Capiert Posted March 31, 2018 Posted March 31, 2018 Kelvin 1900 vs M&M & (K)E is m*(c^2)/2 Wed 2018 03 28 02:48 PS Wi 6.9 C https://en.m.wikipedia.org/wiki/Annus_Mirabilis_papers In 1900, Lord Kelvin, in a lecture titled "Nineteenth-Century Clouds over the Dynamical Theory of Heat and Light",[5] suggested that physics had no satisfactory explanations for the results of the Michelson–Morley experiment and for black body radiation. As introduced, special relativity provided an account for the results of the Michelson–Morley experiments. Einstein's theories for the photoelectric effect extended the quantum theory which Max Planck had developed in his successful explanation of black body radiation. --- "It's about half!": (K)E=m*(c^2)(/2) PS: How did Einstein know the energy was m*(c^2); & NOT KE=m*(c^2)(/2). I say he (=Einstein) did NOT know(!): he only guessed or approximated! (But he=Einstein is very decided in using "2" "halves" of (K)E [#L] in the early part of the paper; & pushing the Lorentz contraction thru to make (up) the (non_sense; non_followable, quantum_jump) so_called following "conclusion?" (i.e. without "K"E). If a body gives off the (kinetic) energy (K)E=m*(c^2)/2 [#L] in the form of radiation, its mass should diminish by (K)E*(2)/c2. The fact that the energy withdrawn from the body becomes (light's_speed's) energy of radiation evidently makes no difference, so that we are led to the more general conclusion that The mass of a body is a measure of its (kinetic) energy_content; if the energy changes by (K)E, (then) the mass changes in the same sense by (K)E*(2)/(9 × 1020), the energy being measured in ergs, and the mass in grammes. (PS: Also typical confusion, preferring Einstein's original units; NOT (promoting) standard (modern) SI units, such as Joules & Kg.) See http://www.fourmilab.ch/etexts/einstein/E_mc2/www/ --- Sat 2018 03 31 04:28.. PS Wi 6.2 C I repeat, I doubt Einstein's calculation(s) because he gave the clue to their basis being the KE's mass, but he pushed (=promoted) his Lorentz contraction 1st with=as priority before mentioning the KE relevance. Instead I would prefer to start with the KE relevance to mass m=KE/(v*va) & work backwards (wrt Einstein), to show he's wrong (when) starting without the half in the 1st place, as his fatal error. Einstein's syntax is also a bit reversed (to my preference for larger postscript having more; but seems (a typical) standard). E.g. Initial_energies=postscripts 0 final_energies=postscript 1 Resting coordinate energy E Moving coordinate energy H. [At 1st I (wrongly) thought (his energy) L could have represented wavelength (or a wave number) but it's NOT true.] Somewhere I've done similar KE comparisons to show KE is (already) relativistic, (e.g. (extrap(olate)able) using simple fractions of c=1, or physicists beta=v/c e.g. let v=0, 1/2, 1/4 (of c). The (non_linear) relativism becomes obvious with the 1/4 vs 3/4 ratios for initial_speed vi & final_speed vf, when v=vf-vi in the form KE=m*((vf^2)-(vi^2))/2. Let vi=0, & vf=1/2 (of c) KE=m*(((1/2)^2)-(0^2))/2 KE=m*((1/4)-(0^2))/2 KE=m/8. Let vi=1/2, & vf=1 (of c) KE=m*((1^2)-((1/2)^2))/2 KE=m*(1-(1/4))/2 KE=m*(3/4))/2 KE=m*3/8. Please notice, both together add to m*4/8 =m/2 i.e. "half" which is the maximum for those cases. That was only a split into 2 producing 1/8th & 3/8th of m. If we split into 4, vf=1/4, 2/4, 3/4, 4/4, then the finals are 1/32, 4/32 9/32 16/32 where (the KE's are (related so:)) m*1/32 (for) vi=0, vf=1/4 m*3/32 vi=1/4, vf=2/4 m*5/32 vi=2/4, vf=3/4 m*7/32 vi=3/4, vf=4/4. Please notice those differ by 1 (wrt 0), 2, 2, 2, 1 of 32 (=(c^2)/2). Important is their sum is m*16/32 =m/2 = half of m, again. (E.g. KE=m*(c^2)/2.) So that is how CO(k)E (conservation of (kinetic) energy) is (non_linearly) guaranteed. I would then have to question what Einstein is doing with [his] coe, because he seems to be in error (when missing the half). I mean, that is how (kinetic) energy is calibrated i.e. related. Getting more than half (m) is abracadabra, i.e. energy creation. =Wishful thinking. Maybe somebody can help point me in the right direction, because I can't confirm him. ?
Bender Posted March 31, 2018 Posted March 31, 2018 On 3/31/2018 at 4:13 AM, Capiert said: Maybe somebody can help point me in the right direction, because I can't confirm him. ? Expand Go to the library and pick up a physics text book. There is no half in the equation because when you derive it properly, it is not there. The E in the equation also does not represent kinetic energy. Your calculations are nothing but numerology and thus meaningless and irrelevant to this topic.
Sensei Posted March 31, 2018 Posted March 31, 2018 (edited) On 3/31/2018 at 4:13 AM, Capiert said: Maybe somebody can help point me in the right direction, because I can't confirm him. ? Expand Suppose so you have Uranium-235 atom. It has mass mu It's decaying to Thorium-231 and Helium-4, with rest-masses mth and mhe. U-235 -> Th-231 + He-4 + 4.68 MeV And it releases approximately 4.68 MeV energy (it's called decay energy). In this case (alpha decay) it (decay energy) is in kinetic energy of newly created particles. (small) part of that energy is taken by Thorium-231, and (larger) part of that energy is taken by highly accelerated Helium-4 nucleus. muc^2 = mthc^2 + mhec^2 + 4.68 MeV Reverse equation: muc^2 - ( mthc^2 + mhec^2 ) = 4.68 MeV Rest-mass of Uranium has been converted to kinetic energies of newly created particles. Rest-mass of Uranium, rest-mass of Thorium, and rest-mass of Helium can be measured independently many different ways. Decay energy can be calculated using above equation. Kinetic energy of newly created particles can be measured (they're (traces) visible by naked eye in e.g. Cloud Chamber). Kinetic energy in Special Relativity is KE = m_0c^2\gamma - m_0c^2 not what you wrote in title! ps. Multiplication by c^2 just coverts units from unit used to express mass to unit used to express energy. ps2. Why am I giving over and over again the same example of radioactive decay process? Because it's easy to see how rest-mass is converted to energy, without bothering much about original Einstein's work. Edited March 31, 2018 by Sensei
Strange Posted March 31, 2018 Posted March 31, 2018 On 3/31/2018 at 4:13 AM, Capiert said: Maybe somebody can help point me in the right direction, because I can't confirm him. ? Expand Learn some basic physics. Learn to write properly.
Capiert Posted March 31, 2018 Author Posted March 31, 2018 On 3/31/2018 at 5:17 AM, Bender said: Go to the library and pick up a physics text book. There is no half in the equation because when you derive it properly, it is not there. The E in the equation also does not represent kinetic energy. Your calculations are nothing but numerology and thus meaningless and irrelevant to this topic. Expand I'm sorry Bender, but I don't think you got my point. (The rest mass energy is in no way correctly calibrated to KE, thus can NOT be energy. Einstein was discussing vis Viva=2*KE; NOT KE!). I know there is no half in Einstein's equation, everybody knows that. But I doubr Einstein got it right because he broke continuity. He broke the rules. (What is for you properly? Starting with an assumption?) Sensei, KE can still calculate the mass, but it's a different value than Einstein's; & perhaps has the guarantee that it's correct because it does NOT break the rules.? Strange, KE is about as basic to physics, as it gets. You guys try to escape (=avoid) that, (basic physics). That's what bluffs me. You prefer 1 guys assumptions, inspite of the hints against it.
Strange Posted March 31, 2018 Posted March 31, 2018 On 3/31/2018 at 10:07 AM, Capiert said: Strange, KE is about as basic to physics, as it gets. Expand All the more reason you should learn about it instead of posting ignorant crap. And, if I were a moderator, I would ban you for failing to write like an adult!
Capiert Posted March 31, 2018 Author Posted March 31, 2018 Strange, maybe I haven't found the right question to make it click for you? What should I be asking you about KE to get you to recognize my perspective?
Sensei Posted March 31, 2018 Posted March 31, 2018 (edited) On 3/31/2018 at 10:07 AM, Capiert said: Sensei, KE can still calculate the mass, but it's a different value than Einstein's; & perhaps has the guarantee that it's correct because it does NOT break the rules.? Expand Difference between Classical Physics kinetic energy KE=\frac{1}{2} * m * v^2 and Special Relativity KE = m_0c^2\gamma - m_0c^2 is only visible at significant velocities (at the moment unavailable for humans, except ions in particle accelerators), while approaching to speed of light in vacuum. Einstein with his 1st wife Mileva, work on Photoelectric effect, they used classical physics kinetic energy equation, not special relativity kinetic energy equation. ps. Probably nobody reading your posts, have any idea, what are you talking about generally.. ps2. You should never replace v by c in classical physics kinetic energy equation! Edited March 31, 2018 by Sensei
Capiert Posted March 31, 2018 Author Posted March 31, 2018 Sensei, the photon energy formula also looks wrong. It's missing the "half"! I.e. The same failure, but in reverse! It really looks like this guy Einstein mixed things up! He makes 1 severe error, & then compesates for it elsewhere. The momentum is mom=h*nu (wavenumber nu; NOT frequency f!) for (Einstein's) Planck's constant h. Energy=mom*va. E=h*f*c/2, where the average_speed va=(vi+vf)/2 is between initial_speed vi=0 m/s & the final_speed vf=c, light's_speed. c=f*lambda for wavelength lambda. -3
Strange Posted March 31, 2018 Posted March 31, 2018 On 3/31/2018 at 10:18 AM, Capiert said: What should I be asking you about KE to get you to recognize my perspective? Expand You should explain what it is you don't understand. You should write your questions in proper English, like an adult, not like a pretentious teenager. On 3/31/2018 at 10:39 AM, Capiert said: It really looks like this guy Einstein mixed things up! Expand The problem is in your failure to understand. Not in well-tested physics.
Sensei Posted March 31, 2018 Posted March 31, 2018 (edited) On 3/31/2018 at 10:39 AM, Capiert said: The momentum is mom=h*nu (wavenumber nu; NOT frequency f!) for (Einstein's) Planck's constant h. Energy=mom*va. E=h*f*c/2, where the average_speed va=(vi+vf)/2 is between initial_speed vi=0 m/s & the final_speed vf=c, light's_speed. c=f*lambda for wavelength lambda. Expand You permanently ignore conversion of units.. and that's just for a start... Edited March 31, 2018 by Sensei
Capiert Posted March 31, 2018 Author Posted March 31, 2018 (edited) Yes Sensei, I wanted to say that formula is still wrong. E=h*nu*c/2 & still won't give you the frequency f, because nu is a photo film distance; NOT the (real) inverse wavelength. lambda=((2*d*nu)^0.5)/m wrt Wiki's diffraction grating syntax slit to slit width d, order m, e.g. 1. Edited March 31, 2018 by Capiert -2
Strange Posted March 31, 2018 Posted March 31, 2018 On 3/31/2018 at 11:00 AM, Capiert said: because nu is a photo film distance; NOT the (real) inverse wavelength Expand Well, if you use the wrong number in the equation then you will get the wrong result. nu (ν) is frequency. lambda (λ) is wavelength. None of these equations use "photo film distance" (whatever that means). Why would the position of a photographic film change the energy of a photon?
Bender Posted March 31, 2018 Posted March 31, 2018 On 3/31/2018 at 10:39 AM, Capiert said: the photon energy formula also looks wrong. Expand I think this is the essence of the problem. Whether something "looks" wrong is irrelevant. What matters is whether it can be used to make predictions which are then verified experimentally. Einstein's theories excell at that. If you make any predictions with your versions, your results will be too small with a factor of about 2. What also matters is communication, such as not Starting (a paragraph) In the middle of a sentence [collection of words]
Sensei Posted March 31, 2018 Posted March 31, 2018 (edited) On 3/31/2018 at 11:00 AM, Capiert said: Yes Sensei, I wanted to say that formula is still wrong. Expand You're "victim of not proper education".. You claim that photon energy formula y E=h v=\omega \hbar is incorrect.. But it used in many places in quantum physics. Starting from photoelectric effect: h f - W = \frac{1}{2} m_e v^2 Photon is disappearing from system, instead free electron is ejected from metal. And its kinetic energy (can be calculated from velocity) is reminder of energy needed to eject it. v = \sqrt{ \frac{ 2 ( h f - W) }{ m_e} } C'mon! This can be proven simply using white light source, prism, and metal such as Cesium, Potassium, Sodium, Lithium.. Then annihilation, then pair production, etc. etc. Edited March 31, 2018 by Sensei
Capiert Posted March 31, 2018 Author Posted March 31, 2018 (edited) On 3/31/2018 at 11:15 AM, Strange said: Well, if you use the wrong number in the equation then you will get the wrong result. nu (ν) is frequency. Expand As you can see above, I think Sensei (also) complained enough about those units NOT being correct. Nu is a wave number, NOT frequency. Quote lambda (λ) is wavelength. None of these equations use "photo film distance" (whatever that means). Expand The (diffracted) distance where the spectral lines are found on film, versus (=wrt) straight thru without diffraction (grating, or prism). Quote Why would the position of a photographic film change the energy of a photon? Expand [Reflected photons can loose some of their energy as a bounce, but your question is wrongly put for what you want to know.] The film does NOT change the photon energy per se, (in what we're interested here). The amount, that the light is bent or deflected (i.e. diffracted) indicates the wavelength. We know c=f*L(ambda); & the energy is proportional to the frequency. Assuming c is constant. So f=c/Lambda; energy is inversely proportional to the deflected distance on the photo film. But it is NOT correct to assume that (either inverse or not inverse) distance nu (which I simply call x (=deflected distance) wrt y_distance (diffraction_grating to film, perpendicular_distance)) is the wavelength. Nu is NOT the wavelength; & it (=nu) is NOT the inverse_wavelength either (which physicists call wavenumber). https://en.m.wikipedia.org/wiki/Wavenumber In spectroscopy, "wavenumber" (=nu_bar) often refers to a frequency which has been divided by the speed of light in vacuum: The historical reason for using this spectroscopic wavenumber rather than frequency is that it proved to be convenient in the measurement of atomic spectra: the spectroscopic wavenumber is the reciprocal of the wavelength of light in vacuum: which remains essentially the same in air, and so the spectroscopic wavenumber is directly related to the angles of light scattered from diffraction gratings and the distance between fringes in interferometers, when those instruments are operated in air or vacuum. Such wavenumbers were first used in the calculations of Johannes Rydberg in the 1880s. The Rydberg–Ritz combination principle of 1908 was also formulated in terms of wavenumbers. A few years later spectral lines could be understood in quantum theory as differences between energy levels, energy being proportional to wavenumber, or frequency. However, spectroscopic data kept being tabulated in terms of spectroscopic wavenumber rather than frequency or energy. Edited March 31, 2018 by Capiert -1
Strange Posted March 31, 2018 Posted March 31, 2018 On 3/31/2018 at 12:23 PM, Capiert said: Nu is a wave number, NOT frequency. Expand Why not (a) learn to write properly and stop behaving like a spoiled teenager and (b) LEARN some basic physics: Quote The energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ): Expand https://en.wikipedia.org/wiki/Photon#Physical_properties On 3/31/2018 at 12:23 PM, Capiert said: In spectroscopy, "wavenumber" (=nu_bar) often refers to a frequency which has been divided by the speed of light in vacuum Expand 1) We are not talking about spectroscopy 2) That is nu-bar, not nu 3) Learn some basic physics 4) Stop writing like a child
Capiert Posted March 31, 2018 Author Posted March 31, 2018 The frequency f=c/lambda (should be) f=c*m/((2*d*x)^0.5). As you can see that's NOT f=c/nu if nu is x (=photo film displacement). You might not be discussing spectroscopy but I am; & a (search for a) consistent (kinetic) energy; NOT a mixture of energy & vis viva. -2
hypervalent_iodine Posted March 31, 2018 Posted March 31, 2018 ! Moderator Note We’re done here. Thread closed.
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