Boost Your Knowledge on Frame of Reference

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 inverted

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?

Having some issues regarding the Euler's angles. Following is the short description of them problem.
In the first step, I determined the Euler's angles to invert my frame of reference that is $X$, $Y$ and $Z$ axes become $-<!--\; -\; -->X$, $-<!--\; -\; -->Y$ and $-<!--\; -\; -->Z$ respectively. I calculated the Euler's angles to be $(135,109.47,45)$ in degrees for $ZXZ$ scheme of transformation.
Now I expect that transforming any vector by same Euler's angles will invert it. For example, if I transform $(1,1,1)$ vector with Euler's angles $(135,109.47,45)$, I should get the vector $(-<!--\; -\; -->1,-<!--\; -\; -->1,-<!--\; -\; -->1)$, but I find that it remains unchanged.

Consider an elevator moving down with uniform velocity. A person standing inside watches an object fall from the ceiling of the elevator to the floor. Say the height of the elevator is $h$. Then the work done by gravity in that frame of reference should be $mgh$. But consider this same event being watched by someone else in the stationary frame of reference. In his reference frame, the object travels a larger distance as it falls from the ceiling to the floor of the elevator because the floor itself is moving downwards (one can calculate this extra distance covered to be $u\sqrt{\frac{2h}{g}}$) and hence the change in kinetic energy should be more in that frame than in the moving frame.
Shouldn't the change in kinetic energy be more in a moving elevator from a stationary frame of reference?

To person A standing on a railway platform, person B on the train travelling past would seems to be aging slower (if such a thing were perceptible) and to person B it would appear that person A was aging faster.
But in the absence of a train, or a platform, or even a planet, the two people would appear to be moving apart. Without a frame of reference it would be difficult to say which is moving and which is not. However, they can't both be appearing to age faster than the other.
So which one would be aging faster and which slower? Or rather, how can one tell which one is moving?

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:
$\begin{array}{}\text{(1)}& a=0⟺<!--\; ⟺\; -->F=0\text{(Newtons First Law)}\end{array}$
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 $9.8\text{m}/{\text{s}}^2$.
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?

Consider two identical charges moving with uniform velocity. There will be a magnetic force of attraction between them as two currents in the same direction attract each other. If I sit on one of the charges, according to me the other charge is not moving. So there wont be any magnetic attraction. How does changing the frame of reference change the outcome of the interaction?

Suppose a neutrino is seen travelling so fast that its Lorentz gamma factor is 100,000. It races past an old, no longer active neutron star, narrowly missing it. As far as the neutrino is concerned, it is the neutron star that is moving at extreme speed, & its mass is 100,000 times larger than 2 solar masses. Therefore, from the speeding neutrino's perspective, the neutron star should appear to be a black hole definitely large enough to trap the neutrino. So how come the speeding neutrino continues its travel right past the old stellar remnant? Is there an agreed name for this question or paradox?

Consider a particle in two inertial reference frames $\mathrm{\Sigma <!--\; \Sigma \; -->}$ and ${\mathrm{\Sigma <!--\; \Sigma \; -->}}^{\prime }$. The reference frame ${\mathrm{\Sigma <!--\; \Sigma \; -->}}^{\prime }$ is moving with uniform velocity $v$ relative to $\mathrm{\Sigma <!--\; \Sigma \; -->}$. The particle is at rest in ${\mathrm{\Sigma <!--\; \Sigma \; -->}}^{\prime }$. Both reference frames have common axes $x$ and $x^{\prime }$. When doing a certain calculation in both reference frames, which one of the obtained results is considered correct? Are they considered both correct or the one obtained by an observer in ${\mathrm{\Sigma <!--\; \Sigma \; -->}}^{\prime }$?

Consider a particle at rest with respect to the rotating frame. Since there is no velocity there is no force felt from $\stackrel{\to <!--\; \to \; -->}{B^{\prime }}$.
$$\begin{gathered} \stackrel{\to <!--\; \to \; -->}{F}=m\stackrel{\to <!--\; \to \; -->}{a}=q\stackrel{\to <!--\; \to \; -->}{E^{\prime }}=q(\stackrel{\to <!--\; \to \; -->}{E}+\stackrel{\to <!--\; \to \; -->}{v}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{B}) \\ \stackrel{\to <!--\; \to \; -->}{E^{\prime }}=\stackrel{\to <!--\; \to \; -->}{E}+\stackrel{\to <!--\; \to \; -->}{v}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{B} \end{gathered}$$
Additionally, considering a stationary particle in the inertial frame of reference, with pseudo forces included:
$$\begin{gathered} \stackrel{\to <!--\; \to \; -->}{F}=m\stackrel{\to <!--\; \to \; -->}{a}+m\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{v}+m\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{v}+m\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{x} \\ =q(\stackrel{\to <!--\; \to \; -->}{E^{\prime }}+\stackrel{\to <!--\; \to \; -->}{v}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{B^{\prime }})=q\stackrel{\to <!--\; \to \; -->}{E} \end{gathered}$$
However if I just plug in $\stackrel{\to <!--\; \to \; -->}{E^{\prime }}$ into the second equation, $\stackrel{\to <!--\; \to \; -->}{B^{\prime }}$ = $\stackrel{\to <!--\; \to \; -->}{B}$ which seems incorrect. Looking at the non-relativistic portions of a Lorentz transformations has $\stackrel{\to <!--\; \to \; -->}{B}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{v}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{E}$

"An inertial frame of reference is one frame where Newton's First Law holds, therefore, a body has a constant velocity or velocity equal zero. And, the sum of all forces equals zero, there is no aceleration etc..."
But, in an inertial frame of reference holds $F=ma$, the second law. It's confuses me, why we can assume $F=ma$ in an inertial frame of reference?

Why Dirac's equation the smallest possible matrices ($2\times <!--\; \times \; -->2$) are used?

How to solve this:
A ${\mathrm{\pi <!--\; \pi \; -->}}^{+}$ decays into a muon and neutrino. Find the pion's energy if
$maxE\nu /minE\nu =100/1;$
$m_{\mathrm{\nu <!--\; \nu \; -->}}=0$
$m_{\mathrm{\pi <!--\; \pi \; -->}}\ast <!--\; \ast \; -->c^2=140\text{ peta-eV}$

Regarding a 1 g rocket that "the amount you accelerate would be less due to relativity". Does that mean that from the crew's time dilated perspective, they would experience less acceleration than we observe in our frame of reference? Could a ship be accelerated at say, 10 g from our frame of reference on Earth, while the crew of the ship only experiences 1 g of acceleration in theirs? If this were possible, how far can we take this, and how quickly?

Relativistic kinetic energy is usually derived by assuming a scalar quantity is conserved in an elastic collision thought experiment, and deriving the expression for this quantity. To me, it looks bodged because it assumes this conserved quantitiy exists in the first place, whereas I'd like a derivation based upon using $KE$$=\frac{1}{2}m{v}^2$ in one frame, and then summing it in another frame say to get the total kinetic energy. Can this or a similar prodecure be done to get the relativistic kinetic energy?

Two frames of references $S$ and $S^{\prime }$ have a common origin $O$ and $S^{\prime }$ rotates with constant angular velocity $\mathrm{\omega <!--\; \omega \; -->}$ with respect to $S$.
A square hoop $ABCD$ is made of fine smooth wire and has side length $2a$. The hoop is horizontal and rotating with constant angular speed $\mathrm{\omega <!--\; \omega \; -->}$ about a vertical axis through $A$. A small bead which can slide on the wire is initially at rest at the midpoint of the side $BC$. Choose axes relative to the hoop and let $y$ be the distance of the bead from the vertex $B$ on the side $BC$. Write down the position vector of the bead in your rotating frame. Show that $\stackrel{¨<!--\; ¨\; -->}{y}-<!--\; -\; -->{\mathrm{\omega <!--\; \omega \; -->}}^{2}y=0$ using the expression for the acceleration. Hence find the time which the bead takes to reach a vertex $C$.
I showed that $\frac{d^2\stackrel{\to <!--\; \to \; -->}{r}}{d{t}^2}=(\frac{d^2\stackrel{\to <!--\; \to \; -->}{r}}{d{t}^2}{)}^{\prime }+2\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}\times <!--\; \times \; -->(\frac{d\stackrel{\to <!--\; \to \; -->}{r}}{dt}{)}^{\prime }+\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}\times <!--\; \times \; -->(\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{r})$ where ′ indicates that it's done in rotating frame. $\stackrel{\to <!--\; \to \; -->}{r}$ is position vector of a point $P$ measured from the origin.
So,
$\stackrel{\to <!--\; \to \; -->}{r}=r\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\stackrel{\to <!--\; \to \; -->}{i}+y\stackrel{\to <!--\; \to \; -->}{j}$
${\stackrel{\to <!--\; \to \; -->}{r}}^{\prime }=(\stackrel{\dot{}<!--\; \dot{}\; -->}{r}cos\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->r\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})\stackrel{\to <!--\; \to \; -->}{i}+\stackrel{\dot{}<!--\; \dot{}\; -->}{y}\stackrel{\to <!--\; \to \; -->}{j}$
${\stackrel{\to <!--\; \to \; -->}{r}}^{″}=(\stackrel{¨<!--\; ¨\; -->}{r}cos\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\stackrel{\dot{}<!--\; \dot{}\; -->}{r}\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}sin\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\stackrel{\dot{}<!--\; \dot{}\; -->}{r}\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->r\stackrel{¨<!--\; ¨\; -->}{\mathrm{\theta <!--\; \theta \; -->}}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->r{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}}^{2}cos\mathrm{\theta <!--\; \theta \; -->})\stackrel{\to <!--\; \to \; -->}{i}+\stackrel{¨<!--\; ¨\; -->}{y}\stackrel{\to <!--\; \to \; -->}{j}$
$\mathrm{\omega <!--\; \omega \; -->}\times <!--\; \times \; -->{\stackrel{\to <!--\; \to \; -->}{r}}^{\prime }=-<!--\; -\; -->\mathrm{\omega <!--\; \omega \; -->}\stackrel{\dot{}<!--\; \dot{}\; -->}{y}\stackrel{\to <!--\; \to \; -->}{i}+(\mathrm{\omega <!--\; \omega \; -->}\stackrel{\dot{}<!--\; \dot{}\; -->}{r}\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\mathrm{\omega <!--\; \omega \; -->}r\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})\stackrel{\to <!--\; \to \; -->}{j}$
$\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}\times <!--\; \times \; -->(\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{r})=-<!--\; -\; -->{\mathrm{\omega <!--\; \omega \; -->}}^{2}r\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\stackrel{\to <!--\; \to \; -->}{i}-<!--\; -\; -->{\mathrm{\omega <!--\; \omega \; -->}}^{2}y\stackrel{\to <!--\; \to \; -->}{j}$

Is pseudo force another form of inertia observed from accelerating frame of reference?

According to the relativity theory, there is no difference between two observers moving with respect to each other, they both experience identical laws of physics. However, does the same go for rotation?
An object gains rotational energy when torque is applied to it, and it also exerts forces on other objects depending on its rotational velocity. This has been used in many sci-fi movies, where the space-ship has a rotating part (as in the Martian for example), and the astronauts are able to stand casually in there, due to the force that the rotating part exerts on them. Obviously, that would not work if that part of the ship wasn't rotating. However, how do we know whether it is in fact rotating or not? If we are able to tell whether it is, isn't there an universal rotational stationary frame of reference, to which we have to compare everything that rotates?

How to transform the stiffness tensor of a rhombohedral crystal from crystallographic frame of reference to laboratory fame of reference
For crystal structures having orthogonal crystallographic axes (like tetragonal or orthorhombic), one can simply use the transformation of axes system (like Euler's angles based transformation matrix) to simply rotate the tensor property to achieve the goal. But I am not sure how to achieve the same with crystal with non orthogonal axes like rhombohedral/trigonal crystals with axes making angles of 60 degrees with each other.