# KJW Mathematics

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1c22ψt22ψx22ψy22ψz2=m2c22ψ

Let
ψ=ψ0exp(iS(t,x,y,z))

2ψt2=t(texp(iS(t,x,y,z)))

$\frac{\partial^2\psi}{\partial t^2} = \psi_0 \frac{\partial}{\partial t}(\frac{\partial}{\partial t} exp(i\frac{S(t,x,y,z)}{\hbar})) = \psi_0 \frac{\partial}{\partial t}(\frac{i}{hbar} exp(i\frac{S(t,x,y,z)}{\hbar}) \frac{\partial S(t,x,y,z)}{\partial t})$

Edited by KJW
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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 $\hbar = 0$, 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 by KJW
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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.

$\frac{\partial^2\psi}{\partial t^2} = \frac{\partial}{\partial t}(\frac{\partial}{\partial t} \psi_0 exp(\frac{i}{\hbar} S)) = \frac{\partial}{\partial t}(\frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial S}{\partial t}) = \frac{i}{\hbar} \psi_0 exp(\frac{i}{\hbar} S) \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi_0 exp(\frac{i}{\hbar} S) (\frac{\partial S}{\partial t})^2 = \frac{i}{\hbar} \psi \frac{\partial^2 S}{\partial t^2} - \frac{1}{\hbar^2} \psi (\frac{\partial S}{\partial t})^2$

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 $\hbar = 0$, 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 by KJW
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• 2 weeks later...

$1 \longleftrightarrow 1$
$2 \longleftrightarrow 4$
$3 \longleftrightarrow 9$
$4 \longleftrightarrow 16$
$5 \longleftrightarrow 25$
$6 \longleftrightarrow 36$
$7 \longleftrightarrow 49$
$8 \longleftrightarrow 64$
$...$
$n \longleftrightarrow n^2$
$...$

$\textstyle \mathbb {N}$

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$\textstyle 2\mathbb {Z}$

$2\textstyle 2\mathbb {Z}$

$\mathbb {ABCDEFGHIJKLMNOPQRSTUVWXYZ}$

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$\documentclass{article} \begin{document} Evaluate the sum \displaystyle\sum\limits_{i=0}^n i^3. \end{document} $

$Evaluate the sum \displaystyle\sum\limits_{i=0}^n i^3.$

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$\displaystyle \lim _{n \to \infty} a_{n}=L$
$\lim _{n \to \infty} a_{n}=L$

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$f(x)\; {\buildrel\rm def\over=} \;x+1$

$\buildrel\rm def\over=$

$\buildrel def \over=$

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$\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk} = \dfrac{1}{1 + x^{n}} \\ y = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk}$

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$\overbrace{\dfrac{d}{dx} \Biggl( \Biggr. x \cdots \dfrac{d}{dx} \Biggl( \Biggr. x}^{n}\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k} \Biggl. \Biggr) \cdots \Biggl. \Biggr) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^n x^{k} = \eta(-n)$ for $x = 1$

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$\times\!\!\!\!\phi^2$

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$(ds)^2 = (c^2 - \dfrac{r^2}{t^2}) (dt)^2 + \color{red} {\dfrac{2r}{t} (dt)(dr)} - (dr)^2 - r^2 ((d\theta)^2 + sin^2 \theta (d\phi)^2)$

For me to complete this, I need to apply a coordinate transformation to remove the part in red😉

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$\buildrel \rm def \over =$

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test bold

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test underline

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