Recent questions in Special 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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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?

RelativityAnswered question

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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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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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?

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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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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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Raiden Barr 2022-10-22

Since time slows down and length contracts, when we travel almost at speed of light, if the speed of light (or EM waves) remains same and the wavelength of light remains same, do we measure the wavelength more than it actually is?

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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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caschaillo7 2022-10-21

In classical (Newtonian) mechanics, every observer had the same past and the same future and if you had perfect knowledge about the current state of all particles in the universe, you could (theoretically) compute the future state of all particles in the universe.

With special (and general) relativity, we have the relativity of simultaneity. Therefore the best we can do is to say that for an event happening right now for any particular observer, we can theoretically predict the event if we know everything about the past light cone of the observer. However, it tachyons (that always travel faster than the speed of light) are allowed, then we cannot predict the future since a tachyon can come in from the space-like region for the observer and can cause an event that cannot be predicted by the past light cone. That is, I believe, why tachyons are incompatible with causality in relativity. Basically, the future cannot be predicted for any given observer so the universe is in general unpredictable - i.e. physics is impossible.

Now in quantum mechanics, perfect predictability is impossible in principle. Instead all we can predict is the probability of events happening. However, Schrodinger's equation allows the future wavefunction to be calculated given the current wavefunction. However, the wavefunction only allows for the predictions of probabilities of events happening. Quantum mechanics claims that this is the calculations of probabilities is the best that can be done by any physical theory.

So the question is: "Is the predictability of the future to whatever extent is possible (based on the present and the past) equivalent to the principle of causality?" Since prediction is the goal of physics and science in general, causality is necessary for physics and science to be possible.

With special (and general) relativity, we have the relativity of simultaneity. Therefore the best we can do is to say that for an event happening right now for any particular observer, we can theoretically predict the event if we know everything about the past light cone of the observer. However, it tachyons (that always travel faster than the speed of light) are allowed, then we cannot predict the future since a tachyon can come in from the space-like region for the observer and can cause an event that cannot be predicted by the past light cone. That is, I believe, why tachyons are incompatible with causality in relativity. Basically, the future cannot be predicted for any given observer so the universe is in general unpredictable - i.e. physics is impossible.

Now in quantum mechanics, perfect predictability is impossible in principle. Instead all we can predict is the probability of events happening. However, Schrodinger's equation allows the future wavefunction to be calculated given the current wavefunction. However, the wavefunction only allows for the predictions of probabilities of events happening. Quantum mechanics claims that this is the calculations of probabilities is the best that can be done by any physical theory.

So the question is: "Is the predictability of the future to whatever extent is possible (based on the present and the past) equivalent to the principle of causality?" Since prediction is the goal of physics and science in general, causality is necessary for physics and science to be possible.

RelativityAnswered question

Antwan Perez 2022-10-20

Integration by parts to derive relativistic kinetic energy

$\underset{0}{\overset{x}{\int}}\phantom{\rule{negativethinmathspace}{0ex}}\frac{d}{dt}{\textstyle [}mv\gamma (v){\textstyle ]}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}dx=v\phantom{\rule{negativethinmathspace}{0ex}}\cdot \phantom{\rule{negativethinmathspace}{0ex}}mv\gamma (v)-\underset{0}{\overset{v}{\int}}\phantom{\rule{negativethinmathspace}{0ex}}mv\gamma (v)\phantom{\rule{thinmathspace}{0ex}}dv$

$\underset{0}{\overset{x}{\int}}\phantom{\rule{negativethinmathspace}{0ex}}\frac{d}{dt}{\textstyle [}mv\gamma (v){\textstyle ]}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}dx=v\phantom{\rule{negativethinmathspace}{0ex}}\cdot \phantom{\rule{negativethinmathspace}{0ex}}mv\gamma (v)-\underset{0}{\overset{v}{\int}}\phantom{\rule{negativethinmathspace}{0ex}}mv\gamma (v)\phantom{\rule{thinmathspace}{0ex}}dv$

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gasavasiv 2022-10-19

There's a voltage difference of 1000 Volts between two points 2 meters apart. An electron starts at the point of lower potential and is left to travel alone in a straight line until it reaches the other point. What speed does the electron have when it reaches the second point?

Whether the retardation effects of the radiation of the electron are important here. I tried doing it using the relativistic formulae for the kinetic energy and got

$v=c\sqrt{1-\left(\frac{{m}_{e}{c}^{2}}{{m}_{e}{c}^{2}+V}\right){}^{2}}$

where ${m}_{e}$ is the electron mass, $c$ the speed of light, $V$ the voltage difference between the two points, and $v$ the final speed of the electron.

Whether the retardation effects of the radiation of the electron are important here. I tried doing it using the relativistic formulae for the kinetic energy and got

$v=c\sqrt{1-\left(\frac{{m}_{e}{c}^{2}}{{m}_{e}{c}^{2}+V}\right){}^{2}}$

where ${m}_{e}$ is the electron mass, $c$ the speed of light, $V$ the voltage difference between the two points, and $v$ the final speed of the electron.

One of the aspects of relativity Physics is taken by the concept of special relativity or the famous Eintsten's Theory that most people already know about. Regardless if you are using a special relativity calculator or working with word problems, it is essential to remember that we are dealing with the theory of gravity. Take a look at the answers and see how gravity can curve or warp in space. The reason why we are using special relativity equations is the relativity itself and the effect of the inertial frames. Don’t forget to check the general relativity concepts as well.