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From: Ross Finlayson <ross.a.finlayson@gmail.com>
Date: Thu, 21 Nov 2024 22:30:21 -0800
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On 11/21/2024 09:55 PM, ProkaryoticCaspaseHomolog wrote:
> On Fri, 22 Nov 2024 2:09:46 +0000, rhertz wrote:
>
>> SOME CALCULATIONS WITH PHOTONS
>>
>> Amount of photons/second INTO the cavity
>>
>> 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)
>>
>> This means an average of 3.00E+09 bounces/sec
>>
>> Reflectivity R = 4999999/5000000 = 0.9999998
>>
>> F = photons/slot_0.33ns = 1.37E+39
>> photons LOST/slot_0.33ns =  2.74E+32
>>
>> 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
>>
>> Photons Slot K = F (Rᴷ + Rᴷ⁻¹ + Rᴷ⁻² + Rᴷ⁻³ + .... + R² + R)
>>
>>
>> If R < 1, it is a geometric series with sum S
>>
>>       S = R (1 - Rᴷ)/(1 - R)
>>
>> Being K = 3,000,000,000 and R = 0.9999998
>>
>>       S = 0.9999998/0.0000002 = 4,999,999
>>
>> So, the energy accumulated PER SECOND is
>>
>>       E = F x S x E(1 photon) = 8.33E-03 Joules
>>
>> In one hour, the energy accumulated is
>>
>>      E(1 hr) = 30 Joules
>>
>> In 72 hours,
>>
>>      E(72 Hr) = 2,160 Joules
>>
>> Its equivalent mass is
>>
>>      M(72 hr) = E(72 Hr)/c² = 2.40E-11 grams
>>
>> Weight W(72 hr) = g . M(72 hr) = 0.235 nanoNewtons
>>
>>
>>
>> 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).
>
> But the photons _do not continually accumulate_ within the cavity.
> They get absorbed after a few thousand or million bounces.
>
>> 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.
>
> Paul was doing a back-of-the-envelope estimate. It's to be expected
> that his estimate was a bit off.
>
> 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)
>
> Let E be the accumulated energy within the cavity
>
> E = P_0 ∫ r^(ct/d) dt
> E = P_0 r^(ct/d) / [ c ln(r) / d ] + constant
>
> c/d = 3.0e9
> ln(r) = -2.0e-7
>
> 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
>
> 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.
>
> 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.
>
> 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.
>
> 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?