Lorentz Boosts in the same direction should form a group. Two boosts along the x axis should produce another boost along the x axis. Is that correct?

Hagman7v
2022-09-26
Answered

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vyhlodatis

Answered 2022-09-27
Author has **14** answers

Yes, it is a one-dimensional subgroup generated by exponentiating an infinitesimal boost. Every one dimensional exponentiation of a generator forms an abelian group, because ${e}^{aG}{e}^{bG}={e}^{(a+b)G}$ there is nothing to not commute. This result is the addition of velocities, you can explicity check that this is associative (it is always manifestly commutative).

asked 2022-07-23

From a Hamiltonian for the Dirac Equation, we can add a potential term to it simply by adjusting the momentum operator so that ${p}^{\mu}\to {p}^{\mu}-{A}^{\mu}$, where ${A}^{\mu}$ is the relevant potential. But how do you calculate ${A}^{\mu}$? For example, what would ${A}^{\mu}$ be for an electron in an electromagnetic field given by the tensor ${F}^{\alpha \beta}$?

asked 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?

asked 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?

asked 2022-05-19

Is there a time + two spatial dimension representation of a Minkowski-space surface which could be constructed within our own (assumed Euclidean) 3D space such that geometric movement within the surface would intuitively demonstrate the “strange" effects of the Lorentz transformation (length contraction, time dilation)?

asked 2022-07-16

is space expansion the same as time dilation ?

asked 2022-09-29

Assume there are two points in spacetime $a=(t,x,y,z)$ and ${a}^{\prime}=({t}^{\prime},{x}^{\prime},{y}^{\prime},{z}^{\prime})$. Let's say that the first one is in the origin of spacetime i.e. $a=(0,0,0,0)$. The point ${a}^{\prime}$ has two possibilities

1. ${a}^{\prime}=(0,10cm,0,0)$ i.e. it's $10cm$ right of $a$

2. ${a}^{\prime}=(1ns,1cm,0,0)$ i.e. it's $1cm$ right of $a$ but $1ns$ ahead of it as well.

Someone want to send a signal from $a$ to ${a}^{\prime}$. Is there a way that these two points(the two ${a}^{\prime}$) receive the signal instantly either in space or time?

1. ${a}^{\prime}=(0,10cm,0,0)$ i.e. it's $10cm$ right of $a$

2. ${a}^{\prime}=(1ns,1cm,0,0)$ i.e. it's $1cm$ right of $a$ but $1ns$ ahead of it as well.

Someone want to send a signal from $a$ to ${a}^{\prime}$. Is there a way that these two points(the two ${a}^{\prime}$) receive the signal instantly either in space or time?

asked 2022-09-27

Einstein's Special Theory of relativity postulates that the speed of light is same for all frames.

Suppose a neutrino is there moving at the speed of light. Then will that neutrino also be flowing with same speed for all frames or is this a special property of EM waves?

Suppose a neutrino is there moving at the speed of light. Then will that neutrino also be flowing with same speed for all frames or is this a special property of EM waves?