Justine Webster

2022-05-19

Suppose we have a sphere of radius $r$ and mass m and a negatively charged test particle at distance d from its center, $d\gg r$. If the sphere is electrically neutral, the particle will fall toward the sphere because of gravity. As we deposit electrons on the surface of the sphere, the Coulomb force will overcome gravity and the test particle will start to accelerate away. Now suppose we keep adding even more electrons to the sphere. If we have n electrons, the distribution of their pairwise distances has a mean proportional to $r$, and there are $n(n-1)/2$ such pairs, so the binding energy is about ${n}^{2}/r$. If this term is included in the total mass-energy of the sphere, the gravitational force on the test particle would seem to be increasing quadratically with $n$, and therefore eventually overcomes the linearly-increasing Coulomb force. The particle slows down, turns around, and starts falling again. This seems absurd; what is wrong with this analysis?

frogoogg31

Beginner2022-05-20Added 12 answers

Suppose we have $N$ electron in the sphere, the electrostatic energy to bring a new electron into the sphere is $Ne/R$. If we add a new electron, only the net electrostatic potential $Ne/R$.

the net potential on far-away electrons is

$N(e-{m}_{e}G)/R$

what is actually quadratic in $N$ is the energy required to bind together $N$ electrons in the sphere, it is actually

$e/R+2e/R+...Ne/R=N(N-1)e/2R$

This also contributes to gravitational weight, but there will be a maximum capacity where the electrons will escape the sphere

the capacitance of a sphere is given by $4\pi \u03f5R$. So in this case the binding energy is bounded by the capacitance of the conductor used for your sphere

the net potential on far-away electrons is

$N(e-{m}_{e}G)/R$

what is actually quadratic in $N$ is the energy required to bind together $N$ electrons in the sphere, it is actually

$e/R+2e/R+...Ne/R=N(N-1)e/2R$

This also contributes to gravitational weight, but there will be a maximum capacity where the electrons will escape the sphere

the capacitance of a sphere is given by $4\pi \u03f5R$. So in this case the binding energy is bounded by the capacitance of the conductor used for your sphere

Marco Villanueva

Beginner2022-05-21Added 6 answers

The statement that the gravitational attraction will eventually dominate the coulomb repulsion as n increases is correct. You probably think the restmass of the electrons will invoke gravitational attraction, but that part is neglible for high electrondensities on the sphere. The gravitational attraction by (the curvature of spacetime caused by) bindingenergy is far greater for high densities.

I don't know where the breakpoint is, but suppose I look way past that limit, i.e. an incredible dense sphere(or shell) of electrons. If I make it dense enough, that could very well become a black hole. Note that this would be a strange black hole, since the 'mass' of such a black hole consists almost entirely of the bindingenergy, and not the restmasses of the electrons. From this point of view it might be more easily imaginable that the gravitional attraction will dominate eventually.

I don't know where the breakpoint is, but suppose I look way past that limit, i.e. an incredible dense sphere(or shell) of electrons. If I make it dense enough, that could very well become a black hole. Note that this would be a strange black hole, since the 'mass' of such a black hole consists almost entirely of the bindingenergy, and not the restmasses of the electrons. From this point of view it might be more easily imaginable that the gravitional attraction will dominate eventually.

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?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?

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 $ds$?