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.
                                                           

— 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.