Newtonian formalism exercises


Exercise 1 A particle of mass {m} moves in the plane {xy}. Its position vector is {\vec{r}=a\cos \omega t \vec{i}+b\sin \omega t \vec{j}} with {a,b,\omega} positive constants and {a>b}. Show that:

  1. The particle’s trajectory is an ellipse.

    For {x=a\cos\omega\ t} and {y=b\sin\omega\ t} it is

    \displaystyle \left( \frac{x}{a} \right)^2+\left( \frac{y}{b} \right)^2=\cos^2\omega t+\sin^2\omega t=1

    Which is the equation of an ellipse

  2. The particle is subject to a central force oriented to the origin

    It is

    {\begin{aligned} \vec{F} &= m\dfrac{d\vec{v}}{dt}\\ &= m\dfrac{d^2\vec{r}}{dt^2}\\ &= m\dfrac{d^2}{dt^2}a\cos \omega t \vec{i}+b\sin \omega t \vec{j}\\ &=-m\omega^2\vec{r} \end{aligned}}

    Which is a central force oriented to the origin (note the minus sign).

Exercise 2 A particle of constant mass {m} is subject to to a force {F}. Suppose that for {t_1} e {t_2} the velocities are {\vec{v}_1} and {\vec{v}_2}, respectively. Show that the work the force does on the particle equals its change in kinetic energy.

{\begin{aligned} W &= \int_{t_1}^{t_2}\vec{F}\cdot d\vec{r}\\ &= \int_{t_1}^{t_2}\vec{F}\cdot\dfrac{d\vec{r}}{dt}dt \\ &= \int_{t_1}^{t_2}\vec{F}\cdot\vec{v}dt\\ &= \int_{t_1}^{t_2}m\dfrac{d\vec{v}}{dt}\cdot\vec{v}dt\\ &=m\int_{t_1}^{t_2}\vec{v}\cdot d\vec{v}\\ &=\dfrac{1}{2} m\int_{t_1}^{t_2}d(\vec{v}\cdot\vec{v})\\ &=1/2m(v_2^2-v_1^2) \end{aligned}}

Exercise 3 For the conditions of exercise 1 calculate:

  1. The particle’s kinetic energy in the positive semi-major axis and on the positive semi-minor axis.

    First note that {\vec{v}=-\omega a\sin \omega t \vec{i}+\omega b \cos \omega t \vec{j}}.

    Now {K=1/2m(\omega^2a^2\sin^2\omega t+\omega^2b^2\cos^2\omega t)}. Let {A} denote the semi-major axis extension and {B} the semi-minor axis extension.

    Hence {K_A=1/2m\omega ^2b^2} and {K_B=1/2m\omega ^2a^2}

  2. The work done by the force field when it moves the particle from the positive semi-major axis to the positive semi-minor axis.

    {\begin{aligned} W &= \int_A^B\vec{F}\cdot d\vec{r}\\ &= \int_A^B(-m\omega ^2\vec{r})\cdot d\vec{r}\\ &= -m\omega ^2\int_A^B\vec{r}\cdot d\vec{r}\\ &= -1/2m\omega ^2\int_A^B d(\vec{r}\cdot\vec{r})\\ &=1/2m\omega ^2(a^2-b^2) \end{aligned}}

  3. The total work done by the force field by moving the particle along the ellipse.

    {W=0}. Do you see why?

Exercise 4 Show that if a particle is subject to {\vec{F}} and {\vec{v}} is its instantaneous velocity then the instantaneous power is

\displaystyle  \mathcal{P}=\vec{F}\cdot\vec{v}

By definition it is {dW=\vec{F}\cdot d\vec{r}}. Hence

{\begin{aligned} \mathcal{P}&=\dfrac{dW}{dt}\\ &= \vec{F}\cdot\dfrac{d\vec{r}}{dt}\\ &= \vec{F}\cdot\vec{v} \end{aligned}}

Exercise 5 Show that the integral {\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}} is independent of a particle’s trajectory if and only if { \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }.

Let {\Gamma=P_1AP_2BP_1} denote a closed curve and admit that {\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}} is path independent. Then

{\begin{aligned} \oint \vec{F}\cdot d\vec{r}&=\int_\Gamma\vec{F}\cdot d\vec{r} \\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}+\int_{P_2BP_1}\vec{F}\cdot d\vec{r}\\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}-\int_{P_1BP_2}\vec{F}\cdot d\vec{r}\\ &=0 \end{aligned}}

Where the last equality follows from our assumption that {\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}} is path independent.

Suppose now that { \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }.

{\begin{aligned} \int_\Gamma\vec{F}\cdot d\vec{r}&=\int_{P_1AP_2}\vec{F}\cdot d\vec{r}+\int_{P_2BP_1}\vec{F}\cdot d\vec{r}\\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}-\int_{P_1BP_2}\vec{F}\cdot d\vec{r}\\ &=0 \end{aligned}}

Where the last equality follows from our { \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }. Hence {\displaystyle\int_{P_1AP_2}\vec{F}\cdot d\vec{r}=\int_{P_1BP_2}\vec{F}\cdot d\vec{r}}.

Exercise 6 A particle of mass {m} moves along {x} subject to to a conservative force field {V(x)}. If the particle’s positions are {x_1} and {x_2} on {t_1} and {t_2}, respectively, show that, if {E} is the total energy then

\displaystyle  t_2-t_1=\sqrt{\frac{m}{2}}\int_{x_1} ^{x_2}\frac{dx}{\sqrt{E-V(x)}}

Write {1/2m\left( \dfrac{dx}{dt} \right)^2+V(x)=E} solve in order to {dt} and integrate.

Exercise 7 Consider a particle of mass {m} that moves vertically on a resistive medium where the retarding force is proportional to the particle’s velocity. Consider that particle initially moves in the downward direction with an initial velocity {v_0} from an height {h}. Derive the particle’s equation of motion {z=z(t)}.

It is {F=m\dfrac{dv}{dt}=-mgg-kmv} with {v<0} and {-kmv>0}. Then solve in order to {dv} and integrate to find {v=\dfrac{dz}{dt}=-\dfrac{g}{k}+\dfrac{kv_0+g}{k}e^{-kt}} (the terminal velocity {v_t} is {v_t=\displaystyle \lim_{t \rightarrow +\infty} v=-g/k}).

Solving in order to {dz} and integrating and it is {z=h-\dfrac{gt}{k}+\dfrac{kv_0+g}{k^2}(1-e^{-kt})}.

Exercise 8 A particle is shot vertically on a region where exists a constant gravitational field with a constant initial velocity {v_0}. Show that, in the presence of resistive force proportional to the square of the particle’s instantaneous velocity, the velocity of the particle when it returns to its initial position is:

\displaystyle  \frac{v_0 v_t}{\sqrt{v_0^2+v_t^2}}

Where {v_t} denotes the particle’s terminal velocity.

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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.

Hence

{\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)

and

\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.

HalfAngleCone01

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}.

Newtonian Mechanics 04

— 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}dx} 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.

Hence

{\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.

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.

Newtonian Mechanics 03

— 1. One Particle Conservation Theorems —

— 1.1. Linear considerations —

Let {\vec{s}} be a constant vector such as {\vec{F}\cdot\vec{s}=0}. Then {\dfrac{d\vec{p}}{dt}\cdot\vec{s}=\vec{F}\cdot\vec{s}=0}. Hence {\vec{p}\cdot\vec{s}} is constant.

The previous derivation shows that if {\vec{F}} is null along a given direction ({\vec{s}}), then the momentum component along that direction is a constant quantity.

— 1.2. Rotational considerations —

Definition 1 Given a reference point one can define the angular momentum of a particle relative to that point.

\displaystyle \vec{L}=\vec{r}\times\vec{p} \ \ \ \ \ (1)

 

The angular momentum is a measure of the amount a rotation that a particle has relative to a given point. For example if a particle moves in a straight line relative to point its angular momentum is {\vec{L}=\vec{r}\times\vec{p}=0} since {\vec{r}} is parallel to {\vec{p}} and the vector product of two parallel vectors is {0} by definition (Do you see why? If not go to this post). see the definition of vector product and prove the previous statement.

Just like we had forces in rectilinear motion to account for the variations of momentum one has the torque in curvilinear motion to account for the variation of angular momentum-

Definition 2 Given a reference point one can define the torque of a particle relative to that point.

\displaystyle \vec{\tau}=\vec{r}\times\vec{F} \ \ \ \ \ (2)

 

Given the definitions of angular momentum it follows that

\displaystyle \frac{d}{dt}\left(\vec{r}\times\vec{p}\right)= \frac{d\vec{r}}{dt}\times\vec{p}+\vec{r}\times\frac{d\vec{p}}{dt}

It is {\dfrac{d\vec{r}}{dt}\times\vec{p}=\dfrac{d\vec{r}}{dt}\times m\vec{v}=m\dfrac{d\vec{r}}{dt}\times\dfrac{d\vec{r}}{dt}=0} by definition.

Hence {\vec{\tau}=\vec{r}\times \dfrac{d\vec{p}}{dt}=\dfrac{d\vec{L}}{dt}}.

Thus if {\vec{\tau}=0}, {\dfrac{d\vec{L}}{dt}=0} and {\vec{L}} is constant in time.

Angular momentum and torque

— 1.3. Energetic considerations —

Let’s consider a particle moves under the action of a force and evolves from mechanical state {1} to mechanical state {2}.

Definition 3 The amount work done by a force against the inertial mass along the trajectory that leads from mechanical state {1} to mechanical state {2} is

\displaystyle W_{12}=\int_1^2\vec{F}\cdot d\vec{r} \ \ \ \ \ (3)

 

It is

{\begin{aligned} \vec{F}\cdot d\vec{r} &= m\frac{d\vec{v}}{dt}\cdot\frac{d\vec{r}}{dt}dt \\ &= m\frac{d\vec{v}}{dt}\cdot\vec{v}dt \\ &= \frac{m}{2}\frac{d}{dt}(\vec{v}\cdot\vec{v})dt \\ &= \frac{m}{2}\frac{d}{dt}(v^2)dt \\ &= d\left(\frac{1}{2}mv^2\right) \end{aligned}}

Hence, the integrand function for {W_{12}} is a total differential (note that we assumed that {\vec{F}} doesn’t depend explicitly on time nor on the velocity). Hence

\displaystyle W_{12}=\frac{1}{2}m(v_2^2-v_1^2)=K_2-K_1

If {W_{12}} depends uniquely on the mechanical states {1} and {2} and not on the trajectory that connects them one says that {\vec{F}} derives from a potential. In this case the force can be written as the negative gradient of a function that is said to be the potential energy function: {\vec{F}=-\nabla U}

It follows

{\begin{aligned} \int_1^2 \vec{F}\cdot d\vec{r} &= -\int_1^2 \nabla U\cdot d\vec{r} \\ &= \int_1^2 \sum_i \frac{\partial U}{\partial x_i}dx_i \\ &= \int_1^2 dU \\ &= U_1-U_2 \end{aligned}}

Let us define

Definition 4 The mechanical, {E}, energy of a mechanic system is the sum of the kinetic energy and the potential energy

\displaystyle E=K+U \ \ \ \ \ (4)

 

Now

{\displaystyle E = T+U \Rightarrow \frac{dE}{dt} = \frac{dT}{dt} + \frac{dU}{dt}}

and

{\displaystyle \vec{F}\cdot d \vec{r}=d\left(1/2mv^2\right)=dT\Rightarrow \frac{dT}{dt}=\vec{F}\cdot \frac{d \vec{r}}{dt} }

For {dU/dt} it is

{\begin{aligned} \frac{dU}{dt} &= \sum_i\frac{\partial U}{\partial x_i}\frac{\partial x_i}{\partial t}+ \frac{\partial U}{\partial t} \\ &= \nabla U \cdot \frac{d \vec{r}}{dt}+ \frac{\partial U}{\partial t} \end{aligned}}

Finally

{\begin{aligned} \frac{dE}{dt} &= \vec{F}\cdot \frac{d \vec{r}}{dt}+\nabla U \cdot \frac{d \vec{r}}{dt}+ \frac{\partial U}{\partial t} \\ &= \left( \vec{F}+\nabla U \right)\cdot \frac{d \vec{r}}{dt}+ \frac{\partial U}{\partial t} \\ &= \frac{\partial U}{\partial t} \end{aligned}}

If {U} isn’t an explicit function of time it follows that {dE/dt=0} and the mechanical system is said to be conservative.

 

Newtonian Mechanics 02

— 3.4. Gravitational Field —

Stating the Law of Universal Gravitation in usual terms one usually says that all particles in the Universe attract eachother with a force that is directly proportional to the masses of the particles and inversely proportional to the square of the distance between them.

Stated in these terms it seems like the force of gravitation is an instantaneous one. Hence one has to resort to field theoretical language in order to describe the action of a gravitational field in terms that are acceptable to us.

Definition 16 Gravitational field: vectorial field, {\vec{g}}, created by a body of mass {m_1} in all points of space (except on the point where the particle is situated) which is responsible for the gravitational interaction.

\displaystyle \vec{g}=G\frac{m_1}{r^2}\hat{r} \ \ \ \ \ (13)

If a particle of mass {m_2} interacts with a gravitic field {\vec{g}} the particle experiences the force {\vec{F}_g}:

\displaystyle \vec{F}_g=\vec{g}m_2=G \frac{m_1 m_2}{r^2}\hat{r} \ \ \ \ \ (14)

 

{\hat{r}} is a unit vector whose direction points from the position of {m_2} to the position of {m_1}.

For the particular case of a body of mass {m} which is suspended from height {h} and interacts with the gravitational field of the Earth the gravitational force is

\displaystyle F_g=G \frac{M_T m}{(R_T+h)^2}

Since {\vec{F}=m\vec{a}} holds for body of constant mass, one can write for the intensity of gravitational acceleration:

\displaystyle g=G\frac{M_T}{(R_T+h)^2}

.

Definition 17

When two bodies of mass {m_1} and mass {m_2} interact via gravity an energy which derives from the gravitational field is established between them. This energy has the name of gravitational potential energy:

\displaystyle U=-G \frac{m_1 m_2}{r} \ \ \ \ \ (15)

 

— 4. Waves and Oscillations —

Definition 18

Period is the minimum time interval necessary for two points in the same oscillatory phenomenon to be i the same mechanical state. It is represented by {T}.

Definition 19

Frequency is the number of cycles of an oscillatory phenomenon that occur in a second. It is represented by {f} and can be calculated by {f=1/T}.

Definition 20

Angular frequency is { \omega = 2\pi/T=2\pi f }

— 4.1. Oscillations —

In this subsection one will study the harmonic motion. This is an important kind of movement since that in first approximation one can always use this model to study oscillatory m otions.

Let us suppose that a particle moves along a straight line under the effect of a force {F}.

Definition 21

A motion is said to be harmonic when in an oscillatory movement the force is proportional to the displacement (initial, also called equilibrium, position is taken as the origin) and as the opposite direction of the movement.

\displaystyle F=-k x

Using Newton’s Axiom 2 and introducing {k/m=\omega^2} one can write:

\displaystyle \frac{\partial ^2 x}{\partial t^2}=-\omega ^2 x \ \ \ \ \ (16)

 

Solutions to this equation can be of the form {x(t)=A\cos (\omega t + \theta)} where {A} is the maximum displacement relative to the equilibrium position and {\theta} is the specific phase which identifies the particle’s initial position.

In the case of harmonic motion the definitions 18 and e 19 Can be written in the form {T=2\pi \sqrt{m/k} } e {f=1/(2\pi) \sqrt{k/m} }.

For an harmonic motion the kinetic and the potential energy are:

  • {K=\dfrac{1}{2} m \omega^2 A^2 \sin^2( \omega t + \theta ) }
  • {U=\dfrac{1}{2} k A^2 \cos^2( \omega t + \theta ) }

Thus the total energy of the system is {E=\dfrac{1}{2}kA^2}

— 4.2. Waves —

Definition 22

A wave is a propagation in a medium that propagates itself transporting energy.

Definition 23

Wavelength, {\lambda} is the minimum distance between two points in a wave that are in the same mechanical state.

Definition 24

The speed of a wave of wavelength {\lambda} and period {T} is {c=\lambda/T=\lambda f}

Definition 25

The wavenumber of a wave is é {k=2\pi/\lambda}

It is possible to demonstrate mathematically that the equation that governs the propagation of a perturbation that moves with constant speed {c}:

\displaystyle \frac{\partial ^2 \phi}{\partial x^2}=\frac{1}{c^2}\frac{\partial ^2 \phi}{\partial t^2} \ \ \ \ \ (17)

 

With the previous definitions it is trivial to see that equations of the form {f_1=A\sin(kx \pm \omega t)} and {f_2=A\cos(kx \pm \omega t)} are solutions of equation 17. These functions are called trignometric functions, {A} is the amplitude and it represents the maximum displacement of the entity that is vibrating.

In general one can say that a progressive wave that propagates to the right is always of the form {f=f(x-ct)}, while a wave that moves to the left is always of the form {g=g(x+ct)}.

Since the wave equation is a linear partial derivative equation any linear combination of solutions of equation 17 is still a solution of the wave equation.

In order for our solutions to have physical sense one has to impose certain conditions that the equations must follow in certain regions of space (and time). These conditions are called boundary conditions and its effect is to restrict the set of values that the solutions might take.

The solutions of the wave equation that follow from boundar conditions are said to be normal modes of vibration.

When is propagating and it encounters the boundary between two different media two things can happen:

  1. Transmission: some of the energy of the wave propagates into the second medium.

    transmissaoonda

  2. Reflection: all of the wave’s energy propagates in the first medium, but with opposite direction.

    reflexaoonda3

When two trigonometric waves of the same amplitude and frequency propagate in the same medium with opposite directions interact thet create a resulting wave whose mathematical expression is given by {f=2A\sin kx \cos \omega t}.

This is the mathematical expression of a stationary wave.

— 4.3. Interference —

When two waves of equal wavelength and constant phase difference interact they are said to interfere.

If the two waves are in the same region of space and are of equal phase the interference is said to be constructive and the amplitude of the resulting wave equals the sum of the individual amplitudes of each original wave.

interferenciaconstrutivapulsoondas

If the two waves are in the same region of space in phase opposition the interference is said to be destructive and the amplitude of the resulting wave is equal to the subtraction of the amplitudes of the two original waves.

interferenciadestrutivapulsoondas

The following diagram is a schematic representation of an experimental realization of an interference pattern:

InterferenciaOndas

— 4.4. Diffraction —

When light of a well defined wavelength passes through a barrier with a slit of width {d} the phenomenom that occurs is called diffraction. Each portion of the slit acts as if it is an independent source of a propagation and as a consequence waves coming from different portions of the slit have different phases. From this interaction one can observe destructive or constructive interference.

The following diagram shows a schematic representation of an experimental realization of the phenomenon of diffraction:

PadraoDifraccao

 

Newtonian Mechanics 01

— 2. Initial Considerations —

Without being too far away from the truth one can say that modern Fundamental Physics rests on these three conceptions

  1. The concept of field
  2. The theory of relativity
  3. Quantum Physics

The concept of field is essential to all of our future discussion hence we’ll define it right away:

Definition 1 Field is an mathematical structure with a defined value in a given set of points.
Definition 2 A field is said to be a vectoral field when it assumes vector values.
Definition 3 A field is said to be a scalar field when it assumes scalar values.

The field equations that we are going to use always represent linear interactions. Hence one can consider each interaction resulting from a field as being independent of all other interactions being analyzed and the resulting interaction is just the sum of all interactions.

Associated to the concept of field we have the concept of potential energy. This energy arises as a result of the result of the interaction of the particle with the field {\vec{A}} and in general it is proportional to {\displaystyle\int_a^b\vec{A}\cdot d\vec{s}} where {d\vec{s}} represents an infinitesimal displacement.

— 3. Mechanics —

This section will be a very brief and shallow introduction to the results and triumphs of the first modern physical thepry. Nevertheless one hopes that its internal elegance and depth can be glimpsed through.

— 3.1. Basic concepts and preliminary definitions —

All mechanical quantities can be expressed in units that derive from the units of the following quantities:

  • Length {L}.
  • Time {T}.
  • Mass {M}. In classical mechanics the mass of a body its a measure of its resistance to alter its state of movement. This characteristic has the name of inertia.

The units that one uses to express the previous quantities are totally conventional. In the following we’ll use the Internation System of Units:

  • {\left[ L \right] =m}
  • {\left[ T \right] =s}
  • {\left[ M \right] = \mathrm{Kg}}
Definition 4

A frame is a set of axes that represent the degrees of freedom of the system that is being studied and an arbitrary point that serves has its origin.

Definition 5 A frame is said to be inertial when it possesses the following properties:

  • Space is homogeneous (all points are equivalent) and isotropic (there is no special orientation)
  • Time is homogeneous (all time instants are equivalent)
Definition 6 Position is the geometric localization of a particle in a frame.
Definition 7 Trajectory is the geometric place of the sucessive positions of a particle in a frame in a given time interval.
Definition 8 Displacement it’s the difference between the final position and the initial position of a particle. The displacement is represented by the symbol {\Delta \vec{x}}.

We know by everyday experience that bodies move by different displacements during different time intervals. The concept that expresses how a particle position changes in a given time interval it’s called velocity.

Definition 9 Average velocity: vectorial quantity that expresses that rate of change in a particles position for a given time interval:

\displaystyle \vec{v}_m=\dfrac{\Delta \vec{x}}{\Delta t} \ \ \ \ \ (4)

Definition 10 Velocity: Vectorial quantity that expresses a particles velocity in a given time instant:

\displaystyle \vec{v}=\lim_{\Delta t\rightarrow 0}\dfrac{\Delta \vec{x}}{\Delta t}=\dfrac{d\vec{x}}{dt} \ \ \ \ \ (5)

Since a particle’s velocity varies in time one can introduce the concept of acceleration.

Definition 11 Average acceleration: Vectorial quantity that expresses the rate of change of velocity for a given time interval.

\displaystyle \vec{a}_m=\dfrac{\Delta \vec{v}}{\Delta t} \ \ \ \ \ (6)

Definition 12 Acceleration: vectorial quantity that expresses the change in velocity in a given time instant.

\displaystyle \vec{a}=\lim_{\Delta t\rightarrow 0}\dfrac{\Delta \vec{v}}{\Delta t}=\dfrac{d\vec{v}}{dt} \ \ \ \ \ (7)

Definition 13 Kinetic energy is the energy associated with the movement of a particle and is defined by:

\displaystyle K=\dfrac{1}{2}m\vec{v}\cdot\vec{v}=\dfrac{1}{2}mv^2=\dfrac{1}{2}m\left( \dfrac{d\vec{x}}{dt}\right)^2 \ \ \ \ \ (8)

Definition 14 Linear momentum : vectorial quantity that is associated with the movement of a particle

\displaystyle \vec{p}=m \vec{v}=m \dfrac{d\vec{x}}{dt} \ \ \ \ \ (9)

Definition 15

The mechanical state of a particle is specified by the simultaneous determination (and with infinite precision) of its coordinates and linear momentum.

— 3.2. Newton’s Axioms —

Thus far we have defined the actors in our stage play but we haven’t defined by which rules do they interact. These rules are given by Newton’s Axioms.

Axiom 1 There is an inertial frame where the linear momentum of a free particle (a particle that experiences no interactions) is constant

Our definition 5 of an inertial frame is, strictly speaking, a mathematical definition. As such nothing in it implies that such a frame exists in our world. As such the function of Newton’s first Axiom is to guarantee that in inertial frame exists in our world.

One important detail is that Axiom 5 only posits the existance of one inertial frame, but we can conclude that there exists an infinite number of inertial frames.

Since an inertial frame space is homogenous and isotropic the point one picks as the origin is arbitrary. Hence one can make a translation of the first origin point and consider the final origin point as being the origin of this new frame, which has to be a new inertial frame.

Moreover one can apply a rotation to our firs inertial frame and obtain a new frame (the time the axes have different orientations). Since space is isotropic this orientation can’t change the nature of the frame and we can conclude that the final frame is also an inertial one.

Additionally one can consider and inertial that moves with constant velocity relatively to an inertial frame. Since space is homegeneous this operation can’t change the nature of the the frame in question and once again it still has to be an inertial frame.

Likewise, since time time is homogeneous one can consider a frame that differs of an inertial frame only in the instant of time that was taken as its origin. Since time is homogeneous this second frame also has to be an inertial one.

Finally let us just notice that any finite or infinite composition of these transformations also produces an inertial frame.

Axiom 2

An in inertial frame the change of linear momentum of a particle is caused by the action of a force {\vec{F}}.

{\vec{F}= \dfrac{d\vec{p}}{dt}}.

When the mass of the particle being considered is constant this axiom becomes {\vec{F}=m\vec{a}}.

In what follows the mass of a particle is to be considered constant unless stated otherwise.

Axiom 3

When two particles interact the force {\vec{F}_{12}} (force that object {1} exerts on object {2}) has an equal intensity as {\vec{F}_{21}} but opposite direction{\vec{F}_{12}=-\vec{F}_{21}}

— 3.3. Kinematics and Dynamics —

On this section we’ll introduce very briefly the concepts that describe (kinematics) and explain (dynamics) the movement of a particle.

— 3.3.1. Equations of movement —

From the definitions of velocity and acceleration that we introduced in section 3 it follows that

\displaystyle d\vec{v}= \vec{a}dt \Rightarrow \int_{t_0}^t d\vec{v}= \int_{t_0}^t \vec{a}dt \Rightarrow \vec{v}(t)-\vec{v}(t_0)=\int_{t_0}^t \vec{a}dt \ \ \ \ \ (10)

 

Since the functional relationship between the acceleration and time is unknown the right hand side of the equality can not be calculated.

It also is

\displaystyle d\vec{x}= \vec{v}dt \Rightarrow \int_{t_0}^t d\vec{x}= \int_{t_0}^t \vec{v}dt \Rightarrow \vec{x}(t)-\vec{x}(t_0)=\int_{t_0}^t \vec{v}dt \ \ \ \ \ (11)

 

Since {\vec{v}(t)} is also an unknown function the calculation has to stop.

If we assume that {\vec{a}} is constant in time (uniformly accelerated motion) we can solve equation 10, { \vec{v}=\vec{v}_0+\vec{a}(t-t_0)}. After substituting in equation 11 it follows

\displaystyle \vec{x}(t)=\vec{x}_0+\vec{v}_0(t-t_0)+\frac{1}{2}\vec{a}(t-t_0)^2 \ \ \ \ \ (12)

 

For the special case {\vec{a}=\vec{0}} the motion is said to be uniform.

— 3.3.2. Galileu Transformations —

After axiom 1 we concluded that there is an infinite number of inertial frames. Hence it makes sense to ask how can calculate the coordinates (velocity) of particle in an inertial frame when knowing its coordinates (velocity) in a first inertial frame.

Let {S} and {S'} be two inertial frames whose origins coincide at {t=0}. Moreover {S'} moves with velocity {\vec{v}_0} relative to {S}.

TransformacaoGalileu

By simple vector addition it is {\vec{v}_0 t+\vec{r}'=\vec{r}} which we can write in component form:

{\begin{aligned} x' & = & x-v_{0x}t\\ y' & = & y-v_{0y}t\\ z' & = & z-v_{0z}t \end{aligned}}

Deriving in order to {t} it is

{\begin{aligned} v'_x & = & v_x-v_{0x}\\ v'_y & = & v_y-v_{0y}\\ v'_z & = & v_z-v_{0z} \end{aligned}}

Galileu transformations are equivalent to the affirmation that the form of the equations of mechanics don’t depend on the inertial frame that one chooses to study the motion of a particle.