Master Special Relativity with Expert Help and Solutions

Up to the work of Riemann and Gauss, this definition would have made clear to me examples of geometrical objects: a line, square, cube, hypercube; each of these possessing geometrical properties such as the number of sides, angles between faces, the dimension of space that contains them etc. Hence a geometrical object was a set of measurements associated with an object using a ruler.
After the work of Einstein and Minkowski who showed that time and space were a part of one another, would it be correct to say:
1. a geometrical object is a set of measurements associated with an object of distance and time using a ruler and a clock?
2. Geometrical objects includes an interval of time, the invariant space-time interval?

What is the current status or acceptance of block time as it relates to Einstein's theory of relativity? Has quantum mechanics ruled it out or is it still the favored view of the world?

Suppose that Standard Model Higgs mechanism broke electromagnetism, by e.g. veving the charged component of the doublet, so that the photon was massive with $m_{\mathrm{\gamma <!--\; \gamma \; -->}}\sim <!--\; \sim \; -->v$. Could such a Universe still have large scale structure? Atoms (i.e. stable electronic orbits)? Life?
Assuming we got past those hurdles, would it have been much more difficult to have discovered special relativity? Would we have been stuck at Galilean invariance, without the invariance of the speed of light from which to build SR? I appreciate that this is speculative.
And, also, the identical question but for a coloured Higgs vacuum that breaks QCD. Would broken QCD still be confining? I guess so - so we could still have nucleons and the resulting chemistry?

Picture a situation where we have two observers, $A$ and $B$, and a system in a certain quantum state. If $B$ makes a measurement of some observable, say energy for example, the state will collapse to one of the possible energy states with a definite probability. If $A$ makes a measurement after this has happened he will observe a precise energy given by the state to which our system has collapsed after the the measurement made by $B$.
Pretty straightforward till now. Now, if both observers make a measurement at the same time they will both measure the same value. But we know from special relativity that simultaneity is relative to the observer, so we may think that from some observer this measurements won't be simultaneous and actually for him 𝐴 will have made the measurement before $B$.
My question is obvious then, how do measurements and relativity of simultaneity marry?

How special relativity leads to Lorentz Invariance of the Maxwell Equations. Differential form will suffice.

The distance between interference lines in the double-slit experiment is:
$w=z\mathrm{\lambda <!--\; \lambda \; -->}/d$
Where:
$w$: Distance between fringes
$z$: Distance from slits to screen.
$\mathrm{\lambda <!--\; \lambda \; -->}$: Wave length of light
$d$: Distance between slits.
Now, looking at the setup from a reference frame traveling in the direction of the distance 𝑧 between slit and screen and moving away from the slit in direction of the screen, the light gets redshifted and the distance 𝑧 gets Lorentz contracted:
$w^{\prime }=\frac{z}{\mathrm{\gamma <!--\; \gamma \; -->}}\mathrm{\gamma <!--\; \gamma \; -->}(1+\frac{v}{c})\mathrm{\lambda <!--\; \lambda \; -->}/d=z(1+\frac{v}{c})\mathrm{\lambda <!--\; \lambda \; -->}/d$
This only holds for the first fringe or so, that aren't too far from the center, because the red-shift formula changes farther out. However, it's obvious that the distance between fringes has increased!
Where is my mistake?

When a free particle move in space with a known momentum and energy then what is the physical process that gives mass to that free (relativistic) particle?
What is role does the Higgs field in that process, if any?

What does a rotating hoop, with each point moving at a velocity close to the speed of light, appear like with respect from a stationary observers perspective. For example how does the shape of the hoop change?

How much does a proton weigh when it is going around the LHC at CERN?
Considering that speed increases weight and the proton is going at almost the speed of light, I would like to know how much a speeding proton would weigh in the LHC.

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?

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

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?

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?

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?

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?

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?

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?