"it can be easily shown that..."

>"it can be easily shown that..."

HURRR DUURR ITS SO LONG IT MUST BE HARD AND ONLY FOR GENIUSES

Go back to twatter you fucking retard.

>"...particle physicists have vomit for brains."

>"we will not insult the intelligence of our reader by providing a proof"

the use of this particular turn of phrase is also known as "proof by intimidation".

They don't actually work all day

They love doing it as it is their passion.

Delet this thread

...

Where has anyone ever said that?

D. Gale - The Game of Hex and the Brouwer Fixed-Point Theorem
(in all fairness, the result in question is indeed rather easy, but still)

Legit kek

math guy here
is it possible that because the standard model isn't complete, that some undiscovered sector would simplify that monstrous equation?

i understand that in ~some models~ the lagrangian is significantly smaller, but i don't really care about those

My diffy q book was full of this shit.

Nagle, Saff, Snider 8th

750 GeV test says no deviation so far

I can just imagine that sentence in the middle of a whole page of dense logic notation.

that's hardly a proof that the standard model is complete, if that's what you're >implyign

this raises another question though
is it possible to determine whether the standard model is complete?

What particular books or studies do you need to follow to be able to understand this?

Stewart Calculus

You would probably need an entire graduate physics curriculum. Think QFT, GR, grad EM, and a grad level particle physics class

Math is awesome. I'm really new to it so sorry if I sound like a newb but is this one whole problem? What is this exactly? This looks really advanced.

It's #6 on the AP Calc BC FRQs this year

it's the standard model of particle physics.

lrn2lrn

(use the built in image search next time)

The entire standard model Lagrangian can be condensed down to [math]\mathcal{L} = \bar{\psi}\mathcal{D}\psi + \frac{1}{4}\int_{M}\operatorname{tr}\mathcal{F}^2[/math] plus some Higgs terms, where [math]\psi[/math] is the spinor containing all matter, [math]\mathcal{D}[/math] is the covariant derivative along the Lie algebra-valued connection one form [math]\mathcal{A} \in \Lambda^1(M) \otimes \operatorname{Lie}SU(5)[/math] and [math]\mathcal{F} = d\mathcal{A}[/math]. The reason why the usual expression for the standard model Lagrangian is so complicated is because physicists explicitly write down the components of the gauge fields and the spinors and do some weird ass Bogolyubov trasnformations that rotates the Hilbert space.

I unironically love it when textbooks insult readers