# 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.

## 13 comments on “Newtonian Mechanics 03”

1. joeschmo26 says:

“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.”

Please bare with me on this ( I am not very good at making my arguments clear in definition, and I feel like I may be overlooking something), but wouldn’t the dot product
${\vec{F}\cdot\vec{s}=0}$ for all ${\vec{F}}$, which in turn would say nothing about ${\vec{p}}$ in the direction of ${\vec{s}}$ ?

• joeschmo26 says:

I guess what I’m trying to say is that ${\vec{F}} = {\vec{0}}$ and/or ${\vec{F}\parallel\vec{s}}$ ?

• ateixeira says:

joeschmo26 I don’t think I fully understand you but let me ask you why do you think that $\vec{F} \cdot \vec{s}=0$ for all $\vec{F}$?

Anyway there is a step that I didn’t show explicitly in my derivation because I wanted to let something for the readerso I’ll just give some pointers.

1 – First I believe that you know that $\frac{d f(x)}{dx}=0$ implies that $f(x)$ is constant
2 – Use the product rule to calculate $\frac{d \left(\vec{p} \cdot \vec{s}\right)}{dt}$ (remember that $\vec{s}$ is a constant vector)
3 – Remember our initial hypothesis about $\vec{s}$

The third step really is the crux of our argument so I’ll just say a little bit more about it. The dot product between two vectors is a mathematical operation that allows one to calculate the projection of a first vector into a second vector.
So when you say that the dot product between two vectors is $0$ it means that the projection between them is just a point (whose length is $0$ and hence the projection is $0$). Which is to say that the two vectors are “perpendicular”

Hence when we’re saying “Let ${\vec{s}}$ be a constant vector such as ${\vec{F}\cdot\vec{s}=0}$“. We’re saying that $\vec{s}$ is a constant vector but we’re also saying that $\vec{s}$ and $\vec{F}$ are perpendicular vectors. Since $\vec{s}$ indicates a direction that means that $\vec{F}$ doesn’t have a component along that direction. By Newton’s Second Axiom that means momentum has to be conserved along the direction of $\vec{s}$.

• joeschmo26 says:

Yeah, the vectors are “perpendicular” when the scalar product is 0 says it all… I was thinking “parallel” …

2. joeschmo26 says:

Sorry for the late comments, but I’m a good bit behind on the material and I know you said answers to these parts would be less frequent, I hope you’ll be patient, because some of the end of this section and Mechanics 04 and beyond is all new to me. I’m going to try to get through as much as I can tonight, and I imagine there will be many questions as I try to dissect their meaning.

• ateixeira says:

joeschmo26 please take your time. I’ve already made a post about Quantum Mechanics, but this weekend I’ll make a final post on the Lagrangian and Hamiltonian formalisms consisting of solved exercises. Do you think this is a good idea?

• joeschmo26 says:

Yes, I would appreciate that very much. At a glance the material looks theoretically and mathematically involved, so the more context I have, the better my chances for survival. (P.s. I have John R. Taylor’s book on classical mechanics, of which I have not studied, but assume will be helpful in filling my knowledge gaps as I attempt to get through these next sections, so your not alone in that task)

• ateixeira says:

I didn’t know John R. Taylor’s book on Classical Mechanics but the reviews about it are pretty much unanimous that it is great. So I guess you’ll be in good hands

3. joeschmo26 says:

In your derivation of the total differential ${ d\left(\frac{1}{2}mv^2\right) }$ from
${\vec{F}\cdot\vec{r}}$ is it supposed to be from the recognition alone that

{\begin{aligned}\frac{d}{dt}(\vec{v}\cdot\{vec{v}) &= frac{d\vec{v}}{dt}\cdot\vec{v} + \vec{v}\cdot\frac{d\vec{v}}{dt} &= 2\frac{d\vec{v}}{dt}\cdot\vec{v}\end{aligned}}

that you move from step 2 to step 3? I ask because I only noticed it when working my way backwards from step 3 to step 2, and I might not have found that without having the result first to work back from. Just wondering if I missed something obvious.

• joeschmo26 says:

Edit:

{\begin{aligned}\frac{d}{dt}(\vec{v}\cdot\vec{v}) &= \frac{d\vec{v}}{dt}\cdot\vec{v} + \vec{v}\cdot\frac{d\vec{v}}{dt} &= 2\frac{d\vec{v}}{dt}\cdot\vec{v}\end{aligned}}

4. joeschmo26 says:

Oh, also, could line 2 of you derivation for ${\frac{dE}{dt}}$ be missing ${\frac{\partial U}{\partial t}}$ ?