~jzck/physics-notes

fe6b680fd367b85811a3d1400efa48b781ad6501 — Jack Halford 1 year, 21 days ago 508d894
mechanical similarities part1
A formularies/LL1/10.1.tex => formularies/LL1/10.1.tex +1 -0
@@ 0,0 1,1 @@
U(\alpha\mathbf{r}_1,\alpha\mathbf{r}_2,\ldots,\alpha\mathbf{r}_n)=\alpha^kU(\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_n)

A formularies/LL1/10.2.tex => formularies/LL1/10.2.tex +1 -0
@@ 0,0 1,1 @@
t'/t=(l'/l)^{1-k/2}

A formularies/LL1/10.3.tex => formularies/LL1/10.3.tex +3 -0
@@ 0,0 1,3 @@
v'/v=(l'/l)^{k/2},\quad
E'/E=(l'/l)^{k},\quad
M'/M=(l'/l)^{1+k/2}\quad

A formularies/LL1/10.4.tex => formularies/LL1/10.4.tex +1 -0
@@ 0,0 1,1 @@
2T=\sum_a\mathbf{p}_a\mathbf{v}_a=\frac{\dd{}}{\dd{t}}(\sum_a\mathbf{p}_a\mathbf{r}_a)-\sum_a\mathbf{r}_a\dot{\mathbf{p}}_a

A formularies/LL1/10.5.tex => formularies/LL1/10.5.tex +1 -0
@@ 0,0 1,1 @@
2\overline{T}=\overline{\sum_a\mathbf{r}_a\cdot\partial U/\partial \mathbf{r}_a}

A formularies/LL1/10.6.tex => formularies/LL1/10.6.tex +1 -0
@@ 0,0 1,1 @@
2\overline{T}=k\overline{U}

A formularies/LL1/10.7.tex => formularies/LL1/10.7.tex +1 -0
@@ 0,0 1,1 @@
\overline{U}=2E/(k+2),\qquad \overline{T}=kE/(k+2)

M md/mechanical-similarity.md => md/mechanical-similarity.md +69 -17
@@ 3,20 3,72 @@ title: mechanical similarity
type: theory
---

\begin{equation} U(\alpha\mathbf{r}_1,\alpha\mathbf{r}_2,\ldots,\alpha\mathbf{r}_n)=\alpha^kU(\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_n) \end{equation}
\begin{equation} t'/t=(l/l)^{1-k/2} \end{equation}

\begin{equation}
v'/v=(l/l)^{k/2},\quad
E'/E=(l/l)^{k},\quad
M'/M=(l/l)^{1+k/2}\quad
\end{equation}

\begin{equation} 2T=\sum_a\mathbf{p}_a\mathbf{v}_a=\frac{\dd{}}{\dd{t}}(\sum_a\mathbf{p}_a\mathbf{r}_a)-\sum_a\mathbf{r}_a\dot{\mathbf{p}}_a \end{equation}
\begin{equation} 2\overline{T}=\overline{\sum_a\mathbf{r}_a\cdot\partial U/\partial \mathbf{r}_a} \end{equation}
\begin{equation} 2\overline{T}=k\overline{U} \end{equation}
\begin{equation} \overline{U}=2E/(k+2),\qquad \overline{T}=kE/(k+2) \end{equation}

## Assumptions
 - The potential energy is a _homogenous_ function of co-ordinates
 - [Classical mechanics](mechanical-system.md)
We have found that the equations of motion are unchanged by multiplication of the [Lagrangian](lagrangian.md) by any constant. This allows us to determine some preperties of motion without necessarily solving the equations.

Let us consider a homogeneous potential $U(q)$

```eq
LL1/10.1
```

where $\alpha$ is any constant and $k$ is the degree of homogeneity of the function.

Let us carry out a transformation where the co-ordinates are changed by a factor $\alpha$ and the time by a factor $\beta$

$$
\mathbf{r}_a\to\alpha\mathbf{r'}_a,\qquad
t\to\beta t'
$$

Then all velocities $\mathbf{v}_a=\dd{\mathbf{r}_a}/\dd{t}$ are changed by a factor $\alpha/\beta$, therefore the kinetic energy by a factor $\alpha^2/\beta^2$. The potential energy is multiplied by a factor $\alpha^k$.

$$
T\to\alpha^2/\beta^2T',\qquad
U\to\alpha^kU'
$$

If these factor are the same, i.e. $\beta=\alpha^{1-k/2}$, then the motion is unchanged. (todo link to paragraph 2 explanation). A change of co-ordinates of the particle by the same factor corresponds to the replacement of the paths by other paths, geometrically similar but differing in size. By definition of $\alpha$ and $\beta$, the times of the motion between corresponding points are in the ratio

```eq
LL1/10.2
```

where $l'/l$ is the ratio of linear dimensions of the two paths. We can find similar relations for other characteristics of motion.

```eq
LL1/10.3
```

## Some consequences

### Small oscillations $k=2$

For $k=2$, we find the motion is `LL1/21.11`, which verifies that the period of oscillations is independant of the amplitude.

### Free fall $k=1$

In a uniform field, the potential energy is a linear function of co-ordinates (see `LL1/5.8`), $k=1$. We then have $t'/t=\sqrt{l'/l}$. Hence, for example, in fall under gravity, the distance fallen goes as the square of the time you've been falling.

### 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.

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

We can see in this log-log plot from wikipedia that the slope is approximately $2/3$, as expected.

## Virial theorem

```eq
LL1/10.4
```
```eq
LL1/10.5
```
```eq
LL1/10.6
```
```eq
LL1/10.7
```


M md/small-oscillations.md => md/small-oscillations.md +4 -2
@@ 48,6 48,8 @@ The general solution the harmonic oscillator is
LL1/21.11
```

where $A=ae^{i\alpha}$ is the complex amplitude, composed of the real amplitude $a$ and the phase $\alpha$ which depend on the initial condition of the system $x(0)$ and $\dot{x}(0)$.
Where $A=ae^{i\alpha}$ is the complex amplitude, composed of the real amplitude $a$ and the phase $\alpha$ which depend on the initial condition of the system $x(0)$ and $\dot{x}(0)$.

we note that the frequency $\omega$ doesn't depend on the inital condition, but only on the parameters of the system $k$ and $m$.
We note that the frequency $\omega$ doesn't depend on the inital condition, but only on the parameters of the system $k$ and $m$.

We also note, that the frequency of the motion is independant on the amplitude, which we have already predicted with `LL1/10.2` for a quadratic potential $k=2$.