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, 20232 yr by KJW

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, 20232 yr by KJW

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, 20232 yr by KJW

[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, 20232 yr by KJW

[math]\textstyle 2\mathbb {Z}[/math]

[math]2\textstyle 2\mathbb {Z}[/math]

[math]\mathbb {ABCDEFGHIJKLMNOPQRSTUVWXYZ}[/math]

Edited December 24, 20232 yr by KJW

[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, 20231 yr by KJW

[math]\displaystyle \lim _{n \to \infty} a_{n}=L[/math]
[math]\lim _{n \to \infty} a_{n}=L[/math]

[math]f(x)\; {\buildrel\rm def\over=} \;x+1[/math]

[math]\buildrel\rm def\over=[/math]

[math]\buildrel def \over=[/math]

Edited December 27, 20231 yr by KJW

[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, 20231 yr by KJW

[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]

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[math]\times\!\!\!\!\phi^2[/math]

 

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[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. 😉

 

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[math]\buildrel \rm def \over =[/math]

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