Deriving the damped oscillator equation from energy conservation and the Larmor formula

[sayebms]

consider energy for a damped electric oscillator . ("$f$" indicates the dipole moment of the oscillator)

in the absence of the damping force

$U= \frac{1}{2}kx^2 +1/2 (\frac{d^2x}{dt^2}) ^2$

and the energy conservation tells us $dU=0$.
but if there is damping force we get the following using larmor formula and energy of a dipole in an electric field, for the conservation of energy:

$\int_t^τ ( \frac{dU}{dt} + \frac{2}{3c^2} (\frac{d^2f}{dt^2})^2 -E \frac{df}{dt} )$

and here is i don't understand: using the fact that $\frac{4π^2 v_0}{3c^3 L} = σ$

and the above conseravtion of energy formula we get to

$Kf+L \frac{d^2f}{dt^2} -2/(3c^3) \frac{d^3f}{dt^3}=E$

i don't really know how we got to this last formula using the above equations. any help is appreciated. and for those who have access to the book The question is from page 184 of the book "Planck's Columbia Lectures".

 

[Simon Bridge]

Check that first equation.

You can't do it with just those equations - you also need to know how that $\sigma$ comes in.

 

[sayebms]

Thank you very much for your time.
actually that is the part i didn't understand. because i don't see any relation between the formula for the conseravation of energy and $σ$ , in the book it just appears suddenly and then the author concludes the final equation (the differential equation) by saying that the constant $σ$ is small.

 

[Simon Bridge]

That sounds... odd.
Either the author has introduced it someplace else that he expects you to have read (maybe quite early in the book) or you are expected to crunch your way through the math and notice a bunch of constants making a mess ... rearrange them into the ratio that he labels with a sigma, realize that it is small...

 

[sayebms]

It is indeed odd. But even more surprisingly i found what it refers to after reading the next chapter.
After some pages it is indicated that the constant sigma actually refers to the damping constant (as the system is considered as damped oscillator since it emits radiation) but still i don't get the relation between this constant beeing small and the last formula which is given:
$Kf+L \frac{d^2f}{dt^2}−2/(3c^3)\frac{d^3f}{dt^3}=E$

but i suspect that it is somehow used to approximate the following:

$2/(3c^3)\frac{d^2f}{dt^2}= −2/(3c^3)\frac{d^3f}{dt^3}$

could it be right??

 

[Simon Bridge]

The approach to take is to consider what the author wants to calculate and where the author starts from.
If this is unclear, then use a different textbook.

 

[Jano L.]

To derive the equation, we may argue this way. Plug-in the expression for $U$ into the integral, differentiate term by term and express $\ddot{f}^2$ as

$$
\frac{d}{dt}\left(\ddot{f}\dot{f}\right) - \dddot{f} \dot{f}.
$$

Collect the terms containing $\dot{f}$. We arrive at

$$
\int_t^{t+\tau} \left( Kf + L\ddot{f} - \frac{2}{3c^3}\dddot{f} - E_z \right)\dot{f} + \frac{2}{3c^3}\frac{d}{dt}\left(\ddot{f} \dot{f}\right) \,dt.
$$

Now since the last terms is total time derivative, its integral is easy to do. It is equal to

$$
\ddot{f} \dot{f} (t + \tau) - \ddot{f} \dot{f} (t).
$$

If we assume that the system oscillates with low damping (low $\sigma$), the values of its amplitude, velocity and acceleration are almost the same as in the previous period, hence this term is small (with respect to $\ddot{f} \dot{f}$.

If we neglect it, we have the condition

$$
\int_t^{t+\tau} \left( Kf + L\ddot{f} - \frac{2}{3c^3}\dddot{f} - E_z \right)\dot{f}\,dt = 0.
$$

One possible way to satisfy this condition is to postulate
$$
Kf + L\ddot{f} - \frac{2}{3c^3}\dddot{f} - E_z = 0.
$$

I think this should be regarded as an heuristic procedure to arrive at some new interesting equation. It should not be regarded as rigorous derivation, since there are those two arbitrary steps.

The equation works approximately (to first approximation) for periodic motions. However, nowadays we know that this equation is not very satisfactory in general, because it leads to all sorts of problems (runaways, acceleration before force is applied, and alike).