Boost Your Knowledge on Frame of Reference

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?

From my understanding, we can do math in an accelerating frame of reference as long as "fictitious" force terms are correctly added. From this point of view, is there anything wrong with viewing the Earth as stationary, and the rest of the universe rotating around it, at least kinematically? And, if so, wouldn't several cosmological objects move faster than light in this frame of reference? How can this be?

The Fourier rate equation of heat conduction states:
$\stackrel{\to <!--\; \to \; -->}{q}=-<!--\; -\; -->k\mathrm{\nabla <!--\; \nabla \; -->}T$
But I'm wondering if this is valid in every frame of reference, because the heat flux $\stackrel{\to <!--\; \to \; -->}{q}$ does obviously change when the area under consideration is moving. The equation is often applied to moving fluids, and because there is no objective way to determine a non-moving frame of reference in a moving substance, I guess that the equation is valid for every inertial frame of reference, but I just wanted to be sure.

"You are traveling in a car going at a constant speed of 100 km/hr down a long, straight highway. You pass another car going in the same direction which is traveling at a constant speed of 80 km/hr. As measured from your car’s reference frame this other car is traveling at -20 km/hr. What is the acceleration of your car as measured from the other car’s reference frame? What is the acceleration of the other car as measured from your car’s reference frame?"

Shouldn't they both appear to have an acceleration of zero, because both velocities are constant? I can imagine sitting in the faster car and watching the slower car, its speed would not appear to change, only its position?

Let's take Faraday's experiments of induce emf where He take a loop and a magnetic field, for instance let's suppose we have an coil and a magnet, Now If we move the magnet towards the coil there will be an induce electric field and Thus the force $\stackrel{\to <!--\; \to \; -->}{F}=q\stackrel{\to <!--\; \to \; -->}{E}$ on charges of loop and thus the induce electric field but Now suppose I'm in frame of reference that is moving along with magnet with the same velocity as the magnet so that magnet is rest in my frame but coil is moving, in this case I will say charge in coil is moving in magnetic field and thus there is an Lorentz force $\stackrel{\to <!--\; \to \; -->}{F}=q(\stackrel{\to <!--\; \to \; -->}{v}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{B})$ and thus the induce emf. So How the same phenomenon is explained from two different Frame of reference with two different principals?

When I moving object in some frame of reference, it has a positive kinetic energy. In the frame of reference where the object is stationary, it has a kinetic energy of 0.
Is the energy difference in some way associated to the frames of reference?

Consider a person standing still on the ground and a car, which is moving with a constant speed $v$ in a straight line. From the frame of reference of the car the person is moving backwards with same speed $v$.
What causes to the person to move backwards in the frame of reference of the car?

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?

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?

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?

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 ?

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.

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?

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?

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?

Imaging that the speed of that spacecraft is almost the speed of light "$c^{-<!--\; -\; -->1}\text{m/s}$". But no acceleration. Meaning it's an inertial frame of reference right? Will you be able to throw a ball in the direction of movement? What about the direction inverse to the movement? Does that means that there is a direction of movement in an inertial frame of reference?

Is this possible to create a inertial frame of reference in the earth? How it is possible?