Learn Relativity Easily with Examples

The perimeter is moving at a speed such that the acceleration is $g=9.81\text{ m/s^2}$. Combining $g={\frac{v^2}{R}}_s$ with $\sqrt{1-<!--\; -\; -->\frac{v^2}{c^2}}$ gives dilation factor
$\sqrt{1-<!--\; -\; -->\frac{g{R}_s}{c^2}}$
Assuming the radius $R_s$ of the space station is $500$ meters, a perimeter clock would lose about $1e-<!--\; -\; -->6$ seconds per year with respect to a clock in the center axis.
Now since the clock at the perimeter is subject to acceleration g, by the equivalence principle it would seem that the gravitational time dilation would apply, which is
$\sqrt{1-<!--\; -\; -->\frac{2R_{e}g}{c^2}}$
where $R_e=6.38\times <!--\; \times \; -->{10}^6\text{ m}$. This would make the perimeter clock slow by about $0.02$ seconds per year.

If we take a source with a frequency of $230Hz$ moving at $30m{s}^{-<!--\; -\; -->1}$ right and the listener moving $25m{s}^{-<!--\; -\; -->1}$ right.Let the source be behind the listener at some distance so that the distance between them decreases.Take the speed of sound as $340m{s}^{-<!--\; -\; -->1}$.
With the speaker as the frame of reference, the listener is moving $5m{s}^{-<!--\; -\; -->1}$ towards the source.
$f_L=\left(\frac{c\pm <!--\; \pm \; -->v_L}{c\pm <!--\; \pm \; -->v_S}\right)f_S$
$=\left(\frac{340+0}{340-<!--\; -\; -->5}\right)230$
$=233.4328Hz$
With the Listener as the frame of reference, the source is moving $5m{s}^{-<!--\; -\; -->1}$ towards the listener.
$f_L=\left(\frac{340+5}{340+0}\right)230$
$=233.3824Hz$
With the system as the frame of reference, the source is moving at $30m{s}^{-<!--\; -\; -->1}$ right and the listener moving $25m{s}^{-<!--\; -\; -->1}$ right.
$f_L=\left(\frac{340-<!--\; -\; -->25}{340-<!--\; -\; -->30}\right)230$
$=233.7097Hz$
Why are the answers all different?

Consider the situation: a body slides down on a wedge with only gravitational force as an external force. The velocity of the body in the wedge's frame using work energy theorem. Now, use this velocity of the body to find the velocity of the wedge in ground frame using conservation of momentum. The point is that can I do this?

Suppose we have an integral $\int <!--\; \int \; -->{\mathrm{d}}^{4}kf(k)$
we want to evaluate and that we're in Minkowski space with some metric $(+,-<!--\; -\; -->,-<!--\; -\; -->,-<!--\; -\; -->)$.
Is it true that:
${\mathrm{d}}^{4}k=\mathrm{d}{k}^0{\mathrm{d}}^3\mathbf{k}=\mathrm{d}{k}^0|\mathbf{k}{|}^2\mathrm{d}|\mathbf{k}|\mathrm{d}(\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->})\mathrm{d}\mathrm{\varphi <!--\; \varphi \; -->}$
just like in ordinary space?

Unruh radiation normally applies to linearly accelerated frames
Is there an equivalent of the Unruh thermal radiation in a frame that is spinning?

A train is travelling towards a tunnel at relativistic velocity. The tunnel has two gates on either side that open/close to admit/reject the train. The train is the same length as the tunnel when it at rest in the tunnel's frame of reference. In the tunnel frame, the train is moving so it is contracted and can fully fit inside the tunnel. Thus, the gates are initially closed, open right before the train enters, close while the train is completely inside the tunnel, open right before the train leaves and closes right after the train has completely left.
How would you describe this sequence of events in the frame of reference of the train?

Does the mass of a moving object in empty space with a constant velocity change within its own frame of reference from its rest mass?

A source of light pulses moves with speed v directly away from an observer at rest in an inertial frame. Let $\mathrm{\Delta <!--\; \Delta \; -->}{t}_e$ be the time between the emission of pulses, and $\mathrm{\Delta <!--\; \Delta \; -->}{t}_o$ be the time between their reception at the observer. Show that $\mathrm{\Delta <!--\; \Delta \; -->}{t}_o=\mathrm{\Delta <!--\; \Delta \; -->}{t}_e+v\mathrm{\Delta <!--\; \Delta \; -->}{t}_e$.
Based on my understanding of special relativity, the space-time interval between two events as measured from two inertial frames of reference should be the same. Therefore,
$\mathrm{\Delta <!--\; \Delta \; -->}{t}_e^2=\mathrm{\Delta <!--\; \Delta \; -->}{t}_o^2-<!--\; -\; -->\mathrm{\Delta <!--\; \Delta \; -->}{x}^2$
$⟹<!--\; ⟹\; -->\mathrm{\Delta <!--\; \Delta \; -->}{t}_e^2=\mathrm{\Delta <!--\; \Delta \; -->}{t}_o^2-<!--\; -\; -->v^2\mathrm{\Delta <!--\; \Delta \; -->}{t}_o^2$
$⟹<!--\; ⟹\; -->\mathrm{\Delta <!--\; \Delta \; -->}{t}_o=(1-<!--\; -\; -->v^2{)}^{-<!--\; -\; -->1/2}\mathrm{\Delta <!--\; \Delta \; -->}{t}_e$
which is not the same relation. What is wrong with my reasoning?

Nothing with mass can travel at the speed of light. Therefor one can not ask what happens when an observer in a car moves with the speed of light. But, is it possible for a frame of reference to move with the speed of light? Is there something preventing that?

How to write $F^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{F}_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}$ in terms of $F_{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}{F}^{\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nu <!--\; \nu \; -->}}$. How to do it?

The kinetic energy is, in the usual form $\frac{m}{\sqrt{-<!--\; -\; -->1-<!--\; -\; -->v^2}}$. As we've seen before, energy can be gained or lost, so if we want to, we can add an arbitrary quantity. Therefore, we can take the energy using $\frac{1}{2}m{v}^2$. (...) But it is better to use the formula as $\frac{m}{\sqrt{1-<!--\; -\; -->v^2}}$, because it is valid for the conservation law.

In classical mechanics we have the notion of inertial frame of reference. If my understanding is correct, such frames are exactly those for which Newton's laws of motion have the usual form. I believe I have also heard, that they are the frames that are "moving at a constant speed" with respect to some distinguished point in space.
So we have to be careful whether our frame is accelerated or not. Which makes me wonder:
Does there exist a set of physical laws (of same descriptive power Newton's laws have, but perhaps expressing relations in some other set of physical quantities) such that the laws from this set look the same, no matter how the frame of reference moves?
Does such a set of physical laws exist if we restrict to polynomially moving frames of reference?

Two person, $A$ and $B$, each holding one end of a long solid rod.
Now person $A$ pushes the rod on one end.
Question: Is it correct that the information that the rod has been pushed will travel to the other end at the speed of light whereas the actual 'push' will travel at the speed of sound in the rod?

We all know that space and time are the fundamental units, means no mathematical expression can express their relation to other variable fundamentally. But as we know that moving rod has a contracted length and dilated time which depend on the velocity of that rod w.r.t. an inertial frame in which the motion is happening. So it's a natural question that space and time measurements are not alone to be measured in any inertial frame but one has to specify velocity also. It sounds like 3d space and 1d time have another dimension (velocity) to be included in their totality. So velocity is also a form of space and time. So what is wrong here?

Let's have a definition of massive particle as an irreucible representation of the Poincare group. Then, let's have a spinor field ${\mathrm{\psi <!--\; \psi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}{\mathrm{\alpha <!--\; \alpha \; -->}}_1...{\mathrm{\alpha <!--\; \alpha \; -->}}_{n-<!--\; -\; -->1}\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}}_1...{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}}_{m-<!--\; -\; -->1}}$, which is equal to $(\frac{m}{2},\frac{n}{2})$ representation of the Lorentz group. There is the hard provable theorem:
${\mathrm{\psi <!--\; \psi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}{\mathrm{\alpha <!--\; \alpha \; -->}}_1...{\mathrm{\alpha <!--\; \alpha \; -->}}_{n-<!--\; -\; -->1}\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}}_1...{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}}_{m-<!--\; -\; -->1}}$ realizes irreducible representation of the Poincare group, if
$({\mathrm{\partial <!--\; \partial \; -->}}^2-<!--\; -\; -->m^2){\mathrm{\psi <!--\; \psi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}{\mathrm{\alpha <!--\; \alpha \; -->}}_1...{\mathrm{\alpha <!--\; \alpha \; -->}}_{n-<!--\; -\; -->1}\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}}_1...{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}}_{m-<!--\; -\; -->1}}=0,$
${\mathrm{\partial <!--\; \partial \; -->}}^{\mathrm{\alpha <!--\; \alpha \; -->}\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}}{\mathrm{\psi <!--\; \psi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}{\mathrm{\alpha <!--\; \alpha \; -->}}_1...{\mathrm{\alpha <!--\; \alpha \; -->}}_{n-<!--\; -\; -->1}\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}}_1...{\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\beta <!--\; \beta \; -->}}}_{m-<!--\; -\; -->1}}=0.$
Can this theorem be interpreted as connection between fields and particles?

Relativistic mass = $m_0\mathrm{\gamma <!--\; \gamma \; -->}$ where $\mathrm{\gamma <!--\; \gamma \; -->}$ is the lorentz factor.
So, if a mass that is $.5$ at rest then it is safe to say that the relativistic mass will be $1$ if it goes at $\frac{\sqrt{3}}{2}c$.
What happens if that $.5$ is actually a radioactive isotope and is decaying while speeding up? Then at what speed will it approximately equal $1$?

Let us assume that we placed lot's of ball touching each other in a hollow cylindrical tube, now if we push one ball at the end the ball at the other end move's instantly. So how do the information from one side of the tube travel's to the other side of the tube instantly. and at what speed at it travel's. If it travels at the speed of light so if make a long tube of the length of about $3\ast <!--\; \ast \; -->{10}^8$ m it would take 1 second to see the effect at the other end assuming that nothing will bend or compress all thing's are ideal?

The mass of a compressed spring is greater than that of an uncompressed spring
according to the equation $E=m{c}^2$

Angular momentum is defined from linear momentum via $\stackrel{\to <!--\; \to \; -->}{L}=\stackrel{\to <!--\; \to \; -->}{r}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{p}$, and is conserved in a closed system. Since energy is the time part of the linear four-momentum, is there a quantity defined from energy that's also conserved?

According to the equivalence principle, gravity and inertial forces are similar. And according to general Relativity, If there’s a large object in space-time, it warps the space time’s geometry and causes the light to bend ( or to be seen as if it was coming from a different source). Now, if there’s an observer in an accelerating frame of reference and there’s no mass, will the observer still see the light to be coming from a different source considering the frame’s acceleration to have the same effect as that of a mass according to the principle of equivalence?