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

On Mon, 18 Nov 2024 19:43:54 +0000, rhertz wrote:

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On Mon, 18 Nov 2024 14:41:26 +0000, gharnagel wrote:

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750,000 Watts?!  That's ridiculous! :-))

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https://www.ligo.caltech.edu/page/ligo-technology
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https://dcc.ligo.org/public/0000/P070082/004/P070082-v4.pdf
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Some insights, that I try to resume for you:
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1) Laser power on each arm is incremented about 300 times,

No.  You misunderstand what is happening.  The beam hits a
different spot on the mirrors each time.

to obtain an effective length of about 1,000 Km.

The 50W beam is spread over 1000 km.  How much energy is
contained in that distance? Delta t = 1000/299792.458 = 3.3 ms.
E = 50*.0033 = 0.167 J.  That's the amount of energy in the
whole 1000 km at any one time, and then it exits the arm on
its way to the detector.

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).

This is your misinterpretation of the LIGO system.  There is
no mention of 750 kW in their blurb, nor of "amplifying" the
signal to more than 50 W.

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 %.

Yes, which is insignificant loss for only 600 reflections:
0.9999998^600 = 0.99988.
But assuming the beam didn't exit but hit the same spots on
the mirrors and built up from 50 W to 750 kW, you would need
30,000 hits. 0.9999998^30000 = 0.994.  Each mirror would have
to absorb .003*750000 = 225 W, which is more than the 50 W
laser can supply.
BUT YOU WANT TO PUT IT IN A SHORT CAVITY.
How many hits if the cavity were only 10 cm?!  Use your
carefulness with numbers to calculate THAT.

This reflectivity means that only 10 μW/hit are absorbed by
the mirrors.

Each hit.  They REALLY add up when you have a short cavity.

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.

Even if you could cool 750 kW, your tiny cavity would be smoked.

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!

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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.

Irrelevant, since all the light would become heat, which wouldn't
be so bad except that the refrigerator, if you could handle that
load, would suck all the energy out that you were trying to detect.

If you don't like the amplification to 750,000 Watts from 50 Watts, go
and comply with LIGO people, not to me.

I guess you mean "complain" Perhaps YOU should ask them if they
ever said anything about 750 kW.  I'm quite sure they didn't.

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.

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I clearly wrote: 2 grams/cavity and 1,000 cm³ each.
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I didn't say that the experiment was going to be cheap. Maybe
it's in the range of ten million USD.

Maybe more like 100 quadrillion, since it's impossible and the
government likes to spend money.  Oh, wait, Musk and Ramaswamy
are looking over their shoulders :-)

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.

But you forgot the superlative issue: poof!