We already know that multiplying the Lagrangian by a constant does not affect the equations of motion. This freedom allows us, in some cases, to have interesting conclusions about motion without integrating the equations.
One of those cases have the potential as an homogeneous function of degree k of the coordinates:
U(α|r₁>,α|r₂>,...) = αᵏU(|r₁>,|r₂>,...).
We can perform a transformation that multiply coordinates by α and time by β, so |rₐ> → α|rₐ> and t → βt.
This implies that velocities |vₐ> = d|rₐ>/dt are changed by a factor α/β, and the kinetic energy by a factor α²/β². The potential energy gets a prefactor αᵏ.
In order to leave equations of motion unaffected, the factor of the kinetic and potential energies must be the same, so α²/β² = αᵏ, which implies β = α¹⁻ᵏ⸍². This condition makes the Lagrangian to be multiplied by αᵏ and the equations of motion are not altered.
If we multiply all the coordinates by the same number, paths are changed in such a way that they change in scale. We keep the geometry of the path but with a different size. Then, if U is homogeneous of degree k, we can obtain geometrically similar paths if time is transformed accordingly, as we have seen. If t' is the time for the transformed path and l' is the transformed version of l, we conclude that
t'/t = (l'/l)¹⁻ᵏ⸍².
Since speeds are also transformed with a factor α/β = αᵏ⸍², v'/v = (l'/l)ᵏ⸍².
Energy transforms as potential or kinetic energy, so E'/E = (l'/l)ᵏ.
Angular momentum is proportional to length and speed, so M'/M = (l'/l)·(v'/v) = (l'/l)¹⁺ᵏ⸍².
Once we know this, we can play with different values of k, according to different particular potential energy functions.
We will see that for small oscillations, U has k=2. This immediately tells us that (t'/t)¹⁻ᵏ⸍² = (t'/t)⁰ = 1. This implies that the period of small oscillations is independent of their amplitude. In other words, the amplitudes can be changes by whatever l'/l, but time ratios will not be affected by such changes.
In a uniform field of force, k=1. Then, (t'/t)¹⸍². For example, falling under uniform gravity, the time of fall is proportional to the square root of the total height difference in the fall.
In a Newtonian gravitational potential or in Coulomb electrostatic potential, k=-1. Then, (t'/t)=(l'/l)³⸍², which is Kepler's third law!
There is another powerful tool worth mentioning here. It is called the *virial theorem*. It works for a potential energy that is homogeneous on coordinates. Motion must also be confined in a finite region of space. When these conditions hold, we can relate the time average values of T and U, i.e. <T> and <E>. Let's develop this theorem.
Applying Euler's theorem for homogeneous functions to T, we know that k=2, and then ∑∂T/∂|vₐ>·|vₐ> = 2T. But we know that ∂T/∂|vₐ> = |pₐ>. Then, we can trivially rewrite 2T as
2T = ∑<vₐ|pₐ> = (d/dt)(∑<vₐ|pₐ>) - ∑<rₐ|ṗₐ>.
Now we average this equation wrt time. For any function f(t), its average value is
<f> = lim{τ→∞} (1/τ)∫f(t)dt {t;0;τ}.
If f(t) is the time derivative of a function F(t), so that f(t)=dF(t)/dt, then we can calculate its mean value as
<f> = lim{τ→∞} (1/τ)∫dF(t)/dt = lim{τ→∞} (F(τ)-F(0))/τ.
Then, if F(t) is a bounded function, we are sure the numerator is always finite, so <f> = 0.
In our case, we want ∑<vₐ|pₐ> bounded, so we need finite velocities and in a finite region. So in our previous equation
2T = ∑<vₐ|pₐ> = (d/dt)(∑<vₐ|pₐ>) - ∑<rₐ|ṗₐ> we can set to 0 the first term of the rhs. In the second term, we know that |ṗₐ> = -∂U/∂|rₐ> according to Newton's 2nd law. For now, we have
2<T> = < ∑<rₐ|∂U/∂|rₐ> >.
Now, if U is homogeneous of degree k in |rₐ>, the rhs can be simplified by Euler's theorem into k<U>. We now have
2<T> = k<U>.
Since E is constant, E=<E>, and then E=<T>+<U>. This allows us to rewrite <T> and <U> as functions of E,
<U> = 2E / (k+2) <T> = kE / (k+2).
For small oscillations (k=2), we get <U> = <T>, a powerful conclusion. And for Newton/Coulomb's potential (k=-1), we get 2<T>=-<U> and E = -<T>. For a uniform potential (k=1) we get <U> = 2<T> and E = 3<T>.
Find the ratio of the times in the same path for particles having different masses but the same potential energy.
If the path is the same, (l₂/l₁)=1.
If the potential energy is the same, (U₂/U₁)=1.
As (U₂/U₁)=(T₂/T₁), (m₂v₂²)/(m₁v₁²)=1.
But v₂/v₁=(l₂/l₁) / (t₂/t₁) = 1 / (t₂/t₁).
Then, (m₂/m₁) / (t₂/t₁)² = 1. This implies (t₂/t₁) = (m₂/m₁)¹⸍².
Find the ratio of the times in the same path for particles having the same mass but potential energies differing by a constant factor.
Again we have the same path, so (l₂/l₁)=1.
Now we also have (m₂/m₁)=1 and (U₂/U₁)=b (b=constant).
Since (U₂/U₁) = (T₂/T₁), (m₂v₂²)/(m₁v₁²) = b = (v₂/v₁)².
We also know that v₂/v₁ = (l₂/l₁) / (t₂/t₁) = 1 / (t₂/t₁).
Then, 1 / (t₂/t₁) = b¹⸍² and then t₂/t₁ = b⁻¹⸍².
Conclusion, t₂/t₁ = (U₁/U₂)¹⸍².