Sincere Garcia

2022-10-08

Imagine a cup stuck to the floor of a falling elevator with the help of some impressive adhesive. In a frame of reference is this cup is being watched through the walls of a transparent elevator, this frame of reference is not accelerating with respect to earth. Now, since earth is an inertial frame of reference, and since this frame of reference is not accelerating with respect to it then it must be the case that this frame of reference in which the cup is viewed is also an inertial frame of reference.

Let's call this frame of reference viewing the cup S since S is an inertial frame of reference, it must follow the law:

$\begin{array}{}\text{(1)}& a=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}F=0\text{}\text{(Newtons First Law)}\end{array}$

The force on the cup is the tension force due to the adhesive and the force due to the Earth, the reaction force, all add up to zero because the cup is seen to not accelerate within the lift.

Okay the so the cup has zero forces acting on it from the observer, the tension force, the gravitational force, the reaction force, all add up to zero. Since $S$ is an inertial frame of reference and $F=0$, the acceleration must be zero from $1$. But according to $S$ it isn't, it is accelerating with $9.8\text{}\text{m}/{\text{s}}^{2}$.

In conclusion:

An inertial frame of reference observes an object on which the total force is zero but still has non-zero acceleration. How can this contradiction be resolved?

Let's call this frame of reference viewing the cup S since S is an inertial frame of reference, it must follow the law:

$\begin{array}{}\text{(1)}& a=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}F=0\text{}\text{(Newtons First Law)}\end{array}$

The force on the cup is the tension force due to the adhesive and the force due to the Earth, the reaction force, all add up to zero because the cup is seen to not accelerate within the lift.

Okay the so the cup has zero forces acting on it from the observer, the tension force, the gravitational force, the reaction force, all add up to zero. Since $S$ is an inertial frame of reference and $F=0$, the acceleration must be zero from $1$. But according to $S$ it isn't, it is accelerating with $9.8\text{}\text{m}/{\text{s}}^{2}$.

In conclusion:

An inertial frame of reference observes an object on which the total force is zero but still has non-zero acceleration. How can this contradiction be resolved?

Dakota Duarte

Beginner2022-10-09Added 7 answers

The problem is in this statement "the tension force, the gravitational force, the reaction force, all add up to zero"

Actually the only force acting on the cup is gravitational force and hence in your "inertial reference frame" the cup is seen to be accelerating downward due to a real force. So there is no contradiction.

Because :

- Glue force acts as friction over here and hence doesn't occur in the absence of relative motion.

- Normal Reaction acts when two bodies try to occupy the same space. But as you may notice that since the cup is just hovering over the surface i.e., it doesn't apply any force to penetrate it and so neither the normal reaction occurs.

Actually the only force acting on the cup is gravitational force and hence in your "inertial reference frame" the cup is seen to be accelerating downward due to a real force. So there is no contradiction.

Because :

- Glue force acts as friction over here and hence doesn't occur in the absence of relative motion.

- Normal Reaction acts when two bodies try to occupy the same space. But as you may notice that since the cup is just hovering over the surface i.e., it doesn't apply any force to penetrate it and so neither the normal reaction occurs.

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:

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

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

How to calculate $ds$?