Boost Your Knowledge on Frame of Reference

Train at a railway station and on the platform there is a glass case (the mass of case is $5kg$ and if it gets hit with a velocity of $10N$ then the glass will break), a frame of reference is attached to me. Now my train starts with an acceleration of $2m/s^2$. Now according to me I'm at rest and glass case is moving backward and so the acceleration on the glass case is $2m/s^2$. Therefore the force acting on the glass case is $F=ma=5\times <!--\; \times \; -->2=10N$ at this the glass case should break, but this is not happening. Whereas according to a frame of reference attached to a person on the platform every thing is fine(i.e. there is no force(horizontal) acting on the glass case).

Is the classical Doopler Effect, for light shift, $1-<!--\; -\; -->v/c$, exact? What is it an approximation of?

Does a photon in vacuum have a rest frame?

Orbital velocity of a circular planet is $\sqrt{aR}$, where a is the centripetal acceleration, and $R$ is radius of the planet. With $v_1$ as the tangential velocity of the rotating planet at the equator.
On the non rotating body, suppose that the orbital velocity is $v_0$, and, for an object launched on the rotating body's "equator", that the orbital velocity will be in the form of $v_1+v_2$ (the body and the object both going counterclockwise). Now, I half-hypothesized $v_0=v_1+v_2=\sqrt{aR}$ and $v_2=\sqrt{a^{\prime }R}$ where $a^{\prime }$ is the "true" rotating body's acceleration, able to be calculated from the rotating frame of reference as
$a^{\prime }=a-<!--\; -\; -->\frac{v_1^2}{R}$
The logic was that from rotating body's reference frame, the object would be traveling at $v_2$, less than $v_0$ because of the centrifugal force, so $v_2$ has to be the orbital velocity if the gravity was "weakened" by centrifugal force.
Tried to solve for $a^{\prime }$ and comparing it to the value, got from rotating frame of reference, ending up with
$a^{\prime }=a-<!--\; -\; -->(\frac{v_1(v_0+v_2)}{R})$
Something's not right, and if I had to choose, I would guess the $v_0=v1+v_2$, that acceleration is not the same on those two planets, but I don't know how it would change, or why.

What is a (spatial) frame of reference in quantum mechanics?
When we make a transformation e.g. in a H-atom from the proton frame of reference to the electron frame of reference. The whole point of QM is that the electron cannot be treated as a dot somewhere in space but that it has an associated wavefunction and only acquires a certain position when I make an appropriate measurement. So how can I transform to the electron frame of reference making the origin to be the position of the electron if I don't know it?

Consider a (classical) system of several interacting particles. Can it be shown that, if the Lagrangian of such a system is Lorenz invariant, there cannot be any space-like influences between the particles?

In the frame of photon does time stop in the meaning that past future and present all happen together?
If we have something with multiple outcomes which is realized viewed from such frame? Are all happening together or just one is possible?
How the communication between two such frame s work meaning is there time delay for the information as $c$ is limited? If there is time delay does it mean that time does not stop?

Since $p^2=E^2-<!--\; -\; -->{\stackrel{\to <!--\; \to \; -->}{p}}^2=m^2$ and
$E=h\mathrm{\nu <!--\; \nu \; -->}=\frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}}$ and
$|\stackrel{\to <!--\; \to \; -->}{p}|=\frac{h}{\mathrm{\lambda <!--\; \lambda \; -->}}$
we have that
$p^2=\frac{h^2{c}^2}{{\mathrm{\lambda <!--\; \lambda \; -->}}^2}-<!--\; -\; -->\frac{h^2}{{\mathrm{\lambda <!--\; \lambda \; -->}}^2}$
If I go to Planck-units ($c=1,h=1$), this becomes zero. Is this a correct thing to do?

Why in accelerated rolling, the center of mass must be at the origin of an inertial frame of reference in order for the second law to be applicable.

Let's say someone who is not moving is seeing a rod which has one edge at $x=0$ and the other at $x=a$ so its length is $a$. An observer moving at constant speed $u$ along the $x$-axis uses lorentz transformation to determine the coordinates of these two points in his frame of refrence.
$x_a^{\prime }=\mathrm{\gamma <!--\; \gamma \; -->}(a-<!--\; -\; -->ut)$, $x_0^{\prime }=\mathrm{\gamma <!--\; \gamma \; -->}(0-<!--\; -\; -->ut)$ so to find the length of the rod we take the difference of these two points and it gives us $x_a^{\prime }-<!--\; -\; -->x_0^{\prime }=\mathrm{\gamma <!--\; \gamma \; -->}a$.
I know that $\mathrm{\gamma <!--\; \gamma \; -->}>1$ so shouldn't the moving reference frame observe the rod to be bigger? Why are we talking about length contraction?

What reference frame is used to measure the velocity, $v=\frac{distance}{time}$?
$\mathbf{p}=\frac{mv}{\sqrt[2]{1-<!--\; -\; -->\frac{v^2}{c^2}}}$
Suppose the momentum of an electron is being observed in a lab. It could be observed in a lab that as the speed of electron increases, it becomes harder to accelerate it. To measure the velocity of electron, we need to measure distance or displacement and time.

If the expansion of the universe is accelerating, doesn't that mean that the entire universe is a non-inertial frame of reference? And if so, what implications does this have?

"The age of the universe is 14 billion years"
Which observers' are these time intervals? According to whom 14 billion years?

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?

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?

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?

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...