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From: hitlong@yahoo.com (gharnagel)
Newsgroups: sci.physics.relativity
Subject: Re: Want to prove =?UTF-8?B?RT1tY8KyPyBVbml2ZXJzaXR5IGxhYnMgc2hvdWxkIHRy?=
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Date: Mon, 18 Nov 2024 03:36:02 +0000
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On Sun, 17 Nov 2024 23:34:24 +0000, rhertz wrote:
>
> On Sun, 17 Nov 2024 21:27:28 +0000, gharnagel wrote:
> >
>> On Sun, 17 Nov 2024 15:42:32 +0000, rhertz wrote:
> > >
> > > I suggest you to read about the cavities used at Berlin
> > > University during 1893-1900 (Wien, Planck - 2 Nobel Prize
> > > on the same matter), and how to measure black body radiation.
> >
> > I have used such cavities many times to calibrate light sources.
> > I suggest you reread what I wrote, particularly how many orders of
> > magnitude a practical system is away from your required energy.
> >
> > > I just thought of the exact opposite use of the small orifice:
> > > to allow the laser beam to ENTER into the perfectly reflecting
> > > cavity, with irregular inner coating, what allow that the laser
> > > beam be dispersed within the small cavity, being confined there.
> > >
> > > The laser beam can't escape,
> >
> > Actually, it can: through the hole.  1 m diameter ball, 1 mm
> > diameter hole ==> 0.0001% loss.  Still many orders of magnitude
> > away.  The light still bounces off the wall of the sphere more
> > often that in a linear cavity.  But you can probably get up to
> > higher energy in the ball this way. 5W continuous input would
> > get you to 5 MW, assuming no losses on the bounces, which is
> > wildly optimistic.
> >
> > > LIGO array use lasers in the kilowatt range, mirrors with almost
> > > 100% reflectivity, etc.
> >
> > "Almost" isn't good enough.  You're still orders of magnitude away
> > from a practical experiment, which is why no one has attempted one.
>
> Obviously, there is a problem understanding the proposed experiment.

Yes, there is.  What do you think the light that enters the cavity
is going to be doing?

It's going to be bouncing off the walls 3x10^8/0.1 = 3x10^9 times
per second.  Each time it bounces, it will lose a minute fraction
of its energy, leaving only 0.999999 of what it had.

What is 0.999999^(3x10^9)?  And that's EACH SECOND!
> *******************************************************************
> 1) ABOUT THE DIFFERENTIAL ELECTROMAGNETIC BALANCE (there are other
> means):
>
> I propose to use TWO DEB, similar to the one of which I posted a link:
>
> Homemade Microgram Electrobalance
>
> https://www.erowid.org/archive/rhodium/chemistry/equipment/scale.html
>
> The output of the two sensors are connected to electronics, which can
> REST, filter and process the electrical output of the EB. The result,
> once calibrated to be ZERO while the laser is off, will be a LINEAR
> relation between weight difference and voltage, in the order of
> nanograms.
>
> The DEB will be working in a vacuum, with a Faraday-like dome, and
> floating on a mercury bed (to filter vibrations, noise and EM
> interferences. It also will be compensated in temperature by means to
> cool it off and measure the heat excess that's eliminated.
>
> *******************************************************************
>
> 2) ABOUT THE CAVITY:
>
> It has not to be a perfect sphere. On the contrary.
> Imagine that you take 2 mg of aluminum foil and build a case with it,
> having dimensions of about 1,000 cm³. The inner coating, made of highly
> reflective material, has artificial irregularities much narrower than
> the laser wavelength (green, 550 nm), such that the laser beam (1 mm
> radius) spread all over the interior of the cavity, with a very small
> percentage escaping through the 2 mm orifice.
>
> If necessary, make the external laser device to spatially oscillate
> slowly a little amount (by mechanical means) to assure that the beam
> is hitting different spots within the cavity. That will spread the
> radiation all over inside its volume.
>
> Two identical cavities are built, being each one placed on the DEB.
> Their weight (a few grams) will be subtracted from their electrical
> outputs, so that (in perfect balance and equilibrium), the output will
> read 0.000 nanograms.
>
> *******************************************************************
>
> 3) ABOUT THE DIFFERENT LOSSES:
>
> Every deviation from a perfect setup, due to different losses and
> perturbations, will be measured and considered in the final result.
>
> The differential output signal [linear function of the weight
> difference] will be processed to filter NOISE, including the electrical
> noise (in the order of nanovolts) generated by the EM mechanisms.
>
> ******************************************************************
>
> 4) ABOUT THE ACCUMULATION OF ENERGY INSIDE THE CAVITY 1:
>
> If the reflected laser beams are almost 100% reflected and only a very
> small fraction of energy dissipates as heat (0.001%), the energy will
> accumulate linearly with time:
>
> 5 W x 1 Hr = 18,000 Joules per hour
>
> By m=E/c², it represents 2.0E-10 gram of mass per hour.
>
> In 72 hours, the accumulated energy within the cavity represents
> 1.44E-08 gram of mass.
>
> This value of mass represents 0.1398 microNewtons in 72 hours.
>
> Such values are perfectly in the range of measurement for current
> technologies (state of the art).
>
> Check my second link of a lab balance ($10,000), capable of measuring
> weight (mass equivalence) in the order of 0.1 microNewtons.
>
> Imagine what can be built investing $ 100,000 or $ 1,000,000 in a
> qualified lab.
>
> *****************************************************************
>
> This experiment allows to probe/disprove E = mc² in the real world,
> not in the fairy land of the quantum/atomic world.

Sorry, but it won't work the way you're thinking.  But there's
another way to think of it:

All the bounce losses will heat up the cavity walls.  Heat is energy.
So just heat up a thick-walled ball, with a laser or whatever?

So how hot would the ball have to be to have 1.3 MJ?

Well, suppose we have a 10 cm dia. aluminum ball with a 1 mm
hole drilled to the center and blackened to absorb the light.
It would weigh about 11 kg and the temperature increase would
be about 128 C.  Radiative heat loss over the surface would be
about 5.4 W, so the laser must stay on -- or extra weight added
for insulation.

11 kg may be a lot to put on a sensitive scale, so let's go to
a 5 cm ball (1.4 kg).  Oops, the temperature goes to 1022 C!
Couldn't use aluminum!  This is quickly getting out of hand.
Forget I said anything :-(

I'm not a metrology expert, but an 11 kg (plus) ball on a scales
that is sensitive to a few ng seems pretty far-fetched to me.