Learn Relativity Easily with Examples

Does the temperature of a body depend on the frame of reference?

Can an article having spin be seen as spinless in a properly choosen frame of reference?
Lets take two-atomic molecule that has spin while the both atoms have zero spin. Can we choose a frame of reference where this molecule has no spin?

Given a point mass, with $\underset{\_<!--\; \_\; -->}{x}$ the position vector, on which acts a force $\underset{\_<!--\; \_\; -->}{F}$ such that it is conservative:
$\underset{\_<!--\; \_\; -->}{F}=-<!--\; -\; -->\mathrm{\nabla <!--\; \nabla \; -->}U(\underset{\_<!--\; \_\; -->}{x}).$
Then if I change frame of reference from a inertial one to a non-inertial one, it is true that the (total) force (meaning the vector sum of all the forces acting w.r.t. the new frame of reference) remains conservative?

According to many authors, a fluid is defined to be incompressible if the material derivative of the density $\frac{D\mathrm{\rho <!--\; \rho \; -->}}{Dt}$ is zero, that is to say, that in an frame of reference following the motion of an air parcel, density doesn't change. This in turn means, according to the continuity equation,
$\frac{D\mathrm{\rho <!--\; \rho \; -->}}{Dt}+\mathrm{\rho <!--\; \rho \; -->}\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{V}=0,$
so that $\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{V}=0$. So far so good.
However, let us consider a simple case in 1D in which the density is of the form $\mathrm{\rho <!--\; \rho \; -->}(x,t)=x-<!--\; -\; -->t$ and $\stackrel{\to <!--\; \to \; -->}{V}=u_x\stackrel{^<!--\; ^\; -->}{\mathrm{ı<!--\; ı\; -->}}$. Both fields satisfy the continuity equation. This is more evident if we use the other form of the continuity equation,
$\frac{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\rho <!--\; \rho \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}+\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->(\mathrm{\rho <!--\; \rho \; -->}\stackrel{\to <!--\; \to \; -->}{V})=0$
Clearly, for the velocity field that I gave, $\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{V}=0$, and the fluid is incompressible, but as we can see, density changes with time and space. Moreover, at a fixed position (i.e. in a stationary frame of reference), density would change with time.
So, does density depend on the frame of reference? What's the real definition of compressibility in fluid mechanics?

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?

Let’s assume we have 2 different observers. Observer 1 sits in space and observer 2 sits in a space lab which is in a free fall state toward the Earth. We further assume that observer 2 in the space lab does not have any information regarding its surrounding.
1. Is there anyway that observer 2 can figure out whether any frame attached to the space lab can be regarded as an inertial or non-inertial frame?
2. Lets assume there is a 1kg ball inside the space lab which is also free falling together with the space lab? Observer 1 (with its true inertial frame) can measure the acceleration of the ball and deduce that 9.8Newton force is acting on the ball. On the other hand, observer 2 measures the acceleration of the ball as 0, and hence, deduces that the net force acting on the ball is 0. As the observer 2 has no way of knowing that space lab is not an inertial frame, he will not have any doubt about his force measurement. Does this mean that definition of force is dependent on the frame of reference?

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?

If the frame of reference moves with the same velocity as the velocity of the object being examined within it, is it truly a frame of reference?

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?

A puck is placed on a friction less disk that is rotating with angular velocity $\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}$. What is the equation of motion with respect to the rotating frame of the puck?
The contradiction that I can not resolve is that it seems really obvious that the puck will rotate with angular velocity $-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{\mathrm{\omega <!--\; \omega \; -->}}$ in the frame of the disk. But if look at the puck from the frame of reference of the disk, we can see that there is a pseudo centrifugal force acting on the puck. This is the only force that acts on the puck in this frame.
Hence, the puck should accelerate radially outward in this frame...

Can an "absolute" frame of reference be determined by measuring the compression of light?
General relativity tells us that there is no absolute frame of reference (actually, it tells us that all frames are relative, which is close but not the same as there is no absolute frame).
Special relativity demonstrates that there is an absolute: the speed of light.
Notwithstanding the impracticality of the issues, is it possible to determine an absolute frame of reference based on minuscule difference in the wave length of light (measured by doppler shift)?
In effect, can we measure our (Earth's) compound frame of reference by measuring the doppler shift / compression of light in our own frame of reference. Or would any relativistic compression be undetectable within that same frame of reference? Or would it be practically infeasible?