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

On Sat, 23 Nov 2024 23:34:40 +0000, rhertz wrote:

EXPLAINING MY APOLOGY:
>
The above calculations are correct in a technical sense, but MY ERROR
was to understand that photon's energies could be accumulated AND STORED
long after the laser was TURNED OFF.
>
This doesn't happen, because the photons stored after T seconds, no
matter how much of them, are ABSORBED by the imperfect inner surface of
the cavity (Reflectivity R < 100.00%).
>
However, it doesn't prevent to measure the weight gain when the laser is
active.
>
Using the best aluminum coating, R = 0.99, which gives a much lower
stored energy than the impractical LIGO-like tech, with R=0.999999998.
>
Using R=0.99, the energy stored after 72 hours is 0.0428 Joules, which
represent m = E/c² of 4.75E-16 grams. A very, very amount of mass.
>
I've learned that, using optomechanical systems (based on micro-quartz
technology) a sensitivity to changes in mass can be measured at scales
as small as 10E-15 grams under ideal conditions.
>
The idea is to measure changes in the mass by measuring the changes in
the frequency of induced vibrations in the "cavity" filled with photons.
>
Of course that these techniques are in the "state of the art" as of
today, but the use of optomechanical resonators is increasing in labs
all over the world, and that its sensitivity has been increased by
10,000 in the last 5 years.
>
The idea is based on several techniques to measure changes, one of them
being using interferometry in the sensors (where the device is placed to
be weighted).
>
But these advanced techniques (there are others) are beyond my interest,
because they are very new and still under the learning curve.
>
That's it.

======================================================================
You are making your calculations FAR more difficult than they need to
be. The important thing to realize is that the temperature of your
chamber does not increase indefinitely, but reaches a temperature at
which the rate at which heat is lost to the environment equals the
power being input to the system.
At this final, steady-state temperature, which I will call T_f as
opposed to the initial temperature T_i, which I presume is also the
temperature of the surrounding environment, the interior cavity of
the chamber will be filled with two types of radiation:
1) Coherent laser light which is reflected and re-reflected within
   the chamber until it is finally absorbed.
2) Black body radiation corresponding to the steady-state temperature
   of the system T_f
The shell of the chamber will have absorbed heat energy corresponding
to the rise in temperature from T_i to T_f.
So the question is, how much does the mass of the internal black
body radiation plus the steady state limit of laser energy within the
chamber plus plus the heat energy of the shell differ between the
initial and final states of the chamber?
I presume that the experiment is being performed in VACUUM. If done
in air, convection currents will mess up any weight measurements, as
well as making computation of the final temperature considerably more
complicated.
I've taken some care to try to avoid the simple arithmetic errors
that I committed in some previous calculations. The emissivity makes
a pretty big difference in the final temperature reached. Starting
from room temperature rather than from 0 C also makes a difference.
However, my record for being able to key things accurately into the
Windows calculator hasn't been very great so far, so I'm prepared for
one of you to point out a silly mistake. :-)
Let T_f = the final, steady-state temperature of the sphere.
T_i = the initial temp of the sphere = temp of environment = 293 K
R = reflectance of the aluminum, assumed to be 0.99
ε = emissivity of aluminum = 0.13
σ = Stefan–Boltzmann constant = 5.67e-8 W⋅m^−2⋅K^−4
a = radiation constant = 4σ/c = 7.57e-16 J m^3 K^-4
P = power input = power output = 5.00 watts
d = density of aluminum = 2.70 g/cm^3 = 2700 kg/m^3
C = specific heat capacity of aluminum = 0.921 J/(g K) = 921 J/kg K)
r = internal radius of sphere = 5.00 cm = 0.0500 m
h = tHickness of the shell = 0.0024 cm = 0.000024 m (heavy duty foil)
V = internal volume of the sphere = 0.0005236 m^3
A = external area of the sphere = 0.03144 m^2
m = mass of the aluminum shell = 2.04 grams = 0.00204 kg
======================================================================
It is implicit from my previous posts that the formula for the steady-
state level of laser energy within the cavity should be
E = -PD / [ c ln(R) ]
where D is the average distance between bounces. I am not sure what
D would be for the presented arrangement. D = 0.07 m might be a
reasonable guess. So at the start and end of the experiment when it
reaches steady state, we have:
E_0 = 0
E_f = -5*(0.07)/[3e8 * ln(0.99)] = 1.16e-7 J
This is quite a bit smaller than our earlier estimates using
unreasonable reflectivities
======================================================================
Continuing on, we calculate the steady-state temperature:
P = ε σ A_e (T_f^4 - T_i^4)
5.00 = 0.13 * 5.67e-8 * 0.03144 * (T_f^4 - 293^4)
2.158e10 = T_f^4 - 7.370e9
T_f^4 = 2.895e10
T_f = 412.5 K
The total black body radiation in the chamber is given by
U = aVT^4
The initial and final black body energies are:
U_i = 2.92e-9 J
U_f = 1.148e-8 J
So that the increase in black body energy from T_i to T_f is
ΔU = U_f - U_i = 8.56e-9 J
======================================================================
The temperature rise from T_i to T_f is 119.5 K
Therefore the increase in thermal energy in the foil is 220 J
======================================================================
Although the laser energy within the shell will reach steady state
almost instantaneously, it will take at least several minutes for
the aluminum foil shell to approach its steady state temperature.
The steady state laser energy and the black body energy within the
sphere are totally insignificant compared with the thermal energy
stored within the heated foil.