8:21 AM
"Bram then additionally says that $|\uparrow ,k \rangle$ is disallowed, because its BRST-momentum is non-vanishing, and its pairing with a genuine physical state results in a delta function. This, for whatever reason, “violates kinematic law.” I don’t understand this."
Not reassuring

8:44 AM
It bothers me that so many sources say that the constraint $p^2 + m^2 = 0$ is natural
It is very much not, that's a choice of gauge!!!
It's the proper time parametrization certainly but that's not the only one possible
That's why you add it to the action, to force the gauge onto the system

0

Like this recent post on Chemistry SE which is the collection for different types of resources for learning Chemistry ( which I find quite useful) do we have any such thing over here at Physics SE? If not, then why not? Is it possible do it( if it isn't already available)? What I think is that ...

9:35 AM
@Slereah but the action is naturally just minimizing proper time, so the proper time parametrization is natural

@bolbteppa I mean sure, but it's certainly not stemming naturally from it
You could decide $$p^2 + m^2 = f(\tau)$$
It is entirely fine
There's no mechanical process which produces $p^2 + m^2 = 0$ from the action
is my point
Is there even a proper way to compute whether a constraint is correct from the get go?
Or do you just have to check the determinant of the hessian of the constrained action

Where does $p^2 + m^2 = f(\tau)$ come from

Well the proper one is $$p^2 + m^2 = [f'(\tau)]^2$$
A change of the parametrization of $x$ produces such a constraint
Or something like that?
I forget

The action is invariant under reparametrizations so idk how this extra factor comes in

The constraint is here to fix the parametrization
Therefore different constraints fix different parametrizations
The proper time parametrization basically fixes $\dot{x}^2 = -1$
But if you switch parametrization, it becomes $$\dot{x}^2 = [f'(\tau)]^2$$
Oh wait
Hm
Am I wrong
I may be

9:44 AM
$S = - m \int \sqrt{-g_{\mu \nu} \frac{dx^{\mu}}{d \tau} \frac{dx^{\nu}}{d \tau}} d \tau = - m \int \sqrt{-g_{\mu \nu} \frac{dx^{\mu}}{d \tau'} \frac{dx^{\nu}}{d \tau'} (\frac{d \tau'}{d \tau})^2 } \frac{d \tau}{d \tau'} d \tau' = m \int \sqrt{-g_{\mu \nu} \frac{dx^{\mu}}{d \tau'} \frac{dx^{\nu}}{d \tau'}} d \tau'$ both will have $p^2 + m^2 = 0$
Hmm

\begin{eqnarray}
p^2 &=& p^\mu p_\mu\\
&=& m^2 \frac{\dot{X}^2}{(\sqrt{\dot{X}^2})^2}\\
&=& m^2 \frac{[f'(\tau)]^2}{ \sqrt{[f'(\tau)]^2}}\\
&=& m^2
\end{eqnarray}
Dang it :V
Nevermind that
How do you fix the gauge
I've seen it done for Polyakov
$$S_G = \int d\tau \lambda(\tau) \ln(e(\tau))$$

Well
I think because $S = - m \int ds$ possesses constraints you have to write it as a constrained action first, and then fixing the Lagrange multiplier is equivalent to fixing a gauge

Hm wait
Let's see

That book with the picture you posted also says setting $\dot{x}^0 = 1$ is a choice of gauge so it's not just fixing the Lagrange multiplier

Well yes that is the obvious gauge

9:51 AM
hmm

the Coulomb gauge of the point particle :p
not very covariant

So fixing a gauge is literally choosing a parametrization, one way to do it is just to set $\dot{x}^0 = 1$, another way is to force the Lagrange multiplier to have a specific form, because it is a function of $\tau$ which changes under parametrizations in a certain way so fixing it fixes the parametrization

Apparently the proper gauge fixation is $$\dot{\lambda} = \xi(x, p, \lambda)$$
Fairly wide gauge
Hm wait, what's the EoM for $\lambda$ anyway
You have $$S = \int \dot{x} p + \lambda (p^2 + m^2)$$

The EOM are $p^2 + m^2 = 0$

So wrt $p$ that's $$\dot{x} + \lambda p = 0$$

10:03 AM
If you find the EOM for $p$ from that action and use them to eliminate $p$ you get the einbein action

That is indeed the parametrization
But it doesn't fix it
I guess it is basically the einbein, yes
So... $$S_G = \int d\tau \alpha(\tau) \ln(\lambda)$$
Maybe???

But now you set $p = - \dot{x}/\lambda$ and plug it into $S = \int \dot{x} p + \lambda(p^2 + m^2)$ to get $S = \frac{1}{2} \int d \tau e (e^{-2} \dot{x}^2 - m^2)$ i.e. a 1D GR action with metric $e$

I'd need to recheck the EoM for $\lambda$ tho
Wait no
That ain't working

This is how you get Polyakov from NG btw if you do all this in 2D but it's pretty complicated writing out the metric

Yeah that seems obvious enough

10:07 AM
So string books usually just say assume the 2D version of $S$ takes that form and use the EM tensor to show you get the NG action back

Just trying to think about how you fix the gauge
Do you just decide it by fiat?
Or can you write down a sufficiently bespoke Lagrangian
You can write down a gauge fixing term for EM

In EM you still have freedom though

True
I hate gauge theory so much

All of physics is basically gauge theory :p
The game is ignoring the subtleties just enough but not too much lest you descend into nlab madness

It's too late
I have caught gauge fever

10:21 AM
Still need to work backwards from EM Gupta-Bleuler to BRST language to see it as natural

BRST seems natural enough using the proof with path integrals
it's a bit trickier for canonical quantization tho
I shudder to think about BRST quantization of the Einstein Hilbert action
spooky metric ghosts
aaaaaaah
Wait I think the gauge fixing is just the einbein constraint
$\alpha(\tau) p_e(\tau)$

10:36 AM
From $S = - m \int ds$ we have $p^2 + m^2 = 0$ is a primary constraint and so $S = \int d \tau [\dot{x} p - H + \frac{1}{2} e(\tau)(p^2 + m^2)] = \int d \tau [\dot{x} p + \frac{1}{2}e(\tau)(p^2 + m^2)]$ is the constrained action, this action must be parametrization invariant which implies $e(\tau)$ has to change in a certain way under reparametrizations. Thus fixing $e(\tau)$ fixes the parametrization i.e. fixes the gauge.
Solving the $p$ EOM for $p$ gives $S = \frac{1}{2} \int d \tau e(\frac{1}{2} e^{-2} \dot{x}^2 - m^2)$ and again fixing $e$ fixes the parametrization.

10:49 AM
I should do something simple to relax, rly
compute apples falling from trees
You know what I haven't done in forever
Optics
Also thermodynamics
Let's do a thermo page on my site
make a little engine
choo choo

11:22 AM
though to be fair
doing it properly requires contact manifolds

11:51 AM
Any body can answer my question which I posted on the site, astronomy.stackexchange.com/questions/33748/…

12:13 PM
What is Dirac Quantization Condition? Roughly, we can say that it is $q_m q_e \in 2\pi \Z$ for eletric charge $q_e$ and magnetic charge $q_m$. But why someone said the existence of monopole implies quantization of electric charge, without any assumption, for instance, "If there exists a quantized monopole.."?
Did I miss something?

The magnetic charge is related to the structure group of the EM principal bundle

Then, Is magnetic charge always quantized?
if exists

If it stems from an EM gauge theory, yes

Okay thanks. I have seen most people omit that kind of hard stuff and just summarize the result without any caution about this. I think.. they need to mention it..

You can do it without bundles by doing the whole Dirac string derivation
Though it's not quite as satisfying

12:24 PM
So, do you mean that the quantization of monopole can be shown from Dirac's original argument?
If yes,,, then I miss something.

Well it's a bit of a fairly specific case
if you want a more general argument it is best to do it via gauge theory yes
stuff like that

12:48 PM
I've never actually worked out how geometric optics stems from EM
The ~Eikonal approximation~

1:01 PM
Landau gets it in 2 ways pretty fast

2 hours later…
2:50 PM
@bolbteppa hello
Is there any easy and quick way to convert Laplace operator to spherical coordinates
?

@bolbteppa Gradient, Curl and Divergence, which one is laplace?

Laplace is 'divergence of a gradient' as explained above 16 where they then state 16 (which is Laplace)

@Slereah is that related?

2:54 PM
$$\Box = g^{\mu\nu} \nabla_\mu \partial_\nu$$
just gonna use the metric tensor of spherical coordinates

@Slereah Tensor is verrrrrrry high-level thing.

@AbhasKumarSinha read that pdf, it sets everything up from the beginning to make it easier and will help prep you for tensors

@bolbteppa Is tensor required? What can I do with it?

When you know tensors very well you can get the formula even faster than as explained in the pdf I sent you

@bolbteppa Oh okay... Can it help me in high school mechanics or mathematics (like matrices or vectors)?

2:58 PM
Vectors are 'rank 1' tensors, matrices are 'rank 2 tensors', you will gain experience as you study more and more advanced things one step at a time

@bolbteppa Okay.... nth rank vectors are multidimensional, so how do you represent them on paper?

A vector can be represented as $v_i$, a rank 1 tensor (one index), a matrix can be represented as $M_{ij}$, this is a rank 2 tensor (2 indices), you can guess how to represent higher rank tensors

@bolbteppa are tensors hard?

It depends on what you want to do with them

@bolbteppa My teacher defined moment of inertia as 7th component of tensor, does that make sense?

3:06 PM
Moment of inertia is a 2nd rank tensor, and it's components give the MOI so I guess that's what they meant

@bolbteppa Well that pdf is well written... I finally understood what is curl
@bolbteppa Probably something like that...

If you want some very bad ways to get the Laplacian I can also do that
using
Dun dun duuun

Pretending $\partial x$ is a quantity

3:08 PM
dun duun duuuun....
@Slereah okay

I mean it's basically the chain rule

@Slereah Those stunts are meant to be performed by the experts only, don't try at home...
@bolbteppa Thanks sir... That method was helpful.

if you have $x$ as a function of $r, \theta, \phi$, then

\frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \phi}{\partial x} \frac{\partial}{\partial \phi} + \frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}
This is just the chain rule
Do this for all coordinates and work it out

@Slereah I know that, that's messy...

It certainly is

3:11 PM
you'll end up doing silly mistakes.

You can also do it the physicist way, where you present some argument of what the derivative will be
doing infinitesimal variations of the coordinates and argue that this is an infinitesimal arc and such

okay...
that's variety of ways.
@Slereah see that, for example, most elegant proof, I've seen today
I dun know how postdocs always manage to think one step ahead and this level of proof is insane.
he's literally controlling maths.
@Slereah What's the most elegant mathematical proof you've come across?

4:08 PM

4:38 PM
@ACuriousMind 9th iteration of yourself? physics.stackexchange.com/users/245302/curiousmind9

@AaronStevens Sure, we call them ACuriousNine ;)

@Slereah that's about 5 minutes working out the first derivatives, now do it again to get second derivatives

4:58 PM
@ACuriousMind There it is

5:22 PM
@bolbteppa that's why you use the tensors

Is $g^{\mu \nu} \nabla_{\mu} \partial_{\nu}$ right?
Well it is, but you need to do simplifications on the Christoffels right
There's a direct formula $\nabla^2 f = \frac{1}{\sqrt{g}} \partial_{\mu} (\sqrt{g} g^{\mu \nu} \partial_{\nu} f)$
Using the scale factors you can simplify it in ortho coordinates

that too

5:37 PM
Where does $A_J(s,t) = -\frac{g^2(-s)^J}{t-m^2}$ come from

Is that the propagator thing

It's a part of a scattering amplitude in Mandelstam variables that motivates going to strings

I vaguely recall it in Schwarz et al
It's the "this is why we can't have spin 2.5" bullshit isn't it

6:10 PM
already seeing posts with things like "my pronouns are var/let". sounds like javascript joke but i guess there's no way to tell if it's legitimate

@SirCumference I saw someone saying their friend used "var/let"; but they also claimed to use "grimble/gromble" as their own personal pronoun, and googling their pronouns + "pronoun" linked right to the SE question, so it seemed like a possible troll scifi.stackexchange.com/posts/221861/revisions

yeah that was the post
maybe i'm overthinking it, though var and let are the only assignment operators in JS. but who knows

@SirCumference Yeah I have severe doubts about that authenticity, the majority of the pronouns they chose plus the word "pronoun" literally gave that question as the top hit in google. Most of the examples they gave are grammatical nightmares, and "grimble/gromble" fits really well with "troll pronouns". The whole post just stinks to me, though proving it's trolling would be difficult.

@JMac clearly a troll; did a similar thing on another site

The CoC requires we refer to the OP as grimble though

6:19 PM
@JMac Fortunately we don't need to "prove" trolls are trolls to simply not engage with the trolls.

@SirCumference I think in this case it would still be within the CoC to just completely disengage instead, and flag as probably trolling

Would it though?
The CoC itself seems really not thought-out

@SirCumference You were neither required to bring up that post nor its OP here, you could've simply chosen to ignore it. Don't overthink it and don't feed the trolls.

Even putting aside the whole coerced speech aspect, if the troll had been more subtle it's not clear what the course of action would be
@ACuriousMind I'm not feeding the troll, the fact is the rules laid out are not written well

@SirCumference Sure. Not addressing the user would make your use of their pronouns irrelevant. I agree that the CoC isn't well thought out; but I think "try to be nice" is still the ultimate goal. If someone has a weird request like that, and you feel like it's trolling and don't want to oblige, just don't engage at all so that things don't get sour if they actually were genuine (or are trying to wind people up).

6:24 PM
It's inevitably going to become a problem elsewhere
@ACuriousMind Not to mention what I brought up was the (probable) JS joke, not the post itself

any astronomers or astrophysicists around?
I'd like some external input on the reliability of a spectroscopic observation from the 70s

6:40 PM
not my forté I'm afraid

never mind

7:09 PM
@Slereah equation 6.13 here (p.12)

2 hours later…
9:24 PM
@DanielSank atomic physics, of course ;-)

10:22 PM
What does integrating over solid angle mean?
Having some troubles understanding why we need to do it
For example, when calculating total luminosities?

140

Higher spin particles have to be coupled to conserved currents, and there are no conserved currents of high spin in quantum field theories. The only conserved currents are vector currents associated with internal symmetries, the stress-energy tensor current, the angular momentum tensor current, a...

Oh look
Again the person banned on this site is the only one who has ever posted about $s^j/(s-M^2)$ propagators on this whole site
Luckily once March '92 comes I could ask him on this site

@JakeRose Solid angle is the 3D generalization of angle
Generally the situations in which you'd want to integrate over solid angle are analogous to 2D situations where you'd integrate over angle
In geometry, a solid angle (symbol: Ω) is a measure of the amount of the field of view from some particular point that a given object covers. That is, it is a measure of how large the object appears to an observer looking from that point. The point from which the object is viewed is called the apex of the solid angle, and the object is said to subtend its solid angle from that point. In the International System of Units (SI), a solid angle is expressed in a dimensionless unit called a steradian (symbol: sr). One steradian corresponds to one unit of area on the unit sphere surrounding the apex,...

10:44 PM
0

In GSW equation (1.1.2) they define the scattering amplitude for a spin $J$ particle at high energies as $$A_J(s,t) = - \frac{g^2(-s)^J}{t-M^2}$$ mentioning it is an asymptotic approximation to a formula involving Legendre polynomials, and in this stackexchange post it is also defined, and in Wil...

@JakeRose see that picture
Note the $\sin \theta d \phi$ part is that red line which comes from noting that the arc length of the circle is found from $s = r \theta$ (not the $\theta$ in the picture, just the way this formula is usually written), but here $r = \sin \theta$, and the angle is $d \phi$, so you have $\sin \theta d \phi$, and the vertical line is $d \theta$ so that the area is $\sin \theta d \theta d \phi$.

11:42 PM
@Slereah no
NO