Recent questions in Relativity

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Heddertyk1j 2023-02-16

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.

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Talia Frederick 2022-11-25

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 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 inverted

RelativityAnswered question

powerinojSs 2022-11-22

In 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?

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

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

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

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

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?

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Bailee Richards 2022-11-22

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.

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.

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Juan Lowe 2022-11-16

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?

In a frame of reference, can we have one reference point or more than one?

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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?

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Hanna Webster 2022-11-13

If time in systems moving with different speed goes differently, does speed of entropy change differ in these systems?

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bruinhemd3ji 2022-11-08

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?

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Jadon Camacho 2022-11-05

The mass of a body on the surface of the Moon is greater than that on Earth

according to the equation $E=m{c}^{2}$

according to the equation $E=m{c}^{2}$

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pighead73283r 2022-11-05

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?

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

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?

RelativityAnswered question

Amy Bright 2022-11-04

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?

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?

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Alvin Parks 2022-11-03

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?

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Aden Lambert 2022-11-02

We know that

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

How to calculate $ds$?

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

How to calculate $ds$?

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cousinhaui 2022-11-01

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?

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Chloe Arnold 2022-10-29

If there were a vast lever floating in free space, a rigid body with length greater than the width of a galaxy, made of a hypothetical material that could endure unlimited internal stress, and this lever began to rotate about its middle like a propeller so that a person looking at the universe would simply see it spinning at a gentle pace like a windmill, would not its ends be moving many, many times the speed of light?

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Kamila Frye 2022-10-29

Why does Lorentz factor not hold for relativistic mass when we apply it to photons?

We know that the photon itself is massless particle ${m}_{0}=0$. But we also know, that the mass of the objects does increase with their energy. And we know that under certain circumstances (gravity, collision with objects) the light does behave like a beam of particles that do have a mass.

Now, this is the equation to get the (relativistic) mass of the object with certain speed:

$m=\frac{{m}_{0}}{\sqrt{1-{v}^{2}/{c}^{2}}}$

We already do know some things: ${m}_{0}=0$, $v=c$.

$m=\frac{0}{\sqrt{1-{c}^{2}/{c}^{2}}}$

$m=\frac{0}{0}$

Expression $\frac{0}{0}$ is not equal to 0. So what is this?

We know that the photon itself is massless particle ${m}_{0}=0$. But we also know, that the mass of the objects does increase with their energy. And we know that under certain circumstances (gravity, collision with objects) the light does behave like a beam of particles that do have a mass.

Now, this is the equation to get the (relativistic) mass of the object with certain speed:

$m=\frac{{m}_{0}}{\sqrt{1-{v}^{2}/{c}^{2}}}$

We already do know some things: ${m}_{0}=0$, $v=c$.

$m=\frac{0}{\sqrt{1-{c}^{2}/{c}^{2}}}$

$m=\frac{0}{0}$

Expression $\frac{0}{0}$ is not equal to 0. So what is this?

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raapjeqp 2022-10-22

Dark energy physically can be interpreted as either a fluid with positive mass but pressure the negative of its density (pressure has units of energy/volume, and energy is mass), or a property of space. If it's a fluid, it should add to the mass of black holes like any form of energy (no hair), and the black hole should grow? However, if dark energy is a property of space, then this won't happen. Is my reasoning correct that we can differentiate (in theory) by looking at black hole's growth rate?

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Seettiffrourfk6 2022-10-22

Having some issues regarding the Euler's angles. Following is the short description of them problem.

In the first step, I determined the Euler's angles to invert my frame of reference that is $X$, $Y$ and $Z$ axes become $-X$, $-Y$ and $-Z$ respectively. I calculated the Euler's angles to be $(135,109.47,45)$ in degrees for $ZXZ$ scheme of transformation.

Now I expect that transforming any vector by same Euler's angles will invert it. For example, if I transform $(1,1,1)$ vector with Euler's angles $(135,109.47,45)$, I should get the vector $(-1,-1,-1)$, but I find that it remains unchanged.

In the first step, I determined the Euler's angles to invert my frame of reference that is $X$, $Y$ and $Z$ axes become $-X$, $-Y$ and $-Z$ respectively. I calculated the Euler's angles to be $(135,109.47,45)$ in degrees for $ZXZ$ scheme of transformation.

Now I expect that transforming any vector by same Euler's angles will invert it. For example, if I transform $(1,1,1)$ vector with Euler's angles $(135,109.47,45)$, I should get the vector $(-1,-1,-1)$, but I find that it remains unchanged.

Understanding relativity is paramount in Physics as it helps to keep one's mind open and unbiased when conducting complex scientific experiments. Still, it is very difficult to find relativity help, especially when it involves equation of logic questions that require written answers. We provide help with both by offering numerous answers that contain relativity problems and solutions provided by your fellow students and volunteers that have faced the same challenges. When you are working with moving objects or designing things, focus on the frame of reference concept to see things from the perspective. Start with the verbal explanations of the relativity concept first to understand the concept bef