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fe6b680fJack Halford mechanical similarities part1 1 year, 1 month ago

#title: mechanical similarity type: theory

We have found that the equations of motion are unchanged by multiplication of the Lagrangian by any constant. This allows us to determine some preperties of motion without necessarily solving the equations.

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

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

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.

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

LL1/10.4
LL1/10.5
LL1/10.6
LL1/10.7