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TBH, I'm indifferent to the issue, it doesn't even crack my top 20 of thing's to give a shit about; why would I waste my emotional energy, spending my daily chip allowance needs deeper problem's to warrant a reasonable return on my investments.

Who thinks he's right and why.

Which is not the photoelectric effect. In an LED you excite electrons to a higher band in a semiconductor, and when they drop back down you get a photon.

I'll start a new topic, you can can ask me there.

Hard to know where to start with this gibberish. Almost everything you say is wrong, almost as if intentionally so. Just to take one point, there is no photoionisation in the retina of the eye. Photons are absorbed by proteins called opsins, which thereby enter an excited state and change from one isomer to another (cis ->trans). The isomerised version then undergoes a chemical reaction with other molecules to start a cascade of biochemistry, resulting in a nerve signal. This is not photoionisation.

The explanation is exactly what the maths says - pick a different path, and you’ll walk a different distance. There’s nothing else to it. No, I showed you that there’s no paradox that needs resolving. It seems to me that you’re wilfully refusing to “get” this. There’s no such thing as “relative to waves”, because light has no rest frame. The speed is always between emitter and receiver. Because it’s he who experiences acceleration locally in his frame. The equivalence principle tells us that uniform acceleration is locally equivalent to a uniform gravitational field; differently put, the accelerated twin sits at a different gravitational potential, which implies frequency shift.

Really? From his article that introduced the theory that later would be named 'Special Relativity': On the electrodynamics of moving bodies I think it is time to close the thread?

yes I did understand that but I'm trying to ascertain your eventual goals with this to provide direction for improvement. If you think about we do much the same with the use of the scale factor under the FLRW However a key point is that G is a constant under the FLRW so your going to have to explain why you feel G would change as a result of change in radius ?

You understand that this math is designed to be factored into change the gravitational variance along a scale that must be adjusted for f(n) each time (n)sin(any deg) changes the radius of a circle along the z plane. With both the degree angle changing and the radius changing so there can be no constant which is why it is variant. A forum I feel is as good a place as any. Accredited or not, if it works to predict the orbital behavior of planetary objects in relation to one another it works. However, it is getting late.

An LED is basically the reverse of photovoltaics, i.e. electroluminescence. As for it's own magnetic moment, whatever the culprit is that doesn't change the fact that magnetization is the point where the two opposing orbitals cross paths as those sides of the atom are adjacent. When I said spin I meant orbital magnetization. As in the port where the fiber optic cable connects to battery a, photoionization occurs and the conductor connecting battery a to battery b adopts one atomic orbital which attracts one end of the cylinder which has a magnet, then the orbitals reverse throughout the conductor as the positive ionized battery a takes back the additional electrons from the negatively ionized batter b, attracting the other end of the cylinder which has an anti-magnet. Your retina are like little photovoltaic batteries, their cells can experience photoinization which is not unlike the process of photosynthesis or why ionized gas in hydrothermal vents are the culprit for the creation of chemical bonds leading to the carbon based dna in our cells. If you have LEDs that are small enough, I'd wage the easiest was to program them to light up in just the right combination to produce a picture through our retina I'd wager the easiest way to program a touch screen interface is what I described here but I can use math to figure out exactly how that interface needs to be set up.

You are right about the brute force. I don’t know how useful it is when the smaller factor is occurring close to zero anyway. But what is given me a glimmer of hope is the first graph of my last post cross the x axis at around 41227. That’s where it appears to cross but computer generated graphics can be deceiving. If anyone could show where y on the graph equal zero that is the x intercept of that fist graph of the last post, it would be well appreciated.

SU(2) \[{\small\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline Field & \ell_L& \ell_R &v_L&U_L&d_L&U_R &D_R&\phi^+&\phi^0\\\hline T_3&- \frac{1}{2}&0&\frac{1}{2}&\frac{1}{2}&-\frac{1}{2}&0&0&\frac{1}{2}&-\frac{1}{2} \\\hline Y&-\frac{1}{2}&-1&-\frac{1}{2}&\frac{1}{6}&\frac{1}{6}& \frac{2}{3}&-\frac{1}{3}&\frac{1}{2}&\frac{1}{2}\\\hline Q&-1&-1&0&\frac{2}{3}&-\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}&1&0\\\hline\end{array}}\] \(\psi_L\) doublet \[D_\mu\psi_L=[\partial_\mu-i\frac{g}{\sqrt{2}}(\tau^+W_\mu^+\tau^-W_\mu^-)-i\frac{g}{2}\tau^3W^3_\mu+i\acute{g}YB_\mu]\psi_L=\]\[\partial_\mu-i\frac{g}{\sqrt{2}}(\tau^+W_\mu^-)+ieQA_\mu-i\frac{g}{cos\theta_W}(\frac{t_3}{2}-Qsin^2\theta_W)Z_\mu]\psi_L\] \(\psi_R\) singlet \[D_\mu\psi_R=[\partial\mu+i\acute{g}YB_\mu]\psi_R=\partial_\mu+ieQA_\mu+i\frac{g}{cos\theta_W}Qsin^2\theta_WZ_\mu]\psi_W\] with \[\tau\pm=i\frac{\tau_1\pm\tau_2}{2}\] and charge operator defined as \[Q=\begin{pmatrix}\frac{1}{2}+Y&0\\0&-\frac{1}{2}+Y\end{pmatrix}\] \[e=g.sin\theta_W=g.cos\theta_W\] \[W_\mu\pm=\frac{W^1_\mu\pm iW_\mu^2}{\sqrt{2}}\] \[V_{ckm}=V^\dagger_{\mu L} V_{dL}\] The gauge group of electroweak interactions is \[SU(2)_L\otimes U(1)_Y\] where left handed quarks are in doublets of \[ SU(2)_L\] while right handed quarks are in singlets the electroweak interaction is given by the Langrangian \[\mathcal{L}=-\frac{1}{4}W^a_{\mu\nu}W^{\mu\nu}_a-\frac{1}{4}B_{\mu\nu}B^{\mu\nu}+\overline{\Psi}i\gamma_\mu D^\mu \Psi\] where \[W^{1,2,3},B_\mu\] are the four spin 1 boson fields associated to the generators of the gauge transformation \[\Psi\] The 3 generators of the \[SU(2)_L\] transformation are the three isospin operator components \[t^a=\frac{1}{2} \tau^a \] with \[\tau^a \] being the Pauli matrix and the generator of \[U(1)_\gamma\] being the weak hypercharge operator. The weak isospin "I" and hyper charge \[\gamma\] are related to the electric charge Q and given as \[Q+I^3+\frac{\gamma}{2}\] with quarks and lepton fields organized in left-handed doublets and right-handed singlets: the covariant derivative is given as \[D^\mu=\partial_\mu+igW_\mu\frac{\tau}{2}-\frac{i\acute{g}}{2}B_\mu\] \[\begin{pmatrix}V_\ell\\\ell\end{pmatrix}_L,\ell_R,\begin{pmatrix}u\\d\end{pmatrix}_,u_R,d_R\] The mass eugenstates given by the Weinberg angles are \[W\pm_\mu=\sqrt{\frac{1}{2}}(W^1_\mu\mp i W_\mu^2)\] with the photon and Z boson given as \[A_\mu=B\mu cos\theta_W+W^3_\mu sin\theta_W\] \[Z_\mu=B\mu sin\theta_W+W^3_\mu cos\theta_W\] the mass mixings are given by the CKM matrix below \[\begin{pmatrix}\acute{d}\\\acute{s}\\\acute{b}\end{pmatrix}\begin{pmatrix}V_{ud}&V_{us}&V_{ub}\\V_{cd}&V_{cs}&V_{cb}\\V_{td}&V_{ts}&V_{tb}\end{pmatrix}\begin{pmatrix}d\\s\\b\end{pmatrix}\] mass euqenstates given by \(A_\mu\) an \(Z_\mu\) \[W^3_\mu=Z_\mu cos\theta_W+A_\mu sin\theta_W\] \[B_\mu= Z_\mu sin\theta_W+A_\mu cos\theta_W\] \[Z_\mu=W^3_\mu cos\theta_W+B_\mu sin\theta_W\] \[A_\mu=-W^3_\mu\sin\theta_W+B_\mu cos\theta_W\] ghost field given by \[\acute{\psi}=e^{iY\alpha_Y}\psi\] \[\acute{B}_\mu=B_\mu-\frac{1}{\acute{g}}\partial_\mu\alpha Y\] [latex]D_\mu[/latex] minimally coupled gauge covariant derivative. h Higg's bosonic field [latex] \chi[/latex] is the Goldstone boson (not shown above) Goldstone no longer applies after spontaneous symmetry breaking [latex]\overline{\psi}[/latex] is the adjoint spinor [latex]\mathcal{L}_h=|D\mu|^2-\lambda(|h|^2-\frac{v^2}{2})^2[/latex] [latex]D_\mu=\partial_\mu-ie A_\mu[/latex] where [latex] A_\mu[/latex] is the electromagnetic four potential QCD gauge covariant derivative [latex] D_\mu=\partial_\mu \pm ig_s t_a \mathcal{A}^a_\mu[/latex] matrix A represents each scalar gluon field Single Dirac Field [latex]\mathcal{L}=\overline{\psi}I\gamma^\mu\partial_\mu-m)\psi[/latex] under U(1) EM fermion field equates to [latex]\psi\rightarrow\acute{\psi}=e^{I\alpha(x)Q}\psi[/latex] due to invariance requirement of the Langrene above and with the last equation leads to the gauge field [latex]A_\mu[/latex] [latex] \partial_\mu[/latex] is replaced by the covariant derivitave [latex]\partial_\mu\rightarrow D_\mu=\partial_\mu+ieQA_\mu[/latex] where [latex]A_\mu[/latex] transforms as [latex]A_\mu+\frac{1}{e}\partial_\mu\alpha[/latex] Single Gauge field U(1) [latex]\mathcal{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}[/latex] [latex]F_{\mu\nu}=\partial_\nu A_\mu-\partial_\mu A_\nu[/latex] add mass which violates local gauge invariance above [latex]\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^2A_\mu A^\mu[/latex] guage invariance demands photon be massless to repair gauge invariance add a single complex scalar field [latex]\phi=\frac{1}{\sqrt{2}}(\phi_1+i\phi_2[/latex] Langrene becomes [latex] \mathcal{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+|D_\mu \phi|^2-V_\phi[/latex] where [latex]D_\mu=\partial_\mu-ieA_\mu[/latex] [latex]V_\phi=\mu^2|\phi^2|+\lambda(|\phi^2|)^2[/latex] [latex]\overline{\psi}=\psi^\dagger \gamma^0[/latex] where [latex]\psi^\dagger[/latex] is the hermitean adjoint and [latex]\gamma^0 [/latex] is the timelike gamma matrix the four contravariant matrix are as follows [latex]\gamma^0=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}[/latex] [latex]\gamma^1=\begin{pmatrix}0&0&0&1\\0&0&1&0\\0&0&-1&0\\-1&0&0&0\end{pmatrix}[/latex] [latex]\gamma^2=\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&i&0&0\\-i&0&0&0\end{pmatrix}[/latex] [latex]\gamma^3=\begin{pmatrix}0&0&1&0\\0&0&0&-1\\-1&0&0&0\\0&1&0&0\end{pmatrix}[/latex] where [latex] \gamma^0[/latex] is timelike rest are spacelike V denotes the CKM matrix usage [latex]\begin{pmatrix}\acute{d}\\\acute{s}\\\acute{b}\end{pmatrix}\begin{pmatrix}V_{ud}&V_{us}&V_{ub}\\V_{cd}&V_{cs}&V_{cb}\\V_{td}&V_{ts}&V_{tb}\end{pmatrix}\begin{pmatrix}d\\s\\b\end{pmatrix}[/latex] [latex]V_{ckm}=V^\dagger_{\mu L} V_{dL}[/latex] the CKM mixing angles correlates the cross section between the mass eigenstates and the weak interaction eigenstates. Involves CP violations and chirality relations. Dirac 4 component spinor fields [latex]\gamma^5=i\gamma_0,\gamma_1,\gamma_2,\gamma_3[/latex] 4 component Minkowskii with above 4 component Dirac Spinor and 4 component Dirac gamma matrixes are defined as [latex] {\gamma^\mu\gamma^\nu}=2g^{\mu\nu}\mathbb{I}[/latex] where [latex]\mathbb{I}[/latex] is the identity matrix. (required under MSSM electroweak symmetry break} in Chiral basis [latex]\gamma^5[/latex] is diagonal in [latex]2\otimes 2[/latex] the gamma matrixes are [latex]\begin{pmatrix}0&\sigma^\mu_{\alpha\beta}\\\overline{\sigma^{\mu\dot{\alpha}\beta}}&0\end{pmatrix}[/latex] [latex]\gamma^5=i{\gamma_0,\gamma_1,\gamma_2,\gamma_3}=\begin{pmatrix}-\delta_\alpha^\beta&0\\0&\delta^\dot{\alpha}_\dot{\beta}\end{pmatrix}[/latex] [latex]\mathbb{I}=\begin{pmatrix}\delta_\alpha^\beta&0\\0&\delta^\dot{\alpha}_\dot{\beta}\end{pmatrix}[/latex] Lorentz group identifiers in [latex](\frac{1}{2},0)\otimes(0,\frac{1}{2})[/latex] [latex]\sigma\frac{I}{4}=(\gamma^\mu\gamma^\nu)=\begin{pmatrix}\sigma^{\mu\nu\beta}_{\alpha}&0\\0&-\sigma^{\mu\nu\dot{\alpha}}_{\dot{\beta}}\end{pmatrix}[/latex] [latex]\sigma^{\mu\nu}[/latex] duality satisfies [latex]\gamma_5\sigma^{\mu\nu}=\frac{1}{2}I\epsilon^{\mu\nu\rho\tau}\sigma_{\rho\tau}[/latex] a 4 component Spinor Dirac field is made up of two mass degenerate Dirac spinor fields U(1) helicity [latex](\chi_\alpha(x)),(\eta_\beta(x))[/latex] [latex]\psi(x)=\begin{pmatrix}\chi^{\alpha\beta}(x)\\ \eta^{\dagger \dot{\alpha}}(x)\end{pmatrix}[/latex] the [latex](\alpha\beta)=(\frac{1}{2},0)[/latex] while the [latex](\dot{\alpha}\dot{\beta})=(0,\frac{1}{2})[/latex] this section relates the SO(4) double cover of the SU(2) gauge requiring the chiral projection operator next. chiral projections operator [latex]P_L=\frac{1}{2}(\mathbb{I}-\gamma_5=\begin{pmatrix}\delta_\alpha^\beta&0\\0&0\end{pmatrix}[/latex] [latex]P_R=\frac{1}{2}(\mathbb{I}+\gamma_5=\begin{pmatrix}0&0\\ 0&\delta^\dot{\alpha}_\dot{\beta}\end{pmatrix}[/latex] Weyl spinors [latex]\psi_L(x)=P_L\psi(x)=\begin{pmatrix}\chi_\alpha(x)\\0\end{pmatrix}[/latex] [latex]\psi_R(x)=P_R\psi(x)=\begin{pmatrix}0\\ \eta^{\dagger\dot{a}}(x)\end{pmatrix}[/latex] also requires Yukawa couplings...SU(2) matrixes given by [latex]diag(Y_{u1},Y_{u2},Y_{u3})=diag(Y_u,Y_c,Y_t)=diag(L^t_u,\mathbb{Y}_u,R_u)[/latex] [latex]diag(Y_{d1},Y_{d2},Y_{d3})=diag(Y_d,Y_s,Y_b)=diag(L^t_d,\mathbb{Y}_d,R_d[/latex] [latex]diag(Y_{\ell 1},Y_{\ell 2},Y_{\ell3})=diag(Y_e,Y_\mu,Y_\tau)=diag(L^T_\ell,\mathbb{Y}_\ell,R_\ell)[/latex] the fermion masses [latex]Y_{ui}=m_{ui}/V_u[/latex] [latex]Y_{di}=m_{di}/V_d[/latex] [latex]Y_{\ell i}=m_{\ell i}/V_\ell[/latex] Reminder notes: Dirac is massive 1/2 fermions, Weyl the massless. Majorona fermion has its own antiparticle pair while Dirac and Weyl do not. The RH neutrino would be more massive than the LH neutrino, same for the corresponding LH antineutrino and RH Neutrino via seesaw mechanism which is used with the seesaw mechanism under MSM. Under MSSM with different Higgs/higglets can be numerous seesaws. The Majorona method has conservation violations also these fermions must be electric charge neutral. (must be antiparticles of themselves) the CKM and PMNS are different mixing angels in distinction from on another. However they operate much the same way. CKM is more commonly used as its better tested to higher precision levels atm. Quark family is Dirac fermions due to electric charge cannot be its own antiparticle. Same applies to the charged lepton family. Neutrinos are members of the charge neutral lepton family Lorentz group Lorentz transformations list spherical coordinates (rotation along the z axis through an angle ) \[\theta\] \[(x^0,x^1,x^2,x^3)=(ct,r,\theta\phi)\] \[(x_0,x_1,x_2,x_3)=(-ct,r,r^2,\theta,[r^2\sin^2\theta]\phi)\] \[\acute{x}=x\cos\theta+y\sin\theta,,,\acute{y}=-x\sin\theta+y \cos\theta\] \[\Lambda^\mu_\nu=\begin{pmatrix}1&0&0&0\\0&\cos\theta&\sin\theta&0\\0&\sin\theta&\cos\theta&0\\0&0&0&1\end{pmatrix}\] generator along z axis \[k_z=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}\] generator of boost along x axis:: \[k_x=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}=-i\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}\] boost along y axis\ \[k_y=-i\begin{pmatrix}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0 \end{pmatrix}\] generator of boost along z direction \[k_z=-i\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0 \end{pmatrix}\] the above is the generator of boosts below is the generator of rotations. \[J_z=\frac{1\partial\Lambda}{i\partial\theta}|_{\theta=0}\] \[J_x=-i\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&1\\0&0&-1&0 \end{pmatrix}\] \[J_y=-i\begin{pmatrix}0&0&0&0\\0&0&0&-1\\0&0&1&0\\0&0&0&0 \end{pmatrix}\] \[J_z=-i\begin{pmatrix}0&0&0&0\\0&0&1&0\\0&-1&0&0\\0&0&0&0 \end{pmatrix}\] there is the boosts and rotations we will need and they obey commutations \[[A,B]=AB-BA\] SO(3) Rotations list set x,y,z rotation as \[\varphi,\Phi\phi\] \[R_x(\varphi)=\begin{pmatrix}1&0&0\\0&\cos\varphi&\sin\varphi\\o&-sin\varphi&cos\varphi \end{pmatrix}\] \[R_y(\phi)=\begin{pmatrix}cos\Phi&0&\sin\Phi\\0&1&0\\-sin\Phi&0&cos\Phi\end{pmatrix}\] \[R_z(\phi)=\begin{pmatrix}cos\theta&sin\theta&0\\-sin\theta&\cos\theta&o\\o&0&1 \end{pmatrix}\] Generators for each non commutative group. \[J_x=-i\frac{dR_x}{d\varphi}|_{\varphi=0}=\begin{pmatrix}0&0&0\\0&0&-i\\o&i&0\end{pmatrix}\] \[J_y=-i\frac{dR_y}{d\Phi}|_{\Phi=0}=\begin{pmatrix}0&0&-i\\0&0&0\\i&i&0\end{pmatrix}\] \[J_z=-i\frac{dR_z}{d\phi}|_{\phi=0}=\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}\] with angular momentum operator \[{J_i,J_J}=i\epsilon_{ijk}J_k\] with Levi-Civita \[\varepsilon_{123}=\varepsilon_{312}=\varepsilon_{231}=+1\] \[\varepsilon_{123}=\varepsilon_{321}=\varepsilon_{213}=-1\] SU(3) generators Gell Mann matrix's \[\lambda_1=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}\] \[\lambda_2=\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}\] \[\lambda_3=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}\] \[\lambda_4=\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}\] \[\lambda_5=\begin{pmatrix}0&0&-i\\0&0&0\\i&0&0\end{pmatrix}\] \[\lambda_6=\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}\] \[\lambda_7=\begin{pmatrix}0&0&0\\0&0&-i\\0&i&0\end{pmatrix}\] \[\lambda_8=\frac{1}{\sqrt{3}}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}\] commutation relations \[[\lambda_i\lambda_j]=2i\sum^8_{k=1}f_{ijk}\lambda_k\] with algebraic structure \[f_{123}=1,f_{147}=f_{165}=f_{246}=f_{246}=f_{257}=f_{345}=f_{376}=\frac{1}{2},f_{458}=f_{678}=\frac{3}{2}\] with Casimer Operator \[\vec{J}^2=J_x^2+J_y^2+j_z^2\]

It would be interesting to imagine what other facts would come to light, as far as the conventional militaries and their future capabilities are concerned. I don't know if the Ukraine war would end or not - it's being fought with conventional weapons. And the military budget would effectively increase for Russia, absent the cost of maintaining and manning a nuclear strike force. I've even wondered if all the nuke talk lately from Putin is just hot air, and they stopped taking care of their missiles some time ago. Could there be Potemkin silos, filled with rusting equipment and non launchable rockets? How good is our intelligence in the West? This is all a bit out there, but I try to remember that Russians are excellent chess players.

Among the things that Pontzer mentions in one of his studies (https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0040503), is that the Ache Indians of Paraguay had low levels of leptin and of testosterone compared to Westerners. Leptin is probably the hormone most related to obesity, while testosterone is related to muscle mass. Although leptin would not be an anabolic hormone itself, the condition known as leptin resistance (common in obese people) would be. It seems that the catabolic tendency in the Ache, possibly related to their long-term cardiovascular physical activity, would lead them to reduce their anabolic processes and maintain an adequate weight without necessarily having to burn more calories after completing 24 hours. Low levels of leptin are an indication that they do not have resistance to leptin and that their anabolic processes related to fat sequestration, therefore, are within the appropriate balance. Similarly, we noticed that relatively low levels of testosterone are related to fewer anabolic processes related to muscle mass gain. In summary, here we have other evidence within a study that maintains that the Hadza, despite spending more calories doing physical activity, do not spend many more calories in 24 hours, that it is anabolic and catabolic processes that make the difference in body weight at a metabolic level and not metabolic acceleration as previously thought.

Obviously Russia doing that and confirming it would end the war in Ukraine pretty quickly and leave them vulnerable to China in their East. China (are they even big three nuclear?) doing it might not change overly vs threats (so Chinese in defense) from the US and Russia but would change their posture with India.(their "nothing more than sticks and rocks" agreement might break down). Their offensiveness in the South Pacific would certainly need rethinking, as would any hope of ultimately claiming Taiwan. US doing it would be throwing the dice.

Every electron has its own magnetic moment. https://en.m.wikipedia.org/wiki/Electron_magnetic_moment#:~:text=In atomic physics%2C the electron,24 J⋅T−1.

Ehrfurcht vor dem leben (3 genders, 4 cases, how do Germans do it?).I hope folks here won't conflate that desire to lay groundwork for global zero with a naive sensibility. We all know how far away a true START agreement is, let alone global zero. As that Brookings fellow pointed out, the latter goal is pragmatic WRT to a longterm winding down of proliferation. Why would all those dictators abandon nuke ambitions so long as the big boys have them? Nothing really happens until Cold War (and hot war) issues are resolved with the big three, and that would require regime change in Russia and liberalization in China to even get parties to the arms reduction table. Well that has been the standard assumption. I think maybe we should test that against the current reality, maybe a thought experiment. What would happen if, right now, one of the big three dismantled all its nukes, fed the fissile material into power plant reactors, and said hey we're done. And then, one assumes, put some of the billions saved into more conventional weapons. (maintaining a large nuclear arsenal is expensive)

Well as an Engineer you certainly know that your field requires mathematical rigor. It's no different for physicist or mathematical theory. So if it's your goal to present some new trig function and have it gain weight in the Professional circles then you will need a mathematical proof. One that doesn't rely on words /pictures or descriptives. Anything less simply wouldn't cut it. I'm sure you can recognize the need fir that no forum has any particular influence in the Professional circles. Forums are useful but mainly to help others learn . Nothing discussed in a forum will ever alter how the scientific or mathematical community does things. That requires a professional peer review paper examined by other experts. The mathematical proof would be needed for that. For example every single physics formula has a corresponding mathematical proof no formula ever gets accepted without one.

Photoelectric effect is basically the same as photoionization of an atom, for a single photon. Photon in, electron out. edit: an LED is not doing this

On what, specifically?

Nuclear weapons are the deterrent to nuclear weapons. Eliminating your nuclear weapons is putting your guard down. Russia would not be in Ukraine if Ukraine had them.

Lol every single spectrograph I've ever examined has some form of redshift. The only time it doesn't is if it was reading some object in the lab. I even had an instructor that was testing the class with a falsified dataset that not only didn't have redshift but also incorrect elements.

Photoelectric effect - As an electron returns to its normal state upon one photon entering and exiting a system, two photons are released. Electroluminescence - Electricity takes grazes through in an atomic orbital, and light is released continually Photovoltaics - A photon passes through an atomic medium and electrons are released Fiber optic cable - Is the photoelectric effects applied to the atoms along the cable Photoionization - A photon enters into a system and an electron is released from the atomic orbital and the atom is now a positive ion Magnetism - The electrons have opposite spin causing the bodies to be magnetized Notice I use the term "grazing through an atomic orbital continually" for electroluminescence now photoionization can also cause this to happen at one positive interval followed by a negative interval. Hence induction motor.

Then what is your position? I am genuinely interested to know. You did poke a few holes for sure and I appreciated it because it does help me figure out how to improve my position and you absolutely provided a lot of good prompting for research, as has MigL. Sorry I can get caught up in the spirit of debate quite a bit but hey, at least I don't get as heated or assholish as I used to get when I first came onto this forum.

But one thing you notice is that such leaders are always around. Before this it was Saddam Hussein, Kim Jong-il, Nicolae Ceaușescu, Pinochet, Idi Amin, Khaddafi, and more, and that’s only going back ~50 years I don’t think I was saying BISS, or really making an argument (or advocating a position) as much as I was poking holes in what you were presenting.