Learn Relativity Easily with Examples

I am asked to show that the matrix
$S=1-<!--\; -\; -->\frac{i}{4}{\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}$
represents the infinitesimal Lorentz transformation
${\mathrm{\Lambda <!--\; \Lambda \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{}_{\mathrm{\nu <!--\; \nu \; -->}}={\mathrm{\delta <!--\; \delta \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{}_{\mathrm{\nu <!--\; \nu \; -->}}+{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{}_{\mathrm{\nu <!--\; \nu \; -->}},$
in the sense that
$S^{-<!--\; -\; -->1}{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}S={\mathrm{\Lambda <!--\; \Lambda \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{}_{\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\nu <!--\; \nu \; -->}}.$
I have already proven that $S^{-<!--\; -\; -->1}={\mathrm{\gamma <!--\; \gamma \; -->}}^0{S}^{†<!--\; †\; -->}{\mathrm{\gamma <!--\; \gamma \; -->}}^0$, so I can begin with the left-hand side of this third equation:
$\begin{array}{rl}{S}^{-<!--\; -\; -->1}{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}S& ={\mathrm{\gamma <!--\; \gamma \; -->}}^0{S}^{†<!--\; †\; -->}{\mathrm{\gamma <!--\; \gamma \; -->}}^0{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}S\\ & ={\mathrm{\gamma <!--\; \gamma \; -->}}^0(1+\frac{i}{4}{\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}){\mathrm{\gamma <!--\; \gamma \; -->}}^0{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}(1-<!--\; -\; -->\frac{i}{4}{\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}})\\ & ={\mathrm{\gamma <!--\; \gamma \; -->}}^0{\mathrm{\gamma <!--\; \gamma \; -->}}^0{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}-<!--\; -\; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^0{\mathrm{\gamma <!--\; \gamma \; -->}}^0{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}\frac{i}{4}{\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}+{\mathrm{\gamma <!--\; \gamma \; -->}}^0\frac{i}{4}{\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\gamma <!--\; \gamma \; -->}}^0{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}-<!--\; -\; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^0\frac{i}{4}{\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathrm{\rho <!--\; \rho \; -->}\mathrm{\tau <!--\; \tau \; -->}}{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\rho <!--\; \rho \; -->}\mathrm{\tau <!--\; \tau \; -->}}{\mathrm{\gamma <!--\; \gamma \; -->}}^0{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}\frac{i}{4}{\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}\\ & ={\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}+\frac{1}{16}{\mathrm{\gamma <!--\; \gamma \; -->}}^0{\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathrm{\rho <!--\; \rho \; -->}\mathrm{\tau <!--\; \tau \; -->}}{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\rho <!--\; \rho \; -->}\mathrm{\tau <!--\; \tau \; -->}}{\mathrm{\gamma <!--\; \gamma \; -->}}^0{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}\\ & ={\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}+\frac{1}{16}\left({\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathrm{\rho <!--\; \rho \; -->}\mathrm{\tau <!--\; \tau \; -->}}{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\rho <!--\; \rho \; -->}\mathrm{\tau <!--\; \tau \; -->}}{\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}\right){\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}.\end{array}$
Remember, we want this to be equal to
${\mathrm{\Lambda <!--\; \Lambda \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{}_{\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\nu <!--\; \nu \; -->}}=({\mathrm{\delta <!--\; \delta \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{}_{\mathrm{\nu <!--\; \nu \; -->}}+{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{}_{\mathrm{\nu <!--\; \nu \; -->}}){\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\nu <!--\; \nu \; -->}}={\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}+{\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{}_{\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\nu <!--\; \nu \; -->}}.$
The first term is there already, but I have no idea how that second term is going to work out to ${\mathrm{\epsilon <!--\; \epsilon \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{}_{\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\nu <!--\; \nu \; -->}}$. Can anyone give me a hint, or tell me what I’ve done wrong?

if $u$ and $u^{\prime }$ are a velocity referred to two inertial frames with relative velocity $v$ confined to the $x$ axis, then the quantities $l$, $m$, $n$ defined by
$(l,m,n)=\frac{1}{|u|}(u_x,u_y,u_z)$
and
$(l^{\prime },m^{\prime },n^{\prime })=\frac{1}{|u^{\prime }|}(u_x^{\prime },u_y^{\prime },u_z^{\prime })$
are related by
$(l^{\prime },m^{\prime },n^{\prime })=\frac{1}{D}(l-<!--\; -\; -->\frac{v}{u},m{\mathrm{\gamma <!--\; \gamma \; -->}}^{-<!--\; -\; -->1},n{\mathrm{\gamma <!--\; \gamma \; -->}}^{-<!--\; -\; -->1})$
and that this can be considered a relativistic aberration formula. The author gives the following definition for $D$, copied verbatim.
$D=\frac{u^{\prime }}{u}(1-<!--\; -\; -->\frac{u_{x}v}{c^2})={[1-<!--\; -\; -->2l\frac{v}{u}+\frac{v^2}{u^2}-<!--\; -\; -->(1-<!--\; -\; -->l^2)\frac{v^2}{c^2}]}^{\frac{1}{2}}$
Why is that better than the second expression?
Also, in case it's not clear, $\mathrm{\gamma <!--\; \gamma \; -->}=1/\sqrt{1-<!--\; -\; -->\frac{v^2}{c^2}}$ and $|u|=|(u_x,u_y,u_z)|=\sqrt{u_x^2+u_y^2+u_z^2}$

Chosing a reference frame in which the Earth is at rest and doesn't rotate
1) Does anybody know of such a publication?
2) I know that even such speeds of ${10}^{18}$ m/s are not in contradiction with relativity because a limiting velocity only exists for exchange of information, which apparantly does not occur.

We have two particles $A$ and $B$ flying with different speed. Now the rest frame for $A$ and $B$ is not the same. But is the inertial frame the same? Or would be the inertial frame the same if we have a third fix point $C$ and we would observe them both from $C$?
For me it seems logical to have $3$ frame of references. From the point of view of $A$, from the point of view of $B$ and from an independent point of view $C$. Is this correct?

What is a homogeneous and isotropic frame of reference?
I have heard that inertial frames of reference in the context of special relativity are both isotropic and homogeneous. I know what isotropic and homogeneous mean in a general context, but what do they mean when relating to a frame of reference?

If a body $A$ is moving with a constant velocity v and an observer on that body $A$ observes another body $B$ to be at rest then the kinetic energy of $B$ is zero. So is energy dependent on the frame of reference if so then how is the conservation of energy stated?

It is said that we can't study quantum gravity because gravity is a weak force. But gravity and acceleration are the same. Why can't we study quantum gravity in a strongly accelerated frame of reference?

The relativistic energy-momentum equation is:
$E^2=(pc{)}^2+(m{c}^2{)}^2.$
Also, we have $pc=Ev/c$, so we get:
$E=m{c}^2/(1-<!--\; -\; -->v^2/c^2{)}^{1/2}.$
Now, accelerating a proton to near the speed of light:
$\begin{array}{}0.990000000000000& c=>& 0.0000000011& J=& 0.01& TeV\\ 0.999000000000000& c=>& 0.0000000034& J=& 0.02& TeV\\ 0.999900000000000& c=>& 0.0000000106& J=& 0.07& TeV\\ 0.999990000000000& c=>& 0.0000000336& J=& 0.21& TeV\\ 0.999999000000000& c=>& 0.0000001063& J=& 0.66& TeV\\ 0.999999900000000& c=>& 0.0000003361& J=& 2.10& TeV\\ 0.999999990000000& c=>& 0.0000010630& J=& 6.64& TeV\\ 0.999999999000000& c=>& 0.0000033614& J=& 20.98& TeV\\ 0.999999999900000& c=>& 0.0000106298& J=& 66.35& TeV\\ 0.999999999990000& c=>& 0.0000336143& J=& 209.83& TeV\\ 0.999999999999000& c=>& 0.0001062989& J=& 663.54& TeV\\ 0.999999999999900& c=>& 0.0003360908& J=& 2,097.94& TeV\\ 0.999999999999990& c=>& 0.0010634026& J=& 6,637.97& TeV\\ 0.999999999999999& c=>& 0.0033627744& J=& 20,991.10& TeV\end{array}$
If the LHC is accelerating protons to $7TeV$ it means they're traveling with a speed of $0.99999999c$.
Is everything above correct?

A rotating frame of reference (since you rotate your BEC to generate these), where the energy is given by:
$\stackrel{~<!--\; ~\; -->}{E}[\mathrm{\Psi <!--\; \Psi \; -->}]=E[\mathrm{\Psi <!--\; \Psi \; -->}]-<!--\; -\; -->L[\mathrm{\Psi <!--\; \Psi \; -->}]\cdot <!--\; \cdot \; -->\mathrm{\Omega <!--\; \Omega \; -->},$
Where $\mathrm{\Omega <!--\; \Omega \; -->}$ is the rotational velocity which you apply to the BEC.
Now the argument that the term $L[\mathrm{\Psi <!--\; \Psi \; -->}]\cdot <!--\; \cdot \; -->\mathrm{\Omega <!--\; \Omega \; -->}$ should be substracted comes from the fact that in a rotating frame your system loses that fraction of rotational energy. Now I was wondering if anyone knew where the form of $L[\mathrm{\Psi <!--\; \Psi \; -->}]\cdot <!--\; \cdot \; -->\mathrm{\Omega <!--\; \Omega \; -->}$ came from?
I know that in a rotating frame of reference you have that $\stackrel{\to <!--\; \to \; -->}{v}={\stackrel{\to <!--\; \to \; -->}{v}}_r+\stackrel{\to <!--\; \to \; -->}{\mathrm{\Omega <!--\; \Omega \; -->}}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{r}$ . If you fill this in into the kinetic energy and use some basic definitions, you get that the extra effect of rotation is given by:
$\frac{1}{2}I{\mathrm{\Omega <!--\; \Omega \; -->}}^2=\frac{1}{2}J\mathrm{\Omega <!--\; \Omega \; -->}.$
This yields half of the value that is used in the book, is there something that I'm missing or not seeing right ?

instead of assuming that the velocity $c$ is a maximal velocity, proving that while assuming $E=m{c}^2$.

What is the coordinate in this system then and how do they all connect to each other? I've read the mentioning of an observer and the observer's state of motion and I don't understand how that relates to a frame of reference.

If the light velocity is a vector quantity, why vector addition cannot be applied to it? Or the light velocity is not a vector quantity?

An inertial frame of reference is a frame of reference which is not accelerating. All laws of physics are the same measured from an inertial frame of reference. A rest frame is a frame of reference where a particle is at rest.
Does this mean that a rest frame could possibly be non-inertial (that is, accelerating), but the particle with respect to his rest frame would have a velocity of $0$? What kind of velocity? And what exactly would it mean to be at rest with respect to a possibly accelerating frame of reference?
What are the differences and relations between rest frame and inertial reference frame?

Take the following gedankenexperiment in which two astronauts meet each other again and again in a perfectly symmetrical setting - a hyperspherical (3-manifold) universe in which the 3 dimensions are curved into the 4. dimension so that they can travel without acceleration in straight opposite directions and yet meet each other time after time.
On the one hand this situation is perfectly symmetrical - even in terms of homotopy and winding number. On the other hand the Lorentz invariance should break down according to GRT, so that one frame is preferred - but which one?
So the question is: Who will be older? And why?
And even if there is one prefered inertial frame - the frame of the other astronaut should be identical with respect to all relevant parameters so that both get older at the same rate. Which again seems to be a violation of SRT in which the other twin seems to be getting older faster/slower...
How should one find out what the preferred frame is when everything is symmetrical - even in terms of GRT

In physics problems, the earth is usually considered to be an inertial frame. The earth has a gravitational field and the second postulate of the general theory of relativity says:
"In the vicinity of any point, a gravitational field is equivalent to an accelerated frame of reference in gravity-free space (the principle of equivalence)."
Does this mean that accelerating frames of reference can be inertial?

Observer A and B are at the same "depth" in a gravity well. Observer B then descends into the well. A will observe B's time as going slower than their own. B will observe A's time as going faster than their own.
What happens if B were to ascend the well back to A's depth, would B's local time speed back up to the same rate as A's, but B would be younger (relative to A)?
What about the paradox caused by relative motion (ignoring gravity)? If A is moving relative to B, A and B will both observe the other's time as going slower. If A and B were together initially, then B moves away and returns, do their clocks agree?

Formula for the Bekenstein bound
$S\le <!--\; \le \; -->\frac{2\mathrm{\pi <!--\; \pi \; -->}kRE}{\mathrm{\hslash <!--\; \hslash \; -->}c}$
where $E$ is the total mass-energy. That seems to imply that the presence of a black hole in the region is dependent on an observer's frame of reference. Yet, my understanding is that the Bekenstein bound is the maximum entropy that any area can withstand before collapsing into a black hole.
Does this mean that the existence of black holes is observer dependent? Or that even if an observer does not report a black hole in their frame, one is guaranteed to form there in the future?

For a car that is accelerating linearly, the non-inertial frame of reference is the driver in the car where from his reference frame, the car is stationary. It is so called stationary because the non-inertial frame of reference has the same acceleration as the car. Is like the car's acceleration "transform" the driver frame of reference into a non-inertial. That's why in the non-inertial frame of reference, there is no force acting on the car.
But when the car is driving in circles at a constant speed, in the non-inertial frame of reference there is a force acting on the car, which is the centripetal force. Why isn't this frame of reference like the above, not having the acceleration found in their each respective inertial reference frame? Why can't we have a non-inertial reference frame(due to rotation) whereby there is no centripetal force, subsequently eradicating the need for a centrifugal force?

Why is it hopeless to view differential geometry as the limit of a discrete geometry?
Classical mechanics can be understood as the limit of relativistic mechanics $R{M}_c$ for $c\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$.
Classical mechanics can be understood as the limit of quantum mechanics $Q{M}_h$ for $h\to <!--\; \to \; -->0$.
As a limit of which discrete geometry ${\mathrm{\Gamma <!--\; \Gamma \; -->}}_{\mathrm{\lambda <!--\; \lambda \; -->}}$ can classical mechanics be understood for $\mathrm{\lambda <!--\; \lambda \; -->}\to <!--\; \to \; -->0$?