Baladdaa9

2022-07-22

Consider a (classical) system of several interacting particles. Can it be shown that, if the Lagrangian of such a system is Lorenz invariant, there cannot be any space-like influences between the particles?

Mireya Hoffman

Beginner2022-07-23Added 14 answers

A counterexample:

1) Take two particles in a laboratory $S$

2) Set the clocks attached to the particles so that at laboratory time $t=0$ both clocks show ${\tau}_{1}=0$, ${\tau}_{2}=0$

3) Leave the particles to evolve under the following lorentz-invariant lagrangian:

$\mathcal{L}={\mathcal{L}}_{1}+{\mathcal{L}}_{2}+{\mathcal{L}}_{\mathrm{i}\mathrm{n}\mathrm{t}}\phantom{\rule{0ex}{0ex}}{\mathcal{L}}_{\mathrm{i}\mathrm{n}\mathrm{t}}={u}_{1}^{\text{}\mu}({\tau}_{1}){u}_{2\mu}({\tau}_{2}),$

where ${\mathcal{L}}_{1},{\mathcal{L}}_{2}$ are free-particle lagrangians. The total lagrangian is lorenz-invariant, but the physical system is not due to its initial setup, which allows the lagrangian to transmit the interactions in a space-like manner.

1) Take two particles in a laboratory $S$

2) Set the clocks attached to the particles so that at laboratory time $t=0$ both clocks show ${\tau}_{1}=0$, ${\tau}_{2}=0$

3) Leave the particles to evolve under the following lorentz-invariant lagrangian:

$\mathcal{L}={\mathcal{L}}_{1}+{\mathcal{L}}_{2}+{\mathcal{L}}_{\mathrm{i}\mathrm{n}\mathrm{t}}\phantom{\rule{0ex}{0ex}}{\mathcal{L}}_{\mathrm{i}\mathrm{n}\mathrm{t}}={u}_{1}^{\text{}\mu}({\tau}_{1}){u}_{2\mu}({\tau}_{2}),$

where ${\mathcal{L}}_{1},{\mathcal{L}}_{2}$ are free-particle lagrangians. The total lagrangian is lorenz-invariant, but the physical system is not due to its initial setup, which allows the lagrangian to transmit the interactions in a space-like manner.

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Due to rectilinear propagation of light

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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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