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

On Thu, 21 Nov 2024 21:33:35 +0000, Paul B. Andersen wrote:

Den 20.11.2024 02:02, skrev rhertz:

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Not bad for my estimation (2 gr) for the weight of the cavity,
with inner surface of 7,850 mm², and thickness of 0.1 mm.
>

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If the cavity is spherical, the diameter will be D = 5 cm.
The time light will use to go from one surface to the opposite
surface is Δt = 16.33 ns
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Even if the cavity may have another shape, I will use
Δt = 16.33 ns as the average time between the reflections
of the laser beam.
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>
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Den 20.11.2024 05:37, skrev rhertz:

THE SYSTEM WILL NEVER REACH INTERNAL EQUILIBRIUM!
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That is IMPOSSIBLE because I'd be pumping ENERGY non-stop, forever if
necessary. Can you get this, please?
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THERE IS NO AMPLIFICATION OF LASER POWER, AND NEVER WAS MEANT TO HAPPEN.
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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.
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When will it happen with the water analogy? Don't know/don't care.
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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.
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Say no more.

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Power of laser P₀ = 5 W
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Let's look at some facts:
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Reflected power after n reflections
   P(n) = P₀⋅Rⁿ    (1)
where R is the reflectivity of the inner walls.
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The  energy stored in the cavity as laser light:
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E = ∑{i = 1 to ∞}P₀⋅ Rⁱ⋅Δt = P₀⋅Δt ⋅∑{i = 1 to ∞} Rⁱ
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∑{i = 1 to ∞} Rⁱ  = R/(1-R) , a converging geometric array
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E = P₀⋅Δt ⋅R/(1-R)   (2)
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Note that this is a constant.
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Calculations with R = 0.99:
-------------------------
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Let us consider that all the laser light is absorbed
when P(i) < 1e-10 W.
 From (1) we find: P(2455) = 9.62e-11
t = 2455⋅Δt = 409 ns
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This means that 409 ns after the laser light enters the cavity,
it will be absorbed by the wall.
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The energy stored as laser light in the cavity will according to (2) be:
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  P₀⋅Δt ⋅R/(1-R) = 8.248e-8 J
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Was it this energy you thought would increase indefinitely?
It won't. It is constant. And tiny.
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Calculations with R = 0.999998:
-------------------------------
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 From (1) we find: P(12350000) = 9.37e-11
t = 12350000⋅Δt = 2.07 ms
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This means that 2.07 ms after the laser light enters the cavity
it will be absorbed by the wall.
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The energy stored as laser light in the cavity will according to (2) be:
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  P₀⋅Δt ⋅R/(1-R) = 4.166e-4 J
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>
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Generally we can say that the laser light will be absorbed
almost immediately after it enters the cavity, and the stored
energy in the form of laser light will be tiny and constant.
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So all the 5 J that enters the cavity every second will heat
the inner wall of the cavity. Since the walls are only 0.1 mm
thick, the outer surface of the cavity will be approximately
the same temperature as the inner surface, so after a short time
the outer wall will radiate 5 W and the system will be in steady state.
>
You could equally well have heated the cavity with a Bunsen burner.
Do you think the mass increase due to the heat energy in the cavity
would be measurable? :-D

Paul, as usual you fucked it up.
CORRECTION 1:
If the cavity is spherical, the diameter will be D = 10 cm (not 5 cm).
NOTE 1: c= 3.0E+10 cm/s
The time light will use to go from one surface to the opposite
surface is Δt = 0.333 ns (NOT 16.33 ns).
NOTE 2: Δt = 0.333 ns IMPLY 3.00E+09 bounces/sec. With LIGO advanced
technology, 1 photon per 5,000,000 photons is lost in every bounce.
This represents a reflectivity R = 0.9999998.
CORRECTION 2 (your fatal error:
The  energy stored in the cavity as laser light:
E = ∑{i = 1 to ∞}P₀⋅ Rⁱ⋅Δt = P₀⋅Δt ⋅∑{i = 1 to ∞} Rⁱ
∑{i = 1 to ∞} Rⁱ  = R/(1-R) , a converging geometric array
E = P₀⋅Δt ⋅R/(1-R)   (2)
Your error here is to use only the energy that exist in the first P₀⋅Δt
interval. This represents ONLY 1.37E+39 photons (550 nm).
You forgot that such quantity is supplied every 0.333 ns, being the
amount of photons/sec 4.12E+48 (3 billions times bigger).
PER SECOND, you have to account the impact of reflectivity on 3 billion
chunks of 1.37E+39 photons EACH, and make the calculations for losses
cumulative with each bounce + new feed of P₀⋅Δt photons.
The above means that you should count losses/sec IN THIS WAY:
Loss 1 = ∑{i = 1 to 3,000,000,000} Rⁱ, for the first pack of photons.
Loss 2 = ∑{i = 1 to 2,999,999,999} Rⁱ, for the 2nd. pack of photons.
Loss 3 = ∑{i = 1 to 2,999,999,998} Rⁱ, for the 3rd. pack of photons.
.....
Loss n = ∑{i = 1 to 3,000,000,001-n} Rⁱ, for the nth. pack of photons.
THEN CALCULATE THE SUM OF LOSSES FOR THE 3 BILLIONS PACKS OF PHOTONS.
You'll understand that such is not an easy task, in particular when
calculating the losses per hour on in 72 hours.
YOUR NUMBERS ARE WAY OFF!! (3,000,000,000 off maybe?).
Now smile, asshole.