Learn Relativity Easily with Examples

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 $\stackrel{\to <!--\; \to \; -->}{B}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{v}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{E}$

How does one tell whether an expression is covariant or not? I get it for a single tensor, but how is it defined when there is no overall up/down index to base it on?

A relativistic train with a rest length of $500.0$ m takes $780$ ns to pass a stationary observer standing on the train platform, as measured by the stationary observer.
(a) What is the speed of the train? (Hint: Remember to account for the Lorentz contraction of the spaceship.)
If $S$ is the inertial frame with respect to the observer and $S^{\prime }$ is the inertial frame with respect to the train, then the front of the train and an observer line up at $t=t^{\prime }=0$ and $x=x^{\prime }=0$,
Then, in the $S^{\prime }$ frame, the front of the train sees itself at $x^{\prime }=500m$ when $t=780\times <!--\; \times \; -->{10}^{-<!--\; -\; -->9}seconds$
The correct answer is:
$2.72\times <!--\; \times \; -->{10}^8\frac{m}{s}=0.907c$

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?

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

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,
$\stackrel{\to <!--\; \to \; -->}{r}=r\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\stackrel{\to <!--\; \to \; -->}{i}+y\stackrel{\to <!--\; \to \; -->}{j}$
${\stackrel{\to <!--\; \to \; -->}{r}}^{\prime }=(\stackrel{\dot{}<!--\; \dot{}\; -->}{r}cos\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->r\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})\stackrel{\to <!--\; \to \; -->}{i}+\stackrel{\dot{}<!--\; \dot{}\; -->}{y}\stackrel{\to <!--\; \to \; -->}{j}$
${\stackrel{\to <!--\; \to \; -->}{r}}^{″}=(\stackrel{¨<!--\; ¨\; -->}{r}cos\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\stackrel{\dot{}<!--\; \dot{}\; -->}{r}\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}sin\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\stackrel{\dot{}<!--\; \dot{}\; -->}{r}\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->r\stackrel{¨<!--\; ¨\; -->}{\mathrm{\theta <!--\; \theta \; -->}}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->r{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}}^{2}cos\mathrm{\theta <!--\; \theta \; -->})\stackrel{\to <!--\; \to \; -->}{i}+\stackrel{¨<!--\; ¨\; -->}{y}\stackrel{\to <!--\; \to \; -->}{j}$
$\mathrm{\omega <!--\; \omega \; -->}\times <!--\; \times \; -->{\stackrel{\to <!--\; \to \; -->}{r}}^{\prime }=-<!--\; -\; -->\mathrm{\omega <!--\; \omega \; -->}\stackrel{\dot{}<!--\; \dot{}\; -->}{y}\stackrel{\to <!--\; \to \; -->}{i}+(\mathrm{\omega <!--\; \omega \; -->}\stackrel{\dot{}<!--\; \dot{}\; -->}{r}\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\mathrm{\omega <!--\; \omega \; -->}r\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\theta <!--\; \theta \; -->}}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})\stackrel{\to <!--\; \to \; -->}{j}$
$\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}\times <!--\; \times \; -->(\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{r})=-<!--\; -\; -->{\mathrm{\omega <!--\; \omega \; -->}}^{2}r\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\stackrel{\to <!--\; \to \; -->}{i}-<!--\; -\; -->{\mathrm{\omega <!--\; \omega \; -->}}^{2}y\stackrel{\to <!--\; \to \; -->}{j}$

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?

What is the minimal voltage between anode and cathode in a linear accelerators to achieve speeds where relativity starts to show?

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?

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?

In momentum and energy is low enough, we end up with the same number of neutrons, protons and electrons after a collision as before it. This can be considered an approximate conservation law. Shouldn't there be a corresponding (approximate) symmetry corresponding to each of these particle groups?

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?

A source of light pulses moves with speed v directly away from an observer at rest in an inertial frame. How will the time of the emission of a light pulse as measured by the emitter be different to the time as measured by the observer (how will it be different than the time of the pulse reaching the observer)? To elaborate slightly, let us say $t_{E_1}$ is the time of emission as measured by the emitter, $t_{E_2}$ is the time of emission as measured by the observer, and $t_O$ is the time of observation of the pulse by the observer. How are these three times different, how would they be measured in practice?

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?

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

In the Lorenz gauge, we have a beautiful relation between the four-current and the four-potential:
$\mathrm{◻<!--\; ◻\; -->}{A}^{\mathrm{\alpha <!--\; \alpha \; -->}}={\mathrm{\mu <!--\; \mu \; -->}}_0{J}^{\mathrm{\alpha <!--\; \alpha \; -->}}$
To get $A$ in terms of $J$, however, we have to use a considerably uglier formula; or, at least, this is the formula presented in textbooks:
$A^{\mathrm{\alpha <!--\; \alpha \; -->}}(t,\mathbf{r})=\frac{{\mathrm{\mu <!--\; \mu \; -->}}_0}{4\mathrm{\pi <!--\; \pi \; -->}}\int <!--\; \int \; -->\frac{J^{\mathrm{\alpha <!--\; \alpha \; -->}}(t-<!--\; -\; -->\frac{1}{c}\Vert <!--\; \Vert \; -->\mathbf{r}-<!--\; -\; -->{\mathbf{r}}^{\prime }\Vert <!--\; \Vert \; -->,{\mathbf{r}}^{\prime })}{\Vert <!--\; \Vert \; -->\mathbf{r}-<!--\; -\; -->{\mathbf{r}}^{\prime }\Vert <!--\; \Vert \; -->}{d}^3{\mathbf{r}}^{\prime }$
The first equation is evidently Lorentz covariant. The second one, on the other hand, doesn't look covariant at all. The integrand has some messy dependence on $(t,\mathbf{r})$ and the integration goes over only the spatial dimensions.
Can we rewrite the second equation in a covariant form? If not, then why not?

consider a particle, $A$ receiving energy from a second one,particle $B$ in a one dimensional collision.
$E^2=p^2+m_0^2$
$EdE=pdp$
For particle $A$:
$E_{A}d{E}_A=p_{A}d{p}_A(1)$
For Particle $B$:
$E_{B}d{E}_B=p_{B}d{p}_B(2)$
Now
$\mid <!--\; \mid \; -->d{E}_A\mid <!--\; \mid \; -->=\mid <!--\; \mid \; -->d{E}_B\mid <!--\; \mid \; -->(3)$
From Conservation of linear momentum we have:
$\stackrel{\to <!--\; \to \; -->}{p_A}+\stackrel{\to <!--\; \to \; -->}{p_B}=\stackrel{\to <!--\; \to \; -->}{K}$
where $\stackrel{\to <!--\; \to \; -->}{k}$ is a constant vector. Now,
$d\stackrel{\to <!--\; \to \; -->}{p_A}+d\stackrel{\to <!--\; \to \; -->}{p_B}=0$
Or,
$d\stackrel{\to <!--\; \to \; -->}{p_B}=-<!--\; -\; -->d\stackrel{\to <!--\; \to \; -->}{p_A}$
Or,
$\mid <!--\; \mid \; -->d\stackrel{\to <!--\; \to \; -->}{p_B}\mid <!--\; \mid \; -->=\mid <!--\; \mid \; -->d\stackrel{\to <!--\; \to \; -->}{p_A}\mid <!--\; \mid \; -->(4)$
Applying relations $(3)$ and $(4)$ to $(1)$ and $(2)$ we have:
$\frac{E_A}{E_B}=\frac{\mid <!--\; \mid \; -->\stackrel{\to <!--\; \to \; -->}{p_A}\mid <!--\; \mid \; -->}{\mid <!--\; \mid \; -->\stackrel{\to <!--\; \to \; -->}{p_B}\mid <!--\; \mid \; -->}(5)$
A pair of particles cannot interact unless relation $(5)$ is satisfied. Can we conclude that that relation $(5)$ to be a restriction for $1D$ collisions?
Now let's move to the general type of $3$d collisions between a pair of particles $A$ and $B$.
A frame is chosen where the particle $B$ is initially at rest in it.
$E_{B}d{E}_B=\stackrel{\to <!--\; \to \; -->}{p_B}\cdot <!--\; \cdot \; -->d\stackrel{\to <!--\; \to \; -->}{p_B}(6)$
If the particle $B$ is initially at rest the RHS of $(6)$ is zero. But the LHS cannot be zero unless $d{E}_B=0$.
How does one get round this problem?

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.

When we observe astronomical objects like distant galaxies there are several complicating factors for estimating the distance:
- Relativistic speed result in length contraction
- Relativistic speed results in Doppler-shifts in the frequency of light
- Simultaneity is skewed due to our differing velocities
With these effects, how are we able to state with decent accuracy how far away all of the distant galaxies are from us? Do the above effects have an impact on the measurements?