$ E = \sum_i \hbar \omega_i = \hbar \sum_i \omega_i = \hbar \Omega $
$ p = \sum_i \hbar k_i = \hbar \sum_i k_i = \hbar K $
$ k_i c = \omeha_i \rightarrow K c = \Omega $
$\rightarrow p = \hbar \frac{\Omega}{c} = \hbar \frac{E}{\hbar c} $
$ p = \frac{1}{c} E $
$ T = \partial_t p = \frac{1}{c} \partial_t E $
$\rightarrow T = \frac{1}{c} P(ower) $
so $ P = 1 GW $, the power output of a nuclear power plant
$ T = \frac{1}{3*10^8 \frac{m}{s}} 10^9 kg \frac{m}{s^2} $
$ T = 3.33.... N $
yeah, I only need the power output of a nuclear power plant to generate a thrust of whole 3.33... N with an emdrive
but to put things into perspective
ion drive
so the Ion drive
BHT8000 can produce 449mN of thrust with a power of 8kW
and em-drive would produce
$ T = \frac{1}{3*10^8 \frac{m}{s}} 8*10^3 kg \frac{m}{s^2} $
$ T = 0.026 mN $
So with the same power input an emdrive can produce 0.0059% thrust of an ion drive. Also why I can't just use a bunch of flash lights to get the same result as this emdrive nobody seems to anwser...
p.s.:
this board needs tex-support