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

On Wed, 20 Nov 2024 4:37:18 +0000, rhertz wrote:

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On Wed, 20 Nov 2024 3:54:10 +0000, ProkaryoticCaspaseHomolog wrote:

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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