Not me. Is ChatGPT that indicates that there are no historical records
about experiments of this kind. Maybe BECAUSE nobody like the results.
Even when it operates near the limits of technology available for the
last 20 years, the concept itself is extremely simple to understand and,
as ChatGPT wrote, it's a novel way to prove/disprove E=mc² at optical
level.
On Sun, 17 Nov 2024 5:22:33 +0000, gharnagel wrote:
On Sun, 17 Nov 2024 4:22:43 +0000, rhertz wrote:
Not me. Is ChatGPT that indicates that there are no
historical records about experiments of this kind.
Maybe BECAUSE nobody like the results.
Even when it operates near the limits of technology
available for the last 20 years, the concept itself
is extremely simple to understand and, as ChatGPT
wrote, it's a novel way to prove/disprove E=mc² at
optical level.
Seems to me that the chatGPT estimate was extremely optimistic.
The maximum energy built up will be limited by the input energy
during each pass of the storage beam. If the cavity were one
meter long, the time per pass would be 1/c. How much energy
does a 5 W laser put out during that time? 1.67e-8 J.
If the end mirror absorbs 0.00001 of the energy per reflection,
the maximum energy in the cavity would reach only .00167 J.
You might get a 10x improvement of reflectivity and make the
cavity 10 m long, but you're still a long way from your dream,
methinks. A scales to weigh a 10 meter pipe seems a bit
unwieldy to me. You think maybe I made a mistake? Check it
out yourself.
You are confusing cavity with something else.
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 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,
LIGO array use lasers in the kilowatt range, mirrors with almost
100% reflectivity, etc.
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.
*******************************************************************
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.
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,
[Blah, blah, blah]
[Blah, blah, blah]
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.
On Mon, 18 Nov 2024 14:41:26 +0000, gharnagel wrote:
750,000 Watts?! That's ridiculous! :-))
https://www.ligo.caltech.edu/page/ligo-technology
https://dcc.ligo.org/public/0000/P070082/004/P070082-v4.pdf
Some insights, that I try to resume for you:
1) Laser power on each arm is incremented about 300 times,
to obtain an effective length of about 1,000 Km.
This power is amplified inside the interferometer using power
recycling and resonant Fabry-Perot cavities, increasing the
effective circulating power in the arms to hundreds of
kilowatts (up to 750 kW in the most recent configurations).
2) The reflectivity of the mirrors is so high that it reflect
all but one of every 5 million photons that hit them. This
means a reflectivity of 99.99998 %.
This reflectivity means that only 10 μW/hit are absorbed by
the mirrors.
Considering that I wrote that the GENERATED HEAT in the entire
cavity is EXTRACTED by the proper use of refrigeration (and also
measured), the environment in which the cavity resides REMAIN IN
THERMAL EQUILIBRIUM.
You are speculating a build up of energy inside your little
ball. With 99.9999% reflectivity each time the light bounces
off the wall, CALCULATE how many times that light must bounce
in 72 hours. What's left will be a skinny zero. CALCULATE
where all that lost energy goes! The only answer is heat.
Your little ball goes poof!
I CLEARLY WROTE: REFRIGERATED, with vibrations cancelled AND
ALMOST IN VACUUM. Both actions are measured. Vacuum can be as
high as one-trillionth that of sea level, which means there are
about 10 million molecules per cubic centimeter.
If you don't like the amplification to 750,000 Watts from 50 Watts, go
and comply with LIGO people, not to me.
I also showed what would happen if you reduced the size of
the ball (poof!). It's still ridiculous because the proposed
scales is too fragile to handle the weight.
I clearly wrote: 2 grams/cavity and 1,000 cm³ each.
I didn't say that the experiment was going to be cheap. Maybe
it's in the range of ten million USD.
The technology exists today, even when not available for anyone.
It is possible an ad-hoc setup, with materials and subsystems
built for this specific purpose.
The proposal is just a sketch, and I cited MAJOR ISSUES only.
On Tue, 19 Nov 2024 0:13:44 +0000, rhertz wrote:
When I look at the screenshot on the following page, taken during an
actual run, https://tinyurl.com/2v3kssvv I see a circulating power of
358.8 KW versus an input power of 63.5 W. This implies an
amplification factor of 5650. This is not a freely adjustable number,
but one dictated by the physics of the Fabry-Perot mirrors. Look down
at the diagram captioned "Basic Michelson interferometer with Fabry
Perot cavities." The far mirrors will be 99.999% reflective, but the
mirrors near the beam splitter must necessarily be partially
transmissive, otherwise power cannot enter the Fabry Perot cavity.
It's an interesting optimization problem. The more transmissive the
near mirrors, the more light enters the FP cavity, but the fewer the reflections. The more reflective the mirrors, the more the power amplification factor, but less light enters the cavities, and the
system's sensitivity to high frequency gravitational waves is also
limited. Once they settled on an optimum transmissivity for the near
mirrors, they could not arbitrarily adjust this parameter.
And about the multilayered coating used in the mirrors, which
also evolve with the pace of time, can be perfectly applied to
the interior of the cavity that I proposed, without restrictions
except money.
Nope. Physics limits their application.
You can achieve 99.999% reflectivity only at one specific angle
(which is dependent on the mirror design). If the mirror reflects
99.999% of light normal to the surface, it won't reflect 99.999% of
the light at other angles.
On Tue, 19 Nov 2024 16:06:10 +0000, gharnagel wrote:
On Tue, 19 Nov 2024 9:41:32 +0000, ProkaryoticCaspaseHomolog wrote:
Nope. Physics limits their application.
You can achieve 99.999% reflectivity only at one specific angle
(which is dependent on the mirror design). If the mirror reflects
99.999% of light normal to the surface, it won't reflect 99.999% of
the light at other angles.
Exactly. I didn't realize how complex the LIGO optical train was,
nor the "power recycling" concept:
https://arxiv.org/pdf/1105.0305
I'm still quite certain, however, that when you throw 750 kW
into a 10 cm ball with walls that are 0.999999 reflective,
the losses will, as you say, cause serious problems.
For one, that's a loss of 0.75 W/bounce, and bounces will happen
c/0.1 = 3x10^9 times per second -- IF one could supply the power
to keep it operating. In which case, the whole thing would make
a beautiful incendiary display. With only 5 W input to drive
the system, however, it would heat up to about 300 C, according
to my radiation slide rule.
No. Your loss per bounce calculation is off.
Think conservation of energy.
At steady state, 5 W input equals 5 W output, which is not
incendiary.
It _is_ warm enough, however, that the whole shebang needs to be
run in ultra-high vacuum to avoid convective effects.
On Wed, 20 Nov 2024 3:54:10 +0000, ProkaryoticCaspaseHomolog wrote:
Given than dielectric coatings don't work here, a realistic
reflection coefficient for aluminum foil would be around 0.85
or so, so you won't have a great build-up of power level
inside your ball.
The system will reach steady-state after a relatively short
period of time. If 5 watts is pumped in, the ball will radiate
5 watts of energy at steady state. The emissivity will affect
the final temperature that the ball will reach, but figures
like 167650K for 0.001" walls are completely unrealistic
numbers.
The accumulated warmth should probably be detectable by a
person with sensitive fingers.
Certainly you won't need to wait 72 hours to get your negative
results.
Most of the time will be spent pumping down your high-vacuum system, evaporating your "getter" to tie up the last bits of gas, etc.
You refuse to accept an elementary fact, clearly stated by me
in this thread. Why? Maybe because you're not the EE/physicist
that you claim to be and you're just an amateur playing with
science as a hobby.
Get this for once:
THE SYSTEM WILL NEVER REACH INTERNAL EQUILIBRIUM!
That is IMPOSSIBLE because I'd be pumping ENERGY non-stop,
forever if necessary. Can you get this, please?
THERE IS NO AMPLIFICATION OF LASER POWER, AND NEVER WAS MEANT
TO HAPPEN.
WHAT HAPPEN IS A CONTINUOUS FEED OF ENERGY!. If you CAN'T SEE
IT, then substitute the 5W laser by a hose POURING WATER INSIDE
THE CAVITY. Some water falls out, but most remain UNTIL THE
CAVITY IS FULL OF IT.
When will it happen with the water analogy? Don't know/don't care.
The only reason by which I used three days to fill the cavity
up is because A LONGER PERIOD would accumulate much more
perturbations and external interferences, complicating the
statistical processing of the electrical signal that IS LINEARLY
PROPORTIONAL to the accumulation of energy inside the cavity.
If you REFUSE to understand this, I advise you to go back to
college or high school, where you could re-learn elementary
logic and arithmetic.
Say no more.
On Wed, 20 Nov 2024 5:29:59 +0000, gharnagel wrote:
On Wed, 20 Nov 2024 4:37:18 +0000, rhertz wrote:
On Wed, 20 Nov 2024 3:54:10 +0000, ProkaryoticCaspaseHomolog wrote:
Given than dielectric coatings don't work here, a realistic
reflection coefficient for aluminum foil would be around 0.85
or so, so you won't have a great build-up of power level
inside your ball.
The system will reach steady-state after a relatively short
period of time. If 5 watts is pumped in, the ball will radiate
5 watts of energy at steady state. The emissivity will affect
the final temperature that the ball will reach, but figures
like 167650K for 0.001" walls are completely unrealistic
numbers.
All three sets of numbers were based upon the heat capacity of Al,
assuming no heat loss.
The accumulated warmth should probably be detectable by a
person with sensitive fingers.
Heat loss can be reduced by lowering the temperature, which can
be accomplished by increasing the mass of the walls.
Certainly you won't need to wait 72 hours to get your negative
results.
With thick walls, and low-emissivity coating, the ball can hold
1.3 MJ of heat without too much loss. The problem will be measuring
the mass increase of a heavier ball.
Most of the time will be spent pumping down your high-vacuum
system, evaporating your "getter" to tie up the last bits of
gas, etc.
You refuse to accept an elementary fact, clearly stated by me
in this thread. Why? Maybe because you're not the EE/physicist
that you claim to be and you're just an amateur playing with
science as a hobby.
If you want to address me, don't delete my words. You have kept
Prok's words, not mine. I don't think Proks background is in
either engineering or physics, IIRC.
Get this for once:
THE SYSTEM WILL NEVER REACH INTERNAL EQUILIBRIUM!
Says the EE playing with physics? :-)
That is IMPOSSIBLE because I'd be pumping ENERGY non-stop,
forever if necessary. Can you get this, please?
How can one explain this to someone who isn't versed in
optics and thermal physics? You have our results. If you
want more explanation, just ask.
“When you talk, you are only repeating what you already know.
But if you listen, you may learn something new.” – Dalai Lama
THERE IS NO AMPLIFICATION OF LASER POWER, AND NEVER WAS MEANT
TO HAPPEN.
No need to shout.
WHAT HAPPEN IS A CONTINUOUS FEED OF ENERGY!. If you CAN'T SEE
IT, then substitute the 5W laser by a hose POURING WATER INSIDE
THE CAVITY. Some water falls out, but most remain UNTIL THE
CAVITY IS FULL OF IT.
Apparently, you don't understand that the cavity has a leak in it,
and the more water is in the cavity the more it leaks. One leak
is the aperture in the cavity to let in the laser energy. I
suggested a interference coated sapphire window. Without it,
the 10 cm diameter ball would leak .01 cm^2/300 cm^2 = 3e-5,
worse than the wall is presumed to absorb/bounce, which is
another assumed to be .000001.
But surely you realize that light has entered the ball continues
to bounce around inside the ball, don't you?
Surely you realize that it's traveling close to the speed of LIGHT,
don't you.
Surely you can calculate how many bounces it will take to
convert most of that light to heat, don't you?
Surely you can estimate that that time is MUCH shorter than
72 hours?
As an EE you have had differential equations, haven't you?
When will it happen with the water analogy? Don't know/don't care.
Did you flunk DE?
The only reason by which I used three days to fill the cavity
up is because A LONGER PERIOD would accumulate much more
perturbations and external interferences, complicating the
statistical processing of the electrical signal that IS LINEARLY PROPORTIONAL to the accumulation of energy inside the cavity.
I say pshaw! Use a more powerful laser. LIGO uses a NdYAG.
If you REFUSE to understand this, I advise you to go back to
college or high school, where you could re-learn elementary
logic and arithmetic.
Pot, kettle, black :-))
Say no more.
“ignorance more frequently begets confidence than does knowledge.”
– Charles Darwin
This is awfully confusing. Sometimes your are answering Richard,
sometimes me. Try to avoid doing this in the future. I don't know
if you have actually confused us two, or if you are merely trying
to multitask by answering us simultaneously. :-(
Current "state of the art" weighting technology (expensive) can measure
about 1 nanogram of MASS.
I asked ChatGPT this question:
"If I use a high precision laser having 5 Watts steady power for 72
hours, injecting the light into an almost perfect reflective cavity
(more than 99% of reflectivity), can I prove that the internal MASS
increased as
m=E/c² states?
ChatGPT answered this:
Conclusion
If the laser operates for 72 hours, the theoretical mass increase inside
the cavity is approximately 14.4 picograms, but in practice, the
steady-state energy inside the cavity will be significantly lower due to losses. Proving this mass increase experimentally would require an exceptionally controlled setup and cutting-edge measurement tools. While
it is a challenging experiment, its successful execution would provide a novel confirmation of m=E/c² in an optical context!
I simplify calculations here:
5 Watts x 72 Hr = 1,296,000 Joules (Kg.m²/s²)
MASS increase in 72 Hr = 1.0368 μgrams of MASS
WEIGHT INCREASE = 1.007 μgrams of WEIGHT ≈ 1.007E-09 Newtons
ChatGPT WARNINGS:
Challenges in Proving the Mass Increase
While the mass increase is theoretically calculable, there are practical hurdles:
1. Energy Retention in the Cavity
Steady-State Limit: A laser does not continuously add energy to a cavity indefinitely. After a certain time, the energy inside the cavity
stabilizes at a steady state determined by:
E(stored) = P(laser)/P(loss) x τ
τ is the cavity’s photon lifetime.
Real cavities lose energy through mirror transmission, scattering, and absorption. Even with highly reflective mirrors (99.999%), steady-state energy may be orders of magnitude smaller than the total energy
delivered by the laser.
2. Balance Sensitivity
Mass Changes: To measure a steady increase up to 1.0368 μgrams of MASS,
you would need an extremely sensitive device like an electromagnetic
balance or other precision mass measurement tools with sub-μgram sensitivity.
Current state-of-the-art balances can detect mass changes at this scale,
but such measurements are delicate and require isolation from
vibrations, air currents, and thermal effects.
3. Environmental Interference
Thermal Effects: Energy absorption by mirrors or cavity components can
cause thermal expansion or convection currents that might mimic a weight change.
Radiation Pressure: The laser light exerts a tiny force on the mirrors,
which might introduce additional complications.
4. Loss Compensation
Not all the laser energy is stored. The difference between the laser's
energy input and the cavity’s steady-state energy is lost as heat, scattered light, or transmitted light. This loss must be carefully
accounted for to isolate the contribution of stored energy to the mass increase.
Experimental Considerations to maximize your chances of success:
High-Q Cavity: Use ultra-high-reflectivity mirrors to maximize energy retention and reduce losses.
Sensitive Instrumentation: Employ a precision electromagnetic balance
capable of detecting picogram-scale changes.
Environmental Isolation: Perform the experiment in a vacuum chamber and
on an isolated platform to reduce noise.
Calibration: Account for all sources of error, including thermal effects
and laser power fluctuations.
***********************************************
BUT WHAT REALLY MATTERS HERE IS THAT THE AI ENGINE CONFIRMED THAT NOBODY
IN THE ENTIRE SCIENTIFIC COMMUNITY HAS EVEN TRIED EXPERIMENTS OF THIS
SORT, WHICH ARE CONCEPTUALLY SIMPLE!
WHY, I ASK? MAYBE THE RESULTS COULD BE SHOCKING!
With a little of ingenuity, you could try this at home, using a
DIFFERENTIAL ELECTROMAGNETIC BALANCE. Here is a guide:
Homemade Microgram Electrobalance
https://www.erowid.org/archive/rhodium/chemistry/equipment/scale.html
You should add a DIFFERENTIAL BALANCE, with two cavities (one with the
laser turned off).
Here is a company that sells microgram balances ($10,000 and up), mainly
for Big Pharma companies.
https://www.mt.com/us/en/home/products/Laboratory_Weighing_Solutions/microbalances/ultra-micro-balance.html
HEY!, THERE IS A CHANCE THAT YOU CAN CONFIRM OR DISPROVE E=mc².E=mc^2 assumes that equal masses contain equal energy regardless of the substance. This seems plainly false to start with, ipso facto.
AND BY DISPROVING IT, RELATIVITY GOES TO THE TRASH BIN. ALL OF IT, AND DERIVATIONS APPLIED ALL OVER THE REALM OF PHYSICS!
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