Newtonian Mechanics 05

— 1. Variational Calculus —

Definition 1 A functional is a mapping from vector spaces to into real numbers.

Let {\displaystyle J = \int _{x_1}^{x^2} f\{y(x),y\prime (x),x\}dx}. Suppose that {x_1} and {x_2} are constants, the functional form of {f} is known.

According to definition 1 {J} is a functional and the goal of the Calculus of Variations is to determine {y(x)} such that the value of {J} is an extremum.

Let {y=y(\alpha, x)} be a parametric representation of {y} such that {y(0,x)=y(x)} is the function that makes {J} an extremum.

We can write {y(\alpha, x)=y(0,x)+ \alpha\eta(x}, where {\eta (x)} is a function of {x} of the class {C^1} (that means that {\eta} is a continuous function whose derivative is also continuous) with {\eta (x_1)=\eta (x_2)=0}.

Now {J} is of the form {\displaystyle J(\alpha) = \int _{x_1}^{x^2} f\{y(\alpha, x),y\prime (\alpha, x),x\}dx}

Therefore the condition for {J} to be an extremum is

\displaystyle \frac{dJ}{d\alpha}(\alpha=0)=0

Example 1 Let {y(x)=x}. Take {y(\alpha, x)= x+ \alpha\sin x} as a parametric representation of {y}. Let {f=\left(dy/dx\right)^2}, {x_1=0} and {x_2=2\pi}.

Given the previous parametric equation find {\alpha} such that {J} is a minimum.

Now {\eta (0)=\eta (2\pi)=0} and {dy/dx=1+\alpha\cos x}.

Hence {\displaystyle J(\alpha)= \int_0^{2\pi}(1+2\alpha\cos x +\alpha^2\cos ^2x)dx=2\pi+\alpha^2\pi}.

By the previous expression {J(\alpha)} it is trivial to see that the minimum value is reached when {\alpha=0}

Exercise 1 Given the points {(x_1,y_1)=(0,0)} and {(x_2,y_2)=(1,0)}, calculate the equation of the curve that minimizes the distance between the points

Now {y(\alpha, x)=y(0,x)+\alpha \eta (x) = 0+\alpha(x^2-x)}.

It is {\eta (x) = x^2-x}, {ds=\displaystyle \sqrt{dx^2+dy^2}=\sqrt{1+(dy/dx)^2}dx}

And it is {s= \displaystyle \int _0^1 \sqrt{1+(dy/dx)^2}} with {dy/dx=\alpha (2x-1)}.

The rest is left as an exercise for the reader.

— 2. Euler Equations —

In the following section we’ll analyze the condition for {J} to be an extremum:

{\begin{aligned} \frac{\partial J}{\partial \alpha} &= \frac{\partial}{\partial \alpha} \int _{x_1}^{x_2}f(y,y\prime,x)dx \\ &= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}+ \frac{\partial f}{\partial y\prime}\frac{\partial y\prime}{\partial \alpha}\right) dx \end{aligned}}

Since it is {\partial y /\partial \alpha = \eta (x)} and {\partial y\prime /\partial \alpha = d\eta/dx} it follows

\displaystyle \frac{\partial J}{\partial \alpha}= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\eta (x)+ \frac{\partial f}{\partial y\prime}\frac{d \eta}{dx}\right) dx

Now {\displaystyle \int _{x_1}^{x_2}\frac{\partial f}{\partial y\prime}\frac{d \eta}{dx}dx=\frac{\partial f}{\partial y\prime}\eta (x)|_{x_1}^{x_2}- \int _{x_1}^{x_2}\frac{d}{dx}\left( \frac{\partial f}{\partial y\prime} \right)\eta (x) dx}.

For the first term it is {\frac{\partial f}{\partial y\prime}\eta (x)|_{x_1}^{x_2}=0} since {\eta (x_1)=\eta (x_2)=0} by hypothesis.


{\begin{aligned} \frac{\partial J}{\partial \alpha} &= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}- \frac{d}{dx}\left( \frac{\partial f}{\partial y\prime} \right) \frac{\partial y}{\partial \alpha}\right)dx \\ &= \int _{x_1}^{x_2}\left( \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y\prime} \right)\eta (x) dx \end{aligned}}

Remembering that {\partial J / \partial\alpha(\alpha=0)=0} and taking into account the fact that {\eta (x)} is an arbitrary function one can conclude that

\displaystyle \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y\prime}=0

The previous equation is known as the Euler’s Equation

Example 2 As an example let us try to derive the equations of motion of a particle that moves in a constant force field starting its movement from the state of rest. The particles starts from point {x_1, y_1} and goes to point {x_2, y_2}.

From the enunciate it follows {K+U=c}. Let us take our original point as being our reference point for the potential. Then it is {k+U=0}.

As always it is {k=1/2mv^2}. For the potential it is {U=-Fx=-mgx}. From the previous equations it follows that {v=\sqrt{2gx}}.

From the definition of velocity it follows that

\displaystyle t=\int _{x_1,y_1}^{x_2,y_2} \frac{ds}{v}=\int _{x_1,y_1}^{x_2,y_2}\frac{\sqrt{dx^2+dy^2}}{\sqrt{2gx}}=\int _{x_1,y_1}^{x_2,y_2}\frac{\sqrt{1+y\prime^2}}{\sqrt{2gx}}dx

Let {f=\sqrt{\frac{1+y\prime^2}{x}}} since {(2g)^{-1/2}} is only a constant factor and can be omitted from our analysis. Given the functional form of {f} it is {df/dy=0} and Euler’s Equation just is:

\displaystyle \frac{d}{dx}\frac{\partial f}{\partial y\prime}=0

From the previous relationship it is

\displaystyle \frac{\partial f}{\partial y\prime}=(2a)^{-1/2}=\mathrm{const}

Hence it is

{\begin{aligned} \frac{y\prime^2}{x(1+y\prime^2)} &= \frac{1}{2a} \Rightarrow\\ y &= \int \frac{x}{\sqrt{2ax-x^2}}dx \end{aligned}}

Making the change of variables {x=a(1-\cos \theta)} it follows {dx=a\sin \theta d\theta}. Hence the expression for {y} is {y=\int a(1-\cos \theta)d\theta\Rightarrow y=a(\theta-\sin \theta)+A}. Since our particle starts from the origin it is {A=0}.

Thus the solution to our initial problem is

{\begin{aligned} x &= a(1-\cos \theta) \\ y &= a(\theta-\sin \theta) \end{aligned}}

Which are the parametric equations of a cycloid.

To close our thoughts on the Euler equation let us say that there also is a second form for the Euler equation. The second form is

\displaystyle f-y\prime\frac{\partial f}{\partial y\prime}= \mathrm{const}

and is used in the cases where {f} doesn’t depend explicitly on {x}.

— 3. Euler Equation for {n} variables —

Let {f} be of the form {f=f\{ y_1(x),y\prime _1(x),y_2(x),y\prime _2(x),\cdots,y_n(x),y\prime _n(x), x \}}.

Now we have {y_i(\alpha, x)= y_i(0,x)+\alpha \eta (x)} and {\displaystyle \int _{x_1}^{x_2}\left( \frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y _i\prime} \right)\eta _i (x) dx} for each of the values of {i}. Since {\eta _i(x)} are independent functions it follows that for {\alpha=0}

\displaystyle \frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y _i\prime}=0

That is to say we have {n} independent Euler equations.

— 4. Hamilton’s Principle —

Minimum principles have a long history in the history of Physics:

In modern Physics one uses a more general extremum principle and the focus of this section will be to state this principle and flesh out its consequences.

Definition 2 The lagrangian (sometimes called the lagrangian function), {L}, of a particle is the difference between its kinetic and potential energies.

\displaystyle L=K-U \ \ \ \ \ (1)

Definition 3 The action, {S}, of a particle’s movement (be it a real or virtual one) is:

\displaystyle \int_{t_1}^{t_2}(K-U)dt \ \ \ \ \ (2)

Axiom 1 Given a collection of paths that a particle can take between points {x_1} and {x_2} in the the time interval {\Delta t= t_2-t_1} the actual path that the particle takes is the one that makes the action stationary

\displaystyle \delta S=\delta \int_{t_1}^{t_2}(K-U)dt=0 \ \ \ \ \ (3)

For rectangular coordinates it is {T=T(x_i)}, {U=U(x_i)}, so {L=T-U=L(x_i,\dot{x}_i)} (where {\dot{x}_i=\dfrac{dx_i}{dt}} is called Newton’s notation).

The function {L} can be identified with the function {f} that we saw on Newtonian Mechanics 04 if one makes the obvious analogies

  • {x \rightarrow t}
  • {y_i(x) \rightarrow x_i(t)}
  • {y\prime_i(x) \rightarrow x\prime_i(t)}
  • {f(y_i(x),y\prime_i (x),x) \rightarrow L(x_i,\dot{x}_i,t)}

In this case the Euler equations are called the Euler-Lagrange equations and it is

\displaystyle \frac{\partial L}{\partial x_i}-\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_i}=0

Example 3 Let us study the harmonic oscillator under Langrangian formalism

\displaystyle L=T-U=1/2m\dot{x}^2-1/2kx^2 \ \ \ \ \ (4)

First it is {\dfrac{\partial L}{\partial x_i}=-kx}.

Then we have {\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{x}_i}=\dfrac{d}{dt}m\dot{x}=m\ddot{x}}.

Hence it is {\dfrac{\partial L}{\partial x_i}-\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{x}_i}=0 \Rightarrow m\ddot{x}+kx=0 \Rightarrow m\ddot{x}=-kx} which is just the harmonic oscillator dynamic equation that we already know.

Example 4 Consider a planar pendulumplanar pendulum write its Lagrangian and derive its equation of motion.

The Lagrangian for the planar pendulum is

\displaystyle L=1/2ml^2\dot{\theta}^2-mgl(1-\cos \theta) \ \ \ \ \ (5)

If we consider {\theta} to be a rectangular coordinate (which it isn’t!) it follows that the equation of motion is:

\displaystyle \ddot{\theta}+g/l\sin \theta=0

This is precisely the equation of motion of a planar pendulum and this result is apparently unexpected since we only analyzed the Lagrangian for rectangular coordinates.

— 5. Generalized coordinates —

Consider a mechanical system constituted by {n} particles. In this case one would need {3n} quantities to describe the position of all particles (since we have 3 degrees of freedom). In the case of having any kind of restraints on the motion of the particles the number of quantities needed to describe the motion of particle is less than {3n}. Suppose that one has {m} restrictions than the degrees of freedom are {3n-m}.

Let {s=3n-m}. These {s} coordinates don’t need to be rectangular, polar, cylindrical nor spherical. These coordinates can be of any kind provided that they completely specify the mechanical state of the system.

Definition 4 The set of {s} coordinates that totally specify the mechanical state of {n} particles is defined to be the set of generalized coordinates.

The generalized coordinates are represented by

\displaystyle q_1,q_2,\cdots,q_s

Since we defined the generalized coordinates of a system of particles one can also define its set of generalized velocities.

Definition 5 The set of {s} velocities of a system of {n} particles described by a set of generalized coordiinattes is defined to be the set of generalized velocities.

The generalized velocities are represented by

\displaystyle \dot{q_1},\dot{q_2},\cdots,\dot{q_s}

Let {\alpha} denote the particle, {\alpha=1,2,\cdots,n}, {i} represent the degrees of freedom {i}, {i=1,2,3} and {j} the number generalized coordinates {j=1,2,\cdots,s}.

\displaystyle x_{\alpha,i}=x_{\alpha,i}(q_1,q_2,\cdots,q_s,t)=x_{\alpha,i}(q_j,t) \ \ \ \ \ (6)

For the generalized velocities it is

\displaystyle \dot{x}_{\alpha,i}=\dot{x}_{\alpha,i}(q_j,t) \ \ \ \ \ (7)

The inverse transformations are

\displaystyle q_j=q_j(x_{\alpha,i},t) \ \ \ \ \ (8)


\displaystyle \dot{q_j}=\dot{q}_j(x_{\alpha,i},t) \ \ \ \ \ (9)

Finally let us note that we also need {m=3n-s} equations of constraint

\displaystyle f_k=f_k(x_{\alpha,i},t) \ \ \ \ \ (10)

with {k=1,2,\cdots,m}.

Example 5 Consider a point particle that moves along the surface of a semi-sphere of radius {R} whose center is the origin of the coordinate system.

The relevant equations are {x^2+y^2+z^2-R^2\geq 0} and {z\geq 0}.

Let {q_1=x/R}, {q_2=y/R} and {q_3=z/R} be our generalized coordinates.

Furthermore we have the condition {q_1^2+q_2^2+q_3^2=1} as a constraint equation. Hence {q_3=\sqrt{1-(q_1^2+q_2^2)}}

Definition 6 Configuration space is the vector space defined by the generalized coordinates

The time evolution of a mechanical system can be represented as a curve in the configuration space.

— 6. Euler-Lagrange Equations in generalized coordinates —

Since {K} and {U} are scalar functions {L} is also a scalar function. Therefore {L} is an invariant for coordinate transformations.

Hence it is

\displaystyle L=K(\dot{x}_{\alpha,i})- U(x_{\alpha,i})=T(q_j,\dot{q}_j,t)-U(q_j,t) \ \ \ \ \ (11)

and {L=L(q_j,\dot{q}_j,t)}.

Hence we can write Hamilton’s Principle (Section 4) in the form

\displaystyle \delta \int_{t_1}^{t_2} L(q_j,\dot{q}_j,t) dt=0 \ \ \ \ \ (12)

That is

  • {x \rightarrow t}
  • {y_i(x) \rightarrow q_j(t)}
  • {y\prime_i(x) \rightarrow q\prime_j(t)}
  • {f(y_i(x),y\prime_i (x),x) \rightarrow L(q_j,\dot{q}_j,t)}

are the analogies to be made now.

Finally the Euler-Lagrange Equations are

\displaystyle \frac{\partial L}{\partial q_j}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}=0 \ \ \ \ \ (13)

for {j=1,2,\cdots,s}

To finalize this section let us note the conditions of validity for the Euler-Lagrange equations:

  • The system is conservative.
  • The equations of constraint have to be functions between the coordinates of the particles and can also be a function of time.
Example 6 Consider the motion of a particle of mass {m} along the surface of a half-angle cone under the action of the force of gravity.


The equations are {z=r\cot\alpha} and {v^2=\dot{r}^2\csc^2\alpha+r^2\dot{\theta}^2}

Now for the potential energy it is {U=mgz=mgr\cot\alpha}. Thus the lagrangian is

\displaystyle L=1/2m(\dot{r}^2\csc^2\alpha+r^2\dot{\theta}^2)-mgr\cot\alpha

Since {\dfrac{\partial L}{\partial \theta}=0} it is {\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{\theta}}=0}. Hence it is {\dfrac{\partial L}{\partial \dot{\theta}}=mr^2\dot{\theta}=\mathrm{const}}.

The angular momentum about the {z} axis is {mr^2\dot{\theta}=mr^2\omega}. Thus {mr^2\omega=\mathrm{const}} expresses the conservation of angular momentum about the axis of symmetry of system.

It is left as an exercise for the reader to find the Euler-Lagrange equation for {r}.


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