Path: ...!weretis.net!feeder9.news.weretis.net!i2pn.org!i2pn2.org!.POSTED!not-for-mail From: tomyee3@gmail.com (ProkaryoticCaspaseHomolog) Newsgroups: sci.physics.relativity Subject: Re: Want to prove =?UTF-8?B?RT1tY8KyPyBVbml2ZXJzaXR5IGxhYnMgc2hvdWxkIHRy?= =?UTF-8?B?eSB0aGlzIQ==?= Date: Sun, 24 Nov 2024 10:04:13 +0000 Organization: novaBBS Message-ID: <98654d26cc4f5fd326f071ea7d4317b8@www.novabbs.com> References: <3cccb55b7c7c451a385b8aad5aac6516@www.novabbs.com> <98a0a1fbdc93a5fcc108882d99718764@www.novabbs.com> <141e19a1c6acd54116739058391ca9f8@www.novabbs.com> <45ed9424edce8c13db24c1dbb8752c26@www.novabbs.com> <7adfc9e5c6884729def0c6a0097c9f37@www.novabbs.com> <092fa494db9895ba52cfac350be5e744@www.novabbs.com> MIME-Version: 1.0 Content-Type: text/plain; charset=utf-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Info: i2pn2.org; logging-data="3898390"; mail-complaints-to="usenet@i2pn2.org"; posting-account="Ooch2ht+q3xfrepY75FKkEEx2SPWDQTvfft66HacveI"; User-Agent: Rocksolid Light X-Spam-Checker-Version: SpamAssassin 4.0.0 X-Rslight-Site: $2y$10$WO68IOzRQLDAaGlg0zTVXexi6cNY/j7cVFzffB7YT4BAxGf5dJUbq X-Rslight-Posting-User: 504a4e36a1e6a0679da537f565a179f60d7acbd8 Bytes: 8047 Lines: 136 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.