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

On Mon, 18 Nov 2024 3:36:02 +0000, gharnagel wrote:

On Sun, 17 Nov 2024 23:34:24 +0000, rhertz wrote:

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On Sun, 17 Nov 2024 21:27:28 +0000, gharnagel wrote:

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On Sun, 17 Nov 2024 15:42:32 +0000, rhertz wrote:

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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.

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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.
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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.
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The laser beam can't escape,

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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.
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LIGO array use lasers in the kilowatt range, mirrors with almost
100% reflectivity, etc.

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"Almost" isn't good enough.  You're still orders of magnitude away
from a practical experiment, which is why no one has attempted one.

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Obviously, there is a problem understanding the proposed experiment.

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Yes, there is.  What do you think the light that enters the cavity
is going to be doing?
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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.
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What is 0.999999^(3x10^9)?  And that's EACH SECOND!

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1) ABOUT THE DIFFERENTIAL ELECTROMAGNETIC BALANCE (there are other
means):
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I propose to use TWO DEB, similar to the one of which I posted a link:
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Homemade Microgram Electrobalance
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https://www.erowid.org/archive/rhodium/chemistry/equipment/scale.html
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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.
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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.
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2) ABOUT THE CAVITY:
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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.
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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.
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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.
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3) ABOUT THE DIFFERENT LOSSES:
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Every deviation from a perfect setup, due to different losses and
perturbations, will be measured and considered in the final result.
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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.
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4) ABOUT THE ACCUMULATION OF ENERGY INSIDE THE CAVITY 1:
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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:
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5 W x 1 Hr = 18,000 Joules per hour
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By m=E/c², it represents 2.0E-10 gram of mass per hour.
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In 72 hours, the accumulated energy within the cavity represents
1.44E-08 gram of mass.
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This value of mass represents 0.1398 microNewtons in 72 hours.
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Such values are perfectly in the range of measurement for current
technologies (state of the art).
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Check my second link of a lab balance ($10,000), capable of measuring
weight (mass equivalence) in the order of 0.1 microNewtons.
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Imagine what can be built investing $ 100,000 or $ 1,000,000 in a
qualified lab.
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This experiment allows to probe/disprove E = mc² in the real world,
not in the fairy land of the quantum/atomic world.

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Sorry, but it won't work the way you're thinking.  But there's
another way to think of it:
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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?
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So how hot would the ball have to be to have 1.3 MJ?
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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.
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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 :-(
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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.

1) You obviously don't know anything about almost perfect reflectivity
of advanced coating materials, like the ones used in the mirrors of the
LIGO instrument, which reflect with ultra-high effectivity 40 Watts
lasers, which are reflected 18,750 times to obtain the cumulative power
of 750,000 watts before the composite beam reach the detector. Make some
research on this.
2) I see that you are in denial of what you read. I wrote CLEARLY that
the weight of each cavity is 2.00 grams. They are done with very thin
composite materials, much more advanced than acrylic mirrors, in the
order of 99.999+ % of reflectivity. Make your calculations again.
I don't know why you are writing about 11 Kg balls, when I wrote that
each cavity weight about 0.002 Kg, and that the output of the
differential electromagnetic balance CANCEL both weights at time ZERO.
The reading of the instrument would be (t = 0) 0.000 micrograms. While
the excited cavity STORE the energy (losses are computed), the companion
cavity is passive, so the balance READ THE DIFFERENCE OF 2 gr + weight
gain - 2 gr (passive cavity).
3) I think that you may learn something about current capabilities by
reading descriptions of LIGO assembly (which are old systems). Then try
to read something about the new systems, built after the couple of LIGO
systems. Maybe, it will be an eye-opener for your stubbornness.
Please, read ALL WHAT I POST, because you don't. I'm an EE, and I don't
write idiocies when dealing with details of experiments involving
technology. I'm very careful with numbers.
If you don't like the idea, write numbers and considerations that make
sense, instead of posting deceiving/incorrect data, which has no support
based on references. I post data WITH REFERENCES, citing links.