2:59 AM
Has anyone figured what the momentum distribution of the universe (FLRW metric) in local thermal equilibrium should be?

6 hours later…
8:59 AM
Morning

@Slereah morn morn

9:49 AM
@JohnRennie Yeah I'm in India and I'm thinking about going to an IIT
I wanted to try my hand at research (physics) and see if I can pursue that later
and also, hello again! I remember you helped me fix my laptop a long time ago

@UmeshKonduru I would ask in the Problem Solving Strategies room. That room is for help with JEE questions and a lot of the students who chat there are about to go to an IIT or are already at an IIT.

sure, thanks
Do you know of any summer research programs offered by universities in USA/Europe that are open for international students?
I had a friend tell me about REUs, but I realised I wouldn't be eligible for them, so I've been looking to find if I can find something similar

4 hours later…
1:43 PM
"Nakanishi-Lautrup field" is a very fancy name for a Lagrange multiplier

2:19 PM
"Although the search for diffeomorphism invariant observables lasts for almost a century, there still seems to be confusion in the subject and claims ranging from denying existence of any observables (except usually for the ADM charges which are defined only in asymptotically flat cases) to claiming that the problem has been solved entirely appear in the literature."

2:42 PM
"diffeomorphism invariant"
::eye twitches::

3:19 PM
Would you prefer the theory to be diffeomorphism variant?

4:00 PM
Can any body help me with this?

Basically, the answer I know is right involves me having an $\omega$ but I have that it doesn’t?
The only thing I can imagine is that I’m not justified with my ansats for some reason
So as far as I can tell, my characteristic equation is wrong

10

In the BRST formalism of gauge theories, the Lautrup-Nakanishi field $B^a(x)$ appears as an auxiliary variable $$\mathcal{L}_\text{BRST}=-\frac{1}{4}F_{\mu\nu}^a F^{a\,\mu\nu}+\frac{1}{2}\xi B^a B^a + B^a\partial_\mu A^{a\,\mu}+\partial_\mu\bar\eta^a(D^\mu\eta)^a,$$ and in the superfield formalis...

nlab calls them a generalization of a Lagrange multiplier

1 hour later…
5:12 PM
could anyone shed light on this, the indices don't even make sense, this is from Tong's Notes

@Charlie what do you mean by "the indices don't even make sense"?
it's an equation with one free index $\mu$, all the rest are properly contracted

oh dear
it's been a while ok go easy on me
:)

I don't even know how to go anything but easy! :P

>:)

but in order to help you untangle this you'll have to explain why you think it doesn't make sense

5:16 PM
$$\partial^{\mu} F_{\mu \nu} = (\partial^{\mu} \partial_{\mu}) A_{\nu} - \partial_{\nu} \partial^{\mu} A_{\mu}$$
It's now one more step to his equation

Oh yeah sorry I see what I was missing lol, ty ;)

2 hours later…
6:55 PM
Can you define an arbitrary gauge with the BRST term?
Do you just add whatever function you want to the auxiliary field term

7:13 PM
If we have two intertial system K and K', and K' moves with a relativistic speed in relation to K, basically the case in which we need to use the lorenz trasformations, when we observer an event, we can have one of the 3 following cases: the event happens in K, or it can happen in K' or it can happen in another reference frame that is none of the above . Is that correct?

What do you mean
The event is gonna happen in all reference frames
Just with different coordinates

the event can be observed in all the frames
but it can happen in one of them
i.e if you have a planet and a rocket, an event can take place in the planet, or in the rocket, or in the space between the two
which translates into 3 different case of where an event can take place

No, the event happens in all frames
Do you mean perhaps that each object has its own frame?

Ofc, an observer in the planet , has the planet as his frame
the observer inside the space ship
has the space ship as his frame

Well, they have their own frames, which may or may not match that of the ship :p
If they're spinning inside the ship, who can tell!

7:18 PM
well
if you observe something
and that rotates when you observe it
while the guy on a intertial frame does not see that=rotation
and they both communicate the datas
the guy in the reference frame that does rotate, should figure that out,no?

The hard part is knowing which one of you is in the inertial frame!
possibly no one

that is another problem
yes
but what I am trying to say is that, you can observe a bouncing ball inside the rocket. for the guy inside of it, the movement observed is different then the guy in the planet
but if the bouncing ball event ,takes place in the planet , the guy that is there observes a different type of event then the guy in the ship

Yes

And ultimately

Jan 8 at 13:24, by ACuriousMind
I'm just saying the phrasing of being "in" a frame is not very helpful

7:21 PM
they are not picture frames indeed

the ball can bounce in another x-planet/ship that is none of the above, then both observers in our initial frames, will observer smth different
if you make a measurement and your location is in earth
aren't you using earth as your reference frame?

a complicated question!
What is Earth's reference frame

for an event happening close to its surface

most experiments do not happen at the earth's core, they are likely rotating wrt it

an intertial reference frame
approximately
or you can consider the sun

7:27 PM
events don't happen in a frame, they happen in all of them and the whole point of relativity is that all inertial frames give equivalent descriptions

how is a bouncing ball=event inside a rocket, happening also in the reference frame of the earths surface?
I can understand that events can be observed in all of them
but happening, taking place in all of them
can't comprehend that

events are just points in spacetime

All those events happen in spacetime
A frame is just a different way of assigning numbers to those points
They're just different functions with the same arguments

assigning numbers to those points, which points?

the events

7:31 PM
so basically in other words reference frame= coordinate system?

the first half of learning math is learning how you can use numbers for everything and the second half is remembering that numbers aren't real
yes

But look at this definition, or distinction between frame and coordinate system
We first introduce the notion of reference frame, itself related to the idea of observer: the reference frame is, in some sense, the "Euclidean space carried by the observer".
In traditional developments of special and general relativity it has been customary not to distinguish between two quite distinct ideas. The first is the notion of a coordinate system, understood simply as the smooth, invertible assignment of four numbers to events in spacetime neighborhoods. The second, the frame of reference, refers to an idealized system used to assign such numbers

Well, a frame is more properly four directions that you generate your coordinates from, but that doesn't change the point much

if you distinguish them, then it's even less clear how an event (a point in spacetime) can "happen" in a frame

that is because

7:35 PM
(btw "the Euclidean space[...]" is obviously not a frame in relativity)

@fqq It happens right at the tip of the vector

I associate a reference frame with an entity (lack of better words) that has physical boundaries while the coordinate system an abstract mathematical tool to asign a set of numbers to an observed event

screams

But maybe,trying to distinguish them, confuses me, and it's better to consider them the same thing

@imbAF I think that's wrong, you cannot be inside/outside of a frame of reference
it's not a region of spacetime

7:36 PM
(unless it's an accelerated frame)

ok
so, an event happening at all frames at the same time
but different montions, of the observers
translate to different coordinates
and via the lorenz transformation, you can transform from the set of coordinates of one observer to the other
is that somehow better as an approach ?
or still faulty ?

somewhat

I can see the simplicity of it
compared to my
spacetime division

@imbAF what time?

in regions
what do you mean ?

7:40 PM
time is one of the coordinates, and it changes between frames

yes i know

It's weird that we believe in Euclidian geometry when we use projective geometry every day
We see objects changing sizes as they get closer and further and we still think it's the same object

isn't it? Mass etc do not change
only our perspective does

7:56 PM
Well the way the eye functions, it has to use projective geometry to figure out how the world is organized

One more thing, why do we use the metric tensor when we want to calculate the square of the absolute value of a 4 vector? Can't we straight up do a scalar product of it with itself ?

You may.

so then why do we use the metric tensor in this case? Is it somehow helpful or necessary in other cases?

The metric tensor isn't super important in special relativity

aha ok

8:02 PM
It gets more important in general relativity

Ok it makes sense now
and regarding the scalar product , when we do that, is there a reason why one vector is in covariant notation and the other in contravariant ?
I am currently looking the notes of my new lecture
and we just take this things, but no explanation behind it

There's a few different ways of considering the scalar product

@imbAF the metric tensor is the scalar product

@fqq True but people don't usually write it out in classical physics!

what you mean by is

8:06 PM
Special relativity delights in using the metric tensor even though most people historically didn't

why would you need to attribute it to a mathematical thing/tool

That's like saying that jets are natural because we use PDEs :p
Who needs to write a differential equation when you can just find the vector field that annihilate that jet section

you lost me here
:D
but i guess my question was dumb

It takes a while to get used to relativity

this is a brief introduction in special relativity, which will soon be connected with the electrodynamics

8:58 PM
@Slereah I am looking at the lorentz transformation , and we have the following eq: $x'^{\mu}=L_v^{\mu}x^v$
what do the indices at $L_v^{\mu}$
the meaning ?

Did you learn about Einstein summation?

yes i know
you are summing over v

Then what do you mean

in the eq. that i wrote
I mean
when you observe $L_v^{\mu}$ what does the superscript and subscript mean

it's the change between two basis
you send one basis to another

9:03 PM
ahaaa
so you sum the two contrary indices
and the result is the index left

10:00 PM
In my notes we write : $a^{\mu}= (ct,\vec x)$ and $a_{\mu}= (ct,\vec -x)$
then here $x'^{\mu}=L_v^{\mu}x^v$ the letters $\mu$ and $v$ are used as indices
for the differential expression we get $\frac{\partial x'^{\mu}}{\partial x'^{v}}=L^{\mu}_v$
$L^{\mu}_v$