Learn Relativity Easily with Examples

Since time slows down and length contracts, when we travel almost at speed of light, if the speed of light (or EM waves) remains same and the wavelength of light remains same, do we measure the wavelength more than it actually is?

In classical (Newtonian) mechanics, every observer had the same past and the same future and if you had perfect knowledge about the current state of all particles in the universe, you could (theoretically) compute the future state of all particles in the universe.
With special (and general) relativity, we have the relativity of simultaneity. Therefore the best we can do is to say that for an event happening right now for any particular observer, we can theoretically predict the event if we know everything about the past light cone of the observer. However, it tachyons (that always travel faster than the speed of light) are allowed, then we cannot predict the future since a tachyon can come in from the space-like region for the observer and can cause an event that cannot be predicted by the past light cone. That is, I believe, why tachyons are incompatible with causality in relativity. Basically, the future cannot be predicted for any given observer so the universe is in general unpredictable - i.e. physics is impossible.
Now in quantum mechanics, perfect predictability is impossible in principle. Instead all we can predict is the probability of events happening. However, Schrodinger's equation allows the future wavefunction to be calculated given the current wavefunction. However, the wavefunction only allows for the predictions of probabilities of events happening. Quantum mechanics claims that this is the calculations of probabilities is the best that can be done by any physical theory.
So the question is: "Is the predictability of the future to whatever extent is possible (based on the present and the past) equivalent to the principle of causality?" Since prediction is the goal of physics and science in general, causality is necessary for physics and science to be possible.

Integration by parts to derive relativistic kinetic energy
$\underset{0}{\overset{x}{\int <!--\; \int \; -->}}\frac{d}{dt}[mv\mathrm{\gamma <!--\; \gamma \; -->}(v)]dx=v\cdot <!--\; \cdot \; -->mv\mathrm{\gamma <!--\; \gamma \; -->}(v)-<!--\; -\; -->\underset{0}{\overset{v}{\int <!--\; \int \; -->}}mv\mathrm{\gamma <!--\; \gamma \; -->}(v)dv$

Consider an elevator moving down with uniform velocity. A person standing inside watches an object fall from the ceiling of the elevator to the floor. Say the height of the elevator is $h$. Then the work done by gravity in that frame of reference should be $mgh$. But consider this same event being watched by someone else in the stationary frame of reference. In his reference frame, the object travels a larger distance as it falls from the ceiling to the floor of the elevator because the floor itself is moving downwards (one can calculate this extra distance covered to be $u\sqrt{\frac{2h}{g}}$) and hence the change in kinetic energy should be more in that frame than in the moving frame.
Shouldn't the change in kinetic energy be more in a moving elevator from a stationary frame of reference?

There's a voltage difference of 1000 Volts between two points 2 meters apart. An electron starts at the point of lower potential and is left to travel alone in a straight line until it reaches the other point. What speed does the electron have when it reaches the second point?
Whether the retardation effects of the radiation of the electron are important here. I tried doing it using the relativistic formulae for the kinetic energy and got
$v=c\sqrt{1-<!--\; -\; -->\left(\frac{m_e{c}^2}{m_e{c}^2+V}\right){}^2}$
where $m_e$ is the electron mass, $c$ the speed of light, $V$ the voltage difference between the two points, and $v$ the final speed of the electron.

To person A standing on a railway platform, person B on the train travelling past would seems to be aging slower (if such a thing were perceptible) and to person B it would appear that person A was aging faster.
But in the absence of a train, or a platform, or even a planet, the two people would appear to be moving apart. Without a frame of reference it would be difficult to say which is moving and which is not. However, they can't both be appearing to age faster than the other.
So which one would be aging faster and which slower? Or rather, how can one tell which one is moving?

Up to the work of Riemann and Gauss, this definition would have made clear to me examples of geometrical objects: a line, square, cube, hypercube; each of these possessing geometrical properties such as the number of sides, angles between faces, the dimension of space that contains them etc. Hence a geometrical object was a set of measurements associated with an object using a ruler.
After the work of Einstein and Minkowski who showed that time and space were a part of one another, would it be correct to say:
1. a geometrical object is a set of measurements associated with an object of distance and time using a ruler and a clock?
2. Geometrical objects includes an interval of time, the invariant space-time interval?

What is the current status or acceptance of block time as it relates to Einstein's theory of relativity? Has quantum mechanics ruled it out or is it still the favored view of the world?

Suppose that Standard Model Higgs mechanism broke electromagnetism, by e.g. veving the charged component of the doublet, so that the photon was massive with $m_{\mathrm{\gamma <!--\; \gamma \; -->}}\sim <!--\; \sim \; -->v$. Could such a Universe still have large scale structure? Atoms (i.e. stable electronic orbits)? Life?
Assuming we got past those hurdles, would it have been much more difficult to have discovered special relativity? Would we have been stuck at Galilean invariance, without the invariance of the speed of light from which to build SR? I appreciate that this is speculative.
And, also, the identical question but for a coloured Higgs vacuum that breaks QCD. Would broken QCD still be confining? I guess so - so we could still have nucleons and the resulting chemistry?

How special relativity leads to Lorentz Invariance of the Maxwell Equations. Differential form will suffice.

Picture a situation where we have two observers, $A$ and $B$, and a system in a certain quantum state. If $B$ makes a measurement of some observable, say energy for example, the state will collapse to one of the possible energy states with a definite probability. If $A$ makes a measurement after this has happened he will observe a precise energy given by the state to which our system has collapsed after the the measurement made by $B$.
Pretty straightforward till now. Now, if both observers make a measurement at the same time they will both measure the same value. But we know from special relativity that simultaneity is relative to the observer, so we may think that from some observer this measurements won't be simultaneous and actually for him 𝐴 will have made the measurement before $B$.
My question is obvious then, how do measurements and relativity of simultaneity marry?

The distance between interference lines in the double-slit experiment is:
$w=z\mathrm{\lambda <!--\; \lambda \; -->}/d$
Where:
$w$: Distance between fringes
$z$: Distance from slits to screen.
$\mathrm{\lambda <!--\; \lambda \; -->}$: Wave length of light
$d$: Distance between slits.
Now, looking at the setup from a reference frame traveling in the direction of the distance 𝑧 between slit and screen and moving away from the slit in direction of the screen, the light gets redshifted and the distance 𝑧 gets Lorentz contracted:
$w^{\prime }=\frac{z}{\mathrm{\gamma <!--\; \gamma \; -->}}\mathrm{\gamma <!--\; \gamma \; -->}(1+\frac{v}{c})\mathrm{\lambda <!--\; \lambda \; -->}/d=z(1+\frac{v}{c})\mathrm{\lambda <!--\; \lambda \; -->}/d$
This only holds for the first fringe or so, that aren't too far from the center, because the red-shift formula changes farther out. However, it's obvious that the distance between fringes has increased!
Where is my mistake?

Imagine a cup stuck to the floor of a falling elevator with the help of some impressive adhesive. In a frame of reference is this cup is being watched through the walls of a transparent elevator, this frame of reference is not accelerating with respect to earth. Now, since earth is an inertial frame of reference, and since this frame of reference is not accelerating with respect to it then it must be the case that this frame of reference in which the cup is viewed is also an inertial frame of reference.
Let's call this frame of reference viewing the cup S since S is an inertial frame of reference, it must follow the law:
$\begin{array}{}\text{(1)}& a=0⟺<!--\; ⟺\; -->F=0\text{(Newtons First Law)}\end{array}$
The force on the cup is the tension force due to the adhesive and the force due to the Earth, the reaction force, all add up to zero because the cup is seen to not accelerate within the lift.
Okay the so the cup has zero forces acting on it from the observer, the tension force, the gravitational force, the reaction force, all add up to zero. Since $S$ is an inertial frame of reference and $F=0$, the acceleration must be zero from $1$. But according to $S$ it isn't, it is accelerating with $9.8\text{m}/{\text{s}}^2$.
In conclusion:
An inertial frame of reference observes an object on which the total force is zero but still has non-zero acceleration. How can this contradiction be resolved?

Consider two identical charges moving with uniform velocity. There will be a magnetic force of attraction between them as two currents in the same direction attract each other. If I sit on one of the charges, according to me the other charge is not moving. So there wont be any magnetic attraction. How does changing the frame of reference change the outcome of the interaction?

When a free particle move in space with a known momentum and energy then what is the physical process that gives mass to that free (relativistic) particle?
What is role does the Higgs field in that process, if any?

What does a rotating hoop, with each point moving at a velocity close to the speed of light, appear like with respect from a stationary observers perspective. For example how does the shape of the hoop change?

Suppose a neutrino is seen travelling so fast that its Lorentz gamma factor is 100,000. It races past an old, no longer active neutron star, narrowly missing it. As far as the neutrino is concerned, it is the neutron star that is moving at extreme speed, & its mass is 100,000 times larger than 2 solar masses. Therefore, from the speeding neutrino's perspective, the neutron star should appear to be a black hole definitely large enough to trap the neutrino. So how come the speeding neutrino continues its travel right past the old stellar remnant? Is there an agreed name for this question or paradox?

Consider a particle in two inertial reference frames $\mathrm{\Sigma <!--\; \Sigma \; -->}$ and ${\mathrm{\Sigma <!--\; \Sigma \; -->}}^{\prime }$. The reference frame ${\mathrm{\Sigma <!--\; \Sigma \; -->}}^{\prime }$ is moving with uniform velocity $v$ relative to $\mathrm{\Sigma <!--\; \Sigma \; -->}$. The particle is at rest in ${\mathrm{\Sigma <!--\; \Sigma \; -->}}^{\prime }$. Both reference frames have common axes $x$ and $x^{\prime }$. When doing a certain calculation in both reference frames, which one of the obtained results is considered correct? Are they considered both correct or the one obtained by an observer in ${\mathrm{\Sigma <!--\; \Sigma \; -->}}^{\prime }$?

"An inertial frame of reference is one frame where Newton's First Law holds, therefore, a body has a constant velocity or velocity equal zero. And, the sum of all forces equals zero, there is no aceleration etc..."
But, in an inertial frame of reference holds $F=ma$, the second law. It's confuses me, why we can assume $F=ma$ in an inertial frame of reference?

How much does a proton weigh when it is going around the LHC at CERN?
Considering that speed increases weight and the proton is going at almost the speed of light, I would like to know how much a speeding proton would weigh in the LHC.