## ~jzck/physics-notes

fe6b680fd367b85811a3d1400efa48b781ad6501 — Jack Halford 1 year, 21 days ago
mechanical similarities part1

9 files changed, 82 insertions(+), 19 deletions(-)

A formularies/LL1/10.1.tex
A formularies/LL1/10.2.tex
A formularies/LL1/10.3.tex
A formularies/LL1/10.4.tex
A formularies/LL1/10.5.tex
A formularies/LL1/10.6.tex
A formularies/LL1/10.7.tex
M md/mechanical-similarity.md
M md/small-oscillations.md

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 @@


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 @@


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}
-\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}
-
-## 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.
+
+
+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 $k=2$.

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