Kareem Mejia

2022-11-14

From eintein's theory of relativity that lets say, a ruler is travelling to a speed if light, then we can say that the ruler (from our view as observers) has shorten. but why, lets say we have a 15 km runway, and we let an electron run through it, in electron's perspective, the length of runway is now only around, very short lets say in centimeters. why is that so?

Kailee Abbott

Beginner2022-11-15Added 14 answers

Amongst the things physicists look for are invarients. That is, quantities that are constant. You often find that if you understand the invarients they tell you a lot about the system you're looking at. In the case of special relativity there is an invarient called the line element, and denoted by ds. If you move a distance $dx$, $dy$ and $dz$ in space, and $dt$ in time, then the line element is given by:

$d{s}^{2}=-{c}^{2}d{t}^{2}+d{x}^{2}+d{y}^{2}+d{z}^{2}$

The significance of ds is that all observers in all frames will agree on the value of $ds$ i.e. it is an invarient.

The fact that ds is invarient is pretty much all you need to know about special relativity. It even encodes the fact that the speed of light is constant. The weird effects like time dilation and length contraction arise from the fact that in the equation of $d{s}^{2}$ the term in $d{t}^{2}$ has a negative sign while all the spatial terms have a positive sign.

I guess your next question is why $ds$ is given by the above equation and why it's an invarient, but those questions I can't answer. That's just the way the universe is.

$d{s}^{2}=-{c}^{2}d{t}^{2}+d{x}^{2}+d{y}^{2}+d{z}^{2}$

The significance of ds is that all observers in all frames will agree on the value of $ds$ i.e. it is an invarient.

The fact that ds is invarient is pretty much all you need to know about special relativity. It even encodes the fact that the speed of light is constant. The weird effects like time dilation and length contraction arise from the fact that in the equation of $d{s}^{2}$ the term in $d{t}^{2}$ has a negative sign while all the spatial terms have a positive sign.

I guess your next question is why $ds$ is given by the above equation and why it's an invarient, but those questions I can't answer. That's just the way the universe is.

Jenny Roberson

Beginner2022-11-16Added 4 answers

In the frame of the electron, the electron is stationary and the runway is rushing past it. So whether a runway is rushing past the electron or a ruler is rushing past the electron, both are length contracted in the direction of motion.

What makes the planets rotate around the sun.

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.

Why is the image formed in a pinhole camera is inverted?

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:

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

where ' indicates a property of the moving frame.For $n$ dimensions, $n+1$ reference points are sufficient to fully define a reference frame. I just want the above line explanation.

In a frame of reference, can we have one reference point or more than one?If time in systems moving with different speed goes differently, does speed of entropy change differ in these systems?

If a rocket ship is traveling at .99c for 1 year, and is streaming a video at 30 frames/sec to earth, how would the earth feed be affected? Would it show the video at a much slower rate, would it remain constant, or would it be sped up?

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

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

If I'm not mistaking one cannot freely change the order of the Dirac spinors $u({\mathbf{k}}_{1},{m}_{{s}_{1}})u({\mathbf{k}}_{2},{m}_{{s}_{2}})\ne u({\mathbf{k}}_{2},{m}_{{s}_{2}})u({\mathbf{k}}_{1},{m}_{{s}_{1}})$ so these expressions seem to be uncompatible. What would the correct expression look like?The mass of a body on the surface of the Moon is greater than that on Earth

according to the equation $E=m{c}^{2}$Is time travel possible?

Time travel - often featured in movies, books, or facetiously in conversation. There are also theories treating time as simply another dimension, which to the layperson might imply forward and backward movement is possible at will. But what do we know scientifically with respect to the possibility or impossibility of controlled time travel? Are there any testable theories on the horizon that may support or eliminate controlled time travel as a possibility?Suppose a particle decays to three other particles. The masses of all particles are assumed to be known and we work in the rest frame of the parent particle. So there are 12 parameters for this because of the 4-momenta of the three daughter particles. Now the constraint of momentum conservation imposes 4 constraints and reduces the number of parameters to 8. Further, the energy-momentum relation for each particle imposes three more constraints and reduces the number of parameters to 5. Are there any other constraints that reduce the number of parameters to 2?

We know that

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

How to calculate $ds$?On a relativistic train a kitchen timer is set to give a signal after 4.95 minutes. You are standing next to the rail road tracks. In your reference frame it takes 8.95 minutes for the kitchen timer to signal. What is the speed of the train?