/README to configure.): m \ddot{\bold{r}} \cdot \delta \bold{r} = m \sum_j \left[ \sum_i \left[  {\mathrm{d} \over \mathrm{d}t}   \left(  \dot{r_i} {\partial r_i \over \partial q_j} \right)  - \dot{r_i} {\mathrm{d} \over \mathrm{d}t}\left(   {\partial r_i \over \partial q_j} \right)      \right] \right] \delta q_j 




Recognizing that Failed to parse (Missing texvc executable; please see math/README to configure.): {\mathrm{d} \over \mathrm{d}t}{\partial r_j \over \partial q_i} = {\partial \dot{r_j} \over \partial q_i}

and Failed to parse (Missing texvc executable; please see math/README to configure.): {\partial r_j \over \partial q_i} = {\partial \dot{r_j} \over \partial \dot{q_i}}

, we obtain:


Failed to parse (Missing texvc executable; please see math/README to configure.): m \ddot{\bold{r}} \cdot \delta \bold{r} = m \sum_j \left[ \sum_i \left[  {\mathrm{d} \over \mathrm{d}t}   \left(  \dot{r_i} {\partial \dot{r_i} \over \partial \dot{q_j}} \right)  - \dot{r_i}   {\partial \dot{r_i} \over \partial q_j}       \right] \right] \delta q_j 




Now, by changing the order of differentiation, we obtain:


Failed to parse (Missing texvc executable; please see math/README to configure.): m \ddot{\bold{r}} \cdot \delta \bold{r} = m \sum_j \left[ \sum_i \left[  {\mathrm{d} \over \mathrm{d}t}  {\partial \over \partial \dot{q_j}} \left( \frac{1}{2} \dot{r_i}^2  \right) -  {\partial \over \partial q_j} \left( \frac{1}{2} \dot{r_i}^2 \right)   \right] \right] \delta q_j 




Finally, we change the order of summation:


Failed to parse (Missing texvc executable; please see math/README to configure.): m \ddot{\bold{r}} \cdot \delta \bold{r} = \sum_j \left[  {\mathrm{d} \over \mathrm{d}t}  {\partial \over \partial \dot{q_j}} \left( \sum_i \frac{1}{2} m \dot{r_i}^2  \right) -  {\partial \over \partial q_j} \left( \sum_i \frac{1}{2} m \dot{r_i}^2 \right) \right] \delta q_j 




Which is equivalent to:

Failed to parse (Missing texvc executable; please see math/README to configure.): m \ddot{\bold{r}} \cdot \delta \bold{r} = \sum_i \left[{\mathrm{d} \over \mathrm{d}t}{\partial T \over \partial \dot{q_i}}-{\partial T \over \partial q_i}\right]\delta q_i

where Failed to parse (Missing texvc executable; please see math/README to configure.): T=\frac{1}{2}m\dot{\bold{r}}\cdot\dot{\bold{r}}

is the kinetic energy of the particle. Our equation for the work done becomes

Failed to parse (Missing texvc executable; please see math/README to configure.): \sum_i \left[{\mathrm{d} \over \mathrm{d}t}{\partial{T}\over \partial{\dot{q_i}}}-{\partial{(T-V)}\over \partial q_i}\right] \delta q_i = 0.

However, this must be true for any set of generalized displacements Failed to parse (Missing texvc executable; please see math/README to configure.): \delta q_i
, so we must have

Failed to parse (Missing texvc executable; please see math/README to configure.): \left[ {\mathrm{d} \over \mathrm{d}t}{\partial{T}\over \partial{\dot{q_i}}}-{\partial{(T-V)}\over \partial q_i}\right] = 0

for each generalized coordinate Failed to parse (Missing texvc executable; please see math/README to configure.): \delta q_i
. We can further simplify this by noting that V is a function solely of r and t, and r is a function of the generalized coordinates and t. Therefore, V is independent of the generalized velocities:

Failed to parse (Missing texvc executable; please see math/README to configure.): {\mathrm{d} \over \mathrm{d}t}{\partial{V}\over \partial{\dot{q_i}}} = 0.

Inserting this into the preceding equation and substituting L = T - V, called the Lagrangian, we obtain Lagrange's equations:

Failed to parse (Missing texvc executable; please see math/README to configure.): {\partial{\mathcal{L}}\over \partial q_i} = {\mathrm{d} \over \mathrm{d}t}{\partial{\mathcal{L}}\over \partial{\dot{q_i}}}.

There is one Lagrange equation for each generalized coordinate qi. When qi = ri (i.e. the generalized coordinates are simply the Cartesian coordinates), it is straightforward to check that Lagrange's equations reduce to Newton's
second law.


The above derivation can be generalized to a system of N particles. There will be 6N generalized coordinates, related to the position coordinates by 3N transformation equations. In each of the 3N Lagrange equations, T is the total kinetic energy of
the system, and V the total potential energy.


In practice, it is often easier to solve a problem using the Euler-Lagrange equations than Newton's laws. This is because appropriate generalized coordinates qi may be chosen to exploit symmetries in the system.

Examples

In this section two examples are provided in which the above concepts are applied. The first example establishes that in a simple case, the Newtonian approach and the Lagrangian formalism agree. The second case illustrates the power of the above formalism, in a case which is hard to solve with Newton's laws.

Falling mass

Consider a point mass m falling freely from rest. By gravity a force F = m g is exerted on the mass (assuming g constant during the motion). Filling in the force in Newton's law, we find Failed to parse (Missing texvc executable; please see math/README to configure.): \ddot x = g

from which the solution

Failed to parse (Missing texvc executable; please see math/README to configure.): x(t) = \frac{1}{2} g t^2

follows (choosing the origin at the starting point). This result can also be derived through the Lagrange formalism. Take x to be the coordinate, which is 0 at the starting point. The kinetic energy is Failed to parse (Missing texvc executable; please see math/README to configure.): T = \frac{1}{2} m v^2

and the potential energy is Failed to parse (Missing texvc executable; please see math/README to configure.): V = - m g x

, hence 

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathcal{L} = T - V = \frac{1}{2} m \dot{x}^2 + m g x

. 
Now we find

Failed to parse (Missing texvc executable; please see math/README to configure.): 0 = \frac{\partial \mathcal{L}}{\partial x} - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial \mathcal{L}}{\partial \dot x} = m g - m \frac{\mathrm{d} \dot x}{\mathrm{d} t}

which can be rewritten as Failed to parse (Missing texvc executable; please see math/README to configure.): \ddot x = g
, yielding the same result as earlier.

Pendulum on a movable support

Consider a pendulum of mass m and length l, which is attached to a support with mass M which can move along a line in the x-direction. Let x be the coordinate along the line of the support, and let us denote the position of the pendulum by the angle θ from the vertical. The kinetic energy can then be shown to be

Failed to parse (Missing texvc executable; please see math/README to configure.): T = \frac{1}{2} M \dot{x}^2 + \frac{1}{2} m \left( \dot{x}_\mathrm{pend}^2 + \dot{y}_\mathrm{pend}^2 \right) = \frac{1}{2} M \dot{x}^2 + \frac{1}{2} m \left[ \left( \dot x + l \dot\theta \cos \theta \right)^2 + \left( l \dot\theta \sin \theta \right)^2 \right],

and the potential energy of the system is

Failed to parse (Missing texvc executable; please see math/README to configure.): V = m g \operatorname{y}_\mathrm{pend} = - m g l \cos \theta .

Image:PendulumWithMovableSupport.svg

Sketch of the situation with definition of the coordinates (click to enlarge)

Now carrying out the differentiations gives for the support coordinate x

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{\mathrm{d}}{\mathrm{d}t} \left[ (M + m) \dot x + m l \dot\theta \cos\theta \right] = 0,

therefore:

Failed to parse (Missing texvc executable; please see math/README to configure.): (M + m) \ddot x + m l \ddot\theta\cos\theta-m l \dot\theta ^2 \sin\theta = 0

indicating the presence of a constant of motion. The other variable yields

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{\mathrm{d}}{\mathrm{d}t}\left[ m( l^2 \dot\theta + \dot x l \cos\theta ) \right] + m (\dot x l \dot \theta + g l) \sin\theta = 0

therefore

Failed to parse (Missing texvc executable; please see math/README to configure.): \ddot\theta + \frac{\ddot x}{l} \cos\theta + \frac{g}{l} \sin\theta = 0

.
These equations may look quite complicated, but finding them with Newton's laws would have required carefully identifying all forces, which would have been much harder and prone to errors. By considering limit cases (Failed to parse (Missing texvc executable; please see math/README to configure.): \ddot x \to 0

should give the equations of motion for a pendulum, Failed to parse (Missing texvc executable; please see math/README to configure.): \ddot\theta \to 0
should give the equations for a pendulum in a constantly accelerating system, etc.) the correctness of this system can be verified.

Hamilton's principle

The action, denoted by Failed to parse (Missing texvc executable; please see math/README to configure.): \mathcal{S}
, is the time integral of the Lagrangian:

Failed to parse (Missing texvc executable; please see math/README to configure.): \mathcal{S} = \int \mathcal{L}\,\mathrm{d}t.

Let q0 and q1 be the coordinates at respective initial and final times t0 and t1. Using the calculus of variations, it can be shown the Lagrange's equations are equivalent to Hamilton's principle:

The system undergoes the trajectory between t₀ and t₁ whose action has a stationary value.

By stationary, we mean that the action does not vary to first-order for infinitesimal deformations of the trajectory, with the end-points (q0, t0) and (q1,t1) fixed. Hamilton's principle can be written as:

Failed to parse (Missing texvc executable; please see math/README to configure.): \delta \mathcal{S} = 0. \,\!

Thus, instead of thinking about particles accelerating in response to applied forces, one might think of them picking out the path with a stationary action.


Hamilton's principle is sometimes referred to as the principle of least action. However, this is a misnomer: the action only needs to be stationary, and the correct trajectory could be produced by a maximum, saddle point, or minimum in the action.


We can use this principle instead of Newton's Laws as the fundamental principle of mechanics, this allows us to use an integral principle (Newton's Laws are based on differential equations so they are a differential principle) as the basis for mechanics. However it is not widely stated that Hamilton's principle is a variational principle only with holonomic constraints, if we are dealing with nonholonomic systems then the variational principle should be replaced with one involving d'Alembert principle of virtual work. Working only with holonomic constraints is the price we have to pay for using an elegant variational formulation of mechanics.

Extensions of Lagrangian mechanics

The Hamiltonian, denoted by H, is obtained by performing a Legendre transformation on the Lagrangian. The Hamiltonian is the basis for an alternative formulation of classical mechanics known as Hamiltonian mechanics. It is a particularly ubiquitous quantity in quantum mechanics (see Hamiltonian (quantum mechanics)).


In 1948, Feynman  invented the path integral formulation extending the principle of least action to quantum mechanics for electrons and photons. In this formulation, particles travel every possible path between the initial and final states; the probability of a specific final state is obtained by summing over all possible trajectories leading to it. In the classical regime, the path integral formulation cleanly reproduces Hamilton's principle, and Fermat's principle in optics.

See also

• Restricted three-body problem
• Hamiltonian mechanics
• Functional derivative
• Nielsen form
• Canonical coordinates
• Generalized coordinates

References

1. ^ ^{a b c d e f g h i j k l m n o p q r s t u v w} Torby, Bruce (1984). "Energy Methods", Advanced Dynamics for Engineers, HRW Series in Mechanical Engineering (in English). United States of America: CBS College Publishing. ISBN 0-03-063366-4. 

• Goldstein, H. Classical Mechanics, second edition, pp.16 (Addison-Wesley, 1980)
• Moon, F. C. Applied Dynamics With Applications to Multibody and Mechatronic Systems, pp. 103-168 (Wiley, 1998).

Further reading

• Landau, L.D. and Lifshitz, E.M. Mechanics, Pergamon Press.
• Gupta, Kiran Chandra, Classical mechanics of particles and rigid bodies (Wiley, 1988).

External links

• Tong, David, Classical Dynamics Cambridge lecture notes
• Principle of least action interactive Excellent interactive explanation/webpage
• Aerospace dynamics lecture notes on Lagrangian mechanics
• Aerospace dynamics lecture notes on Rayleigh dissipation functionar:ميكانيك لاغرانج

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