Lagrangian and Hamiltonian Mechanics Week 1

Landau & Lifshitz, Mechanics, pages 1-3 and 96. Kibble & Berkshire, Classical Mechanics, 5th edition, page 231.

The module is mostly just heavy mathematics applied to mechanics.

Physical notes in book, too.

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Recap of Derivatives

Notation for a full derivative involving $x$…

$$y(x)=x^2⟹\frac{dy}{dx}=y^{\prime }=2x$$

Notation for a full derivative involving $t$…

$$y(t)=t^3⟹\frac{dy}{dt}=\dot{y}=3t^2$$

Types of derivative…

$$\frac{d}{d(...)}:\text{full derivative}$$

or…

$$\frac{\partial }{\partial (...)}:\text{partial derivative}$$

When differentiating a variable, the other variable(s) are treated as constant:

$$\begin{array}{rl}f(x,y)=x^{2}y⟹& \frac{\partial f}{\partial x}=2xy,\\ & \frac{\partial f}{\partial y}=x^2\end{array}$$

In terms of $t$…

$$\begin{array}{rl}f(x,t)=xt⟹& \frac{\partial f}{\partial x}=t,\\ & \frac{\partial f}{\partial t}=x\end{array}$$

But what would be the total derivative of $f(x,t)$ with respect to $t$? Keep in mind that whilst $x$ is a variable, it can be expressed in terms of $t$ (and hence as a function $x=x(t)$). Total derivative = rate of change if all variables can change.

$$\frac{df}{dt}\equiv \frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial t}$$

This equivalence is essentially made up of partial derivatives treating each thing as a variable whilst keeping another constant, but ensuring the result still stays in the form $\frac{(...)f}{(...)t}$. It’s like you’re adding all the changes up for every variable in order to calculate how much change the entire function has, with respect to a single variable.

$$\begin{array}{rl}\frac{df}{dt}& =\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\cdot \dot{x}\\ & =x+t\dot{x}\end{array}$$

Note: How do we tell the difference between partial and total derivatives when the notation is in the form $\dot{x}$?

$$f(x,y,t)⟹\frac{df}{dt}=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\cdot \frac{\partial y}{\partial t}$$

This pattern for a full derivatives extends continually for additional variables. Note: What other notation is there in different languages, like French?

$$\begin{array}{rl}f(x,\dot{x})=x\cdot {\dot{x}}^2⟹& \frac{\partial f}{\partial x}={\dot{x}}^2\cdot \frac{\partial x}{\partial x}={\dot{x}}^2\\ & \frac{\partial f}{\partial \dot{x}}=x\frac{\partial ({\dot{x}}^2)}{\partial x}=x\cdot 2\dot{x}\end{array}$$ $$\begin{array}{rl}f(x,\dot{x})=x\cdot \dot{x}⟹\frac{df}{dt}& =\frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial t}+\frac{\partial f}{\partial \dot{x}}\cdot \frac{\partial \dot{x}}{\partial t}\\ & =\dot{x}\cdot \dot{x}+x\cdot \ddot{x}\\ & ={\dot{x}}^2+x\cdot \ddot{x}\end{array}$$

(the first term with respect to $t$ is omitted, since it’s equal to $0$) Note: figure out a better way to type partial derivatives in LaTeX - it’s so annoying

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Recap of Newton Mechanics

Energy $E=T+U$: the sum of the kinetic and potential energy, respectfully.

Kinetic Energy (scalar and vector)… for a particle of mass $m$: $T=\frac{m{v}^2}{2}$ for two particles of masses $m_1,m_2$: $T=T_1+T_2=\frac{m_1{v}_1^2}{2}+\frac{m_2{v}_2^2}{2}$

for a particle of mass $m$ moving along $x$ (where everything is moving →, including a unit vector $\hat{i}:|\hat{i}|=1$): $\vec{v}\equiv \dot{x}\hat{i}$, $T=\frac{m{\dot{x}}^2|\hat{i}{|}^2}{2}=\frac{m{\dot{x}}^2}{2}$. for two particles of masses $m_1,m_2$ moving along $x$ (where everything is moving →, including a unit vector $\hat{i}:|\hat{i}|=1$): $T=T_1+T_2=\frac{m_1{\dot{x_1}}^2}{2}+\frac{m_2{\dot{x}}_2^2}{2}$.

Potential Energy (gravity and spring)… for a mass $m$ a distance $h$ from the Earth’s surface: $U=mgh:g\text{ is a constant}$. for masses $m_1,m_2$ distances $h_1,h_2$ from the Earth’s surface: $U=U_1+U_2=m_{1}g{h}_1+m_{2}g{h}_2$.

for a spring compressed with a change in length $\Delta x$: $U=\frac{1}{2}k(\Delta x{)}^2,\text{ where }k\text{ is a constant}$, or if $\Delta x$ is not given, then it can be calculated by calculating the original length of the spring and subtracting its final length ($(x_2-x_1)-l$): $U=\frac{1}{2}k(x_2-x_1-l{)}^2$.

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Generalised co-ordinates and degrees of freedom. By definition, if co-ordinates completely define a position then they are generalised co-ordinates. Often denoted by $q_i$, where $i$ is the index. Equal to the number $N$ of degrees of freedom, $\text{d.f.}$

For a particle in one-dimension, there is one degree of freedom ($1\text{d.f.}$) since it can only move along one axis, $x$, hence $1\text{g.c.}$ For a particle in two-dimensions (on a plane), there are two degrees of freedom ($1\text{d.f.}$) since it can move along two axes, $x$ and $y$, hence $2\text{g.c.}$. Note: the generalised co-ordinates here are $x$ and $y$, but they’re also often denoted simply as $q_1$ and $q_2$. This allows for higher dimensions than $x,y,z$ allows - again, this is hinting towards the mathematical nature of this module. For a particle in three-dimensions (in a volume)… you get the idea. Three degrees of freedom ($3\text{d.f.}$) along $q_1,q_2,q_3$.

These co-ordinates do not need to be Cartesian, for example they could be polar co-ordinates: in two dimensions, $(r,\theta )$ have $2\text{d.f.}$ and hence $2\text{g.c.}:q_1,q_2$. in three dimensions (spherical co-ordinates), $(r,\theta ,\varphi )$ have $3\text{d.f.}$ and hence $3\text{g.c.}:q_1,q_2,q_3$. insert funny haha about modelling the physics of spherical horses

These definitions hold for all co-ordinate systems, for $N$ dimensions, where the co-ordinate system will have $(N)\text{d.f.}$ and $(N)\text{g.c.}$ with $q_1,q_2,\dots ,q_i$. Note that this is for a single particle - for $n$ particles, the system would have $(nN)\text{d.f.}$.

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Constraints - particle limitations on position or velocity. Holonomic constraints - can be expressed by the equation $f(q_1,\dots ,q_s,t)=0$, i.e. a function dependent on generalised co-ordinates and time which is equal to zero. If constraints can not be expressed in this way, they are non-holonomic. Note that we focus on holonomic, I assume because of the complexity increase when solving differential equations that are equal to a non-zero function.

For example… a one-dimensional rod stretching between $x_1$ and $x_2$ (hence with length $l=x_2-x_1$) can be expressed as $x_2-x_1-l=0$ and thus as $f(x_1,x_2)=0$. Therefore we can conclude the constraint is holonomic.

Another example… a two-dimensional rod stretching between $(x_1,y_1)$ and $(x_2,y_2)$ (hence with length $l^2=(x_2-x_1{)}^2+(y_2-y_1{)}^2$) can be expressed as $(x_2-x_1{)}^2+(y_2-y_1{)}^2-l^2=0$ and thus as $f(x_1,x_2,y_1,y_2)=0$. Therefore we can again conclude the constraint is holonomic.

Note: the absence of time is okay, it’s simply equal to $0$.

$N$ particles with $p$ holonomic constraints has degrees of freedom ($\text{d.f.}$): one-dimension: $n=N-p$ two-dimensions: $n=2N-p$ three-dimensions: $n=3N-p$ etc.

This makes sense extending the logic before, as each constraint restricts the movement of a particle by one “axis”. Remember, the degrees of freedom is just a numerical value of how much the particles can move in a system.