Sujet : Re: Want to prove E=mc²? University labs should try this!
De : hertz778 (at) *nospam* gmail.com (rhertz)
Groupes : sci.physics.relativityDate : 22. Nov 2024, 17:31:06
Autres entêtes
Organisation : novaBBS
Message-ID : <092fa494db9895ba52cfac350be5e744@www.novabbs.com>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
User-Agent : Rocksolid Light
I owe an apology to everyone that participated in this thread.
The experiment described in the OP is ILL CONCEIVED (by me. Thanks,
ChatGPT, which encouraged me to follow since the OP). But it's my fault
entirely.
The source of my HUGE ERROR was to NOT UNDERSTAND that ONLY with a
perfect reflectivity of 100%, this experiment could be realized. Even a
slightly lower reflectivity (say 1 photon lost every 1,000,000,000,000
photons) causes that ALL THE PHOTONS bouncing back and forth be
dissipated as heat.
The accumulation of photons at the K-th Slot is correctly given by
N = F (Rᴷ + Rᴷ⁻¹ + Rᴷ⁻² + Rᴷ⁻³ + .... + R² + R)
where F is the amount of photons per 0.33 nsec slot.
With a 5 Watts laser, F = 4.12E+48 photons/sec x 0.333 nsec = 1.37E+39
If R = 0.9999998 (LIGO, 1 photon lost every 5,000,000), the survival
time of each pack F is 1.67 msec. In that time, all the photons are
wasted as heat.
As an example, if R = 0.9999999998 (1 photon lost/5 billion photons),
the survival time of each pack would be about 1.67 seconds.
After K bounces in the cavity (K = 3.00E+09 bounces/sec), IT WAS WRONG
TO CALCULATE accumulated energy in this way:
E(1 second) = F x [R(1-Rᴷ)/(1-R)] x E(1 photon)
because I'm not contemplating the extinction rate of each pack F.
So, in the end, and no matter if there were seconds or hours, NO PHOTONS
remain within the cavity.
It was fun, but the experiment is impossible to be implemented.
BTW: Do you all understand that I was trying to weight LIGHT? Can it be
possible? Maybe the formula E=mc² works only in one way: mass to energy.