odcinaknr

2022-09-30

Consider a particle at rest with respect to the rotating frame. Since there is no velocity there is no force felt from $\overrightarrow{{B}^{\prime}}$.

$\overrightarrow{F}=m\overrightarrow{a}=q\overrightarrow{{E}^{\prime}}=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})\phantom{\rule{0ex}{0ex}}\overrightarrow{{E}^{\prime}}=\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B}$

Additionally, considering a stationary particle in the inertial frame of reference, with pseudo forces included:

$\overrightarrow{F}=m\overrightarrow{a}+m\overrightarrow{\omega}\times \overrightarrow{v}+m\overrightarrow{\omega}\times \overrightarrow{v}+m\overrightarrow{\omega}\times \overrightarrow{\omega}\times \overrightarrow{x}\phantom{\rule{0ex}{0ex}}=q(\overrightarrow{{E}^{\prime}}+\overrightarrow{v}\times \overrightarrow{{B}^{\prime}})=q\overrightarrow{E}$

However if I just plug in $\overrightarrow{{E}^{\prime}}$ into the second equation, $\overrightarrow{{B}^{\prime}}$ = $\overrightarrow{B}$ which seems incorrect. Looking at the non-relativistic portions of a Lorentz transformations has $\overrightarrow{B}-\overrightarrow{v}\times \overrightarrow{E}$

$\overrightarrow{F}=m\overrightarrow{a}=q\overrightarrow{{E}^{\prime}}=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})\phantom{\rule{0ex}{0ex}}\overrightarrow{{E}^{\prime}}=\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B}$

Additionally, considering a stationary particle in the inertial frame of reference, with pseudo forces included:

$\overrightarrow{F}=m\overrightarrow{a}+m\overrightarrow{\omega}\times \overrightarrow{v}+m\overrightarrow{\omega}\times \overrightarrow{v}+m\overrightarrow{\omega}\times \overrightarrow{\omega}\times \overrightarrow{x}\phantom{\rule{0ex}{0ex}}=q(\overrightarrow{{E}^{\prime}}+\overrightarrow{v}\times \overrightarrow{{B}^{\prime}})=q\overrightarrow{E}$

However if I just plug in $\overrightarrow{{E}^{\prime}}$ into the second equation, $\overrightarrow{{B}^{\prime}}$ = $\overrightarrow{B}$ which seems incorrect. Looking at the non-relativistic portions of a Lorentz transformations has $\overrightarrow{B}-\overrightarrow{v}\times \overrightarrow{E}$

Jarrett Pearson

Beginner2022-10-01Added 4 answers

Let there be an electromagnetic bivector $f=\frac{E}{c}\wedge {e}^{t}+{e}^{xyz}B$. Let's check this against a current $j=\rho c{e}_{t}$:

$F=f\cdot j=(\frac{E}{c}\rho c\wedge {e}^{t})\cdot {e}_{t}=E\rho $

so that checks, sign-wise.

We define the primed frame by the following transformations:

${x}^{\prime}=x\mathrm{cos}\omega t-y\mathrm{sin}\omega t,\phantom{\rule{1em}{0ex}}{y}^{\prime}=y\mathrm{cos}\omega t+x\mathrm{sin}\omega t,\phantom{\rule{1em}{0ex}}{z}^{\prime}=z,\phantom{\rule{1em}{0ex}}{t}^{\prime}=t$

Collectively, let $s=x{e}_{x}+y{e}_{y}+z{e}_{z}+ct{e}_{t}$, and let ${s}^{\prime}=g(s)$. We find the Jacobian of the transformation:

$\begin{array}{rl}\underset{\_}{g}({e}_{t})& =\frac{1}{c}{\mathrm{\partial}}_{t}{s}^{\prime}={e}_{t}+\frac{\omega}{c}(-{y}^{\prime}{e}_{x}+{x}^{\prime}{e}_{y})\\ \underset{\_}{g}({e}_{x})& ={\mathrm{\partial}}_{x}{s}^{\prime}={e}_{x}\mathrm{cos}\omega t+{e}_{y}\mathrm{sin}\omega t\\ \underset{\_}{g}({e}_{y})& ={\mathrm{\partial}}_{y}{s}^{\prime}={e}_{y}\mathrm{cos}\omega t-{e}_{x}\mathrm{sin}\omega t\\ \underset{\_}{g}({e}_{z})& ={\mathrm{\partial}}_{z}{s}^{\prime}={e}_{z}\end{array}$

Currents will transform according to the Jacobian, so you can easily choose a stationary particle in the stationary frame and map it to its four-current in the primed frame. To consider a stationary particle in the primed frame, however, you will need to compute the inverse.

To compute the transformation of the Faraday field, we'll need the adjoint transformation-this is kinda like a transpose, but slightly different due to the metric. The adjoint is more fundamental, however--this is the same reason conjugate transposes are used in QM, instead of plain transposes.

$\begin{array}{rl}\overline{g}({e}^{t})& ={e}^{t}\\ \overline{g}({e}^{x})& =-{y}^{\prime}\frac{\omega}{c}{e}^{t}+{e}^{x}\mathrm{cos}\omega t-{e}^{y}\mathrm{sin}\omega t\\ \overline{g}({e}^{y})& ={x}^{\prime}\frac{\omega}{c}{e}^{t}+{e}^{x}\mathrm{sin}\omega t+{e}^{y}\mathrm{cos}\omega t\\ \overline{g}({e}^{z})& ={e}^{z}\end{array}$

Luckily, the determinant of this operator is $1$, so the inverse is equal to the adjoint. That actually simplifies things considerably! This is, as a result, an orthogonal transformation--it is a position dependent rotation. Not a surprise; we designed it that way. That means we can map from the original, unprimed space to the primed space using only $\underset{\_}{g}$ for all objects.

There are some immediate consequences of this. First, you should see that ${B}_{z}$ remains invariant-the $xy$ plane is mapped to itself, and that is exactly the component that corresponds to ${f}_{xy}$. (Edit: ${E}_{z}$ however does not remain invariant, as $tz$ gets mixed with $xz$ and $yz$, so some electric effects appear magnetic in the rotating frame.)

One insight from relativity is that we should always think about the EM field as planes, and knowing which planes are left invariant by a transformation tells us immediately which components stay the same and which change.

$F=f\cdot j=(\frac{E}{c}\rho c\wedge {e}^{t})\cdot {e}_{t}=E\rho $

so that checks, sign-wise.

We define the primed frame by the following transformations:

${x}^{\prime}=x\mathrm{cos}\omega t-y\mathrm{sin}\omega t,\phantom{\rule{1em}{0ex}}{y}^{\prime}=y\mathrm{cos}\omega t+x\mathrm{sin}\omega t,\phantom{\rule{1em}{0ex}}{z}^{\prime}=z,\phantom{\rule{1em}{0ex}}{t}^{\prime}=t$

Collectively, let $s=x{e}_{x}+y{e}_{y}+z{e}_{z}+ct{e}_{t}$, and let ${s}^{\prime}=g(s)$. We find the Jacobian of the transformation:

$\begin{array}{rl}\underset{\_}{g}({e}_{t})& =\frac{1}{c}{\mathrm{\partial}}_{t}{s}^{\prime}={e}_{t}+\frac{\omega}{c}(-{y}^{\prime}{e}_{x}+{x}^{\prime}{e}_{y})\\ \underset{\_}{g}({e}_{x})& ={\mathrm{\partial}}_{x}{s}^{\prime}={e}_{x}\mathrm{cos}\omega t+{e}_{y}\mathrm{sin}\omega t\\ \underset{\_}{g}({e}_{y})& ={\mathrm{\partial}}_{y}{s}^{\prime}={e}_{y}\mathrm{cos}\omega t-{e}_{x}\mathrm{sin}\omega t\\ \underset{\_}{g}({e}_{z})& ={\mathrm{\partial}}_{z}{s}^{\prime}={e}_{z}\end{array}$

Currents will transform according to the Jacobian, so you can easily choose a stationary particle in the stationary frame and map it to its four-current in the primed frame. To consider a stationary particle in the primed frame, however, you will need to compute the inverse.

To compute the transformation of the Faraday field, we'll need the adjoint transformation-this is kinda like a transpose, but slightly different due to the metric. The adjoint is more fundamental, however--this is the same reason conjugate transposes are used in QM, instead of plain transposes.

$\begin{array}{rl}\overline{g}({e}^{t})& ={e}^{t}\\ \overline{g}({e}^{x})& =-{y}^{\prime}\frac{\omega}{c}{e}^{t}+{e}^{x}\mathrm{cos}\omega t-{e}^{y}\mathrm{sin}\omega t\\ \overline{g}({e}^{y})& ={x}^{\prime}\frac{\omega}{c}{e}^{t}+{e}^{x}\mathrm{sin}\omega t+{e}^{y}\mathrm{cos}\omega t\\ \overline{g}({e}^{z})& ={e}^{z}\end{array}$

Luckily, the determinant of this operator is $1$, so the inverse is equal to the adjoint. That actually simplifies things considerably! This is, as a result, an orthogonal transformation--it is a position dependent rotation. Not a surprise; we designed it that way. That means we can map from the original, unprimed space to the primed space using only $\underset{\_}{g}$ for all objects.

There are some immediate consequences of this. First, you should see that ${B}_{z}$ remains invariant-the $xy$ plane is mapped to itself, and that is exactly the component that corresponds to ${f}_{xy}$. (Edit: ${E}_{z}$ however does not remain invariant, as $tz$ gets mixed with $xz$ and $yz$, so some electric effects appear magnetic in the rotating frame.)

One insight from relativity is that we should always think about the EM field as planes, and knowing which planes are left invariant by a transformation tells us immediately which components stay the same and which change.

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

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

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

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

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

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,

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