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

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Sujet : Re: Want to prove E=mc²? University labs should try this!
De : hertz778 (at) *nospam* gmail.com (rhertz)
Groupes : sci.physics.relativity
Date : 24. Nov 2024, 00:34:40
Autres entêtes
Organisation : novaBBS
Message-ID : <afe961104287110aab310b0cc3b5f8ef@www.novabbs.com>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
User-Agent : Rocksolid Light
On Fri, 22 Nov 2024 16:31:06 +0000, rhertz wrote:

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.
EXPLAINING MY APOLOGY:
The above calculations are correct in a technical sense, but MY ERROR
was to understand that photon's energies could be accumulated AND STORED
long after the laser was TURNED OFF.
This doesn't happen, because the photons stored after T seconds, no
matter how much of them, are ABSORBED by the imperfect inner surface of
the cavity (Reflectivity R < 100.00%).
However, it doesn't prevent to measure the weight gain when the laser is
active.
Using the best aluminum coating, R = 0.99, which gives a much lower
stored energy than the impractical LIGO-like tech, with R=0.999999998.
Using R=0.99, the energy stored after 72 hours is 0.0428 Joules, which
represent m = E/c² of 4.75E-16 grams. A very, very amount of mass.
I've learned that, using optomechanical systems (based on micro-quartz
technology) a sensitivity to changes in mass can be measured at scales
as small as 10E-15 grams under ideal conditions.
The idea is to measure changes in the mass by measuring the changes in
the frequency of induced vibrations in the "cavity" filled with photons.
Of course that these techniques are in the "state of the art" as of
today, but the use of optomechanical resonators is increasing in labs
all over the world, and that its sensitivity has been increased by
10,000 in the last 5 years.
The idea is based on several techniques to measure changes, one of them
being using interferometry in the sensors (where the device is placed to
be weighted).
But these advanced techniques (there are others) are beyond my interest,
because they are very new and still under the learning curve.
That's it.

Date Sujet#  Auteur
17 Nov 24 * Want to prove E=mc²? University labs should try this!93rhertz
17 Nov 24 +* Re: Want to prove E=mc²? University labs should try this!85LaurenceClarkCrossen
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