KJW Posted December 8, 2023 Share Posted December 8, 2023 (edited) 1c2∂2ψ∂t2−∂2ψ∂x2−∂2ψ∂y2−∂2ψ∂z2=−m2c2ℏ2ψ Let ψ=ψ0exp(iS(t,x,y,z)ℏ) ∂2ψ∂t2=∂∂t(∂∂texp(iS(t,x,y,z)ℏ)) \[\frac{\partial^2\psi}{\partial t^2} = \psi_0 \frac{\partial}{\partial t}(\frac{\partial}{\partial t} exp(i\frac{S(t,x,y,z)}{\hbar})) = \psi_0 \frac{\partial}{\partial t}(\frac{i}{hbar} exp(i\frac{S(t,x,y,z)}{\hbar}) \frac{\partial S(t,x,y,z)}{\partial t})\] Edited December 8, 2023 by KJW Link to comment Share on other sites More sharing options...

KJW Posted December 8, 2023 Author Share Posted December 8, 2023 (edited) Klein-Gordon equation: \[\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2} - \frac{\partial^2\psi}{\partial x^2} - \frac{\partial^2\psi}{\partial y^2} - \frac{\partial^2\psi}{\partial z^2} = -\frac{m^2c^2}{\hbar^2}\psi \] \[\psi = \psi_0 exp(\frac{i}{\hbar} S(t,x,y,z))\] where S(t,x,y,z) is the action \[\frac{\partial^2\psi}{\partial t^2} = \frac{\partial}{\partial t}(\frac{\partial}{\partial t} \psi_0 exp(\frac{i}{\hbar} S)) = \frac{\partial}{\partial t}(\frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial S}{\partial t}) = \frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi_0 exp(\frac{i}{\hbar} S) (\frac{\partial S}{\partial t})^2 = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial t})^2\] \[-i\hbar (\frac{1}{c^2}\frac{\partial^2 S}{\partial t^2} - \frac{\partial^2 S}{\partial x^2} - \frac{\partial^2 S}{\partial y^2} - \frac{\partial^2 S}{\partial z^2}) + (\frac{1}{c^2}(\frac{\partial S}{\partial t})^2 - (\frac{\partial S}{\partial x})^2 - (\frac{\partial S}{\partial y})^2 - (\frac{\partial S}{\partial z})^2) = m^2c^2\] In the classical limit of [math]\hbar = 0[/math], the linear second-order Klein-Gordon equation becomes the non-linear first-order Hamilton-Jacobi equation: \[(\frac{1}{c^2}(\frac{\partial S}{\partial t})^2 - (\frac{\partial S}{\partial x})^2 - (\frac{\partial S}{\partial y})^2 - (\frac{\partial S}{\partial z})^2) = m^2c^2\] Edited December 8, 2023 by KJW Link to comment Share on other sites More sharing options...

KJW Posted December 8, 2023 Author Share Posted December 8, 2023 (edited) Klein-Gordon equation: \[\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2} - \frac{\partial^2\psi}{\partial x^2} - \frac{\partial^2\psi}{\partial y^2} - \frac{\partial^2\psi}{\partial z^2} = -\frac{m^2c^2}{\hbar^2}\psi \] \[\psi = \psi_0 exp(\frac{i}{\hbar} S(t,x,y,z))\] where [math]S(t,x,y,z)[/math] is the action. \[\frac{\partial^2\psi}{\partial t^2} = \frac{\partial}{\partial t}(\frac{\partial}{\partial t} \psi_0 exp(\frac{i}{\hbar} S)) = \frac{\partial}{\partial t}(\frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial S}{\partial t}) = \frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi_0 exp(\frac{i}{\hbar} S) (\frac{\partial S}{\partial t})^2 = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial t})^2\] Similarly: \[\frac{\partial^2\psi}{\partial x^2} = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial x^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial x})^2 \;\;;\;\; \frac{\partial^2\psi}{\partial y^2} = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial y^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial y})^2 \;\;;\;\; \frac{\partial^2\psi}{\partial z^2} = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial z^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial z})^2\] Thus: \[-i\hbar (\frac{1}{c^2}\frac{\partial^2 S}{\partial t^2} - \frac{\partial^2 S}{\partial x^2} - \frac{\partial^2 S}{\partial y^2} - \frac{\partial^2 S}{\partial z^2}) + (\frac{1}{c^2}(\frac{\partial S}{\partial t})^2 - (\frac{\partial S}{\partial x})^2 - (\frac{\partial S}{\partial y})^2 - (\frac{\partial S}{\partial z})^2) = m^2c^2\] In the classical limit of [math]\hbar = 0[/math], the linear second-order Klein-Gordon equation becomes the non-linear first-order Hamilton-Jacobi equation: \[\frac{1}{c^2}(\frac{\partial S}{\partial t})^2 - (\frac{\partial S}{\partial x})^2 - (\frac{\partial S}{\partial y})^2 - (\frac{\partial S}{\partial z})^2 = m^2c^2\] Edited December 8, 2023 by KJW Link to comment Share on other sites More sharing options...

KJW Posted December 19, 2023 Author Share Posted December 19, 2023 (edited) [math]1 \longleftrightarrow 1[/math] [math]2 \longleftrightarrow 4[/math] [math]3 \longleftrightarrow 9[/math] [math]4 \longleftrightarrow 16[/math] [math]5 \longleftrightarrow 25[/math] [math]6 \longleftrightarrow 36[/math] [math]7 \longleftrightarrow 49[/math] [math]8 \longleftrightarrow 64[/math] [math]...[/math] [math]n \longleftrightarrow n^2[/math] [math]...[/math] [math]\textstyle \mathbb {N}[/math] Edited December 19, 2023 by KJW Link to comment Share on other sites More sharing options...

KJW Posted December 24, 2023 Author Share Posted December 24, 2023 (edited) [math]\textstyle 2\mathbb {Z}[/math] [math]2\textstyle 2\mathbb {Z}[/math] [math]\mathbb {ABCDEFGHIJKLMNOPQRSTUVWXYZ}[/math] Edited December 24, 2023 by KJW Link to comment Share on other sites More sharing options...

KJW Posted December 25, 2023 Author Share Posted December 25, 2023 (edited) [math] \documentclass{article} \begin{document} Evaluate the sum $\displaystyle\sum\limits_{i=0}^n i^3$. \end{document} [/math] [math]Evaluate the sum $\displaystyle\sum\limits_{i=0}^n i^3$.[/math] Edited December 25, 2023 by KJW Link to comment Share on other sites More sharing options...

KJW Posted December 26, 2023 Author Share Posted December 26, 2023 [math]\displaystyle \lim _{n \to \infty} a_{n}=L[/math] [math]\lim _{n \to \infty} a_{n}=L[/math] Link to comment Share on other sites More sharing options...

KJW Posted December 27, 2023 Author Share Posted December 27, 2023 (edited) [math]f(x)\; {\buildrel\rm def\over=} \;x+1[/math] [math]\buildrel\rm def\over=[/math] [math]\buildrel def \over=[/math] Edited December 27, 2023 by KJW Link to comment Share on other sites More sharing options...

KJW Posted December 30, 2023 Author Share Posted December 30, 2023 (edited) [math]\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk} = \dfrac{1}{1 + x^{n}} \\ y = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk}[/math] Edited December 30, 2023 by KJW Link to comment Share on other sites More sharing options...

KJW Posted January 4 Author Share Posted January 4 (edited) [math]\overbrace{\dfrac{d}{dx} \Biggl( \Biggr. x \cdots \dfrac{d}{dx} \Biggl( \Biggr. x}^{n}\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k} \Biggl. \Biggr) \cdots \Biggl. \Biggr) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^n x^{k} = \eta(-n)[/math] for [math]x = 1[/math] Edited January 4 by KJW Link to comment Share on other sites More sharing options...

KJW Posted January 5 Author Share Posted January 5 (edited) deleted Edited January 5 by KJW Link to comment Share on other sites More sharing options...

KJW Posted January 12 Author Share Posted January 12 (edited) deleted Edited January 12 by KJW Link to comment Share on other sites More sharing options...

KJW Posted January 13 Author Share Posted January 13 (edited) deleted Edited January 13 by KJW Link to comment Share on other sites More sharing options...

KJW Posted January 17 Author Share Posted January 17 (edited) [math]\times\!\!\!\!\phi^2[/math] Edited January 17 by KJW Link to comment Share on other sites More sharing options...

KJW Posted January 27 Author Share Posted January 27 (edited) deleted Edited January 27 by KJW Link to comment Share on other sites More sharing options...

KJW Posted March 30 Author Share Posted March 30 (edited) [math](ds)^2 = (c^2 - \dfrac{r^2}{t^2}) (dt)^2 + \color{red} {\dfrac{2r}{t} (dt)(dr)} - (dr)^2 - r^2 ((d\theta)^2 + sin^2 \theta (d\phi)^2)[/math] For me to complete this, I need to apply a coordinate transformation to remove the part in red. 😉 Edited March 30 by KJW Link to comment Share on other sites More sharing options...

KJW Posted March 30 Author Share Posted March 30 (edited) deleted Edited March 30 by KJW Link to comment Share on other sites More sharing options...

KJW Posted March 30 Author Share Posted March 30 (edited) deleted Edited March 30 by KJW Link to comment Share on other sites More sharing options...

KJW Posted March 31 Author Share Posted March 31 (edited) deleted Edited March 31 by KJW Link to comment Share on other sites More sharing options...

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