A friend asked me, rather innocently, "how long will it take to learn string theory", I gave an estimate number, "perhaps in the next 6 years if we're lucky"

They said "that's too long, I need to learn it in the next 6 days"

Apparently their physics lab (thermodynamics lab) TA was asking them about Stephen Hawking beckenstein formula, going off on a tangent from Stefan boltzmann law 😂

Safe to assume that TA lost it that day. They were rambling about black hole thermodynamics to 2nd year kids in a thermodynamics lab

@Obliv why dont you just write it down and avoid all the integration business?

6 years!!!

premed students expect to be doctors in that amount of time

(if they're lucky)

I have been using $S=m \int d \tau$ for a while now but never wondered about its spin unless some article presented it as the action for a spinless massive particle, so I wonder what the "spinfull" version is

You basically get an extra term for a little direction of your particle

arxiv.org/pdf/2307.13644.pdf

for instance

@user85795 going by a macroscopic analysis: most students start studying at 18 and get their PhDs, on avg by 28. 6 yrs was actually little. You keep learning after that too

@Sanjana Actually, you should be looking at it as a categorisation based upon the spin. These are the "representations of the Poincaré group" that keep appearing as the definition of elementary particles in QFT. Spin 0 follows Klein-Gordon equation, and you need the Klein-Gordon action for that. Spin half follows Dirac equation, etc. Spin 1 follows Proca action, which are closely related to photons.

This is one of the gruppenpest parts. The reason why this works, is that we do not just want to know specific wave equations, hoping that they happen to work well. Instead, now that we accept that SR is correct, we want to find all possible wave equations that are invariant under the Poincaré group. That leads to a study between Lie groups, Lie algebras, and group theory, and the result is this classification of all possible Poincaré group representations under different spins.

Hello fellas

Hello Cracked Pot

You may have helped me right meow

I'm trying to remember an old saying and your message lit up a candle

om nom

i have finally overcome the flu

Ditto is finally about to om nom nom? Does ditto feel like a cute slime?

@SirCrackpot ????

Hey everyone

Not a candle :P

Making lunch here

miao miao should go get some om nom nom too

but it is like, what just happened??

@naturallyInconsistent you good

miaoooo

purr purr

popi popi

(i am a clown)

boop

youtube.com/shorts/mSzPkxIdyx8?feature=share

@Loong I hope this was a deliberate choice by someone. I'd treat them to a beer. Juvenile fun is so good.

@naturallyInconsistent There was also the famous Finnair Flight 666 to HEL.

@nickbros123 I would not assume that. It is fun to excite eager students with fun stuff

@Loong Hell sinky?

miehehehe

Reminds me of one of my favourite journals

@naturallyInconsistent that would be fine, until the TA starts asking questions from Black Hole thermodynamics, and unironically grades us based on that. That I would classify as having lost it

Hell's Physics

@nickbros123 Ah, yes, totally unhinged and needs disciplinary action.

@Loong why is this? I know it is seriously sad to study but why Hell?

brb

@naturallyInconsistent I humored my poor friend who loves grades by recommending him this: youtube.com/playlist?list=PL3PVFGnaPl_sCp2A87NVD8GT5Z8Oqw3Yr

@naturallyInconsistent only works for German speakers

I told him he should watch it at 2x speed

funny thing is he might have taken it seriously

@Loong :(

Apparently only this was sent

But I said that I meant a bulb, not a candle

The Sailor's Child problem, introduced by Radford M. Neal, is somewhat similar. It involves a sailor who regularly sails between ports. In one port there is a woman who wants to have a child with him, across the sea there is another woman who also wants to have a child with him.

The sailor cannot decide if he will have one or two children, so he will leave it up to a coin toss. If Heads, he will have one child, and if Tails, two children. But if the coin lands on Heads, which woman would have his child? He would decide this by looking at The Sailor's Guide to Ports and the woman in the port that appears first would be the woman that he has a child with.

You are his child. You do not have a copy of The Sailor's Guide to Ports. What is the probability that you are his only child, thus the coin landed on Heads (assume a fair coin)?

what do u think about this?

@nickbros123 what if you friend turns out to be an unrivaled genius and your joke turns him into a world leading expert in ST? :P

@SirCrackpot Plead to the lord for forgiveness for sending a genius down the ST path

@SirCrackpot bing?

@Slereah @Sanjana any idea what the extra terms in (4.33) after $p_a dx^a$ are?

@bolbteppa There's a few explicit examples given later in the paper

$\phi_C$ is the Casimir invariant for the momentum, it's just $m^2$

$\phi_I$ is the same but for spin

@Slereah what a great naming choice

$A(p_a p^a - \phi_C)$ is just the Polyakov constraint term

With $A$ some Lagrange multiplier

The other two A terms are similarly constraints for spin stuff but I'm not as sure how to interpret them

Presumably this part of it is old and better explained elsewhere

Presumably the $A^{\alpha\beta}$ term is to keep the particle's spin constant

And $A^\alpha$ is to keep it... the appropriate angle with the momentum?

$\phi_I$ is the coadjoint orbit element related to the momentum, so it is the "type" of the momentum

A massless spinning particle (with local supersymmetry) has $S = \int d \tau (p_{\mu} \dot{x}^{\mu} - \frac{1}{2} \chi_{\mu} i \dot{\chi}^{\mu} - \frac{1}{2} e p^2 - \frac{i}{2} \psi \chi^{\mu} p_{\mu})$ where the last term is basically the supercurrent from global susy, if this is what they are trying to cover...

Well this is only for integer spin stuff I think

But presumably the terms are somewhat analogous

Section 5.2 is on spinning particles but I'm not sure if it's integer spin

It's spin 1

They're only covering the "normal" kinematic groups

But yeah if you compare it to that it does seem an analogous case

Except the last term lacking

What's the 2nd term in 4.33

Not entirely sure

Seems to be equivalent to the $\chi i \dot{\chi}$ term

That $\Sigma_a^b$ thing appears right below (4.33) making it look like a metric or a group transformation or something... - yes it is an orthogonal transformation...

This is pretty crazy

It's the method of coadjoint orbits

where you write the Lagrangian of a point particle as an action on the kinematic group

and then project it down to the manifold, ie the homogeneous space associated to it

The Sigma is generators of rotations

You select a coadjoint vector from the kinematic group to represent the "type" of particle, ie the translational part is gonna be (E,0,0,0) or something

And the rotational part for the spin

Then under applying the homogeneous part of the group to it you get the full "configuration space"

Like the mass hyperboloid

Or the sphere for the spin

do i need to download mathematica/maple or something to do calculations with latex

wolfram is kinda slow and kills the vibeeee man

What do you mean by doing calculations with LaTeX?

Like if there was a way to directly input my algebraic expressions into a calculator to solve them. It's okay, I'll just use a physical calculator for now. It's faster than messing with wolfram

even though the level of precision is pretty bad for thermo lol (too many large/small numbers)

There is sympy

You should not be needing algebraic calculations that are so big as to require Mathematica.

@bolbteppa Where can I find this action? Where did you learn these things from? The best I know is Polyakov form of the action for spinless particle...

@Sanjana String theory notes/books, e.g. section 8 here

is it not available on arxiv?

How come we can omit the factor $$\frac{1}{\sqrt{2\pi q(q+N)/N}}$$ from $$\Omega(N,q) = \frac{\left(\frac{q+N}{q}\right)^q\left(\frac{q+N}{N}\right)^N}{{\sqrt{2\pi q(q+N)/N}}} \approx \left(\frac{q+N}{q}\right)^q\left(\frac{q+N}{N}\right)^N$$ for very large q,N

That's the best link I can find right now, there is one somewhere else for sure

@bolbteppa its okay ill look it up myself if i need

@bolbteppa I didn't know Townsend had string theory notes. Thanks...but if it's not too much can you say which equation in section 8 am I looking for?

is $\ln\left(\frac{1}{\sqrt{2\pi q(q+N)/N}}\right)$ really negligible or something

Oh yeah I guess since it's still in the log, a very large number / a small number is still a very large number or whatever

It's 9.5 of the version I have, that one looks like a different version and I can't see those pages, I see another different version on there too, basically that will point you in the right direction

ok ok thanks

@Sanjana it's even in 'Fields'

ncatlab.org/nlab/show/spinning+particle

you can generally find this stuff looking for wordline quantization or wordline formalism or stuff like that

Thanks @bolbteppa and @ekardnam_

it happens that for string theory we basically only know the worldsheet formalism whose analogue is the wordline formalism in qft, so often you find these in string theory books i guess

With that ncatlab link, I got tons of references

@Obliv That is just $\Omega\propto q^{-1/2}\left(1+\frac Nq\right)^q\left(1+\frac qN\right)^{N-1/2}$ and those halves are too small to matter.