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

On Tue, 19 Nov 2024 9:41:32 +0000, ProkaryoticCaspaseHomolog wrote:

On Tue, 19 Nov 2024 0:13:44 +0000, rhertz wrote:

<snip>

At steady state, the absorbed power equals the input power. So the
cavity will emit IR in all directions, distorting your weight
measurements since the emitted IR carries momentum.

<snip>
The discussion was distorted when I used LIGO to cite examples of
current advancements in technology.
The proposed cavity IS NOT accumulating power, but continuously storing
energy. Quite a different story.
Grossly speaking, and with a perfect 100% reflectivity, the cavity
stores
18,000 J/Hr, or 1,296,000 Joules in 72 hours.
But, as Caltech doc. stated, 1 photon out of 5,000,000 is lost at every
reflection. So, I made calculations with photons instead. Maybe I'm
fatally wrong, but these are the numbers:
h = 6.62607E-34 J.sec
λ = 5.50E-07 m
c = 3.00E+08 m/s
E(1 photon) = 1.21E-48 Joules
5 W = 5 Joules x sec = 4.12E+48 photons x sec
Number of 3.14 mm² spots within the 5cm radius cavity = 2,500
bounces/sec in the cavity = 3.00E+09 bounces/sec
photon hits/sec in the cavity 1.37E+39
photons hit/sec x spot 5.4879643E+35
photons reflected/sec x spot 5.4879632E+35
photons lost/sec x spot 1.10E+29
energy lost/sec x spot 1.33E-19 J
energy lost/sec in the cavity 3.33E-16 J
energy lost/Hr in the cavity 1.20E-12 J
energy lost/Hr in the cavity 2.87E-13 cal/Hr
********************************************
This tiny result is quite different if I use 18,000 J/Hr stored in a
perfect cavity, and produce a loss of 2.0E-07 due to a reflectivity of
99.99998 %, which gives a loss of 0.0036 Joules/Hr = 0.00086 cal/Hr.
********************************************
Enjoy destroying the calculations with photons. The heat generated seems
to be extremely low for my taste.
I don't know if I made a huge mistake in the first part. The second part
(18,000 J/Hr) surely is wrong.
Finally, don't forget that the cavity should increase the number of
photons inside at a rate of 1,48E+52 photons/Hr.
Not all of them suffer 3.00E+09 bounces/sec. The new wave of photons/sec
should have a discount in the number of hits/sec (or hour), so not all
of them are affected by the same loss in 72 hours.