The equation that allows us to calculate stuff in special relativity!

Take two observers with identical rules and stopwatch, and aligned axes, but one is on a car moving at towards the $+x$ direction at speed $v$.

TODO image.

When both observe an event, if we denote:It is of course arbitrary who is standing and who is moving, we will just use the term "standing" for the one without primes.

- $(t,x,y,z)$ the observation of the standing observer
- $(t_{′},x_{′},y_{′},z_{′})$ the observation of the ending observer on a car

Then the coordinates of the event observed by the observer on the car are:
where:

$t_{′}x_{′}y_{′}z_{′} =γ(t−c_{2}vx )=γ(x−vt)=y=z $

$γ=1−(cv )_{2} 1 $

Note that if $cv $ tends towards zero, then this reduces to the usual Galilean transformations which our intuition expects:

$t_{′}y_{′}z_{′} =tx_{′}=y=z =x−vt$

This explains why we don't observe special relativity in our daily lives: macroscopic objects move too slowly compared to light, and $cv $ is almost zero.

Same motivation as Galilean invariance, but relativistic version of that: we want the laws of physics to have the same form on all inertial frames, so we really want to write them in a way that is Lorentz covariant.

This is just the relativistic version of that which takes the Lorentz transformation into account instead of just the old Galilean transformation.

Basically a synonym of Lorentz covariance?

OK, so let's verify the main desired consequence of the Lorentz transformation: that everyone observes the same speed of light.

Observers will measure the speed of light by calculating how long it takes the light going towards $+x$ cross a rod of length $L=x_{2}−x_{1}$ laid in the x axis at position $X1$.

TODO image.

Each observer will observe two events:

- $(t_{1},x_{1},y_{1},z_{1})$: the light touches the left side of the rod
- $(t_{2},x_{2},y_{2},z_{2})$: the light touches the right side of the rod

Supposing that the standing observer measures the speed of light as $c$ and that light hits the left side of the rod at time $T1$, then he observes the coordinates:

$t_{1}x_{1}t_{2}x_{2} =T1=X1=cL =X1+L $

Now, if we transform for the moving observer:
and so the moving observer measures the speed of light as:

$t_{1}x_{1}t_{2}x_{2} =γ(t_{1}−c_{2}vx_{1} )=γ(x_{1}−vt_{1})=γ(t_{2}−c_{2}vx_{2} )=γ(x_{2}−vt_{2}) $

$c_{′} =t_{2}−t_{1}x_{2}−x_{1} =(t_{2}−c_{2}vx_{2} )−(t_{1}−c_{2}vx_{1} )(x_{2}−vt_{2})−(x_{1}−vt_{1}) =(t_{2}−t_{1})−c_{2}v (x_{2}−x_{1})(x_{2}−x_{1})−v(t_{2}−t_{1}) =1−c_{2}v t_{2}−t_{1}x_{2}−x_{1} t_{2}−t_{1}x_{2}−x_{1} −v =1−c_{2}v cc−v =cc−v c−v =c $

Suppose that a rod has is length $L$ measured on a rest frame $S$ (or maybe even better: two identical rulers were manufactured, and one is taken on a spaceship, a bit like the twin paradox).

Question: what is the length $L_{′}$ than an observer in frame $S_{′}$ moving relative to $S$ as speed $v$ observe the rod to be?

The key idea is that there are two events to consider in each frame, which we call 1 and 2:Note that what you visually observe on a photograph is a different measurement to the more precise/easy to calculate two event measurement. On a photograph, it seems you might not even see the contraction in some cases as mentioned at en.wikipedia.org/wiki/Terrell_rotation

- the left end of the rod is an observation event at a given position at a given time: $x_{1}$ and $t_{1}$ for $S$ or $x_{1}$ and $t_{1}$ for $S_{′}$
- the right end of the rod is an observation event at a given position at a given time : $x_{2}$ and $t_{2}$ for $S$ or $x_{2}$ and $t_{2}$ for $S_{′}$

Measuring a length means to measure the $x_{2}−x_{1}$ difference for a single point in time in your frame ($t2=t1$).

So what we want to obtain is $x_{2}−x_{1}$ for any given time $t_{′}2=t_{′}1$.

In summary, we have:

$LL_{′} =x_{2}=x_{2} −x_{1}−x_{1}t_{2}=t_{1} $

By plugging those values into the Lorentz transformation, we can eliminate $t_{2}andt_{1}$, and conclude that for any $t_{2}=t_{1}$, the length contraction relation holds:

$L_{′}=γL $

The key question that needs intuitive clarification then is: but how can this be symmetric? How can both observers see each other's rulers shrink?

And the key answer is: because to the second observer, the measurements made by the first observer are not simultaneous. Notably, the two measurement events are obviously spacelike-separated events by looking at the light cone, and therefore can be measured even in different orders by different observers.

What you would see the moving rod look like on a photo of a length contraction experiment, as opposed as using two locally measured separate spacetime events to measure its length.

One of the best ways to think about it is the transversal time dilation thought experiment.

Light watch transverse to direction of motion. This case is interesting because it separates length contraction from time dilation completely.

Of course, as usual in special relativity, calling something "time dilation" leads us to mind boggling ideas of "symmetry breaking": if both frames have a light watch, how can both possibly observe the other to be time dilated?

And the answer to this, is the usual: in special relativity time and space are interwoven in a fucked up way, everything is just a spacetime event.

In this case, there are three spacetime events of interest: both clocks start at same position, your beam hits up at x=0, moving frame hits up at x>0.

Those two mentioned events are spacelike-separated events, and therefore even though they seem simultaneous to you, they are not going to be simultaneous to the moving observer!

If little clock one meter away from you tells you that at the time of some event (your light beam hit up) the moving light watch was only 50% up, this is just a number given by your one meter away watch!

The key question is: why is this not symmetrical?

One answer is: because one of the twin accelerates, and therefore changes inertial frames.

But the better answer is: understand what happens when the stationary twin sends light signals at constant time intervals to each other. When does the travelling twin receives them?

By doing that, we see that "all the extra aging happens immediately when the twin turns around":

- on the out trip, both twins receive signals at constant intervals
- when the moving twin turns around and starts to accelerate through different inertial frames, shit happens:
- the moving twin suddenly notices that the rate of signals from the stationary twin increased. They are getting older faster than us!
- the stationary twin suddenly notices that the rate of signals from the moving twin decreased. They are getting older slower than us!

- then when the moving twin reaches the return velocity, both see constant signal rates once again

Another way of understanding it is: you have to make all calculations on a

*single*inertial frame for the entire trip.Supposing the sibling quickly accelerates out (or magically starts moving at constant speed), travels at constant speed, and quickly accelerates back, and travels at constant speed setup, there are three frames that seem reasonable:

- the frame of the non-accelerating sibling
- the outgoing trip of the accelerating sibling
- the return trip of the accelerating sibling

If you do that, all three calculations give the exact same result, which is reassuring.

Another way to understand it is to do explicit integrations of the acceleration: physics.stackexchange.com/questions/242043/what-is-the-proper-way-to-explain-the-twin-paradox/242044#242044 This is the least insightful however :-)

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