In accordance with classical electrodynamic an electron moving with acceleration `W` loses its energy due to radiation as
`(dE)/(dt)= -(2e^(2))/(3c^(3))W^(2)`,
Where `e` is the electron, `c` is the velocity of light. Estimate the time during which the energy of an electron performing alomist harmonic oscillations with frequency `omega= 5.10^(15)s^(-1)` will decrease `eta= 10 time s`

Explanation

The formula in `MKS` units is
`(dE)/(dt)= (mu_(0)e^(2))/(6pic)w^(rarr_2)`
For an electron performing (linear) harmonic vibrations `vec(w)` is the some definite direction with
`w_(x)=-w^(2)x` say.
Thus `(dE)/(dt)=-(mu_(0)e^(2)omega^(4))/(6pic)x^(2)`
If the radiation loss is small (i.e., if `omega` is no to large), then the motion of the electron is always close to simple harmonic wiht slowly decreasing amplitude. Then we can write
`E=(1)/(2)m omega^(2)a^(2)`
and `x=a cos omegat`
and avarage the above equation ignoring the variation of `a` in any cycle. Thus we get the equation, on using `lt x^(2) ge(1)/(2)a^(2)`
`(dE)/(dt)=-(mu_(0)e^(2)omega^(4))/(6pic).(1)/(2)a^(2)=-(mu_(0)e^(2)omega^(2))/(6pi mc)E`
since `E=(1)/(2)momega^(2)a^(2)` for a harmonic oscillator.
This equation intergates to
`E=E_(o)e^(-t//T)`
where `T= 6pi mc//e^(2)omega^(2)mu_(0)`
It is than see that energy decreases `eta` times in
`t_(0)=T In eta=(6pi mc)/(e^(2)omega^(2)mu_(0))In eta= 14.7ns`