Brenton Dixon

2022-07-22

Orbital velocity of a circular planet is $\sqrt{aR}$, where a is the centripetal acceleration, and $R$ is radius of the planet. With ${v}_{1}$ as the tangential velocity of the rotating planet at the equator.

On the non rotating body, suppose that the orbital velocity is ${v}_{0}$, and, for an object launched on the rotating body's "equator", that the orbital velocity will be in the form of ${v}_{1}+{v}_{2}$ (the body and the object both going counterclockwise). Now, I half-hypothesized ${v}_{0}={v}_{1}+{v}_{2}=\sqrt{aR}$ and ${v}_{2}=\sqrt{{a}^{\prime}R}$ where ${a}^{\prime}$ is the "true" rotating body's acceleration, able to be calculated from the rotating frame of reference as

${a}^{\prime}=a-\frac{{v}_{1}^{2}}{R}$

The logic was that from rotating body's reference frame, the object would be traveling at ${v}_{2}$, less than ${v}_{0}$ because of the centrifugal force, so ${v}_{2}$ has to be the orbital velocity if the gravity was "weakened" by centrifugal force.

Tried to solve for ${a}^{\prime}$ and comparing it to the value, got from rotating frame of reference, ending up with

${a}^{\prime}=a-(\frac{{v}_{1}({v}_{0}+{v}_{2})}{R})$

Something's not right, and if I had to choose, I would guess the ${v}_{0}=v1+{v}_{2}$, that acceleration is not the same on those two planets, but I don't know how it would change, or why.

On the non rotating body, suppose that the orbital velocity is ${v}_{0}$, and, for an object launched on the rotating body's "equator", that the orbital velocity will be in the form of ${v}_{1}+{v}_{2}$ (the body and the object both going counterclockwise). Now, I half-hypothesized ${v}_{0}={v}_{1}+{v}_{2}=\sqrt{aR}$ and ${v}_{2}=\sqrt{{a}^{\prime}R}$ where ${a}^{\prime}$ is the "true" rotating body's acceleration, able to be calculated from the rotating frame of reference as

${a}^{\prime}=a-\frac{{v}_{1}^{2}}{R}$

The logic was that from rotating body's reference frame, the object would be traveling at ${v}_{2}$, less than ${v}_{0}$ because of the centrifugal force, so ${v}_{2}$ has to be the orbital velocity if the gravity was "weakened" by centrifugal force.

Tried to solve for ${a}^{\prime}$ and comparing it to the value, got from rotating frame of reference, ending up with

${a}^{\prime}=a-(\frac{{v}_{1}({v}_{0}+{v}_{2})}{R})$

Something's not right, and if I had to choose, I would guess the ${v}_{0}=v1+{v}_{2}$, that acceleration is not the same on those two planets, but I don't know how it would change, or why.

edgarovhg

Beginner2022-07-23Added 12 answers

You had some problems of frames of reference when making your hypothesis. Say that the non rotating planet is $A$ and the other is $B$. The orbit velocity is $v$. So an object orbiting $A$ have a velocity with respect to the center of $A$ and to the surface is the same, and is ${v}_{0}=v$. For $B$, say the object's velocity with respect to $B$'s surface is ${v}_{1}$, and $B$'s surface with respect to $B$'s center(the velocity of $B$'s rotation at its equator), ${v}_{2}$. Then the object's velocity with respect to $B$'s center is ${v}_{1}+{v}_{2}$.

And since $A$ and $B$ are otherwise identical, ${v}_{0}={v}_{1}+{v}_{2}=v$.

And since $A$ and $B$ are otherwise identical, ${v}_{0}={v}_{1}+{v}_{2}=v$.

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Light travels at a very high speed

Due to rectilinear propagation of light

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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:

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

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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

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

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How to calculate $ds$?