Re: Want to prove E=mc²? University labs should try this!

SOME CALCULATIONS WITH PHOTONS
Amount of photons/second INTO the cavity
E(1 photon) =  1.21E-48 Joules
5 Joules/sec  =  4.12E+48  photons/sec
Time between hits =  3.333E-10 seconds (10 cm distance)
This means an average of 3.00E+09 bounces/sec
Reflectivity R = 4999999/5000000 = 0.9999998
F = photons/slot_0.33ns = 1.37E+39
photons LOST/slot_0.33ns =  2.74E+32
Photons Slot 1 = F R
Photons Slot 2 = F R² + F R
Photons Slot 3 = F R³ + F R² + F R
Photons Slot 4 = F R⁴ + F R³ + F R² + F R
Photons Slot 5 = F R⁵ + F R⁴ + F R³ + F R² + F R
Photons Slot K = F (Rᴷ + Rᴷ⁻¹ + Rᴷ⁻² + Rᴷ⁻³ + .... + R² + R)
If R < 1, it is a geometric series with sum S
      S = R (1 - Rᴷ)/(1 - R)
Being K = 3,000,000,000 and R = 0.9999998
      S = 0.9999998/0.0000002 = 4,999,999
So, the energy accumulated PER SECOND is
      E = F x S x E(1 photon) = 8.33E-03 Joules
In one hour, the energy accumulated is
     E(1 hr) = 30 Joules
In 72 hours,
     E(72 Hr) = 2,160 Joules
Its equivalent mass is
     M(72 hr) = E(72 Hr)/c² = 2.40E-11 grams
Weight W(72 hr) = g . M(72 hr) = 0.235 nanoNewtons
This value is far from my calculations on the OP of this thread by a
factor of about 1,000, but still feasible of being measured (maybe
letting the experiment run for months).
At any case, very far from the 4.166E-4 Joules calculated by Paul, or
the mean life of 2.07 ms before the "5 Watts" are DISSIPATED.