Payton Rasmussen

2022-10-08

Consider two identical charges moving with uniform velocity. There will be a magnetic force of attraction between them as two currents in the same direction attract each other. If I sit on one of the charges, according to me the other charge is not moving. So there wont be any magnetic attraction. How does changing the frame of reference change the outcome of the interaction?

Bestvinajw

Beginner2022-10-09Added 15 answers

You're right; in the frame moving with the charges there will be no magnetic attraction. There will however be an electrostatic repulsion, so you will see the two particles move away from each other.

Consider another frame (the rest frame) where both particles move side by side in the $x$ direction. The particles still repel each other electrostatically, but now they attract one another magnetically as well.

This may seem like a problem, but it isn't because even though the force measured in the rest frame is weaker since the electrostatic repulsion is partially canceled by the magnetic attraction, the time experienced by moving observers is also different due to the witchcraft of special relativity. If the frame moving with the charges is primed, and the rest frame is unprimed, then

$\begin{array}{rl}{t}^{\mathrm{\prime}}& =\gamma t\\ F& =q(E+v\times B)=q(E-vB)\\ {F}^{\mathrm{\prime}}& =q{E}^{\mathrm{\prime}}\end{array}$

What we seek to show is that since force is inversely proportional to the square of the time, $F$ and ${F}^{\prime}$ are related by a factor of ${\gamma}^{2}$ in which case observers in either frame will calculate matching trajectories using Maxwell's equations.

Since for a point charge

$E=\frac{1}{4\pi \u03f5}\frac{q}{{r}^{2}}$

$B=\frac{\mu}{4\pi}\frac{qv\times \hat{r}}{{r}^{2}}=\frac{\mu}{4\pi}\frac{qv}{{r}^{2}}$

Now we have all we need. Plugging $B$ in to $F$ and using $\mu =1/({c}^{2}\u03f5)$ and $\gamma =1/\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$

$F=q(E-vB)=q(E-\frac{{v}^{2}}{{c}^{2}}\frac{1}{4\pi \u03f5}\frac{q}{{r}^{2}})=q{\gamma}^{2}E$

${F}^{\mathrm{\prime}}=q{E}^{\mathrm{\prime}}=qE=\frac{1}{{\gamma}^{2}}F$

(noting that dy is unaffected by a boost in the $x$ direction)

To solve more complicated problems of this sort, you usually want to talk about faraday tensors, $4$-currents, and einstein notation, but I hope this simple example gave you some insight into how electromagnetism and special relativity are related. You may have wondered what if the particles go so fast that the magnetic force exceeds the electrostatic repulsion, but indeed you can see that to do so you must exceed the speed of light!

Consider another frame (the rest frame) where both particles move side by side in the $x$ direction. The particles still repel each other electrostatically, but now they attract one another magnetically as well.

This may seem like a problem, but it isn't because even though the force measured in the rest frame is weaker since the electrostatic repulsion is partially canceled by the magnetic attraction, the time experienced by moving observers is also different due to the witchcraft of special relativity. If the frame moving with the charges is primed, and the rest frame is unprimed, then

$\begin{array}{rl}{t}^{\mathrm{\prime}}& =\gamma t\\ F& =q(E+v\times B)=q(E-vB)\\ {F}^{\mathrm{\prime}}& =q{E}^{\mathrm{\prime}}\end{array}$

What we seek to show is that since force is inversely proportional to the square of the time, $F$ and ${F}^{\prime}$ are related by a factor of ${\gamma}^{2}$ in which case observers in either frame will calculate matching trajectories using Maxwell's equations.

Since for a point charge

$E=\frac{1}{4\pi \u03f5}\frac{q}{{r}^{2}}$

$B=\frac{\mu}{4\pi}\frac{qv\times \hat{r}}{{r}^{2}}=\frac{\mu}{4\pi}\frac{qv}{{r}^{2}}$

Now we have all we need. Plugging $B$ in to $F$ and using $\mu =1/({c}^{2}\u03f5)$ and $\gamma =1/\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$

$F=q(E-vB)=q(E-\frac{{v}^{2}}{{c}^{2}}\frac{1}{4\pi \u03f5}\frac{q}{{r}^{2}})=q{\gamma}^{2}E$

${F}^{\mathrm{\prime}}=q{E}^{\mathrm{\prime}}=qE=\frac{1}{{\gamma}^{2}}F$

(noting that dy is unaffected by a boost in the $x$ direction)

To solve more complicated problems of this sort, you usually want to talk about faraday tensors, $4$-currents, and einstein notation, but I hope this simple example gave you some insight into how electromagnetism and special relativity are related. You may have wondered what if the particles go so fast that the magnetic force exceeds the electrostatic repulsion, but indeed you can see that to do so you must exceed the speed of light!

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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

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

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