The h Bar

Silly Goose

00:41

Let's say I am dealing with a factorized Hilbert space $\mathcal{H}_1 \otimes \mathcal{H}_2$. I want to generate the Lie group $SU(2) \otimes SU(2)$ consisting of all unitaries of the form $(U_1: \mathcal{H}_1 \rightarrow \mathcal{H}_1) \otimes (U_2: \mathcal{H}_2 \rightarrow \mathcal{H}_2)$. Can this be done by taking the direct product $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$?

I feel I am not using standard notation... does $SU(2) \otimes SU(2)$ usually refer to local operators of the form $U_1 \otimes U_2$?

also I mean to continue...taking the direct product of the lie algebras and exponentiating elements in the resulting direct sum of lie algebras?

Other than using $\otimes$ instead of $\times$, I think I am using standard notation actually based on the wikipedia for direct product.

123

02:24

Hello Everyone..

Physics Meta

03:29

0

Q: Should we flag a duplicate question if it has more engagements comparing the original post?

Look at this question: Does the universe have a center?. It is the original one. (original, unpopular) Now, look at this one: Did the Big Bang happen at a point?. (duplicate, famous) If you notice, the duplicate post has a comment by a moderator that redirects to the original post: Now, the orig...

naturallyInconsistent

05:03

Been looking at prices of coal power, solar power, wind power, utility electricity, and holy fuck prices have been skyrocketing.

123

Hello @JohnRennie Sir

DanielSank

05:24

Just finished a draft of a paper.

First one I've written myself in seven years.

:-p

naturallyInconsistent

Ohh, what is it about?

nickbros123

05:54

My neglect of mathematics all through the semester has finally caught up to me, now I gotta do the first 3 chapters of Hoffman kunze in 25 days

naturallyInconsistent

06:17

@nickbros123 First 2 chapters should be fine, chap 3 is where you should be sinking some time into.

123

Bob attached to the string and rotating in circular motion. Is this an example of conical pendulum?

Silly Goose

the notation of H&K kills me

@Relativisticcucumber honk

naturallyInconsistent

@SillyGoose h o n k

nickbros123

06:36

Yes, I started now, 1st chap is a breeze

I guess it's one of those books that exponentially rise in difficulty

Qmechanic

07:23

@Relativisticcucumber : Yes.

Slereah

07:54

I don't know how to use Tikz properly so I am just moving shapes millimeter by millimeter until they fit

Silly Goose

lol

i wonder if there is an ai tool to draw manifold diagrams

would be nice

inkscape manually is a little bit inconsistent

Slereah

You can just input functions in Tikz directly

But that's some pretty heavy stuff for the compiler

I try to keep things light

Silly Goose

i should make an effort to utilize tikZ :P i guess i haven't the need yet though

started the day with linear ending the day with lie theory woop

Slereah

It is a bit of a pain but a well illustrated concept is always nice

Currently illustrating angle measurement techniques for astronomy in the Ibak tribes of Borneo

The most important illustration

naturallyInconsistent

08:58

@Slereah What did you want to do, that you have to do this? You know that TikZ gives you freedom to use general vector coördinates, right, not just relative coördinates of objects?

Slereah

Trying to have a few cylinders inside each others

BUT

They are rotated cylinders

The rotation was making things a bit hard to predict

naturallyInconsistent

TikZ can do the rotation for you...

Slereah

Probably but I couldn't figure out how to do it properly

I could only find how to rotate the objects individually and if I did that their relative positioning got messed up

naturallyInconsistent

09:22

hmm, maybe you should open a LaTeX question about it, hehehe

Slereah

Probably quicker to just hack something together rly

Mad

09:36

question about dirac notation.
if i want to write something like
$ <A^\dagger v, B v> $ do i write $ <v \vert A^\dagger B \vert v> $ or$ <v \vert B A^\dagger\vert v> $

naturallyInconsistent

No, it is $\left<v\mid AB\mid v\right>$

Mad

thats one way

i am trying to show that (AB)^\dagger = Bdagger A dagger

which is EXTREMLY trivial in the standard maths notation

but i cant do it in dirac

Slereah

$(A | \psi \rangle)^\dagger = \langle \psi | A^\dagger$

roughly speaking

Mad

this requires knowledge of the dual space

i want to show it with only the attribute and the definition of the scalarproduc

this is the proof in standard notation

i am trying to translate this into diracs

naturallyInconsistent

Mad

09:46

Well, you are inserting the operators into the kets

which is something that the lecture didnt do

naturallyInconsistent

it seems like you also did that.

Mad

Yes but i am not using dirac notation

But okay

naturallyInconsistent

It is the same thing. Dirac notation is scalar product notation.

Mad

No its not

anyways, the way you wrote the first product

should be tho AB^dagger?

naturallyInconsistent

what I wrote is correct.

Mad

09:49

I am not saying its not

the issue is , look, you are using two notations

naturallyInconsistent

$\left<AB\psi|\phi\right>=\left<\psi|(AB)^\dagger|\phi\right>$

Mad

the dirac notation i know contains always 3 bars if operators are involved

(the right side of the equation)

the left side, you are just using standard notation

it doesnt look right at all

naturallyInconsistent

That is simply not necessarily the case. $\left<\psi|A|\phi\right>\equiv\left<\psi|A\phi\right>$

Mad

Yes but we did not define this like this!!

Nowhere have i ever read this definition

And you cant even define it like that

naturallyInconsistent

I know you are coming from maths, but this is how physics is usually going to be.

Mad

09:52

because the left side also implies <A^\dagger \psi, \phi>

I am not arguing if this is useful or not

i am just asking to translate what i wrote into what physicsts define

naturallyInconsistent

$\left<\psi|A\phi\right>=\left<A^\dagger\psi|\phi\right>$

Mad

i think you are not understanding

my lecture is based on sakurai, my tutor refuses any kind of notation that is not 100% equal to what he has on his solutions paper. this is simply not using dirac notation with the three bars (as Sakurai uses it)

which includes 3 bars

naturallyInconsistent

If you want the more complete version, A is an operator representable by a matrix, $\left|\phi\right>$ is a ket vector that is usually representable by a column vector, and the action of a matrix on a column vector is a column vector, and so $A\left|\phi\right>=\left|A\phi\right>$ always makes sense: matrix multiply vector equals vector.

And it is 1 or 2 bars, never 3.

Mad

Thank you

what you wrote would have been very useful

if i did not have a degree in mathematics

naturallyInconsistent

Look, what you wrote as the usual mathematical proof, is not even correct. It should be

If you want to go into greater detail, you can complex conjugate basically everything there, so then you can express them with the 2 bars.

But as you already wrote down in maths notation, this thing kinda requires the 1 bar version, not the 2 bar version.

Relativisticcucumber

12:21

@Mad related: physics.stackexchange.com/questions/502606/… (+ please use \langle and \rangle before my eyes bleed)

@Slereah i have a follow up question to something you said a few days ago. so iirc, you said that $\sqrt{-g_{\mu \nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}}$ is a valid lagrangian when attempting to use lagrangian mech to get the geodesic equation. however, i see that the einstein-hilbert action is defined by $S_H = \int \sqrt{-g}Rd^nx$ so why, in the first case, do we not have to include $R$? [...]

[...] in that case i believe we are solving for geodesics of free, non-interacting particles, but this is still in a curved space-time, right?

Slereah

If you include both terms at once, what you have is the couple equations between a point particle and the metric

(it's the Schwarzschild metric if you solve it)

Relativisticcucumber

sorry what do you mean "both terms"

Slereah

For the geodesic equation you need to do it in the test limit, where the energy of the particle doesn't couple to gravity

The action for a point particle and the EH action

Relativisticcucumber

@Slereah and what is "it" here

@Qmechanic ok thank you!

Slereah

@Relativisticcucumber The metric should not be dynamical, ie you don't vary with respect to g

Relativisticcucumber

12:33

im not really understanding. so R is the curvature essentially, right? so im trying to see why the first action doesnt have an R in it as the einstein-hilbert action does.

Slereah

Matter doesn't interact with the curvature directly usually

but with the metric

Relativisticcucumber

but the metric is nothing physical, right? its just a description of the manifold?

hm wait as is R

bah

@Slereah but then wouldnt we never need an R..?

Slereah

It is similar to how particles couple with the electromagnetic potential A, but the kinetic term of EM is the EM tensor

Or how classically the acceleration depends on the derivative of the potential, but the EoM is the Poisson equation, with the Laplacian of the potential

ACuriousMind

13:06

@Relativisticcucumber I think we first need be clearer that these are actions for two completely different physical systems

the "geodesic action" is an action $S_G[\gamma]$ on the space of paths $\gamma: I\to M$, the dynamical variable is the path of a particle, the metric is a fixed background here

the Einstein-Hilbert action $S_{EH}[g]$ is an action on the space of all metrics on a manifold, the dynamical variable is the metric itself, and there aren't any particles here

so the question "why does one of these include $R$ and the other doesn't?" is about the same as asking "why does the action for the trajectory of a particle in an electromagnetic field not involve the field strength tensor, while the action for electromagnetism with the EM field being the dynamical variable does?"

Relativisticcucumber

okay i think i need to go back and reread this variational method for finding stationary points of the action

i will do so

then i will reexamine your answers. thank you @ACuriousMind @Slereah

Slereah

13:56

How does any other field do it without ArXiv

Ryder Rude

hi

Slereah

14:18

Why is every object used in experimental GR all the objects that are the worst things to model in GR

Why can't we be made of pure energy

Ryder Rude

@Slereah what objects do u mean?

i agree that fields are better than rigid bodies

Slereah

A stick, for instance

Worst object

Ryder Rude

people study sticks using GR?

Slereah

It's the classic example of a rigid object

Ryder Rude

i just think of the universe as fields

but engineers have to study this ugly stuff

@Slereah tbf everything is a field at the fundamental level

a rigid body isnt even allowed in relativity, even in principle. am i right?

because a rigid body has to instantaneously communicate forces

Slereah

14:27

It depends what you mean by rigid

ACuriousMind

@Slereah I'm fairly sure there are several cults claiming we actually are pure energy; the GR sticks are probably part of the conspiracy to suppress the truth :P

Slereah

Proclus was right

Ryder Rude

@ACuriousMind it's not a cult. everything is a field according to QFT

Slereah

He took the extra step of saying that we are all one to remove the problems due to high cardinality

Ryder Rude

some spirituality people may also say this stuff tho

@Slereah what is a rigid body in relativity? can it be deformed?

ok but energy isnt even conserved. doesnt make sense to say we are pure energy

Slereah

14:31

It is a body such that the projection operator projecting to a local spacelike distribution is invariant under the motion of the object

Ryder Rude

everything is a field is the right thing to say

Slereah

So $$\mathcal{L}_{u} (u \otimes u - g) = 0$$

where $u$ is the velocity field of the object

Ryder Rude

@Slereah is deformation allowed?

Slereah

What's a deformation

Ryder Rude

a body changes its shape under forces

Slereah

14:35

If the body had no force acting on it it would just explode into particles moving along geodesics

Ryder Rude

yeh. i meant external forces

non-science people think of energy as a substance rather than a property of fields

there are popular sayings like matter is concentrated energy

it has gotten very popular because of E=mc2

there is a thing called the cosmological argument which claims to logically conclude a God

but it suffers from the problem of induction according to Hume

ACuriousMind

14:52

@RyderRude Checkmate: I'm a Weinbergian and believe in the primacy of particles.

Neither QM nor QFT in their formalisms really come with an ontology attached, stop claiming they say such ill-defined things like "everything is a field"

@Slereah the ultra-est of finitists?

123

Hello Everyone

ACM give me your brain. So i can save my time in understanding physics.

Ryder Rude

@ACuriousMind how do u explain that particles are only defined asymptotically?

Weinberg builds stuff from particles, but we have to note that particles in QFT are not well defined in the interacting theory

ACuriousMind

I don't have to explain anything

the QFT formalism works regardless of what ontology, if any, you ascribe to an interacting state

"is" a proton three quarks? or "is" it a mass of sea quarks and gluons? or "is" it an "excitation in a quantum field"? You can argue that the answer to all of these questions is "Yes", but I find the last description the most unenlightening

Ryder Rude

is there any particle interpretation of the interaction region of scattering experiments?

@ACuriousMind the field ontology is unenlightening because it's more reductive, more fundamental. the particle description is best for the special cases

but when u r in the interaction region, there is no particle interpretation as even an option

ACuriousMind

I don't know what that even means

what do you need an "interpretation" for, here?

neither the LSZ formula specifically nor the time evolution operator in general cares about what you "think happens" during an interaction

Ryder Rude

15:08

yes, we can do physics without ever talking about an ontology. but if we r going to choose an ontology, the field ontology is the one that applies to all situations.

so i am just saying that it is superior to the particle ontology. im not saying that u r forced to think of an ontology

i can describe the interaction region's ontology using fields

while the particle ontology ceases to be an option

More Anonymous

@ACuriousMind about your comments on my question (now deleted). I can choose any amount of time to move the blade through say a chunk of metal. There are lambda primes but they cancel on both sides. As for point 2. You need a finite force you have $1/ r \to \infty$ , the only chance you have is $ r \dot \to 0$

What I do know is the y and x or r relation

ACuriousMind

@MoreAnonymous my point was that the path of the knife can just be modeled as the trajectory of the knife point $t\mapsto (h(t),r(t))$. I do not understand what $\lambda(t)$ is supposed to represent

More Anonymous

@ACuriousMind Yes but newtons second law has $d p /dt $ and not $d p /dh(t)$

ACuriousMind

hm? the second law would just be $m \ddot{h} = F_h$ and $m\ddot{r} = F_r$, where $m$ is the mass of the knifepoint

More Anonymous

Ah I see your point

All the $\lambda(t)$ did was try to account for the variation in speed one may apply while moving the knife

ACuriousMind

15:22

but what is $\lambda(t)$ supposed to be

the "variation in speed" is just you being free to choose $\dot{h}(t)$ and $\dot{r}(t)$ within the constraints of the shape

More Anonymous

I think your right there was no need to write lambda(t)

ACuriousMind

also you're right that you need $\dot{r}\to 0$ at $x=0$ for the forces to work (this is just the observation that the particle moving up the dome will reach the top with zero velocity), but what does this have to do with "infinite velocity"?

Ryder Rude

i posted this answer about equivalence principle in QM. Is it correct

i will delete it if it's incorrect

the cosmological argument claims that since everything we know has a cause, the universe must too

but this suffers from the problem of induction

Mad

15:44

Question, does this equation, hold only if the operator is unitary?

Or maybe theres another manipulation i am not seeing

(trace of V^dagger A V = Trace A

Slereah

Feyerabend's book came in the mail

It's time to end it all and destroy science

Relativisticcucumber

16:11

@Slereah what do you think about knot theory if anything

sorry to ping you you neednt reply if u do not want

Slereah

All I know about knot theory is what little I read on Baez' site

Relativisticcucumber

thats where i just found out about it XD

Slereah

well you know as much as I

Relativisticcucumber

Spaghettification

In astrophysics, spaghettification (sometimes referred to as the noodle effect) is the vertical stretching and horizontal compression of objects into long thin shapes (rather like spaghetti) in a very strong, non-homogeneous gravitational field. It is caused by extreme tidal forces. In the most extreme cases, near a black hole, the stretching and compression are so powerful that no object can resist it. Within a small region, the horizontal compression balances the vertical stretching so that a small object being spaghettified experiences no net change in volume. Stephen Hawking described the flight...

i love it

Slereah

see also

Nuclear pasta

In astrophysics and nuclear physics, nuclear pasta is a theoretical type of degenerate matter that is postulated to exist within the crusts of neutron stars. If it exists, nuclear pasta would be the strongest material in the universe. Between the surface of a neutron star and the quark–gluon plasma at the core, at matter densities of 1014 g/cm3, nuclear attraction and Coulomb repulsion forces are of comparable magnitude. The competition between the forces leads to the formation of a variety of complex structures assembled from neutrons and protons. Astrophysicists call these types of structures...

Relativisticcucumber

16:21

omg

strongest material in the universe

yes

Slereah

16:32

Never eat the antispaghettis

Mad

17:12

why is it okay to pull the translation operator here from the ket alpha

and why does he choose the translation to be equal to the integration variable

(he is trying to represent the impuls operator in x basE)

ACuriousMind

the translation variable is not the integration variable, $\Delta x'$ and $x'$ are different objects

and I don't know what you mean by "pulling" the operator

Mad

on the left is P alpha

he inserts the completion relation

int x> <x,Palpha>

he pulled p out

he inserted the completion relation in the middle i see

ok nvm

Slereah

18:15

The weak equivalence principle as viewed by experimentalists

Mad

i am practicing some homework as exam preperation

one of the questions i met was " write the impulse eigenfunction in the x space"

$\vert p \rangle = \int dx \vert x \rangle \langle x\vert p \rangle = 1/(2 \pi) \hbar)^{0.5} \int dx \vert x \rangle exp [i p x / \hbar]$

Amit

Is there a way to check if a singularity of the metric is a coordinate singularity?

Mad

Does that look good?

Slereah

@Amit Depends how broad you understand the term

A fairly broad method is to compute the components of the Riemann tensor along some parallel transported frame of an observer

although a simpler method is to just change coordinates so that it's not singular, if you can find such a coordinate

Amit

@Slereah along meaning, covariant derivative?

Slereah

18:20

yes

Amit

I see, thanks

@Slereah oh btw, if it is a "true" singularity, the riemann tensor will be undefined too at those points?

Slereah

It will not converge, yes

Which usually means that it diverges

Amit

So it can't also have a coordinate type singularity?

Slereah

What do you mean

Amit

I mean is it a conclusive test

A sufficient condition

That it diverges

Slereah

18:25

As I said, depends what you mean by a singularity

There are singularities for which this does not work

It is for a type of singularities called scalar singularities

some singularities don't even have any badly behaved tensors

Amit

I only divide it to those that go away via a coordinate transformation vs those that don't

Slereah

I'm only talking about coordinate invariant singularities here

The usual tricky example is the cone

The cone minus its apex is a manifold, and the apex is a singularity

But it is also everywhere flat

Amit

Yes, I was asking if similar to the metric, the riemann tensor can also exhibit a coordinate dependent singularity

Slereah

Sure, you can just make up some fake singularities

One easy trick you can use for example is to bring the infinity within finite distance

Like if you take some coordinate $r' = \arctan(r)$, infinity will be at $r' = \pm \pi/2$

Amit

Ahhh, okay, nice

Slereah

18:30

So if your Riemann tensor isn't a bounded function at infinity, it will appear divergent there

Amit

To hold infinity at the palm of your tensor!

Slereah

You can also make Minkowski space divergent easily enough

Pick some coordinate like $t' = 1/t$

Amit

Coordinates are evil, if only Descrates knew the mischief he had done, lol

Slereah

In the 70's they just made giant lists of weird singular spacetimes

There are plenty of them

Amit

Prog rock and singular spacetimes

Interesting, thanks

Slereah

18:43

Worst one imo is the singularity that is only singular coming from specific directions

Slereah

19:56

Apparently the PPN formalism is specifically made with perfect fluids in mind

Why must they always omit such things

Slereah

20:18

"We assume throughout that the matter composing the solar system can be idealized as a perfect fluid. For the purpose of most solar system experiments in the coming decades, this is an adequate assumption (see, however, section 9.2)"

Oh no

God there's a truckload of conditions for the classical limit

Amit

21:14

The sky is the $\lim$

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