~jzck/physics-notes

535ed88d417794c01cb3a29b2ff1a9e92b17c4dd — Jack Halford 7 months ago a9992e5
started second book
M Makefile => Makefile +12 -2
@@ 9,18 9,25 @@ SVG	=	$(subst md/,www-data/,$(MD:.md=.svg))
TEX	=	$(subst md/,www-data/,$(MD:.md=.tex))
HTML	=	$(subst md/,www-data/,$(MD:.md=.html)) www-data/index.html www-data/style.css

ifeq ($(PROD),yes)
PROD_OPT=	"-V prod:yes"
endif

all: $(HTML) $(SVG) www-data/graph.svg

PROD = yes
www-data/%.html: md/%.md | www-data/colors.css www-data
> @printf '$@\n'
> pandoc	-s --filter tools/pf-filter.py \
> pandoc	\
		-s --filter tools/pf-filter.py \
		-H tools/header.html \
		--template tools/template.html \
		--mathjax=https://cdn.jsdelivr.net/npm/mathjax@3.1/es5/tex-mml-chtml.js \
		-f markdown \
		-f markdown+pipe_tables \
		--resource-path .:formularies \
		-V type:$(shell tools/get type $<) \
		-V slug:$(shell tools/get slug $<) \
		$(PROD_OPT) \
		tools/math.tex \
		$< \
		-o $@ &


@@ 48,8 55,11 @@ clean:
re: clean all

deploy:
> $(MAKE) re PROD=yes
> pass otp dev/aws.amazon.com | aws-mfa --profile default
> aws s3 sync --content-type 'text/html;charset=utf-8' www-data s3://0x5.be/physics-notes \
	--exclude "*" --include "*.html"
> aws s3 sync www-data s3://0x5.be/physics-notes --exclude "*.html"
> $(MAKE) clean

.PHONY: clean all re sync

A formularies/LL1/14.10.tex => formularies/LL1/14.10.tex +2 -0
@@ 0,0 1,2 @@
\delta\phi=2\int_{r_\text{min}}^{r_\text{max}}
\frac{M\dd{r}/r^2}{\sqrt{2m(E-U)-M^2/r^2}}

A formularies/LL1/14.11.tex => formularies/LL1/14.11.tex +1 -0
@@ 0,0 1,1 @@
\left[r^2U(r)\right]_{r\to 0}<-M^2/2m

M formularies/LL1/14.2.tex => formularies/LL1/14.2.tex +1 -1
@@ 1,1 1,1 @@
\frac{\partial L}{\partial \dot{\phi}}=mr^2\dot{\phi}=\text{constant} (=M_z=p_\phi)
\frac{\partial L}{\partial \dot{\phi}}=mr^2\dot{\phi}=\text{constant} (=M_z)

M formularies/LL1/14.5.tex => formularies/LL1/14.5.tex +1 -1
@@ 1,1 1,1 @@
\dot{\mathbf{r}}=\frac{\dd{r}}{\dd{t}}=\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}
\dot{r}\equiv\frac{\dd{r}}{\dd{t}}=\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}

A formularies/LL1/14.8.tex => formularies/LL1/14.8.tex +1 -0
@@ 0,0 1,1 @@
U_{\text{eff}}=U(r)+M^2/2mr^2

A formularies/LL1/14.9.tex => formularies/LL1/14.9.tex +1 -0
@@ 0,0 1,1 @@
U(r)+M^2/2mr^2=E

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

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

A formularies/LL2/1.1.tex => formularies/LL2/1.1.tex +1 -0
@@ 0,0 1,1 @@
c=2.99793\text{ cm/s}

A formularies/LL2/2.1.tex => formularies/LL2/2.1.tex +1 -0
@@ 0,0 1,1 @@
c^2(t_2-t_1)-(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2=0

A formularies/LL2/2.2.tex => formularies/LL2/2.2.tex +1 -0
@@ 0,0 1,1 @@
c^2(t_2'-t_1')-(x_2'-x_1')^2+(y_2'-y_1')^2+(z_2'-z_1')^2=0

A formularies/LL2/2.3.tex => formularies/LL2/2.3.tex +1 -0
@@ 0,0 1,1 @@
s_{12}=\sqrt{c^2(t_2-t_1)-(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

A formularies/LL2/2.4.tex => formularies/LL2/2.4.tex +1 -0
@@ 0,0 1,1 @@
\dd{s}^2=c\dd{t}^2-\dd{x}^2-\dd{y}^2-\dd{z}^2

A formularies/LL2/2.5.tex => formularies/LL2/2.5.tex +1 -0
@@ 0,0 1,1 @@
\frac{a(V_2)}{a(V_1)}=a(V_{12})

A formularies/LL2/2.6.tex => formularies/LL2/2.6.tex +1 -0
@@ 0,0 1,1 @@
t_{12}'=\frac{1}{c}\sqrt{c^2t_{12}^2-l_{12}^2}=\frac{s_{12}}{c}

A formularies/LL2/2.7.tex => formularies/LL2/2.7.tex +1 -0
@@ 0,0 1,1 @@
l_{12}'=\sqrt{l_{12}^2-c^2t_{12}^2}=is_{12}

A formularies/LL2/3.1.tex => formularies/LL2/3.1.tex +1 -0
@@ 0,0 1,1 @@
\dd{t'}=\frac{\dd{s}}{c}=\dd{t}\sqrt{1-\frac{v^2}{c^2}}

A formularies/LL2/3.2.tex => formularies/LL2/3.2.tex +1 -0
@@ 0,0 1,1 @@
t_2'-t_1'=\int_{t_1}^{t_2}\dd{t}\sqrt{1-\frac{v^2}{c^2}}

A formularies/LL2/4.1.tex => formularies/LL2/4.1.tex +4 -0
@@ 0,0 1,4 @@
x=x'+Vt,\qquad
y=y',\qquad
z=z',\qquad
t=t'

A formularies/LL2/4.2.tex => formularies/LL2/4.2.tex +2 -0
@@ 0,0 1,2 @@
x=x'\ch\psi+ct'\sh\psi,\qquad
ct=x'\sh\psi+ct'\ch\psi

A formularies/LL2/4.3.tex => formularies/LL2/4.3.tex +4 -0
@@ 0,0 1,4 @@
x=\frac{x'+Vt'}{\sqrt{1-(V/c)^2}},\qquad
y=y',\qquad
z=z',\qquad
t=\frac{t'+Vx'/c^2}{\sqrt{1-(V/c)^2}}

A formularies/LL2/6.1.tex => formularies/LL2/6.1.tex +4 -0
@@ 0,0 1,4 @@
A^0=\frac{A'^0+\frac{V}{c}A'^1}{\sqrt{1-\frac{V^2}{c^2}}},\qquad
A^1=\frac{A'^1+\frac{V}{c}A'^0}{\sqrt{1-\frac{V^2}{c^2}}},\qquad
A^2=A'^2,\qquad
A^3=A'^3

A formularies/LL2/6.10.tex => formularies/LL2/6.10.tex +6 -0
@@ 0,0 1,6 @@
A^{ik}=\begin{pmatrix}
\phantom{-}0 & \phantom{-}p_x & \phantom{-}p_y & \phantom{-}p_z \\
-p_x & \phantom{-}0 & -a_z & \phantom{-}a_y \\
-p_y & \phantom{-}a_z & \phantom{-}0 & -a_x \\
-p_z & -a_y & \phantom{-}a_x & \phantom{-}0 \\
\end{pmatrix}

A formularies/LL2/6.2.tex => formularies/LL2/6.2.tex +4 -0
@@ 0,0 1,4 @@
A_0\equiv A^0,\qquad
A_1\equiv-A^1,\qquad
A_2\equiv-A^2,\qquad
A_3\equiv-A^3

A formularies/LL2/6.3.tex => formularies/LL2/6.3.tex +1 -0
@@ 0,0 1,1 @@
\delta_i^k A^i=A^k

A formularies/LL2/6.4.tex => formularies/LL2/6.4.tex +5 -0
@@ 0,0 1,5 @@
\delta_i^k=
\begin{cases}
1, & \text{if} & i=k \\
0, & \text{if} & i\neq k
\end{cases}

A formularies/LL2/6.5.tex => formularies/LL2/6.5.tex +7 -0
@@ 0,0 1,7 @@
g^{ik}=g_{ik}=
\begin{pmatrix}
1 & \phantom{-}0 & \phantom{-}0 & \phantom{-}0\\
0 & -1 & \phantom{-}0 & \phantom{-}0\\
0 & \phantom{-}0 & -1 & \phantom{-}0\\
0 & \phantom{-}0 & \phantom{-}0 & -1
\end{pmatrix}

A formularies/LL2/6.6.tex => formularies/LL2/6.6.tex +3 -0
@@ 0,0 1,3 @@
A^iA_i=
g_{ik}A^kA^i=
g^{ik}A_kA_i

A formularies/LL2/6.8.tex => formularies/LL2/6.8.tex +1 -0
@@ 0,0 1,1 @@
e^{0123}=+1

A formularies/LL2/6.9.tex => formularies/LL2/6.9.tex +1 -0
@@ 0,0 1,1 @@
e^{iklm}e_{iklm}=-24

A formularies/LL2/7.1.tex => formularies/LL2/7.1.tex +1 -0
@@ 0,0 1,1 @@
u^i=\frac{\dd{x}^i}{\dd{s}}

M md/angular-momentum.md => md/angular-momentum.md +80 -7
@@ 5,31 5,104 @@ type: theory

Here we find that the conservation of angular momentum arises from the _isotropy of space_

the _isotropy of space_ means that the mechanical properties of a system do not vary when rotated as a whole.

<hr>

[lagrangian](lagrangian.md)

Let us consider an infinitesimal rotation $\delta\v{\phi}$ of a system in space. The resulting displacement from this rotation, is `LL1/9.1` from figure 5.

```eq
LL1/9.1
```

The velocities are also changed accordingly,

```eq
LL1/9.2
```

```fig
LL1/5
```

We now write the condition that the [Lagrangian](lagrangian.md) is unchanged by rotiation

$$
\delta L=\sum_a \left(
\frac{\partial L}{\partial \v{r}_a}\cdot\delta\v{r}_a +
\frac{\partial L}{\partial \v{v}_a}\cdot\delta\v{v}_a
\right) = 0
$$

$$
\delta L=\sum_a \left(
\v{\dot{p}}_a\cdot\delta\v{\phi}\times\v{r}_a+
\v{p}_a\cdot\delta\v{\phi}\times\v{v}_a
\right) = 0
$$

we use the permutation rule $a\cdot b\times c=c\cdot a\times b =b\cdot c\times a$

$$
\delta\v{\phi}\sum_a \left(
\v{r}_a\times\v{\dot{p}}_a +
\v{v}_a\times\v{p}_a \right)
= \delta\v{\v{\phi}}\frac{\dd{}}{\dd{t}}\sum_a \v{r}_a \times \v{p}_a
= 0
$$

Since $\delta\v{\phi}$ is arbitrary, we conclude that the vector

```eq
LL1/9.3
```

called the _angular momentum_ is conserved in the motion of a closed system. Like the linear momentum, it is additive, whether or not the particles in the system interact.

There are no other additive integrals of the motion. Every closed system has seven: energy, three components of momentum and three omponents of angular momentum.

### Change of origin
Since the angular momentum $\v{M}$ `LL1/9.3` involves the radius vectors $\v{r}_a$ of the particles, its value depends in general on the choice of origin. If we change the origin to $\v{r}_a=\v{r}_a'+\v{a}$ we have

$$
\v{M}=\sum_a\v{r}_a'\times\v{p}_a+\v{a}\times\sum_a\v{p}_a
$$

```eq
LL1/9.4
```

### Change between two internal frames

we take two internal references frames $K$ and $K'$. We let the origin coincide at an instant. Their relative velocity is $\v{V}$ such that $\v{v}_a=\v{v}_a'+\v{V}$.

We have

$$
\v{M}
=\sum_a m_a\v{r}_a\times\v{v}_a
=\sum_a m_a\v{r}_a\times\v{v}_a'+\sum_a m_a\v{r}_a\times\v{V}
$$

Where the first sum is $\v{M}'$, the angular momentum in $K'$, and the second sum is the position of the center of mass. Thus we obtain
```eq
LL1/9.5
```

This is the law of transfation of angular momentum, corresponding the `LL1/8.1` for momentum and `LL1/8.5` for energy.

If $K'$ is the frame of rest, then $\mu\v{V}$ is the total momentum $\v{P}$ relative to $K$, and

```eq
LL1/9.6
```
```eq
LL1/9.7
```
```eq
LL1/9.8
```

In other words, the angular momentum $\v{M}$ consists of an "intrinsic angular momentum" in a frame at rest, and the angular momentum $\v{R}\times\v{P}$ due to its motion as a whole.


## Conservation in non-closed systems

The components of angular momentum along an axis about which the field is symmetrical is always conserved, since the mechanical properties of the system are unaltered by any rotation about that axis. Here the angular momentum must, of course, be defined relative to an origin lying on the axis.

Some examples include central fields, field homogeneous in one direction.

M md/central-field.md => md/central-field.md +40 -3
@@ 7,9 7,13 @@ Some simplifications arise when dealing with a potential depends only on distanc

<hr>

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

We have shown that the [angular-momentum](angular-momentum.md) $\mathbf{M}=\mathbf{r}\times\mathbf{p}$ of a system in a central field is conserved. Since $\mathbf{M}$ is perpendicular to $r$, the constancy of $\mathbf{M}$ show that the radius vector $\mathbf{r}$ must remain must remain in the plane perpendicular to $\mathbf{M}$.
## Conservation of $M_z$

We have shown that the [angular-momentum](angular-momentum.md) $\v{M}=\v{r}\times\v{p}$ of a system in a central field is conserved, because of the symmetry of the field around the center.

Since $\v{M}$ is perpendicular to $\v{r}$, the constancy of $\v{M}$ show that the radius vector $\v{r}$ must remain must remain in the plane perpendicular to $\v{M}$. (like planets in the solar system, rotate each on a single plane)

We use polar coordinates $r,\phi$ to describe the Lagrangian in this plane



@@ 23,7 27,11 @@ As this function does not have the co-ordinate $\phi$ explicitly (Lagrangian is 
LL1/14.2
```

We know from polar coordinates that $\dd{A}=\mfrac{1}{2}\mathbf{r}^2\dd{\phi}$ is the area element of the trajectory. We can write this as
We immediately deduce that $\phi$ varies monotonically, as it can never change sign according to `LL1/14.2`.

## Kepler's second law

We know from polar coordinates that $\dd{A}=\mfrac{1}{2}\v{r}^2\dd{\phi}$ is the area element of the trajectory (TODO: math note). We can write this as

```eq
LL1/14.3


@@ 31,6 39,8 @@ LL1/14.3

in other words, in equal times the radius vector of the particle sweeps out equal areas (Kepler's second law).

## General solution

To find the laws of motion, we start from the conservation of energy

```eq


@@ 54,3 64,30 @@ writing `LL1/14.2` as $\dd{\phi}=\frac{M}{mr^2}\dd{t}$ we also find
```eq
LL1/14.7
```

Formulae `LL1/14.6` and `LL1/14.7` give the general solution of the problem.

## Properties of the motion

The expression `LL1/14.4` can be re-interpreted, as a motion in one dimension where the "effective potential" is

```eq
LL1/14.8
```
We call the quantity $M^2/2mr^2$ the _centrifugal energy_.

In one dimension the limits of the motion are at $T=E+U=0$, which is
```eq
LL1/14.9
```

Theses points determines the limits of $r$. At these points the radial velocity is zero . However the particle is not at rest as in one dimension because the angular velocity is not zero. Instead, $\v{dot{r}}=0$  indicated a _turning point_ of the radius.

If the range is limited only by $r>r_\text{min}$ then the motion is infinite. If the range has two limits $r_\text{min}$ and $r_\text{max}$ then the path lies completely within an annulus bounded by theses circles.

```eq
LL1/14.10
```
```eq
LL1/14.11
```

A md/einstein-relativity.md => md/einstein-relativity.md +45 -0
@@ 0,0 1,45 @@
---
title: Einstein's relativity
type: model
---

We can deduce a whole new theory of mechanics by changing one assumption: interactions between particles doesn't happen instantaneously.

<hr>

## Velocity of propagation of interaction

The interaction of particles is described in ordinary mechanics by means a of potential energy of interaction, which is a function of only the coordinates of the particles interacting. This manner of decription contains the assumption of instantaneous propagation of interaction.

However, experiment shows that instantaneous interactions do not exist in nature. It seems natural to consider that a change in one body, will influence other bodies only after a certain lapse of time. We obtain the interaction velocity by dividing the distance between the two bodies by the this time interval.

This velocity is calculated from the amount of time the bodies _begin_ interacting, therefore it is strictly speaking the _maximum_ velocity of interaction. From this, we conclude that motions bodies with greater velocities aren't possible, as they would produce interactions with velocities larger than the maximum velocity of interaction.

From the principle of relativity, it follows that the velocity of propagation of interactions is the _same_ in _all_ internal systems of reference. This it is a universal constant.

(TODO: show that this speed is also the speed of light in empty space)

```eq
LL2/1.1
```

The large value of this velocity shows that Galileo's is sufficiently accurate in most cases. The formal passage to Galileo's relativity, in which interaction is instantaneous is done by passing to the limit $c\to\infty$.

## Consequences



## Summary of differences with Galilean relativity

$K$ and $K'$ are two inertial frames of reference  
  
_relative_: is not necessarily the same in $K$ as in $K'$.  
_absolute_: is the same in $K$ and in $K'$.

<!-- --- | Classical mechanics | Relativistic mechanics -->
<!-- --- | --- | --- -->
<!-- point in space $x$ | relative | relative -->
<!-- point in time $t$ | absolute | relative -->
<!-- space interval $\dd{x}$ | absolute | relative (contracted) -->
<!-- time interval $\dd{t}$ | absolute | relative (dilated) -->
<!-- spacetime interval $\dd{s}$ | absolute | absolute -->

A md/four-vectors.md => md/four-vectors.md +251 -0
@@ 0,0 1,251 @@
---
title: four vectors
type: theory
---

The coordinates of an event $(ct,x,y,z)$ can be considered as the components of a four-dimensional radius vector. We denote its components by $x^i$.

$$
x^0=ct,\qquad
x^1=x,\qquad
x^2=y,\qquad
x^3=z
$$

We generalize the concept to any four quantities $A^0,A^1,A^2,A^3$ which transform according to the [Lorentz transform](lorentz-transformation.md)

```eq
LL2/6.1
```

The square magnitude is defined analagously to the square of the radius four-vector

$$
(A^0)^2-(A^1)^2-(A^2)^2-(A^3)^2
$$

### covariant notation

To make notation simpler we introduce two types of components

```eq
LL2/6.2
```

$A_i$ are called the _covariant_ components  
$A^i$ are called the _contravariant_ components

Then the square of the four-vector is simpler to write:

$$
\sum_{i=0}^3 A^iA_i=A^0A_0+A^1A_1+A^2A_2+A^3A_3
$$

We write simply $A^iA_i$, ommiting the summation sign.

In analogy with the square of a four-vector, we form the scalar product of two different four-vectors:

$$
A^iB_i=A^0B_0+A^1B_1+A^2B_2+A^3B_3
$$

The rules of `LL2/6.2` generalize to higher order tensors, i.e. when we raise or lower _any_ indice, the sign changes if that indice is $1,2,3$, the sign doesn't change if the indice is 0.

For a a rank-2 four-tensor, if we represent the tensor with a $4\text{x}4$ matrix, where the first indice is the rows and the second is the columns, we have the following sign changes

$$
A^{ik}=\begin{pmatrix}
+ & + & + & +\\
+ & + & + & +\\
+ & + & + & +\\
+ & + & + & +
\end{pmatrix}
,\qquad
{A^i}_k=\begin{pmatrix}
+ & - & - & -\\
+ & - & - & -\\
+ & - & - & -\\
+ & - & - & -
\end{pmatrix}
$$
$$
{A_i}^k=\begin{pmatrix}
+ & + & + & +\\
- & - & - & -\\
- & - & - & -\\
- & - & - & -
\end{pmatrix}
,\qquad
A_{ik}=\begin{pmatrix}
+ & - & - & -\\
- & + & + & +\\
- & + & + & +\\
- & + & + & +
\end{pmatrix}
$$

### tensor contraction (trace)

We can form a scalar from a tensor $A^{ik}$ with the following operation

$$
{A^i}_i\equiv {A^0}_0+{A^1}_1+{A^2}_2+{A^3}_3
$$


### tensor symmetry

$A^{ik}$ is said to be symmetric is $A^{ik}=A^{ki}$. Then the components ${A^i}_k$ and ${A_k}^i$ must coincide, we shall then write simply $A^i_k$. A symmetric tensor 

$A^{ik}$ is said to be antisymmetric is $A^{ik}=-A^{ki}$. The diagonal components of antisymmetric tensors are zero.

All tensors can be written as the sum of a symmetric tensor and an anti-symmetric tensor.

### substitution tensor

We introduce the four-tensor $\delta_i^k$ as

```eq
LL2/6.3
```

the components have to be

```eq
LL2/6.4
```

and its trace is $\delta^i_i=4$.

### metric tensor

By raising or lowering on index in $\delta_i^k$, we can obtain the tensors $g^{ik}$ or $g_{ik}$, which we call the metric tensor. These tensors have identical components, which can be written as a matrix (where $i$ labels the rows and $k$ the columns)

```eq
LL2/6.5
```

It is clear that

```eq
LL2/6.6
```

We can also write the scalar product in the form

```eq
LL2/6.7
```

### One last invariant tensor

$\delta_k^i$, $g_{ik}$ and $g^{ik}$ all have the property that their components are the same in all coordinate systems. The _completely antisymmetric unit tensor_ of rank four, $e^{iklm}$ also has this property.

This is the tensor whose components change sign on interchange of any pair of indices. From the antisymmetry it follows that all components in which two indices are the same are zero. We set

```eq
LL2/6.8
```

hence $e_{0123}=-1). Thus all other components of $e^{iklm}$ are $+1$ or $-1$, depending on whether the numbers $i,k,l,m$ can be brought to an arrangement $0,1,2,3$ in an even of odd number of transpositions. The number of components is $4!=24$. Thus,

```eq
LL2/6.9
```

for reference we have the following formulas

$$
e^{iklm}e_{prst}=
\begin{vmatrix}
\delta_p^i & \delta_r^i & \delta_s^i & \delta_t^i \\
\delta_p^k & \delta_r^k & \delta_s^k & \delta_t^k \\
\delta_p^l & \delta_r^l & \delta_s^l & \delta_t^l \\
\delta_p^m & \delta_r^m & \delta_s^m & \delta_t^m \\
\end{vmatrix}
,\qquad
e^{iklm}e_{prsm}=
\begin{vmatrix}
\delta_p^i & \delta_r^i & \delta_s^i \\
\delta_p^k & \delta_r^k & \delta_s^k \\
\delta_p^l & \delta_r^l & \delta_s^l \\
\end{vmatrix}
$$

$$
e^{iklm}e_{prlm}=-2
\begin{vmatrix}
\delta_p^i & \delta_r^i \\
\delta_p^k & \delta_r^k \\
\end{vmatrix}
,\qquad
e^{iklm}e_{pklm}=-6\delta_p^i
$$

As a consequence of these formulas we have

$$
\begin{aligned}
e^{prst}A_{ip}A_{kr}A_{ls}A_{mt} &= -\det{(A)}e_{iklm}\\
e^{iklm}e^{prst}A_{ip}A_{kr}A_{ls}A_{mt} &= -24\det{(A)}
\end{aligned}
$$

blablabla about $e$ being a pseudotensor because it doesn't change under reflection.  

blablabla about $e^{iklm}e^{prst}$ being a true four-tensor of rank 8.  

blablabla about $A^{*ik}=\mfrac{1}{2}e^{iklm}A_{lm}$ being a "dual" tensor, need to do some algebra to get this.

TODO logic about linking 3d vectors to spatial components of four-tensors

```eq
LL2/6.10
```

$$
A^{ik}=(\v{p},\v{a})
$$

$$
A_{ik}=(-\v{p},\v{a})
$$

### differentiation of four-tensors

The covariant four-gradient of a scalar $\phi$ is the four-vector

$$
\frac{\partial\phi}{\partial x^i}=\left(\frac{1}{c}\frac{\partial\phi}{\partial t},\nabla\phi\right)
$$

The differential of the scalar

$$
\dd{\phi}=\frac{\partial\phi}{\partial x^i}\dd{x^i}
$$

is also a scalar.

We may note for simplicity

$$
\partial^i=\frac{\partial}{\partial x^i},\qquad
\partial_i=\frac{\partial}{\partial x_i}
$$

### integration of four-tensors

In three-dimension space, one can extend integrals over a volume, a surface or a curve. In four-dimensional space there are four types of integrations

1) integral over a curve in four-space

the element of integration is the four-vector $\dd{x^i}$

2) integral over a two-dimension surface in four-space.

blabla

3) integral over a hypersurface, i.e. over a three-dimensional manifold.

4) integral over a four-dimensional volume

A md/four-velocity.md => md/four-velocity.md +12 -0
@@ 0,0 1,12 @@
---
title: four velocity
type: theory
---

From the ordinary three-dimensional velocity vecotr one can form a [four-vector](four-vectors.md). We define the four-velocity as

```eq
LL2/7.1
```

and `LL2/3.1`

A md/intervals.md => md/intervals.md +119 -0
@@ 0,0 1,119 @@
---
title: intervals
type: theory
---

In classical mechanics we deal with point in space (3 scalars or 1 vector), and points in time (1 scalar) separately. Here we call an _event_ a point in spacetime (4 scalars or 1 four-vector).

Let us evaluate the interval between two events in two different reference frames. $K$ moves with constant velocity along the $x$ axis relative to $K'$. The first event is sending a signal with velocity $c$ from the event $(t_1, x_1, y_1, z_1)$, the second event is the arrival of this signal at $(t_2, x_2, y_2, z_2)$. The distance covered by the signal is $c(t_2-t_1)$. The distance is also $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. Therefore we can write the following relationship

```eq
LL2/2.1
```

In the second frame $K'$, the velocity of light is the same $c$, this is a postulate of [relativistic mechanics](einstein-relativity.md). We have similarly

```eq
LL2/2.2
```

We call the expresion the _interval_ between two events

```eq
LL2/2.3
```

We also have, for two infinitely close events an infinitesimal interval

```eq
LL2/2.4
```

This expression looks like a regular square, but with varying signs, we think of this geometrically as spacetime being "pseudo-euclidian".

## Intervals are absolute

From the reasoning above, we see that if $s_{12}=0$ in one reference frame, then the interval has to be zero in any other reference frame, $s_{12}'=0$. We also know that $\dd{s}$ and $\dd{s'}$ are infinitesimals of the same order, from these two conditions it follows that the two infinitesimals must be proportional:

$$
\dd{s}=a\dd{s}'
$$

The only differentiator between $K$ and $K'$ is the magnitude of their relative velocity, so $a$ must be a function of only this velocity $a=a(V)$.

We find the function of a by taking three frames $K, K_1, K_2$, when then have

$$
\dd{s}^s=a(V_1)\dd{s_1}^2,\qquad
\dd{s}^s=a(V_2)\dd{s_2}^2,\qquad
\dd{s_1}^s=a(V_{12})\dd{s_2}^2
$$

where $V_12$ is the $K_2$ relative to $K_1$. comparing the relation we have

```eq
LL2/2.5
```

However $V_{12}$ not only depends on $V_1$ and $V_2$, but also on the angle between them, but this angle doesn't appear on the left side, this formula can only be correct if $a(V)=1$. Thus

$$
\dd{s}^2=\dd{s'}^2
$$

and from the equality of infinitesimals we also have the equality of finite intervals: $s=s'$. In other words, intervals are the same in all reference frames, they are absolute.

## Timelike intervals

Again, we consider two events $(t_1, x_1, y_1, z_1)$, $(t_2, x_2, y_2, z_2)$.

We put, to simplify notation

$$
t_2-t_1=t_{12},\qquad
(x_2-x_1)^2-(y_2-y_1)^2-(z_2-z_1)^2=l_{12}^2
$$

From the invariance of intervals we have

$$
c^2t_{12}^2-l_{12}^2= c^2t_{12}'^2-l_{12}'^2
$$

Can we find another frame in which these events happen the same point in space? this frame would have to verify $l_{12}'=0$, then

$$
s_{12}^2=c^2t_{12}^2-l_{12}^2= c^2t_{12}'^2\gt0
$$

We conclude that we can only find the desired frame of reference iff $s_{12}^2\gt0$, i.e. the interval is a real number. We call these types of intervals _timelike_.

The time that elapses between the two events is

```eq
LL2/2.6
```

A single body always follows a timelike path, since the velocity of motion is always smaller than $c$

$$
l_{12}\lt ct_{12}
$$

## Spacelike intervals

Similarly to before, we ask if we can find a frame $K'$ in which the two events happen at the same instant, mathematically we ask for $t_{12}'=0$, or

$$
s_{12}^2=-l_{12}'^2\lt 0
$$

We find that this is only possible if the interval between the two events is an imaginary number. We call these intervals _spacelike_.

The distance between the two points where the events occur will be

```eq
LL2/2.7
```

Because intervals are absolute, their division into spacelike and timelike intervals is also absolute.

M md/keplers-problem.md => md/keplers-problem.md +1 -1
@@ 3,7 3,7 @@ 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) which has $k=-1$ (Newtonian or Coulombian potential)
Kepler's problem is a special case of the [two body problem](two-body-problem.md) with a $k=-1$ potential (e.g. Newtonian or Coulombian)

<hr>


A md/lorentz-transformation.md => md/lorentz-transformation.md +69 -0
@@ 0,0 1,69 @@
---
title: Lorentz transformation
type: theory
---

Let us consider a frame $K'$ moving with constant velocity $V$ relative to $K$.

The question arises of how do we transform from coordinates of one frame of reference $K$ to another frame of reference $K'$ and vice-versa. In classical mechanics, the times being equal we have

```eq
LL2/4.1
```

In the context of special relativity, we obtain the transformation between two frames of references by using the fact that the [interval](intervals.md) $s$ between two events must be the same in every inertial frame.

The interval can be regarded as a distance of two points in a four dimensional system of coordinates.

We neglect parallel displacements because these operations leave the distances unchanged, since the interval depends only on relative coordinates, not on the point of reference.

We are left with rotations in the four-dimensional space $(x,y,z,ct)$. We can resolve these into 6 planar rotations $xy,zy,xz,tx,ty,tz$. The first three corresponds to spatial rotation that we already know.

Let us consider the $tx$ rotation, in this rotation $y,z$ don't change. This rotation must leave $(ct)^2-x^2$ unchanged, the square of the "distance" of the point (ct,x) from the origin. The general form of the rotation is


```eq
LL2/4.2
```

where $\psi$ is the hyperbolic angle of rotation. A simple check shows that $(ct)^2-x^2=(ct')^2-x'^2$.

Let's consider a frame $K$ moving with velocity $V$ relative to $K'$ along the axis $x$. The angle $\psi$ may only depends on the relative velocity $V$. We consider the motion of the origin of $K'$, in which $x'=0$, then `LL2/4.2` takes the form

$$
x=ct'\sh\psi,\qquad
ct=ct'\ch\psi
$$

or dividing one by the other,

$$
\frac{x}{ct}=\th\psi
$$

But $x/t$ is the velocity $V$ of the $K'$ system relative to $K$. So

$$
\th\psi=\frac{V}{c}
$$

And we immediately can deduce,

$$
\sh\psi=\frac{V/c}{\sqrt{1-(V/c)^2}},\qquad
\ch\psi=\frac{1}{\sqrt{1-(V/c)^2}}
$$

Substituting in `LL2/4.2`, we find

```eq
LL2/4.3
```

The inverse formulas are obtained by changing $V$ to $-V$, since the $K$ system moves with velocity $-V$ relative to $K'$.

We see that the Lorentz transform `LL2/4.3` becomes Galileo's transform `LL2/4.1` when we take the limit $c\to\infty$

# Proper length

# Transformation of velocities

A md/proper-time.md => md/proper-time.md +72 -0
@@ 0,0 1,72 @@
---
title: proper time
type: theory
---

Let us observe clock which are moving relative to us arbitrarily. In the course of time $\dd{t}$, we can attach an inertial reference frame to these clocks, and they will have gone a distance $\sqrt{\dd{x}^2+\dd{y}^2+\dd{z}^2}$. In the coordinate system of the moving clock, the clock is at rest $\dd{x'}=\dd{y'}=\dd{z'}=0$. Because of the invariance of [intervals](intervals.md)

$$
\dd{s}^2=c^2\dd{t}^2-\dd{x}^2-\dd{y}^2-\dd{z}^2=c^2\dd{t'}^2
$$

from which

$$
\dd{t'}^2=\dd{t}\sqrt{1-\frac{\dd{x}^2+\dd{y}^2+\dd{z}^2}{c^2\dd{t}^2}}
$$

But

$$
\frac{\dd{x}^2+\dd{y}^2+\dd{z}^2}{\dd{t}^2}=v^2
$$

where $v$ is the velocity of the moving clocks relative to the observer, therefore

```eq
LL2/3.1
```

Integrating this expression, we obtain the amount of time elapsed on the moving clock during the time $t_2-t_1$ for the observer.

```eq
LL2/3.2
```

The time read by a clock attached to an object is called the _proper time_ of this object. Formulas `LL2/3.1` and `LL2/3.2` express the proper time in terms of the time for another frame of reference.

As we see from `LL2/3.1` and `LL2/3.2`, the proper time of a moving object is always less than the corresponding interval in the rest system.

## Duration perspective

Similar to how two distant people both see each other's size as smaller, two moving frames both see the other's time going slower.

Let us convince ourselves that there is no contradiction here. From a frame $K$, we observe a frame $K'$ moving ouwards.

1) At the first instant, the clock in $K'$ passes the clock in $K$, at that moment the readings coincide.

2) To compare the rate after a moment, we must compare the same clock in $K'$ with a new clock in $K$ at the same position. Similarly to the reason we can't measure with a ruler from a distance.

3) We conclude, to compare rates of clock in two reference frame, we need one clock in $K'$ and multiple clocks in $K'$, this is an assymetrical process.

#### What if the movement of $K'$ varies and circles back to coincide with $K$ ? (twins paradox)

If $K'$ describes a closed path returning to the starting point, from what we deduced the moving clock appears to lag behind relative to the one at rest. The converse reasoning, in which the moving clock is considered at rest, is not possible since the clock describing a closed trajectory does not carry out a uniform rectilinear motion, it's reference frame is not intertial.

Since the laws of nature are the same only for intertial reference frames, the frames linked to the clock at rest (inertial frame) and to the moving clock (non-inertial) have different properties, and the argument which leads to the results that the clock at rest must lag is not valid.

## Straight line is the longest path

The time interval read by a clock is equal to the integral

$$
\frac{1}{c}\int_a^b\dd{s}
$$

If the clock is at rest, then its world line is clearly parallel to the $t$ axis. If the clock carries an arbitrary motion and returns to the starting point, then its world line will be a curve passing through the two points $a$ and $b$. We saw that the clock at rest always indicates a greater time interval than the moving one. Thus we arrive at the result that the integral

$$
\int_a^b\dd{s}
$$

has its maxmimum value if it is taken along the straight world line joining $a$ to $b$. We have of course assumed only curves for which every element \dd{s} along the curves are timelike. In euclidian space, the integral would be a minimum along the straight line.

M md/two-body-problem.md => md/two-body-problem.md +16 -1
@@ 3,19 3,34 @@ title: two body problem
type: model
---

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

<hr>

The potential energy of interaction of two particles depends only on the distance between them, therefore we can write the Lagrangian `LL1/5.1` as

```eq
LL1/13.1
```

We note this relative position vector $\v{r}\equiv\v{r}_1-\v{r}_2$, and we set the origin be at the center of mass, i.e. $m_1\v{r}_1+m_2\v{r}_2=0$. These two equations give

```eq
LL1/13.2
```

Substitution in `LL1/13.1` gives

```eq
LL1/13.3
```

where

```eq
LL1/13.4
```

is called the _reduced mass_.

The function `LL1/13.3` is formally identical with the Lagrangian of a particle of mass m moving in a central field $U(\v{r})$.

A minkowski-template/P Grid small.png => minkowski-template/P Grid small.png +0 -0
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A minkowski-template/minkowski.png => minkowski-template/minkowski.png +0 -0
A minkowski-template/minkowski.py => minkowski-template/minkowski.py +51 -0
@@ 0,0 1,51 @@
#!/usr/bin/env python
import math
import numpy as np
import matplotlib.pyplot as plt

fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)

# Move left y-axis and bottim x-axis to centre, passing through (0,0)
# ax.spines['left'].set_position('center')

# Eliminate upper and right axes
ax.spines['right'].set_color('lightgrey')
ax.spines['top'].set_color('none')
ax.spines['left'].set_color('none')
ax.spines['bottom'].set_color('lightgrey')

x = np.linspace(-10,10,100)
plt.plot(x,x, linewidth=1, color='lightgrey')
plt.plot(x,-x, linewidth=1, color='lightgrey')
for a in (1,2,3,4,5,6,7,8,9):
    plt.plot(x,np.sqrt(a**2+x**2), linewidth=1, color='lightgrey')
    plt.plot(x,-np.sqrt(a**2+x**2), linewidth=1, color='lightgrey')
    plt.plot(np.sqrt(a**2+x**2), x, linewidth=1, color='lightgrey')
    plt.plot(-np.sqrt(a**2+x**2), x, linewidth=1, color='lightgrey')
# plt.plot(x,x, color='lightgrey')

# Show ticks in the left and lower axes only
ax.xaxis.set_ticks_position('bottom')
ax.yaxis.set_ticks_position('left')

#
plt.tick_params(bottom=False, left=False)
plt.rc('grid', linestyle="-", color='grey')
plt.xticks(range(-10,11))
plt.yticks(range(-10,11))
ax.set_xticklabels([])
ax.set_yticklabels([])
ax.set_aspect('equal')
plt.grid(True, color='lightgrey')
plt.ylim([-10,10])
plt.xlim([-10,10])


# fig = plt.figure()


plt.savefig(fname='minkowski.svg', format='svg', dpi=fig.dpi)
plt.savefig(fname='minkowski.png', format='png', dpi=fig.dpi)

plt.show()

A minkowski-template/minkowski.svg => minkowski-template/minkowski.svg +4272 -0
@@ 0,0 1,4272 @@
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