Master Special Relativity with Expert Help and Solutions

on the Dirac equation, expands the ${\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\partial <!--\; \partial \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}}$ term as:
${\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\partial <!--\; \partial \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}}={\mathrm{\gamma <!--\; \gamma \; -->}}^0\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}+\stackrel{\to <!--\; \to \; -->}{\mathrm{\gamma <!--\; \gamma \; -->}}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{\mathrm{\nabla <!--\; \nabla \; -->}}$
where $\stackrel{\to <!--\; \to \; -->}{\mathrm{\gamma <!--\; \gamma \; -->}}=({\mathrm{\gamma <!--\; \gamma \; -->}}^1,{\mathrm{\gamma <!--\; \gamma \; -->}}^2,{\mathrm{\gamma <!--\; \gamma \; -->}}^3)$, but to my knowledge,
${\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\partial <!--\; \partial \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}}={\mathrm{\gamma <!--\; \gamma \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\eta <!--\; \eta \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\partial <!--\; \partial \; -->}}^{\mathrm{\nu <!--\; \nu \; -->}}={\mathrm{\gamma <!--\; \gamma \; -->}}^0\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{\mathrm{\gamma <!--\; \gamma \; -->}}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{\mathrm{\nabla <!--\; \nabla \; -->}}$
using the convention ${\mathrm{\eta <!--\; \eta \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}=\mathrm{diag}<!--\; \; -->(+,-<!--\; -\; -->,-<!--\; -\; -->,-<!--\; -\; -->)$.

When an electron emits a photon from changing energy levels, the frequency of the photon depends on the difference between the energy levels.
But if someone is moving with respect to the atom, the frequency will be apparently red shifted or blue shifted.
Does this mean that the energy levels of the orbits of the atom look to have different values if you are moving with respect to them? But, the apparent energy difference can be different depending on whether the photon is emitted towards you or away from you.
What's going on? Aren't energy levels of orbits supposed to be a fixed value given the kind of atom?

Let $A$ be a rocket moving with velocity $v$.
Then the slope of its worldline in a spacetime diagram is given by $c/v$.
Since it is a slope, $c/v=\tan <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$ for some $\mathrm{\theta <!--\; \theta \; -->}$ $>45$ and theta $<90$.
Does this impose a mathematical limit on $v$?
If so what is it?
As in, we know $\tan <!--\; \; -->(89.9999999999)=572957795131.$
And $c=299792458.$
Using $\tan <!--\; \; -->(89.9999999999)$ as our limit of precision, the smallest $v$ we can use is:
$c/v=\tan <!--\; \; -->(89.9999999999)$
$\Rightarrow <!--\; \Rightarrow \; -->299792458/v=572957795131$
Therefore, $v=1911.18m/s$
What is the smallest non zero value of $v$?

Relativistic Lagrangian transformations, it's $L=-<!--\; -\; -->m{c}^2\sqrt[2]{1-<!--\; -\; -->\frac{|u{|}^2}{c^2}}$ I need to study the translation, boost and rotation symmetry. I say it doesn't depend of the position, so it has translation symmetry and the momentum will conserve. It's rotation invariant because depends only of the modulus of the speed $|u|$ (What is the conserved quantity derived by this symmetry?)

The book said that the mass of a microscopic oscillator (what is that?) is not continuous, but discrete and the difference between states is an energy quanta:
$\mathrm{\epsilon <!--\; \epsilon \; -->}=h\mathrm{\nu <!--\; \nu \; -->}=E_k-<!--\; -\; -->E_i$
And since $m=\frac{h\mathrm{\nu <!--\; \nu \; -->}}{c^2}$ then the (relativistic) mass of the photon is
$m=\frac{h\mathrm{\nu <!--\; \nu \; -->}}{c^2}$
How did they deduce that?

Suppose a spring with stiffness $k$, is strained by constant forces on each end.
In a frame where the strained spring moves at a constant velocity, what's the total relativistic energy of the moving strained string as $k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$?

Suppose $\mathrm{\varphi <!--\; \varphi \; -->}(x)$ a scalar field, $v^{\mathrm{\mu <!--\; \mu \; -->}}$ a $4$-vector. According to my notes a quantity of form $v^{\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\partial <!--\; \partial \; -->}}_{\mathrm{\mu <!--\; \mu \; -->}}\mathrm{\varphi <!--\; \varphi \; -->}(x)$ will not be Lorentz invariant.
But explicitly doing the active transformation the quantity becomes
${\mathrm{\Lambda <!--\; \Lambda \; -->}}_{\mathrm{\nu <!--\; \nu \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{v}^{\mathrm{\nu <!--\; \nu \; -->}}({\mathrm{\Lambda <!--\; \Lambda \; -->}}^{-<!--\; -\; -->1}{)}_{\mathrm{\mu <!--\; \mu \; -->}}^{\mathrm{\rho <!--\; \rho \; -->}}{\mathrm{\partial <!--\; \partial \; -->}}_{\mathrm{\rho <!--\; \rho \; -->}}\mathrm{\varphi <!--\; \varphi \; -->}(y)=v^{\mathrm{\nu <!--\; \nu \; -->}}{\mathrm{\partial <!--\; \partial \; -->}}_{\mathrm{\nu <!--\; \nu \; -->}}\mathrm{\varphi <!--\; \varphi \; -->}(y)$
where $y={\mathrm{\Lambda <!--\; \Lambda \; -->}}^{-<!--\; -\; -->1}x$ and the partial differentiation is w.r.t. $y$. This seems to suggest that the quantity is a Lorentz scalar, so could be used to construct a Lorentz invariant first order equation of motion.
I'm clearly making a mistake here. But I don't see what I've done wrong. Am I wrong to think that $v$ transforms nontrivially under the active transformation?

Is there an easy way to show that $x^2-<!--\; -\; -->t^2=1/g^2$ for a (relativistic) body undergoing acceleration $g$?

Consider two observer in a tree-torus space of size $L$. Observer $A$ is at rest, while observer $B$ moves in the $x$-direction with constant velocity $v$. $A$ and $B$ began at the same event, and while $A$ remains still, $B$ moves once around the universe and comes back to intersect the worldline of $A$ without ever having to accelerate (since the universe is periodic).
Obtained proper time measured by $A$ (because observer $A$A is at rest, so the trajectory of $A$ is characterized by $v=0$):
$\mathrm{\tau <!--\; \tau \; -->}=\int <!--\; \int \; -->d\mathrm{\tau <!--\; \tau \; -->}=\int <!--\; \int \; -->\sqrt{1-<!--\; -\; -->v^2}dt=t.$
For particle $B$, it moves at constant velocity $v$:
$\mathrm{\tau <!--\; \tau \; -->}=\int <!--\; \int \; -->d\mathrm{\tau <!--\; \tau \; -->}=\int <!--\; \int \; -->\sqrt{1-<!--\; -\; -->v^2}dt=t\sqrt{1-<!--\; -\; -->v^2}.$
This result is inconsistent with my understanding of Lorentz invariance because the times don’t match and no one was accelerated and this is not the usual intuition for Lorentz invariance. Is this really consistent with Lorentz invariance?

Three events $A,B,C$ are seen by observer $O$ to occur in the order $ABC$. Another observer $O\prime$ sees the events to occur in the order $CBA$. Is it possible that a third observer sees the events in the order $ACB$?
I could draw spacetime diagram for three arbitrary-ordered events for $O$ and $O\prime$. But I couldn't draw spacetime diagram for third observer.

Why do we say that irreducible representation of Poincare group represents the one-particle state?
Only because
1. Rep is unitary, so saves positive-definite norm (for possibility density),
2. Casimir operators of the group have eigenvalues $m^2$ and $m^{2}s(s+1)$, so characterizes mass and spin, and
3. It is the representation of the global group of relativistic symmetry,
yes?

Can one explain the relativistic energy transformation formula:
$E=\mathrm{\gamma <!--\; \gamma \; -->}{E}^{\prime },$
where the primed frame has a velocity $v$ relative to the unprimed frame, in terms of relativistic time dilation and the quantum relation $E=h\mathrm{\nu <!--\; \nu \; -->}$?
Imagine a pair of observers, A and B, initially at rest, each with an identical quantum system with oscillation period $T$.
Now A stays at rest whereas B is boosted to velocity 𝑣.
Just as in the "twin paradox" the two observers are no longer identical: B has experienced a boost whereas A has not. Both observers should agree on the fact that B has more energy than A.
From A's perspective B has extra kinetic energy by virtue of his velocity $v$. Relativistically A should use the energy transformation formula above.
But we should also be able to argue that B has more energy from B's perspective as well.
From B's perspective he is stationary and A has velocity $-<!--\; -\; -->v$. Therefore, due to relativistic time dilation, B sees A's oscillation period $T$ increased to $\mathrm{\gamma <!--\; \gamma \; -->}T$.
Thus B finds that his quantum oscillator will perform a factor of $\mathrm{\gamma <!--\; \gamma \; -->}T/T=\mathrm{\gamma <!--\; \gamma \; -->}$ more oscillations in the same period as A's quantum system.
Thus B sees that the frequency of his quantum system has increased by a factor of $\mathrm{\gamma <!--\; \gamma \; -->}$ over the frequency of A's system.
As we have the quantum relation, $E=h\mathrm{\nu <!--\; \nu \; -->}$, this implies that B observes that the energy of his quantum system is a factor of $\mathrm{\gamma <!--\; \gamma \; -->}$ larger than the energy of A's stationary system.
Thus observer B too, using his frame of reference, can confirm that his system has more energy than observer A's system.
Is this reasoning correct?

is space expansion the same as time dilation ?

Special relativity says that anything moving (almost) at the speed of light will look like its internal clock has (almost) stopped from the perspective of a stationary observer. How do we see light as alternating electric and magnetic fields?

If an object has very energetic particles in it like that of the sun then wouldn't its mass be higher hence making its gravity greater than that of the still state ones ?

If there is a non-zero expectation value for the Higgs boson even in "vacuum", since the Higgs boson has a mass unlike photons, then I would expect it to have a rest frame.
So why doesn't a non-zero expectation value for the Higgs boson not only break electroweak symmetry, but also break Lorentz symmetry?

Special relativity theory says simultaneity is relative, meaning that different observers will not agree on what happened first and what second. Does it then make sense to say that looking at distant stars, we see them how they looked "billions of years ago" and not how they look now? Does it make sense to talk about what these stars look like now? How do we define this "now" if simultaneity is relative?

If you are traveling at $x$ speed the time will pass for you slower than to an observer that is relatively stopped. That's all just because a photon released at the $x$ speed can't travel faster than the $c$ limit.
What happens if you have two bodies, $A$ and $B$ moving towards each other. If $A$ releases a light beam, and $B$ measures it (the speed of the photons), the speed measured is still the same? The only difference will be the wave length? And if we have the opposite case, $A$ and $B$ are moving away from each other, we get the red shift, but the speed measured will be still the same?

If the velocity is a relative quantity, will it make inconsistent equations when applying it to the conservation of energy equations?
For example:
In the train moving at $V$ relative to ground, there is an object moving at $v$ relative to the frame in the same direction the frame moves. Observer on the ground calculates the object kinetic energy as $\frac{1}{2}m(v+V{)}^2$. However, another observer on the frame calculates the energy as $\frac{1}{2}m{v}^2$.

It is experimentally known that the equation of motion for a charge $e$ moving in a static electric field $\mathbf{E}$ is given by:
$\frac{\mathrm{d}}{\mathrm{d}t}(\mathrm{\gamma <!--\; \gamma \; -->}m\mathbf{v})=e\mathbf{E}$
Is it possible to show this using just Newton's laws of motion for the proper frame of $e$, symmetry arguments, the Lorentz transformations and other additional principles?