Ciro Santilli @cirosantilli 34

When we particles particles, the action is obtained by integrating the Lagrangian over time:

In the case of field however, we can expand the Lagrangian out further, to also integrate over the space coordinates and their derivatives.

Since we are now working with something that gets integrated over space to obtain the total action, much like density would be integrated over space to obtain a total mass, the name "Lagrangian density" is fitting.

E.g. for a 2-dimensional field $f(x,y,t)$:

Of course, if we were to write it like that all the time we would go mad, so we can just write a much more condensed vectorized version using the gradient with $x\in {\mathbb{R}}^2$:

And in the context of special relativity, people condense that even further by adding $t$ to the spacetime Four-vector as well, so you don't even need to write that separate pesky $t$.

The main point of talking about the Lagrangian density instead of a Lagrangian for fields is likely that it treats space and time in a more uniform way, which is a basic requirement of special relativity: we have to be able to mix them up somehow to do Lorentz transformations. Notably, this is a key ingredient in a/the formulation of quantum field theory.

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 $L^{\prime }$ than an observer in frame $S^{\prime }$ 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

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 $x_2^{\prime }-x_1^{\prime }$ for any given time $t^{\prime }2=t^{\prime }1$.

In summary, we have:

By plugging those values into the Lorentz transformation, we can eliminate $t_{2}and{t}_1$, and conclude that for any $t_2^{\prime }=t_1^{\prime }$, the length contraction relation holds:

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.

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.

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:

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:

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

Why should I care when I can calculate new x and new time with Lorentz transformation?

Answer: it can give some qualitative intuition on what is larger/smaller happens before/after based only on arguably more intuitive geometric considerations, without requiring you to do any calculations, see e.g. Figure "Spacetime diagram illustrating how faster-than-light travel implies time travel".