~jzck/physics-notes

af8a21b686e01fdd86f06784acc93522f4f6ecbe — Jack Halford 1 year, 7 days ago e1407e6
new things
M Makefile => Makefile +6 -3
@@ 16,7 16,8 @@ www-data/%.html: md/%.md | www-data/colors.css www-data
> pandoc	-s --filter tools/pf-filter.py \
		-H tools/header.html \
		--template tools/template.html \
		--mathjax -f markdown \
		--mathjax=https://cdn.jsdelivr.net/npm/mathjax@3.1/es5/tex-mml-chtml.js \
		-f markdown \
		--resource-path .:formularies \
		-V type:$(shell tools/get type $<) \
		-V slug:$(shell tools/get slug $<) \


@@ 46,7 47,9 @@ clean:

re: clean all

sync:
> aws s3 sync www-data s3://jack-halford-website/physics-notes
deploy:
> aws s3 sync --content-type 'text/html;charset=utf-8' www-data s3://www.0x5.be/physics-notes \
	--exclude "*" --include "*.html"
> aws s3 sync www-data s3://www.0x5.be/physics-notes --exclude "*.html"

.PHONY: clean all re sync

M formularies/LL1/6.1.tex => formularies/LL1/6.1.tex +1 -0
@@ 1,1 1,2 @@
E\equiv\sum_i\dot{q_i}\frac{\partial L}{\partial \dot{q_i}}-L
=\sum_i\dot{q}_i p_i-L

A formularies/LL1/7.1.tex => formularies/LL1/7.1.tex +1 -0
@@ 0,0 1,1 @@
\sum_a\partial L/\partial \mathbf{r}_a=0

A formularies/LL1/7.2.tex => formularies/LL1/7.2.tex +1 -0
@@ 0,0 1,1 @@
\mathbf{P}\equiv\sum_a\partial L/\partial \mathbf{v}_a

A formularies/LL1/7.3.tex => formularies/LL1/7.3.tex +1 -0
@@ 0,0 1,1 @@
\mathbf{P}=\sum_am_a\mathbf{v}_a

A formularies/LL1/7.4.tex => formularies/LL1/7.4.tex +1 -0
@@ 0,0 1,1 @@
\sum_a\mathbf{F}_a=0

A formularies/LL1/7.5.tex => formularies/LL1/7.5.tex +1 -0
@@ 0,0 1,1 @@
p_i=\partial L/\partial \dot{q}_i

A formularies/LL1/7.6.tex => formularies/LL1/7.6.tex +1 -0
@@ 0,0 1,1 @@
F_i=\partial L/\partial q_i

A formularies/LL1/7.7.tex => formularies/LL1/7.7.tex +1 -0
@@ 0,0 1,1 @@
\dot{p}_i=F_i

A formularies/LL1/8.1.tex => formularies/LL1/8.1.tex +1 -0
@@ 0,0 1,1 @@
\mathbf{P}=\mathbf{P}'+\mathbf{V}\sum_a m_a

A formularies/LL1/8.2.tex => formularies/LL1/8.2.tex +1 -0
@@ 0,0 1,1 @@
\mathbf{V}=\mathbf{P}/\sum_a m_a=\sum_a m_a\mathbf{v}_a/\sum_a m_a

A formularies/LL1/8.3.tex => formularies/LL1/8.3.tex +1 -0
@@ 0,0 1,1 @@
\mathbf{R}\equiv\sum_a m_a\mathbf{r}_a/\sum_a m_a

A formularies/LL1/8.4.tex => formularies/LL1/8.4.tex +1 -0
@@ 0,0 1,1 @@
E=\frac{1}{2}\mu V^2+E_i

A formularies/LL1/8.5.tex => formularies/LL1/8.5.tex +1 -0
@@ 0,0 1,1 @@
E=E'+\mathbf{V}\cdot\mathbf{P}'+\frac{1}{2}\mu V^2

A formularies/LL1/9.1.tex => formularies/LL1/9.1.tex +1 -0
@@ 0,0 1,1 @@
\delta \mathbf{r}=\delta\mathbf{ϕ}\times\mathbf{r}

A formularies/LL1/9.2.tex => formularies/LL1/9.2.tex +1 -0
@@ 0,0 1,1 @@
\delta \mathbf{v}=\delta\mathbf{ϕ}\times\mathbf{v}

A formularies/LL1/9.3.tex => formularies/LL1/9.3.tex +1 -0
@@ 0,0 1,1 @@
\mathbf{M}\equiv\sum_a\mathbf{r}_a\times\mathbf{p}_a

A formularies/LL1/9.4.tex => formularies/LL1/9.4.tex +1 -0
@@ 0,0 1,1 @@
\mathbf{M}=\mathbf{M}'+a\times\mathbf{P}

A formularies/LL1/9.5.tex => formularies/LL1/9.5.tex +1 -0
@@ 0,0 1,1 @@
\mathbf{M}=\mathbf{M}'+\mu \mathbf{R}\times\mathbf{V}

A formularies/LL1/9.6.tex => formularies/LL1/9.6.tex +1 -0
@@ 0,0 1,1 @@
\mathbf{M}=\mathbf{M}'+\mathbf{R}\times\mathbf{P}

A formularies/LL1/9.7.tex => formularies/LL1/9.7.tex +1 -0
@@ 0,0 1,1 @@
M_z=\sum_a\frac{\partial L}{\partial \dot{\phi_a}}

A formularies/LL1/9.8.tex => formularies/LL1/9.8.tex +1 -0
@@ 0,0 1,1 @@
M_z=\sum_a m_ar_a^2\dot{\phi_a}

M md/angular-momentum.md => md/angular-momentum.md +30 -8
@@ 3,11 3,33 @@ title: angular momentum
type: theory
---

$$ \delta \mathbf{r}=\delta\mathbf{ϕ}\times\mathbf{r} $$
$$ \delta \mathbf{v}=\delta\mathbf{ϕ}\times\mathbf{v} $$
$$ \mathbf{M}\equiv\sum_a\mathbf{r}_a\times\mathbf{p}_a $$
$$ \mathbf{M}=\mathbf{M}'+a\times\mathbf{P} $$
$$ \mathbf{M}=\mathbf{M}'+\mu \mathbf{R}\times\mathbf{V} $$
$$ \mathbf{M}=\mathbf{M}'+\mathbf{R}\times\mathbf{P} $$
$$ M_z=\sum_a\frac{\partial L}{\partial \dot{\phi_a}} $$
$$ M_z=\sum_a m_ar_a^2\dot{\phi_a} $$
Here we find that the conservation of angular momentum arises from the _isotropy of space_

<hr>

[lagrangian](lagrangian.md)

```eq
LL1/9.1
```
```eq
LL1/9.2
```
```eq
LL1/9.3
```
```eq
LL1/9.4
```
```eq
LL1/9.5
```
```eq
LL1/9.6
```
```eq
LL1/9.7
```
```eq
LL1/9.8
```

M md/central-field.md => md/central-field.md +4 -2
@@ 3,7 3,9 @@ title: central field
type: theory
---

We saw that we can reduce a [two body problem](two-body-problem.md) into a system of one particle and a potential $U(r)$ that only depends on the distance $r$ from some fixed point. This is called a central field.
Some simplifications arise when dealing with a potential depends only on distance

<hr>

This potential generates a radial force $\mathbf{F}=-\frac{\partial U(r)}{\partial\mathbf{r}}=-\frac{\dd{U}}{\dd{r}}\frac{\mathbf{r}}{r}$



@@ 47,7 49,7 @@ by integrating
LL1/14.6
```

writing $(14.2)$ as $\dd{\phi}=\frac{M}{mr^2}\dd{t}$ we also find
writing `LL1/14.2` as $\dd{\phi}=\frac{M}{mr^2}\dd{t}$ we also find

```eq
LL1/14.7

D md/centre-of-mass.md => md/centre-of-mass.md +0 -7
@@ 1,7 0,0 @@
---
title: centre of mass
type: theory
---

## Model
 - [Classical mechanics](mechanical-system.md)

M md/energy.md => md/energy.md +30 -3
@@ 3,21 3,48 @@ title: energy
type: theory
---

We write the total time derivative of the [Lagrangian](lagrangian.md) as
Here we find that the conservation of energy arises from the _homogeneity of time_.

<hr>

Let us suppose a closed system, we can then write the total time derivative of the [Lagrangian](lagrangian.md) as

$$
\frac{\dd{L}}{\dd{t}}=\sum_i\frac{\partial L}{\partial q_i}q_i+\sum_i\frac{\partial L}{\partial \dot{q}_i}\dot{q}_i
\frac{\dd{L}}{\dd{t}}=\sum_i\frac{\partial L}{\partial q_i}\dot{q}_i+\sum_i\frac{\partial L}{\partial \dot{q}_i}\ddot{q}_i
$$

which doesn't have a $\partial L/\partial t$ because $L$ doesn't depend explicitly on time. We apply Lagranges equation `LL1/2.6`
which doesn't have a $\partial L/\partial t$ because $L$ doesn't depend explicitly on time because it is a closed system, and we have supposed the _homogeneity of time_. We apply Lagrange's equation `LL1/2.6`

$$
\frac{\dd{L}}{\dd{t}}=-\sum_i\frac{\dd{}}{\dd{t}}\frac{\partial L}{\partial \dot{q}_i}\dot{q}_i+\sum_i\frac{\partial L}{\partial \dot{q}_i}\ddot{q}_i
$$
$$
\frac{\dd{L}}{\dd{t}}-\frac{\dd{}}{\dd{t}}\left(\sum_i\frac{\partial L}{\partial \dot{q}_i}\dot{q}_i\right)
=-\sum_i\frac{\partial L}{\partial \dot{q}_i}\ddot{q}_i
+\sum_i\frac{\partial L}{\partial \dot{q}_i}\ddot{q}_i
=0
$$

Hence we find the conserved quantity

```eq
LL1/6.1
```

This property doesn't hold for a system with a potential dependant on time.

As we know, the Lagrangian has the form $L=T(q,\dot{q})-U(q)$, where $T$ is a quadratic function of velocities. Using Euler theorem on homogenous functions

$$
\sum_i\dot{q_i}\frac{\partial L}{\partial \dot{q_i}}=
\sum_i\dot{q_i}\frac{\partial T}{\partial \dot{q_i}}=2T
$$

from which we find immediately
```eq
LL1/6.2
```
In Cartesian co-ordinates,
```eq
LL1/6.3
```

M md/galileo-relativity.md => md/galileo-relativity.md +0 -1
@@ 11,4 11,3 @@ We call _inertial_, a frame in which
 - space is _homogenous_, all regions of space behave similarly.  
 - space is _isotropic_, all directions of space behave similarly.  
 - time is _homogenous_, past and future behave similarly.  


M md/hermitian-eigenfunctions.md => md/hermitian-eigenfunctions.md +2 -2
@@ 5,8 5,7 @@ title: hermitian eigenfunctions

We have seen that real physical quantities are represented by hermitian linear integral [operators](quantum-operator.md) $(3.15)$. Show that the eingenfunctions of these operators are orthogonal.

## solution

:::solution
Let $f_n$ and $f_m$ be two different eigenvalues of the quantity $f$, and $\Psi_n$, $\Psi_m$ the corresponding eigenfunctions

$$ \hat{f}\Psi_n=f_n\Psi_n,\qquad \hat{f}\Psi_m=f_m\Psi_m $$


@@ 20,3 19,4 @@ integrating over q, and using the fact that $\hat{f}$ is Hermitian $\hat{f}\equi
$$ (f_n-f_m)\int\Psi_n\Psi_m\vphantom{f}^*\dd{q}=0 $$

and since $f_n\neq f_m$, we obtain the orthogonality property.
:::

A md/integrals-of-motion.md => md/integrals-of-motion.md +10 -0
@@ 0,0 1,10 @@
---
title: integrals of motion
type: theory
---

During the motion of a system, we can find functions of the co-ordinates which are constant, we call them _integrals of the motion_.

The number of integrals of motion for a system wit $s$ degrees of freedom is $2s-1$.

TODO p6

M md/keplers-problem.md => md/keplers-problem.md +3 -1
@@ 3,6 3,8 @@ title: Kepler's problem
type: model
---

Kepler's problem is a special case of the [two body problem](two-body-problem.md) in a [central field](central-field.md)
Kepler's problem is a special case of the [two body problem](two-body-problem.md) in a [central field](central-field.md) which has $k=-1$ (Newtonian or Coulombian potential)

<hr>

TODO LL1 p15

M md/lagrangian.md => md/lagrangian.md +3 -4
@@ 3,8 3,7 @@ title: Lagrangian
type: theory
---

Every mechanical system is characterized by a definite function
$=L(\mathbf{q},\dot{\mathbf{q}},t)$, this _Lagrangian_ contains only $q$ and $\dot{q}$ because, from experience, we know that the mechanical state of the system is completely defined by these, without need of the higher derivatives $\ddot{q}, \dddot{q}, \ldots$.
Every mechanical system is characterized by a definite function $L(\mathbf{q},\dot{\mathbf{q}},t)$, this _Lagrangian_ contains only $q$ and $\dot{q}$ because, from experience, we know that the mechanical state of the system is completely defined by these, without need of the higher derivatives $\ddot{q}, \dddot{q}, \ldots$.

Let the system have a single degree of freedom, which occupies two positions $q_1$ and $q_2$ at times $t_1$ and $t_2$, the condition that the system moves along a certain path is that the _action_ integral



@@ 167,7 166,7 @@ Substituting these equation in the cartesian Lagrangian $L=\mfrac{1}{2}\sum m_a(
```eq
LL1/5.5
```
where each particles has 6 functions $a_{i,k}$ that depend only on co-ordinates. The kinetic energy is still quadratic in valocities, it may now depend on co-ordinates also.
where each particle has 6 functions $a_{i,k}$ that depend only on co-ordinates. The kinetic energy is still quadratic in valocities, it may now depend on co-ordinates also.

## Open systems



@@ 185,7 184,7 @@ LL1/5.7

### Uniform field

A field such that the same force $\mathbf{F}$ act ona  particle at any point is said to be uniform, the potential energy for such a field is
A field such that the same force $\mathbf{F}$ acting on a particle at any point is said to be uniform, the potential energy for such a field is

```eq
LL1/5.8

M md/mechanical-similarity.md => md/mechanical-similarity.md +1 -1
@@ 51,7 51,7 @@ In a uniform field, the potential energy is a linear function of co-ordinates (s

### Kepler's third law $k=-1$

the Newtonian attraction of two masses, of the Coulomb attraction of two charges, the potential energy is inversely proportional to the distance apart $k=-1$. Then $(t'/t)^2=(l'/l)^3$. Hence, the square of the time of revolution in the orbit goes as the cube of the size of the orbit.
For the Newtonian attraction of two masses, or the Coulomb attraction of two charges, the potential energy is inversely proportional to the distance apart $k=-1$. Then $(t'/t)^2=(l'/l)^3$. Hence, the square of the time of revolution in the orbit goes as the cube of the size of the orbit.

<img src="https://upload.wikimedia.org/wikipedia/commons/5/52/Solar_system_planets_a%28AU%29_vs_period%28terrestrial_years%29.svg">


M md/momentum.md => md/momentum.md +72 -9
@@ 3,12 3,75 @@ title: momentum
type: theory
---

[lagrangian](lagrangian.md)

$$ \sum_a\partial L/\partial \mathbf{r}_a=0 $$
$$ \mathbf{P}\equiv\sum_a\partial L/\partial \mathbf{v}_a $$
$$ \mathbf{P}=\sum_am_a\mathbf{v}_a $$
$$ \sum_a\mathbf{F}_a=0 $$
$$ p_i=\partial L/\partial \dot{q}_i $$
$$ F_i=\partial L/\partial q_i $$
$$ \dot{p}_i=F_i $$
Here we find that the conservation of momentum arises from the _homogeneity of space_.
<hr>

The homogeneity of space tells us that if we displace all the particles in a closed system by the same amount $\epsilon$, the mechanical properties should be conserved. Note that this displacement will change the Lagrangian in the case of an open system because the external field will be relatively displaced.

We can write this change in [Lagrangian](lagrangian.md) as

$$
\delta L=\sum_a \frac{\partial L}{\partial \mathbf{r}_a}\delta\mathbf{r}_a
=\epsilon\cdot\sum_a \frac{\partial L}{\partial \mathbf{r}_a}
$$

This must be true for an arbitrary $\epsilon$, so, the condition $\delta L=0$ is equivalent to

```eq
LL1/7.1
```
From Lagrange's equations `LL1/5.2`, we have

$$
\sum_a\frac{\partial L}{\partial \mathbf{r}_a}=
\sum_a\frac{\dd{}}{\dd{t}}\frac{\partial L}{\partial \mathbf{v}_a}=0
$$

We have found a conserved vector that we call momentum of the system.

```eq
LL1/7.2
```

Differentiating `LL1/5.1` we find that

```eq
LL1/7.3
```

The additivity of the momentum is evident. Unlike the Energy, the momentum of the system is equal to the sum of the momentum of the particles $\mathbf{p}_a=m_a\mathbf{v}_a$, whereas the energy also depends on the interactions between the particles.

By applying the same displacement logic in a single direction, we find that the individual Cartesian components of the momentum $r_i$ may be conserved in an open system, only if the external field doesn't depend on $r_i$, $\partial U/\partial r_i=0$.

e.g., in a uniform field in the $z$-direction, the $x$ and $y$ components of the momentum $p_x$, $p_y$ are conserved.

## Force

Because we are using Cartesian co-ordinates, the kinetic energy $T$ doesn't depend on co-ordinates, i.e. we have `LL1/7.1` which becomes $\partial L/\partial\mathbf{r}_a=\partial U/\partial\mathbf{r}_a$, which we know from `LL1/5.4` is the force. Thus `LL1/7.1` signifies that the sum of the forces on all the particles in a closed system is zero

```eq
LL1/7.4
```

This is _Newton's third law_, the equality of action and reaction.

## Generalised co-ordinates

We call generalised momenta the components

```eq
LL1/7.5
```
And generalised forces the components
```eq
LL1/7.6
```
In this notation, Lagrange's equation `LL1/5.1` become
```eq
LL1/7.7
```
However, in generalised co-ordinates, $p_i$ do not reduce to $mv_i$, because the 3 directions may not be linearly independant as we have seen in `LL1/5.5`

## Centre of mass



A md/natural-reference-frame.md => md/natural-reference-frame.md +62 -0
@@ 0,0 1,62 @@
---
title: natural reference frame
type: theory
---

We generalise concepts of positon, velocity and energy to systems with multiple particles.

<hr>


## Rest frame

The [momentum](momentum.md) of a system takes different values in different frames of reference.

Let's consider a frame $K'$ that moves with velocity $\mathbf{V}$ relative to a frame $K$, then we have the velocity of a particle $\mathbf{v}_a=\mathbf{v}'_a+\mathbf{V}$. We can then relate the momenta $\mathbf{P}$ and $\mathbf{P}'$.

$$
\mathbf{P}=\sum_a m_a\mathbf{v}_a=\sum_a m_a\mathbf{v}'_a+\mathbf{V}\sum_a m_a
$$

or

```eq
LL1/8.1
```


In particular, we can always find a frame in which $\mathbf{P}'=0$, the velocity of this frame is then
```eq
LL1/8.2
```

In this frame, we say that the system is _at rest_.

### Velocity of the system

The velocity of the rest frame found in `LL1/8.2` is a natural definition for the "velocity of the system".

### Position of the system (centre of mass)

From `LL1/8.2`, we can find a special point in space, called the _centre of mass_.

```eq
LL1/8.3
```

$\mathbf{R}$ is a natural definition for the "position of the system".

### Energy of the system (rest energy)

We call the [energy](energy.md) in the rest frame the _internal energy_ $E_i$ of thesystem. The energy of a system moving as a whole can be written

```eq
LL1/8.4
```

with the following proof

```eq
LL1/8.5
```


M md/quantum-operator.md => md/quantum-operator.md +4 -4
@@ 79,13 79,13 @@ where the kernel function $K(q,q')$ is
LL3/3.11
```

Thus, for every physical quantity there is a definite linear integral operator. From $(3.9)$ we see that if $\Psi$ is one the the eigenfunction $\Psi_n$, then
Thus, for every physical quantity there is a definite linear integral operator. From `LL3/3.9` we see that if $\Psi$ is one the the eigenfunction $\Psi_n$, then

```eq
LL3/3.12
```

The values taken by physical quantities are necessarily real, hence the mean must also be real, in any state. And if the mean is real, then all of the eigenvalues must be real, because the mean values coincidence with the eigenvalues in the states $\Psi_n$. Because the mean is real, we can equate $(3.8)$ to it's conjugate form
The values taken by physical quantities are necessarily real, hence the mean must also be real, in any state. And if the mean is real, then all of the eigenvalues must be real, because the mean values coincidence with the eigenvalues in the states $\Psi_n$. Because the mean is real, we can equate `LL3/3.8` to it's conjugate form

```eq
LL3/3.13


@@ 99,7 99,7 @@ For an arbitrary $\qop{f}$ we can find the transposed operator $\qop{f}\qop[t]{f
LL3/3.14
```

where $\Psi$ and $\Phi$ are different, if we take $\Phi=\Psi^*$ then from $(3.13)$ we have
where $\Psi$ and $\Phi$ are different, if we take $\Phi=\Psi^*$ then from `LL3/3.13` we have

```eq
LL3/3.15


@@ 195,7 195,7 @@ And we notice that if $\{\qop{f},\qop{h}\}=0$ and $\{\qop{g},\qop{h}\}=0$, it do

### The continuous spectrum

We can generalize the above results for operator an operator $\qop{f}$ with a continuous spectrum. We shall denote it's eigenvalues $f$, without suffix, because they take a continuous range of values. We note $\Psi_f$ the eigenfunction corresponding to the eigenvalue $f$. Just as $\Psi$ can be exanded in a series $(3.2)$ of eingenfunctions, it can also be expanded in terms of of the complete set of eigenfunctions of a quantity with a continuous spectrum as an integral.
We can generalize the above results for operator an operator $\qop{f}$ with a continuous spectrum. We shall denote it's eigenvalues $f$, without suffix, because they take a continuous range of values. We note $\Psi_f$ the eigenfunction corresponding to the eigenvalue $f$. Just as $\Psi$ can be exanded in a series `LL3/3.2` of eingenfunctions, it can also be expanded in terms of of the complete set of eigenfunctions of a quantity with a continuous spectrum as an integral.

```eq
LL3/5.1

M md/translation-operator.md => md/translation-operator.md +3 -1
@@ 5,4 5,6 @@ type: problem

Express the [operator](quantum-operator.md) $\hat{T_a}$ of a parallel displacement over a finite distance $a$ in terms of the momentum operator.

## solution
:::solution

:::

M md/two-body-problem.md => md/two-body-problem.md +2 -6
@@ 3,13 3,9 @@ title: two body problem
type: model
---

a system of two [particles](particle.md)
We show that a two body problem can be reduced to one body in a [central field](central-field.md)

\begin{equation} \mathbf{P}=\mathbf{P}'+\mathbf{V}\sum_a m_a \end{equation}
\begin{equation} \mathbf{V}=\mathbf{P}/\sum_a m_a=\sum_a m_a\mathbf{v}_a/\sum_a m_a \end{equation}
\begin{equation} \mathbf{R}\equiv\sum_a m_a\mathbf{r}_a/\sum_a m_a \end{equation}
\begin{equation} E=\frac{1}{2}\mu V^2+E_i \end{equation}
\begin{equation} E=E'+\mathbf{V}\cdot\mathbf{P}'+\frac{1}{2}\mu V^2 \end{equation}
<hr>

```eq
LL1/13.1