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Chapter 7 - Lagrange's Equations
Section 7.1 - Lagrange's Equations for Unconstrained Motion
Consider a particle experiencing unconstrained motion in three dimensions and subject to some conservative net force \mathbf{F}(\mathbf{r}). Then, we see that
T = \frac{1}{2}mv^2 = \frac{1}{2}m\dot{\mathbf{r}}^2 = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)
Naturally, U = U(\mathbf{r}) = U(x, y, z).
Definition. The Lagrangian function or Lagrangian is then \mathcal{L} = T - U.
We can calculate that \frac{\partial \mathcal{L}}{\partial x} = -\frac{\partial U}{\partial x} = F_x, and which by Newton's second law, F_x = \dot{p}_x. Additionally, we can see that \frac{\partial \mathcal{L}}{\partial \dot{x}} = m\dot{x} = p_x, so then \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q_i}} = \dot{p}_x. An important note is that these derivatives are taken holding the higher and lower order derivatives constant. Thus, we see that
\frac{\partial \mathcal{L}}{\partial x} = \frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot{x}}
This holds true in y and z as well. This tells us that Newton's second law is equivalent to the Lagrange equations.
Theorem. Hamilton's Principle. The actual path in which a particle flows between points 1 and 2 in a given time t_1 to t_2 is such that the action integral S is stationary on the given path. Here,
S = \int_{t_1}^{t_2} \mathcal{L} dt
It is important to note that if we switch to a coordinate system with coordinates q_1, q_2, q_3, in order for this criterion to hold, the Lagrangian must be written in an inertial reference frame (that is, F_{q_i} = \dot{p_i}).
Definition. The generalized force of a system can be expressed in terms of the Lagrangian, where \frac{\partial \mathcal{L}}{\partial q_i} is the $i$-th component of generalized force.
Definition. The generalized momentum of a system can be expressed in terms of the Lagrangian, where \frac{\partial \mathcal{L}}{\partial \dot{q}_i} is the $i$-th component of generalized momentum.
Notably, when using polar or cylindrical coordinates, the $\phi$-component of generalized momentum is the angular momentum of the system.
With N unconstrained particles, we see 3N equations.
Section 7.2 - Constrained Systems, an Example
Consider a pendulum with mass m, constrained by \sqrt{x^2 + y^2} = l. Now, we can clearly see that T = \frac{1}{2} m v^2 = \frac{1}{2}l^2\dot{\phi}^2, and U = mgh = mgl(1 - \cos \phi). Then,
\mathcal{L} = T - U = \frac{1}{2}l^2\dot{\phi}^2 - mgl(1 - \cos \phi)
Then, we see that solving the \phi component of Lagrange's equations,
-mgl \sin \phi = \frac{d}{dt}(ml^2 \dot{\phi})
We know that ml^2 = I is the pendulum's moment of inertia, (mg) * (-l)\sin\theta is the torque on the pendulum, and $\ddot{\phi}$$ is the angular acceleration \alpha, the equation then becomes
\Gamma = I \alpha
Section 7.3 - Constrained Systems in General
Consider a system with N particles, indexed by \alpha = 1, \ldots, N, and positions \mathbf{r}_\alpha. We can then say that q_1, \ldots, q_n are a set of generalized coordinates for the system if and only if each r_\alpha can be expressed as a function of the generalized coordinates and time, so that
\mathbf{r}_\alpha = \mathbf{r}_\alpha(q_1, \ldots, q_n, t)
Conversely, we can write each q_i in terms of the positions and time, so that
q_i = q_i(\mathbf{r}_1, \ldots, \mathbf{r}_N, t)
Additionally, we require n to be minimal. Then, this implies that n \leq 3N.
Definition. We define a set of coordinates q_1, \ldots, q_n as natural if the transformation between the set and cartesian coordinates is time-independent.
Definition. The number of degrees of freedom of a system is the number of coordinates that can be independently varied in a small displacement. That is, it is the number of variables that are free parameters of the system.
Definition. A constrained $N$-particle system is one in which the number of degrees of freedom of the system is less than 3N.
Definition. A holonomic system is one in which the number of degrees of freedom is the same as the number of generalized coordinates.
Consider a ball on a surface that is free to roll but not slide or spin. Then, it has two degrees of freedom (x and y). However, to describe the system we must indicate rotation as well, so that it requires five generalize coordinates.
Section 7.4 - Proof of Lagrange's Equations with Constraints
Let us consider the case of one particle constrained to a surface so that it can be described as a holonomic system with two generalized coordinates q_1 and q_2. Then, we can see that there are two kinds of forces on the particle: the force of constraint, and any non-constraint forces.
For this problem, we will denote the non-constraint forces as \mathbf{F} such that \mathbf{F} = -\nabla U(\mathbf{r}, t). Then, we can define the Lagrangian as usual, so that \mathcal{L} = T - U.
Now, consider any two points \mathbf{r}_1 and \mathbf{r}_2, at which the particle passes through at times t_1 and t_2 respectively. Then, we can denote the "correct" path as \mathbf{r}(t) and any other path as \mathbf{R}(t) = \mathbf{r}(t) + \mathbf{\epsilon}(t). By the endpoint constraints, we see that \mathbf{\epsilon}(t_1) = \mathbf{\epsilon}(t_2) = 0
Now, we can define the action integral S as
S = \int_{t_1}^{t_2} \mathcal{L}(\mathbf{R}, \dot{\mathbf{R}}, t) dt
If we let S_0 be the value of S when \mathbf{R} = \mathbf{r}, we see that then for some path, we can write \delta S = S - S_0 for some difference in paths \mathbf{\epsilon}(t).
We can then substitute to see that \delta \mathcal{L} = \mathcal{L}(\mathbf{R}, \dot{\mathbf{R}, t}) - \mathcal{L}(\mathbf{r}, \dot{\mathbf{r}}, t). Then, substituting the formulas for the Lagrangian and for any arbitrary path, we see that, neglecting any \mathbf{\epsilon}^2 terms,
\delta \mathcal{L} = \frac{1}{2}m [(\dot{\mathbf{r}} + \dot{\mathbf{\epsilon}})^2 - \dot{\mathbf{r}}^2] - [U(\mathbf{r} + \mathbf{\epsilon}, t) - U(\mathbf{r}, t)] = m \dot{\mathbf{r}} \cdot \dot{\mathbf{\epsilon}} - \mathbf{\epsilon} \cdot \nabla U
We can then see that
\delta S = \int_{t_1}^{t_2} \delta \mathcal{L} dt = \int_{t_1}^{t_2} [m \dot{\mathbf{r}} \cdot \dot{\mathbf{\epsilon}} - \mathbf{\epsilon} \cdot \nabla U] dt
This can be solved to see that \delta S = -\int_{t_1}^{t_2} \mathbf{\epsilon} \cdot[m\ddot{\mathbf{r}} + \nabla U] dt. We see then that m \ddot{\mathbf{r} } = F_{tot} = F_{constraint} + F, which implies that \delta S = -\int_{t_1}^{t_2} \mathbf{\epsilon} \cdot \mathbf{F}_{constraint}. As \mathbf{\epsilon} is in the plane of motion and the constraint force is normal to the plane, we see that \delta S = 0.
Section 7.5 - Examples of Lagrange's Equations
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Section 7.6 - Generalized Momenta and Ignorable Coordinates
We can write the generalized equations for force and momentum as
F_i = \frac{d}{dt}p_i
In general, the generalized force and momenta are not the same as the usual, as the coordinate system is not usually cartesian.
Definition. If the Lagrangian is independent of a coordinate q_i, then the coordinate's momentum is conserved, and the coordinate is said to be ignorable or cyclic.
The connection between the invariance of the Lagrangian and certain conservation laws is known as Noether's Theorem.
Section 7.7 - Conclusion
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Section 7.8 - More about Conservation Laws
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Section 7.9 - Lagrange's Equations for Magnetic Forces
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Section 7.10 - Lagrange Multipliers and Constraint Forces
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