Learn Relativity Easily with Examples

How does translational kinetic energy of a system depend on frame of reference where as rotational kinetic energy does not depend on reference frame?

Massless particle (such as a photon) cannot have a frame of reference. But the electromagnetic field does have mass; does it have a frame of reference? If so, if one takes a charge as "setting up" its own EMF, then the field's frame of reference is the charges, so the distinction doesn't matter. But if we take the EMF to pervade all of space, and charges merely to alter the value of the field at any given location, then it makes sense to ask about the frame of reference of the field per se. However the EMF varies from point to point; how could a volume of the EMF, which is not homogenous, have a single frame of reference?

In string theory, If an open string obeys the Neumann boundary condition, then in the static gauge, one can show that the end points move at the speed of light. The derivation is straightforward, but how can this apply to the massive string?

In Schwarzschild coordinates ${\displaystyle (t,r,\mathrm{\theta <!--\; \theta \; -->},\mathrm{\varphi <!--\; \varphi \; -->})}$ the Schwarzschild metric (or equivalently, the line element for proper time) has the form
${\displaystyle g=-<!--\; -\; -->c^2{d\mathrm{\tau <!--\; \tau \; -->}}^2=-<!--\; -\; -->(1-<!--\; -\; -->\frac{r_{\mathrm{s}}}{r})c^{2}d{t}^2+{(1-<!--\; -\; -->\frac{r_{\mathrm{s}}}{r})}^{-<!--\; -\; -->1}d{r}^2+r^2{g}_{\mathrm{\Omega <!--\; \Omega \; -->}},}$
where $g_{\Omega }$ is the metric on the two sphere, i.e. ${\displaystyle g_{\mathrm{\Omega <!--\; \Omega \; -->}}=(d{\mathrm{\theta <!--\; \theta \; -->}}^2+{\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}d{\mathrm{\phi <!--\; \phi \; -->}}^2)}.$
Isn't $(t,r,\mathrm{\theta <!--\; \theta \; -->},\mathrm{\varphi <!--\; \varphi \; -->})$ representing the spherical coordinates of an observer, looking at the black-hole from infinity?

"A high-speed train is traveling at a constant $150$ m/s (about $300$ mph) on a straight horizontal track across the south pole. Find the angle between a plumb line suspended from the ceiling inside the train and another inside a hut on the ground. In what direction is the plumb line on the train deflected?"
we're looking for is a relationship $\tan <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})=\frac{a_n}{a_t}$, because for the plumb line in the hut $a_t=1$ but not for the plumb line in the train due to the Coriolis force.
However,
1. What is the frame of reference here? Is it rotating, fixed so that the train is standing still? Is it rotating with the earth?
2. Since the train is on the south pole, isn't that the same to say that the earth's rotation doesn't affect it?
3. Can you show me how to get the right answer? The right answer is supposed to be $0.13$ degrees. In which direction?

If we have a light ray $x^{\mathrm{\mu <!--\; \mu \; -->}}$ with velocity $c$, what is $c^0$ (the time component)?

If an object in free fall near earth, then an inertial frame of reference and I see the earth accelerating towards me with no force acting upon it. What causes that acceleration?

Have any known experiments ruled out travelling faster than the speed of light? Or is this just a widely accepted theory?

The argument is often given that the early attempts of constructing a relativistic theory of quantum mechanics must not have gotten everything right because they led to the necessity of negative energy states. What's so wrong with that? Why can't we have negative energy states?

Cosmic microwave background temperature is about $2.7K$. But what temperature we will measure in space using a simple Kelvin thermometer in the shadow? Can it be lower than $2.7K$?

Is it allowed to have the zeroth-component of a four-velocity be negative?
This is referring to $V^0$ for a curved space metric with signature $-+++$.

On Einstein original paper "On the electrodynamics of moving bodies", on section 2 of the first part (Kinematics), the following thought experiment is described: a rod is imparted constant speed $v$ in the direction of the $x$$x$ axis (growing $x$). At both ends of the rod, A and B, there is a clock, both clocks are synchronous in the stationary system. Then, a ray of light is sent from A at time $t_A$, which is reflected at B at time $t_B$, and arrives at A at time $t_A^{\prime }$. It is then stated that "taking into consideration the principle of the constancy of the velocity of light", we get:
$t_B-<!--\; -\; -->t_A=\frac{r_{AB}}{c-<!--\; -\; -->v}\text{and }{t}_A^{\prime }-<!--\; -\; -->t_B=\frac{r_{AB}}{c+v}$
where $r_{AB}$ is the length of the moving rod, measured in the stationary system. Now, if besides the rod's length, both time intervals refer to time in the stationary system, so must speed. Given the previous assumption of its constancy, why are terms like $c+v$ and $c-<!--\; -\; -->v$ in the above equations?

If light travels at the speed of light, and anything with rest mass will experience relativistic effects based on the Lorentzian equations, why doesn't light experience these kinds of effects?
For example, relativistic mass and rest mass are related via
$m = \frac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}$
Shouldn't light therefore have no rest mass (since the Lorentzian is $0$)?

If some kind of source was able to supply an infinite amount of energy, does that imply that it also must have an infinite mass? Is the contrary also true?

Manipulating an $n\times <!--\; \times \; -->n$ metric where $n$ is often $>4$, depending on the model. The $00$ component is always $\mathrm{\tau <!--\; \tau \; -->}$*constant, as in the Minkowski metric, but the signs on all components might be $+$ or $-<!--\; -\; -->$ , depending on the model. Can I call this metric a Minkowski metric? Or what should I call it?

Why do the principles of physics get more complex when the frame of reference undergoes acceleration?

$\mathrm{\rho <!--\; \rho \; -->}={\mathrm{\varphi <!--\; \varphi \; -->}}^{\ast <!--\; \ast \; -->}\mathrm{\varphi <!--\; \varphi \; -->}$
$\mathrm{\rho <!--\; \rho \; -->}={\mathrm{\varphi <!--\; \varphi \; -->}}^{\ast <!--\; \ast \; -->}\mathrm{\varphi <!--\; \varphi \; -->}$
......
$J=-<!--\; -\; -->\frac{i\mathrm{\hslash <!--\; \hslash \; -->}}{2m}({\mathrm{\varphi <!--\; \varphi \; -->}}^{\ast <!--\; \ast \; -->}\mathrm{\nabla <!--\; \nabla \; -->}\mathrm{\varphi <!--\; \varphi \; -->}-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}\mathrm{\nabla <!--\; \nabla \; -->}{\mathrm{\varphi <!--\; \varphi \; -->}}^{\ast <!--\; \ast \; -->})$
with the conservation of probability current and density following from the Schrödinger equation:
$\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->J+\frac{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\rho <!--\; \rho \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}=0.$
The fact that the density is positive definite and convected according to this continuity equation, implies that we may integrate the density over a certain domain and set the total to 1, and this condition will be maintained by the conservation law. A proper relativistic theory with a probability density current must also share this feature. Now, if we wish to maintain the notion of a convected density, then we must generalize the Schrödinger expression of the density and current so that the space and time derivatives again enter symmetrically in relation to the scalar wave function. We are allowed to keep the Schrödinger expression for the current, but must replace by probability density by the symmetrically formed expression
$\mathrm{\rho <!--\; \rho \; -->}=\frac{i\mathrm{\hslash <!--\; \hslash \; -->}}{2m}({\mathrm{\psi <!--\; \psi \; -->}}^{\ast <!--\; \ast \; -->}{\mathrm{\partial <!--\; \partial \; -->}}_t\mathrm{\psi <!--\; \psi \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->}{\mathrm{\partial <!--\; \partial \; -->}}_t{\mathrm{\psi <!--\; \psi \; -->}}^{\ast <!--\; \ast \; -->}).$
which now becomes the 4th component of a space-time vector, and the entire 4-current density has the relativistically covariant expression
$J^{\mathrm{\mu <!--\; \mu \; -->}}=\frac{i\mathrm{\hslash <!--\; \hslash \; -->}}{2m}({\mathrm{\psi <!--\; \psi \; -->}}^{\ast <!--\; \ast \; -->}{\mathrm{\partial <!--\; \partial \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}\mathrm{\psi <!--\; \psi \; -->}-<!--\; -\; -->\mathrm{\psi <!--\; \psi \; -->}{\mathrm{\partial <!--\; \partial \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}{\mathrm{\psi <!--\; \psi \; -->}}^{\ast <!--\; \ast \; -->})$
1. What exactly are ${\mathrm{\partial <!--\; \partial \; -->}}_t$ and ${\mathrm{\partial <!--\; \partial \; -->}}^{\mathrm{\mu <!--\; \mu \; -->}}$?
2. Are they tensors?
3. If they are, how are they defined?

If I float motionless in outer space, I will measure spacetime to be approximately flat. By the equivalence principle, I will get flat spacetime if I am free falling near earth. Is this right? otherwise I can distinguish between the two scenarios.
Let say I declare myself (while free falling) to be at rest in an inertial frame of reference under GR. I see the earth accelerating towards me. What causes that acceleration? It's not the curvature of spacetime, right?

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).