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

On Wed, 20 Nov 2024 0:48:23 +0000, gharnagel wrote:

Possibly, covering the outside of the ball with a low-emissivity
coating might give a faint possibility of weighing the heat
energy in the ball.
>
https://en.wikipedia.org/wiki/Low_emissivity
>
Actually, they have a value of 0.03 for Al foil itself.  But
Al would melt before it was hot enough.  In fact, EVERYTHING
would melt:
>
Wall thickness: .001" 167650K
Wall thickness: .100" 1676K (weight: 862 gm)
Wall thickness: .200" 838K (weight: 8620 gm)
>
Might as well stick with Al and use a thicker wall.  That
puts all the pressure on the scales though, but I think that's
a much easier problem than weighing actual 455 nm light.  Just
keep pumping 5W into the cavity for 72 hours, and the hole
for introducing the 5W could be a sapphire disc with a 455 nm
transmission band, otherwise, too much energy would be lost.
>
What do you think, still out of bounds?  Yeah, I think so.

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.