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

On Tue, 19 Nov 2024 16:41:15 +0000, ProkaryoticCaspaseHomolog wrote:

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On Tue, 19 Nov 2024 16:06:10 +0000, gharnagel wrote:

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On Tue, 19 Nov 2024 9:41:32 +0000, ProkaryoticCaspaseHomolog wrote:

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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.

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Exactly.  I didn't realize how complex the LIGO optical train was,
nor the "power recycling" concept:
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https://arxiv.org/pdf/1105.0305
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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.
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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.

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No. Your loss per bounce calculation is off.
Think conservation of energy.

Prok, Prok, Prok!  Did you not read the above sentence that begins
with "-- IF"?

At steady state, 5 W input equals 5 W output, which is not
incendiary.
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It _is_ warm enough, however, that the whole shebang needs to be
run in ultra-high vacuum to avoid convective effects.

The 300 degree temperature rise I calculated was under such
condition, based upon 300 cm^2 of surface area radiating with
no convection.
The point is, though, that with 3x10^9 bounces/second, it would
take much less than a second to whittle 5 W down to nothing:
1E5 bounces: 0.905Po
1E6 bounces: 0.368Po
2E6 bounces: 0.135Po
3E6 bounces: 0.050Po (in the first msec)
4E6 bounces: 0.018Po