Ciro Santilli @cirosantilli 35

Authentication is one of the hardest parts of cryptography, because the only truly secure way to do it is by driving to the other party yourself to establish a pre-shared key so you can do message authentication code. Or to share your public key with them if you are satisfied with the safety of post-quantum cryptography.

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Ising model by

Ciro Santilli 35  Updated 2025-01-10  +Created 1970-01-01

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Toy model of matter that exhibits phase transition in dimension 2 and greater. It does not provide numerically exact results by itself, but can serve as a tool to theorize existing and new phase transitions.

Each point in the lattice has two possible states: TODO insert image.

As mentioned at: stanford.edu/~jeffjar/statmech/intro4.html some systems which can be seen as modelled by it include:

the spins direction (up or down) of atoms in a magnet, which can undergo phase transitions depending on temperature as that characterized by the Curie temperature and an externally applied magnetic field
Neighboring spins like to align, which lowers the total system energy.
the type of atom at a lattice point in a 2-metal alloy, e.g. Fe-C (e.g. steel). TODO: intuition for the neighbour interaction? What likes to be with what? And aren't different phases in different crystal structures?

Also has some funky relations to renormalization TODO.

Bibliography:

stanford.edu/~jeffjar/statmech/intro4.html

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Length contraction by

Ciro Santilli 35  Updated 2025-01-10  +Created 1970-01-01

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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:

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 $S^{'}$

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 (

t 2 = t 1

So what we want to obtain is

x_{2}^{'} - x_{1}^{'}

for any given time

t^{'} 2 = t^{'} 1

In summary, we have:

L L^{'} = x_{2} = x_{2}^{'} - x_{1} - x_{1}^{'} t_{2}^{'} = t_{1}^{'}

(1)

By plugging those values into the Lorentz transformation, we can eliminate

t_{2} a n d t_{1}

, and conclude that for any

t_{2}^{'} = t_{1}^{'}

, the length contraction relation holds:

L^{'} = \frac{L}{γ}

(2)

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.

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Indefinite orthogonal group by