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

On 11/21/2024 09:55 PM, ProkaryoticCaspaseHomolog wrote:

On Fri, 22 Nov 2024 2:09:46 +0000, rhertz wrote:
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SOME CALCULATIONS WITH PHOTONS
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Amount of photons/second INTO the cavity
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E(1 photon) =  1.21E-48    Joules
5 Joules/sec  =  4.12E+48  photons/sec
Time between hits =  3.333E-10    seconds (10 cm distance)
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This means an average of 3.00E+09 bounces/sec
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Reflectivity R = 4999999/5000000 = 0.9999998
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F = photons/slot_0.33ns = 1.37E+39
photons LOST/slot_0.33ns =  2.74E+32
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Photons Slot 1 = F R
Photons Slot 2 = F R² + F R
Photons Slot 3 = F R³ + F R² + F R
Photons Slot 4 = F R⁴ + F R³ + F R² + F R
Photons Slot 5 = F R⁵ + F R⁴ + F R³ + F R² + F R
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Photons Slot K = F (Rᴷ + Rᴷ⁻¹ + Rᴷ⁻² + Rᴷ⁻³ + .... + R² + R)
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If R < 1, it is a geometric series with sum S
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      S = R (1 - Rᴷ)/(1 - R)
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Being K = 3,000,000,000 and R = 0.9999998
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      S = 0.9999998/0.0000002 = 4,999,999
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So, the energy accumulated PER SECOND is
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      E = F x S x E(1 photon) = 8.33E-03 Joules
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In one hour, the energy accumulated is
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     E(1 hr) = 30 Joules
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In 72 hours,
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     E(72 Hr) = 2,160 Joules
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Its equivalent mass is
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     M(72 hr) = E(72 Hr)/c² = 2.40E-11 grams
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Weight W(72 hr) = g . M(72 hr) = 0.235 nanoNewtons
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This value is far from my calculations on the OP of this thread by a
factor of about 1,000, but still feasible of being measured (maybe
letting the experiment run for months).

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But the photons _do not continually accumulate_ within the cavity.
They get absorbed after a few thousand or million bounces.
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At any case, very far from the 4.166E-4 Joules calculated by Paul, or
the mean life of 2.07 ms before the "5 Watts" are DISSIPATED.

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Paul was doing a back-of-the-envelope estimate. It's to be expected
that his estimate was a bit off.
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Let's use your figures of
P_0 = 5.0 watt input,
c = speed of light
d = 0.10 m distance between bounces,
r = 0.9999998 reflectivity (much higher than anything Paul assumed)
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Let E be the accumulated energy within the cavity
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E = P_0 ∫ r^(ct/d) dt
E = P_0 r^(ct/d) / [ c ln(r) / d ] + constant
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c/d = 3.0e9
ln(r) = -2.0e-7
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Calculate the accumulated energy for the interval 0 to 1 second
E(0,1) = 5.0/[3.0e9 * 2.0e-7]
E(0,1) = 0.0083 Joules
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If we use Paul's figure for reflectivity, I compute
E(0,1) = 0.00083 Joules, only a factor of two higher than Paul's
back-of-the-envelope estimate.
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Letting the system run longer doesn't buy you any extra energy
accumulated within the cavity, but the shell does keep getting
warmer up to a point, accumulating mass-energy until the temperature
reaches a steady-state.
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Now, I've been writing about incredible power densities within the
shell. But it's the _same damn photons_ bouncing back and forth many
thousands of times. We aren't really getting that much accumulation
of mass-energy.
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And remember, in a realistic experiment, reflectivity is going to be
much closer to 0.9 than it will be to 0.9999998

How about put a Weber bar next to a cyclotron and turn it on and off?
Have you tried turning it on and off?
How about you put a linac through a cyclotron,
and variously turn them on and off?
How about one of those hand helicopters?
Or just one of those tree seeds that wafts its way down?