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We reviewed Galilean Relativity and introduced the fact that the predictions for the speed of light from Electromagnetism do not depend on the relative velocities of observers.
While electromagnetism was seen as a great triumph, experiments were beginning to find deviations from predictions and some physicists such as Lorentz came up with mathematical patches to try to "fix" classical Physics.
Maxwell's Equations predict light has a constant speed $c=3.0 \times 10^8 m/s$ no matter how fast the observer's velocity is.
Einstein achieved a breakthrough by accepting what the mathematics and physics was telling the world: that the constancy of the speed of light, $c$, was so important to the universe that time and space would be "warped".
Let's say your birthday twin travels on an airplane, when they land they are a tiny little bit younger than you because moving clocks tick slower, i.e. their experience of time proceeded slower than yours due to them traveling faster than you.
While your friend is in the air on the airplane, your friend's measurement of the airplane is not consistent with your measurement. You would find the airplane to be just very slightly shorter than your friend measured it.
If your friend, on the plane in the air, saw that lightening had struck the tip and the back of the plane simultaneously, you would most definitely not see them strike simultaneously. In your spacetime experience, they did not strike simultaneously, and you are both right.
The laws of physics are the same in all inertial frames of reference (all observers not accelerating but may be traveling at different velocities relative to each other.)
All such observers will measure the speed of light to be $c$ no matter how fast they are going relative to another observer.
$$ L = \frac{L_p}{\gamma} $$
$$ \gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} \geq 1 $$
Longest length is measured in the proper frame of the object
being measured.
$$ \Delta t = \gamma \Delta t_p $$
Shortest time interval is measured in proper frame
of event being timed.
e.g. muon decay, we detect more muons from space on the ground than would be expected if time dilation did not exist. Because they are going so fast they effectively get to live longer in our spacetime experience.
$$ x^{\prime} = \gamma \left( x - ut \right) $$ $$ y^{\prime} = y $$ $$ z^{\prime} = z $$ $$ t^{\prime} = \gamma \left(t-\frac{ux}{c^2}\right) $$
Tip: Use Galilean Transforms to orient yourself between reference frames. For example, imagine a truck being the $S^{\prime}$ frame and yourself on the ground as the $S$ frame. You measure a sign on the road to be five meters ahead of you. In the trucks coordinate frame, the sign is $5 - ut$ behind it assuming it has passed it already. If the truck is going $10 m/s$ for $2s$, then the sign is $5 - 20 = -15 m$ or 15 meters behind it already.
We will try to consistently use $S$ as the frame at rest relative to Earth and $S^{\prime} as the moving frame. However, in relativity, it doesn't really matter as it is all relative.
Relativity has us think about spacetime rather than space and time, but in a way, we already do so...
Suppose a truck and observer on the ground start out at $t=0$ at $x=0$ and the truck travels $u=30m/s$ in the $+x$ direction. The ground observer shoots a rocket in the $+x$ direction and let's pretend it doesn't get pulled down by gravity. If the ground observer measures the rocket to be going at $v_x = 100 m/s$ then how fast will the truck observe the rocket to be going?
$$ v^{\prime}_x = \frac{ v_x - u}{1 - \frac{u v_x}{c^2}} $$
$$ v^{\prime}_y = \frac{ v_y }{\gamma \left( 1 -\frac{v_x u}{c^2}\right)} $$
$$ v^{\prime}_z = \frac{ v_z }{\gamma \left( 1 -\frac{v_x u}{c^2}\right)} $$
$v^{\prime}_x =$ the velocity of the rocket as measured by the truck
$v_x =$ the velocity of the rocket as measured by ground observer
$u =$ the relative velocity of truck/ground frames (how fast the truck is going relative to the ground observer or how fast the ground observer is receding relative to the truck observer.)
$$ v_x = \frac{ v^{\prime}_x + u}{1 + \frac{u v^{\prime}_x}{c^2}} $$
Suppose an observer is moving towards, at speed $u$, a wave source whose wave speed is $c$, then classically the wavelength they would measure would be $\lambda = (c - u)T_0$ where $T_0$ is the period of the wave.
The detected frequency is
$$ f = \frac{c}{\lambda^{\prime}} = \frac{c}{(c-u)T_0} $$
However, from relativity we know that $$ T = \gamma T_0 $$
Some algebra work yields: $$ f = \sqrt{ \frac{c + u }{c - u} }f_0 $$ where $f$ is the frequency observed by the traveler, and $f_0$ is the frequency the source would measure itself emitting.
The relativistic Doppler effect is used to measure of shifts in the frequency of light emitted by a moving astronomical object such as a galaxy. Light emitted by atoms and normally found in the extreme violet region of the spectrum is shifted toward the red end of the spectrum for atoms in other galaxies, indicating that these galaxies are receding from us. Edwin Hubble (1889–1953) found that most galaxies are moving away from us, indicating that the Universe is expanding. This surprised Einstein.
Relativity does not do away with Classical Physics, but rather, enhances it to be consistent with Electrodynamics.
One of the key features of Classical Physics, something you should always remember, is conservation of momentum and conservation of energy.
Conservation of momentum and relativity lead to the finding that:
$$ \vec{p} = \frac{ m \vec{v}}{\sqrt{ 1- \frac{v^2}{c^2}}} $$
When $v$ is much smaller than $c$, $p \approx mv$.
Because Newton's Second Law says that $F = dp/dt$, $$ F = \frac{ma}{\left( 1 - \frac{v^2}{c^2}\right)^{\frac{3}{2}}} $$
$$ a = \frac{F}{m} \left( 1 - \frac{v^2}{c^2}\right)^{\frac{3}{2}} $$
As $v \rightarrow c$, $a \rightarrow 0$.
The faster you travel, the less effective your applied force is.
It is impossible for anything to exceed the speed of light.
Sorry science fiction fans, we will never explore the stars in person$^{*}$.
$^*$ Actually, there is still a little bit of hope with General Relativity.
A further implication of the refinement of the definition of momentum is that energy is also redefined.
Work-Energy theorem says $$K = W = \int^{x_2}_{x_1} F dx = \left(\gamma - 1 \right) mc^2$$
Using the binomial theorem, $(1+x)^n = 1 + nx + n(n-1)x^2/ + \cdots$, $$ K = \frac{1}{2} mv^2 + \frac{3}{8} \frac{mv^4}{c^2} + \cdots$$ So when $v$ is much smaller than $c$, $K = \frac{1}{2} mv^2$ works fine.
The form for relativistic kinetic energy has a component that is not dependent on speed: $$ K = \left(\gamma - 1 \right) mc^2 = \gamma mc^2 - mc^2 $$
We call this the "Rest Energy". It is an energy inherent in all objects that have mass. $E_0 = mc^2$.
The implication is now that Energy and Matter separately aren't conserved, but that together they are; and they can be converted back and forth.
The total energy of a massive particle that isn't in a potential field, i.e. only has kinetic energy and the inherent rest energy is: $$ E_{TOT} = K + E_0 = \gamma mc^2 $$
Combining this with the momentum equation $p$ and with some algebra, $$ E^2 = \left( mc^2\right)^2 + \left( pc\right)^2 $$ For particles without mass such as light, $E = pc$, and for particles with mass that are at rest, $E = mc^2$.
$E = mc^2$ implies that there is something special about mass.
Mercury wasn't behaving as expected:
In our Solar System, the largest perihelion shift is 560 arcsec/century for
Mercury. Most of the shift can be attributed to perturbations from other planets, but
after all the known planets are taken into account a shift of 43 arcsec/century remains
unexplained. Historically, some people speculated that there might be another planet
closer to the Sun than Mercury that caused the additional perihelion shift. The
hypothetical planet was called Vulcan.
Einstein extended special relativity to general relativity, which finds that gravity can be better seen as a warping of space and time.
Principal of equivalence: gravitational mass is equivalent to inertial mass.
Stated another way, acceleration due to non-gravitational force in
a gravity free part of space
and acceleration due to gravitation are effectively equivalent.
Equivalence principal:
Time and space are warped by massive bodies (gravity):
Time and space are warped by massive bodies (gravity):
These arcs are actually the lensed light of galaxies that lie
behind the cluster of normal looking galaxies in the pictures.
The huge gravity well of this galaxy cluster "lenses" light around
it, providing us views of dozens of galaxies (albeit a warped view)
that we otherwise wouldn't be able to see.
$$ \frac{\Delta f}{f} = \frac{\Delta U_g}{mc^2} $$
This has consequences for GPS satellites...we must apply the above adjustment as well as time dilation adjustment from special relativity or else your GPS will send you into the ocean.
The gravitational redshift: electromagnetic radiation originating from a source that is in a gravitational field is reduced in frequency (redshifted) when observed in a region at a higher gravitational potential (further away from the gravitational source[s]).
The gravitational blueshift: electromagnetic radiation originating from a source that is far away gravitational field is increased in frequency (blueshifted) when observed in a region with higher gravitational field. e.g. the frequency of a GPS satellite.
October 1971: Joseph Hafele and Richard Keating fly atomic clocks on
airplanes around the Earth. Results in $ns$:
If gravity warps spacetime and alters the trajectory of light, then a dense enough gravitational source should be able to "trap" light. GR predicts such "black holes".
7 minutes $\rightarrow$ 10 minutes in (feel free to watch entire):
A problem we will return to: GR predicts a singularity; quantum mechanics does not. Both theories perform spectacularly well in their own domains. Where they intersect, in a black hole, we only have?
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