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

To make things more realistic and feasible, I changed the material of
the cavity for aluminum, which has a reflectivity of 99%. I figure out
the weight of each cavity as being 2 grams.
The aluminum coating on the inside has, on purpose, an irregular surface
in the order of nanometers, so the laser beam will disperse in each
bounce.
I suggest that the laser output be dispersed by a vibrating lens, prior
to its entry to the cavity. In this way, the beam will be continuously
dispersed. Even more, 4 or 5 different lasers (total power = 5 W) be
placed in such a way that they cover the entire volume of the cavity,
accelerating the process of energy spread in its volume.
New calculations, using aluminum:
h = 6.62607E-34 J.sec
λ = 5.50E-07 m
c = 3.00E+08 m/s
E(photon) = 1.21E-48 Joules
5 W = 5 Joules/sec = 4.12E+48 photons/sec
5 Joules/hour = 1.48E+52 photons/hour
Inner surface of cavity = 7.85E+03 mm²
Area of laser spot  = 3.14159 mm²
                 2,500 spots
Time between hits = 3.333E-10 seconds (10 cm avg. trajectory)
                3.00E+09 bounces/sec
photon hits/sec in the cavity = 1.37E+39
photons hit/sec x spot = 5.4879643E+35
Reflectivity with aluminum = 99.0%
photons reflected/sec x spot = 5.4330846E+35
photons absorbed/sec x spot = 5.49E+33
energy lost/sec x spot = 6.67E-15 J
energy lost/sec in the cavity = 1.67E-11 J
energy lost/Hr in the cavity = 6.00E-08 J
energy lost/Hr in the cavity = 1.43E-08 cal/Hr
energy lost in 72 Hr, in the cavity = 1.032E-06 calories
Even if I made errors in the order of 1,000:1, still the heat absorbed
is insignificant (0.00103 calories).
AS A REFERENCE:
1 calorie is the heat that raises 1"C to one gram of water.
With this new arrangement and materials, proving/disproving E=mc² should
be much easier and cheap. The weight increment is in the order of 9 μN
(1.0368 μgrams of MASS) can be measured with EM balances of $ 10,000.