Learn Relativity Easily with Examples

What kind of significant impacts have originated from $E=m{c}^2$.
Generally, it is regarded as the most famous equation of all time. Except for nuclear energy (fission and fusion) I do not know any other way in which this equation has made an impact on the world.
Can somebody list some developments and impacts based on this equation?

Consider me sitting on the top of a train which is travelling near to the speed of light, will I be able to look my image on a mirror which I am holding in my hand?

Trying to derive the infinitesimal time dilation relation $dt=\mathrm{\gamma <!--\; \gamma \; -->}d\mathrm{\tau <!--\; \tau \; -->}$, where $\mathrm{\tau <!--\; \tau \; -->}$ is the proper time, 𝑡 the coordinate time, and $\mathrm{\gamma <!--\; \gamma \; -->}=(1-<!--\; -\; -->v(t{)}^2/c^2{)}^{-<!--\; -\; -->1/2}$ the time dependent Lorentz factor. The derivation is trivial if one starts by considering the invariant interval $d{s}^2$, but it should be possible to obtain the result considering only Lorentz transformations. So, in my approach I am using two different reference frames $(t,x)$ will denote an intertial laboratory frame while $(t^{\prime },x^{\prime })$ will be the set of all inertial frames momentarily coinciding with the observed particle, i.e. the rest frame of the particle. These frames are related by
$t^{\prime }=\mathrm{\gamma <!--\; \gamma \; -->}(t-<!--\; -\; -->\frac{Vx}{c}),x^{\prime }=\mathrm{\gamma <!--\; \gamma \; -->}(x-<!--\; -\; -->Vt),$
where $V$ is some nonconstant (i.e. time dependent) parameter which is, hopefully, the velocity of the particle in the laboratory frame. Treating $x$, $t$ and $V$ as independent variables (for now) and taking the differential of the above relations,
$d{t}^{\prime }=\mathrm{\gamma <!--\; \gamma \; -->}(dt-<!--\; -\; -->\frac{Vdx}{c})-<!--\; -\; -->\frac{{\mathrm{\gamma <!--\; \gamma \; -->}}^3}{c^2}(x-<!--\; -\; -->Vt)dV,$ and
$d{x}^{\prime }=\mathrm{\gamma <!--\; \gamma \; -->}(dx-<!--\; -\; -->Vdt)-<!--\; -\; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^3(t-<!--\; -\; -->\frac{Vx}{c})dV.$
Imposing either the definition of the rest frame $d{x}^{\prime }=0$ or (what should be equivalent) $dx=Vdt$, the only way in which i obtain $dt=\mathrm{\gamma <!--\; \gamma \; -->}d{t}^{\prime }$ is if $dV=0$. So, the derivation breaks badly at some point or I must be wrong in using some of the above equations. Which one is it?

Suppose we have a train moving. When the origin of train's frame coincides with the origin of observers frame; the the time is set to zero. At that very instant, a photon is emitted from train towards the direction train is moving. After time $t$ measured by observer at rest the photon will be at a distance $ct$ and the train at a distance $vt$; but the photon is at a distance $ct-<!--\; -\; -->vt$ from the train, now the time that the passenger on train measures for the photon to reach that point is the distance divided by the velocity of light ( which of course is $c$) so
time measured by observer on train is $t^{\prime }=\frac{ct-<!--\; -\; -->vt}{c}=(1-<!--\; -\; -->\frac{v}{c})t$ . where $t$ is time measured by observer at rest.
Where is the mistake in this reasoning?

The electric dipole moment for a charge distribution is:
$\stackrel{\to <!--\; \to \; -->}{P}={\int <!--\; \int \; -->}_V\mathrm{\rho <!--\; \rho \; -->}(\stackrel{\to <!--\; \to \; -->}{x})\stackrel{\to <!--\; \to \; -->}{x}{d}^{3}x$
If we imagine that $\mathrm{\rho <!--\; \rho \; -->}=const$ on a spherical volume then it's easy to see that $\stackrel{\to <!--\; \to \; -->}{P}$ depends by the origin of the frame of reference, it can be zero or something totally different. For this reason it's very difficult for me to give a physical meaning at this quantity (because it's strange that depends by the origin).

How do Vectors transform from one inertial reference frame to another inertial reference frame in special relativity.
A bound vector in an inertial reference frame ($x$,$ct$) has its line of action as one of the space axis in that frame and is described by $x$*$i$*,then what would it be in form of new base vectors ($\mathbf{a}$) and ($\mathbf{b}$) in a different inertial system ($x‘<!--\; ‘\; -->$,$ct‘<!--\; ‘\; -->$) moving with respect to the former inertial system with v*i* velocity.Let (i) and (j) be the two bounded unit vectors with the line of action as co-ordinate axis($x$) and ($ct$) respectively and senses in the positive side of co-ordinates and similarly ($\mathbf{a}$) and ($\mathbf{b}$) are defined for co-ordinates ($x‘<!--\; ‘\; -->$) and ($ct‘<!--\; ‘\; -->$) respectively.

34 years have passed since Voyager I took off and it's just crossing the solar system, being approximately at 16.4 light-hours away. How much time have passed for itself, though?

$E=(t{)}^0(m{)}^1(c{)}^2$. Here, $m$ = mass of the body. $c$ = velocity of light. Is $t$ the time?

Since frequency of a wave is a function of time, then for a particular ray of the wave, why would the frequency remain the same when observed from a moving reference frame? The frequency should change according to time dilation. No?

1) From inside the room I would view the light moving north at the speed of light
2) From outside the room, observing from a stationary position, would the light exiting the window move at twice the speed of light? or would it change speed?

An object at the end of a rope is attached to the ceiling of an elevator. If the elevator is allowed to free-fall and hit the ground, at impact the tension of the rope is greater than if the elevator was standing still.
Why in the equation below, the acceleration is positive for $m\stackrel{\to <!--\; \to \; -->}{a}$
$T-<!--\; -\; -->mg=ma$
Is this because to the object at the end of the rope, the elevator appears to be accelerating upward? Why is the acceleration not negative, since the elevator was free-falling?

Do we assume that the rest mass of a fundamental particle is constant in all inertial reference frames? i.e. is the rest mass of an electron if it is travelling at constant velocity c/2 (relative to the distant stars) the same as the rest mass of the electron if it is travelling at velocity 0 relative to the distant stars?

Imagine there are 2 digital count up timers, A and B, separated by 100 light days in the same frame of reference. There is a person at each timer. Both timers are turned on at the same time (this could be done via a signal at a point C which is the same distance away from both A and B).
At this point, a person at A will see the timer at B being 100 days less than the timer at A (due to the time light takes to travel from A to B). Vice versa for a person at B.

Let's say a body with m=2kg falls from 100 meters. Obviously it's speed would be far lower than the speed of light so the change in mass (if it exists) would be very tiny. However, if the speed increases, its mass would increase too. That's because its kinetic energy would become bigger. On the other hand, its potential energy would decrease in the same amount that KE has increased.
Does this suggest that either the mass is not going to change (due to the conservation of energy) or it would become slightly bigger, but never less?

Imagine you are one lightyear away from a photon sensitive (light sensitive) switch. So it is obvious that light would take one year to reach to the switch. Now I have a one lightyear long plank. I simply point the plank towards the switch and press it. Now I just did work which light would take 1 year to do in a matter of seconds.
Now the question is, did I break the laws of physics?

From atomic transitions to clockwork gears. So, does this property/constraint makes every clock available to use inherently susceptible to effects of relativistic motion?
For example, if you travel at 99% of speed of light with a clock, then there'd be some particle in the clock whose motion, w.r.t you, is also in the same direction, and when added up, it'll surpass 100%.
For above situation to be avoided, every particle that makes up the clock will have to be:
1. Stationary w.r.t observer, but if this happens, the (frozen) clock can't tell time to the observer.
2. Move in a direction other than that of observer's motion, If this happens, clock can't travel with observer.

The momentum of a particle will change if it is moving at speed close to light speed. In the general case, the wavelength is given as
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$
and
$p=\frac{mv}{\sqrt{1-<!--\; -\; -->v^2/c^2}}$
when $v\to <!--\; \to \; -->c$, $p\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, so is it say that the wavelength is ZERO?

struggling to derive Lorentz transformations for a sine wave, which is traveling at random direction. I started by prooving that phase $\mathrm{\varphi <!--\; \varphi \; -->}$ is invariant for relativity and that equation $\mathrm{\varphi <!--\; \varphi \; -->}={\mathrm{\varphi <!--\; \varphi \; -->}}^{\prime }$ holds.
By using the above equation i am now trying to derive Lorentz transformations for angular frequency $\mathrm{\omega <!--\; \omega \; -->}$, and all three components of the wave vector $k$, which are $k_x$, $k_y$ and $k_z$.
This is my attempt:
$\begin{array}{rl}{\mathrm{\varphi <!--\; \varphi \; -->}}^{\prime }& =\mathrm{\varphi <!--\; \varphi \; -->}\\ {\mathrm{\omega <!--\; \omega \; -->}}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}{t}^{\prime }+k^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}{r}^{\prime }& =\mathrm{\omega <!--\; \omega \; -->}\mathrm{\Delta <!--\; \Delta \; -->}t+k\mathrm{\Delta <!--\; \Delta \; -->}r\\ {\mathrm{\omega <!--\; \omega \; -->}}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}{t}^{\prime }+[{k_x}^{\prime },{k_y}^{\prime },{k_z}^{\prime }][\mathrm{\Delta <!--\; \Delta \; -->}{x}^{\prime },\mathrm{\Delta <!--\; \Delta \; -->}{y}^{\prime },\mathrm{\Delta <!--\; \Delta \; -->}{z}^{\prime }]& =\mathrm{\omega <!--\; \omega \; -->}\mathrm{\Delta <!--\; \Delta \; -->}t+[k_x,k_y,k_z][\mathrm{\Delta <!--\; \Delta \; -->}x,\mathrm{\Delta <!--\; \Delta \; -->}y,\mathrm{\Delta <!--\; \Delta \; -->}z]\\ {\mathrm{\omega <!--\; \omega \; -->}}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}{t}^{\prime }+{k_x}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}{x}^{\prime }+{k_y}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}{y}^{\prime }+{k_z}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}{z}^{\prime }& =\mathrm{\omega <!--\; \omega \; -->}\mathrm{\Delta <!--\; \Delta \; -->}t+k_x\mathrm{\Delta <!--\; \Delta \; -->}x+k_y\mathrm{\Delta <!--\; \Delta \; -->}y+k_z\mathrm{\Delta <!--\; \Delta \; -->}z\end{array}$
$\begin{array}{c}{\mathrm{\omega <!--\; \omega \; -->}}^{\prime }\mathrm{\gamma <!--\; \gamma \; -->}(\mathrm{\Delta <!--\; \Delta \; -->}t-<!--\; -\; -->\mathrm{\Delta <!--\; \Delta \; -->}x\frac{u}{c^2})+{k_x}^{\prime }\mathrm{\gamma <!--\; \gamma \; -->}(\mathrm{\Delta <!--\; \Delta \; -->}x-<!--\; -\; -->u\mathrm{\Delta <!--\; \Delta \; -->}t)+{k_y}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}y+{k_z}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}z\\ \mathrm{\gamma <!--\; \gamma \; -->}({\mathrm{\omega <!--\; \omega \; -->}}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}t-<!--\; -\; -->{\mathrm{\omega <!--\; \omega \; -->}}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}x\frac{u}{c^2})+\mathrm{\gamma <!--\; \gamma \; -->}({k_x}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}x-<!--\; -\; -->{k_x}^{\prime }u\mathrm{\Delta <!--\; \Delta \; -->}t)+{k_y}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}y+{k_z}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}z\\ \mathrm{\gamma <!--\; \gamma \; -->}({\mathrm{\omega <!--\; \omega \; -->}}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}t-<!--\; -\; -->{k_x}^{\prime }cc\mathrm{\Delta <!--\; \Delta \; -->}t\frac{u}{c^2})+\mathrm{\gamma <!--\; \gamma \; -->}({k_x}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}x-<!--\; -\; -->\frac{{\mathrm{\omega <!--\; \omega \; -->}}^{\prime }}{c}u\frac{\mathrm{\Delta <!--\; \Delta \; -->}x}{c})+{k_y}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}y+{k_z}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}z\\ \mathrm{\Delta <!--\; \Delta \; -->}t\mathrm{\gamma <!--\; \gamma \; -->}({\mathrm{\omega <!--\; \omega \; -->}}^{\prime }-<!--\; -\; -->{k_x}^{\prime }u)+\mathrm{\Delta <!--\; \Delta \; -->}x\mathrm{\gamma <!--\; \gamma \; -->}({k_x}^{\prime }-<!--\; -\; -->{\mathrm{\omega <!--\; \omega \; -->}}^{\prime }\frac{u}{c^2})+{k_y}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}y+{k_z}^{\prime }\mathrm{\Delta <!--\; \Delta \; -->}z\end{array}$
From this I can write down the Lorentz transformations.
$\begin{array}{rl}\mathrm{\gamma <!--\; \gamma \; -->}({\mathrm{\omega <!--\; \omega \; -->}}^{\prime }-<!--\; -\; -->{k_x}^{\prime }u)& =\mathrm{\omega <!--\; \omega \; -->}\\ \mathrm{\gamma <!--\; \gamma \; -->}({k_x}^{\prime }-<!--\; -\; -->{\mathrm{\omega <!--\; \omega \; -->}}^{\prime }\frac{u}{c^2})& =k_x\\ {k_y}^{\prime }& =k_y\\ {k_z}^{\prime }& =k_z\end{array}$
What am i doing wrong?

A proton accelerated with electric field gives off E.M. radiation and therefore should lose mass. Larmor's formula gives us a value for the power emitted (varies as acceleration squared). However, as the proton picks up speed, it also gains mass. Now, say I set up an immense electric field which provides an immense acceleration to the proton. In the initial moments of motion, even though its acceleration is extremely high, its velocity is low. In those moments, does the proton lose mass faster than it gains?

Is there any way in which a bound state could consist only of massless particles? If yes, would this "atom" of massless particles travel on a light-like trajectory, or would the interaction energy cause it to travel on a time-like trajectory?