Paxton Hoffman

2022-07-22

on the Dirac equation, expands the ${\gamma}^{\mu}{\mathrm{\partial}}_{\mu}$ term as:

${\gamma}^{\mu}{\mathrm{\partial}}_{\mu}={\gamma}^{0}\frac{\mathrm{\partial}}{\mathrm{\partial}t}+\overrightarrow{\gamma}\cdot \overrightarrow{\mathrm{\nabla}}$

where $\overrightarrow{\gamma}=({\gamma}^{1},{\gamma}^{2},{\gamma}^{3})$, but to my knowledge,

${\gamma}^{\mu}{\mathrm{\partial}}_{\mu}={\gamma}^{\mu}{\eta}_{\mu \nu}{\mathrm{\partial}}^{\nu}={\gamma}^{0}\frac{\mathrm{\partial}}{\mathrm{\partial}t}-\overrightarrow{\gamma}\cdot \overrightarrow{\mathrm{\nabla}}$

using the convention ${\eta}_{\mu \nu}=\mathrm{diag}(+,-,-,-)$.

${\gamma}^{\mu}{\mathrm{\partial}}_{\mu}={\gamma}^{0}\frac{\mathrm{\partial}}{\mathrm{\partial}t}+\overrightarrow{\gamma}\cdot \overrightarrow{\mathrm{\nabla}}$

where $\overrightarrow{\gamma}=({\gamma}^{1},{\gamma}^{2},{\gamma}^{3})$, but to my knowledge,

${\gamma}^{\mu}{\mathrm{\partial}}_{\mu}={\gamma}^{\mu}{\eta}_{\mu \nu}{\mathrm{\partial}}^{\nu}={\gamma}^{0}\frac{\mathrm{\partial}}{\mathrm{\partial}t}-\overrightarrow{\gamma}\cdot \overrightarrow{\mathrm{\nabla}}$

using the convention ${\eta}_{\mu \nu}=\mathrm{diag}(+,-,-,-)$.

Kali Galloway

Beginner2022-07-23Added 16 answers

Yes. You are missing the fact that he is using the convention

$\mathrm{\nabla}=({\mathrm{\partial}}_{1},{\mathrm{\partial}}_{2},{\mathrm{\partial}}_{3})$

as opposed to

$\mathrm{\nabla}=({\mathrm{\partial}}^{1},{\mathrm{\partial}}^{2},{\mathrm{\partial}}^{3})$

The first convention is by far the most common in my experience.

$\mathrm{\nabla}=({\mathrm{\partial}}_{1},{\mathrm{\partial}}_{2},{\mathrm{\partial}}_{3})$

as opposed to

$\mathrm{\nabla}=({\mathrm{\partial}}^{1},{\mathrm{\partial}}^{2},{\mathrm{\partial}}^{3})$

The first convention is by far the most common in my experience.

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What is the distance from the Earth's center to a point in space where the gravitational acceleration due to the Earth is 1/24 of its value at the Earth's surface.

What is proper motion.

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Light travels at a very high speed

Due to rectilinear propagation of light

Light can reflect from a surface

Screen of the pinhole camera is invertedIn order to calculate the cross-section of an interaction process the following formula is often used for first approximations:

$\sigma =\frac{2\pi}{\hslash \phantom{\rule{thinmathspace}{0ex}}{v}_{i}}{\left|{M}_{fi}\right|}^{2}\varrho \left({E}_{f}\right)\phantom{\rule{thinmathspace}{0ex}}V$

${M}_{fi}=\u27e8{\psi}_{f}|{H}_{int}|{\psi}_{i}\u27e9$

${M}_{fi}=\u27e8{\psi}_{f}|{H}_{int}|{\psi}_{i}\u27e9$

$\varrho \left({E}_{f}\right)=\frac{\mathrm{d}n\left({E}_{f}\right)}{\mathrm{d}{E}_{f}}=\frac{4\pi {{p}_{f}}^{2}}{{\left(2\pi \hslash \right)}^{3}}\frac{V}{{v}_{f}}$

The derivation of this equation in the context of the non relativistic Schrödinger equation. Use this formula in the relativistic limit: ${v}_{i},{v}_{f}\to c\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}{p}_{f}\approx {E}_{f}/c$

Very often books simply use this equation with matrix element derived from some relativistic theory, e.g. coupling factors and propagators from the Dirac equation or Electroweak interaction. How is this justified?Why does the azimuthal angle, $\varphi $, remain unchanged between reference frames in special relativity?

I think this comes from the aberration formula, showing dependence only on the polar angle, $\theta $.

The aberration formula is:

$\mathrm{tan}\theta =\frac{{u}_{\perp}}{{u}_{||}}=\frac{{u}^{\prime}\mathrm{sin}{\theta}^{\prime}}{\gamma ({u}^{\prime}\mathrm{cos}{\theta}^{\prime}+v)}$

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Suppose we want to construct a wave function for 2 free (relativistic) fermions. As we are dealing with fermions the total wave function has to be antisymmetric under interchange of the coordinates,Suppose we want to construct a wave function for 2 free (relativistic) fermions. As we are dealing with fermions the total wave function has to be antisymmetric under interchange of the coordinates,

$\mathrm{\Psi}({x}_{1},{x}_{2})=-\mathrm{\Psi}({x}_{2},{x}_{1})$

If we assume that we can factorize the wave function in terms of single particle wave functions we can write

$\mathrm{\Psi}({x}_{1},{x}_{2})={\psi}_{1}({x}_{1}){\psi}_{2}({x}_{2})-{\psi}_{1}({x}_{1}){\psi}_{2}({x}_{2})$

which fulfills the anti-symmetry requirement. The plane wave single particle states are given by,

${\psi}_{\mathbf{k},{m}_{s}}(x)={u}_{\mathbf{k},{m}_{s}}(s)\varphi (\mathbf{k}\cdot \mathbf{r})$

So expect the total wavefunction to be

$\begin{array}{rl}\mathrm{\Psi}({x}_{1},{x}_{2})& ={u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{1})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-{u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{2})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})\\ & ={u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{1}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{2})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-{u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{2}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{1})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})\end{array}$

However

$u({\mathbf{k}}_{1},{m}_{{s}_{1}})u({\mathbf{k}}_{2},{m}_{{s}_{2}})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-u({\mathbf{k}}_{2},{m}_{{s}_{2}})u({\mathbf{k}}_{1},{m}_{{s}_{1}})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})$

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$d{s}^{2}={g}_{\mu \nu}d{x}^{\mu}d{x}^{\nu},$

How to calculate $ds$?