Ciro Santilli @cirosantilli  40

Given the function $\psi$:the operator can be written in Planck units as:often written without function arguments as:Note how this looks just like the Laplacian in Einstein notation, since the d'Alembert operator is just a generalization of the laplace operator to Minkowski space.

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.

Some sources say that this is just the part that says that the norm of a $L^2$ function is the same as the norm of its Fourier transform.

Others say that this theorem actually says that the Fourier transform is bijective.

The comment at math.stackexchange.com/questions/446870/bijectiveness-injectiveness-and-surjectiveness-of-fourier-transformation-define/1235725#1235725 may be of interest, it says that the bijection statement is an easy consequence from the norm one, thus the confusion.

TODO does it require it to be in $l^1$ as well? Wikipedia en.wikipedia.org/w/index.php?title=Plancherel_theorem&oldid=987110841 says yes, but courses.maths.ox.ac.uk/node/view_material/53981 does not mention it.

The Kibble balance is so precise and reproducible that it was responsible for the 2019 redefinition of the Kilogram.

It relies rely on not one, but three macroscopic quantum mechanical effects:How cool is that! As usual, the advantage of those effects is that they are discrete, and have very fixed values that don't depend either:One downside of using some quantum mechanical effects is that you have to cool everything down to 5K. But that's OK, we've got liquid helium!

The operating principle is something along:Then, based on all this, you can determine how much the object weights.

Bibliography:

An overview of what you can do with a pre-shared key with tradeoffs can be found at: quantumcomputing.stackexchange.com/questions/142/advantage-of-quantum-key-distribution-over-post-quantum-cryptography/25727#25727 The options are:

The fact that you cannot have trailing commans in lists or dicts as in 3, at:is one of the most infuriating design choices of all time!!!

Maximum current that can flow across a Josephson junction, as can be directly seen from the Josephson equations.

Is a fixed characteristic value of the physical construction of the junction.

A function $\mathbb{R}\to \mathbb{R}$ defined by the second of the Josephson equations plus initial conditions.

It represents an internal state of the junction.