i am a little bit confused about how to obtain the second line.

i have that $\kappa([A \wedge \delta A] \wedge A) = \kappa([\delta A \wedge A] \wedge A)$ by antisymmetry of $\wedge$ and $[,]$

but i get that $\kappa([A \wedge A] \wedge \delta A) = -\kappa(\delta A \wedge [A \wedge A]) = -\kappa([\delta A \wedge A] \wedge A)$ by cyclicity of $\kappa$ and antisymmetry of $\wedge$ and $[,]$

but this would make the term in question $\kappa(\frac{2}{3} \delta A \wedge A \wedge A)$, which I don't think is correct

in indices i also get this result

okay i think the above expressions are just wrong (the variations i mean)

wait nvm...

ahh i misunderstood the cyclicity proeprty of the killing form

Since earth's gravitational field is independent of object mass all bodies fall at same rate but when we throw a feather and a ball they don't fall at same rate. Is this has to do something with air resistance?

Yes

So what does air resistance depends on I read somewhere it depends on area and in others it depends on velocity. Why does this force does not have a mathematical equation

The problem is that except at very slow speeds the air flow is turbulent.

It does

And turbulence is currently impossible to treat analytically so we have to resort to approximations and numerical methods.

Oh so air flow is different at each point in space in direction and is continuously changing is that why?

Typically you write the equation as $$m\ddot{z} = mg - \rho \dot{z}$$

The friction term is $\rho \dot{z}$

$\rho$ is some friction coefficient and $\dot{z}$ the velocity of the falling object

If you divide by $m$, you get $$\ddot{z} = g - \frac{\rho}{m} \dot{z}$$

@alam The reasons why turbulence is hard to describe are quite involved. Basically it's a chaotic system.

So unlike the frictionless case, it does depend on the mass

Ok thanks

@Slereah That's linear drag, which is easy to describe. Sadly above speeds of a few mm/s the flow becomes turbulent and the linear drag equation no longer applies.

@JohnRennie I tried keeping it simple :p

i have finally completed by study of chern-simons theory to a sufficient degree...for now..!

@SillyGoose Did you follow any particular lecture note?

@Sanjana i only ultimately stationized the non-abelian classical chern-simons action on a trivial bundle and also computed how it changes under arbitrary gauge transformations. so, i didn't actually do a lot of chern-simons theory content-wise. i used nakahara mainly to learn (or perhaps just familiarize) some differential geometry and then looked at D. S. Freed's notes on classical chern-simons part 1

but the Freed notes are a little bit turgid to me now :P maybe to someone more well versed in the mathematics would see it as clear

and then asking questions here to ACM mainly and also some mathoverflow and physics stack answers

i found the answer by josé here particularly helpful

@vengaq Yes. When we talk about Hawking radiation we mean that the radiation escapes to infinity, so it can be observed by an observer far from the horizon.

isnt this inaccurate

@nickbros123 it's not even an integer, an goes to zero for N->\infty

there was a claim (not in the book, but in another SE post) that it seems reasonable for large G and N

the actual counting I suppose would be ${N+G-1}\choose{G-1}$, Approximations still give some factors of $\sqrt{2\pi}$ here and there

notes from David Tong dont even deal with this formula when speaking about entropy.

@ACuriousMind in its unwavering attempts to become more like the US, education in India for the most part isnt cheap, unless you study @ a government institution, wherein, post scholarship, youre still paying some $250 a year

@SillyGoose i think the first few chapters r good

@Semiclassical i expected him to give up hidden variables for a richer philosophy. but i got the impression that he was pushing hidden variables as the explicate order and the pilot wave as the implicate order, but i havent read it fully because i got bored

"the action of a group on a set is a functor from the group to $Set$"

how does this make sense? what is the map between the objects

suppose we take the action of $SO(2)$ on $R^2$

then we hav to map the objects. what is the map?

do we hav to map $S^1$ to $R^2$?

it's not making sense to me

@Slereah

do we simply say $F(SO(2))=R^2$ for the map between objects

@RyderRude Because a group is a groupoid with one object, commonly called *.

The functor sends * to X in Set, which is the set that it is acting on

The group elements, which are automorphisms of *, naturally become automorphisms of the set X.

Hope this clears things up for you

What do we call the solution to the Helmholtz equation?

Can we have in real life, only linear polarized light? Wouldn't that be impossible, as a linearly polarized light, implies one frequency i.e a plane wave, and plane waves are not physical ?

@imbAF It depends on the context

what do you mean?

You have vibrating membrane interpretations, Laplace equations (which give electrostatic field), quantum mechanical free particle interpretations, etc

plus, depending on the experiment it may not matter if the light isn't "perfectly" linearly polarized

it just has to be close enough to linearly-polarized for it to interact with other objects in interesting ways

Schweg

idk why i was watching this because I literally have no idea what they're saying but here ya go youtube.com/watch?v=PZ_qnBwSejE

@ACuriousMind Afaik elementary education is compulsory and free even in the USA. Its quality, well we could talk about :-) But things are not fine in this sense even in Germany, imho

@imbAF wdym plane waves are not physical?

@Obliv Because they have infinite extent

It's difficult to construct a device that produces an infinitely-wide coherent wavefront

yeah. plane waves are a useful approximation in certain cases but

I don't follow, why is circular/elliptical polarization physical but not plane waves?

i'm not sure the intention was to pick out linear as compared with circular. just having a fixed polarization in general?

@Obliv Polarization is the direction of oscillation. The wavefronts are the surfaces where the wave is at maximum.

They are different things.

there is somewhat of a difference between linearly and circularly polarized light, insofar as you can relate left/right circular polarized light to photons having spin angular momentum

til ed witten's father is 103 years old ...

imagine being 72 years old and your father is still alive