The h Bar

0celo7

19:00

@Semiclassical are your parents wondering when you’ll have kids yet?

Or is that only a thing for women

Semiclassical

not really

i think they're not holding out much optimism there :P

maybe more with my sister

0celo7

My half sister gets asked constantly by her mother and she despises her for it

Semiclassical

she got married last October

however, both her and her husband are actors

0celo7

Is Gromov married?

Sid

@Semiclassical ...It's only been an year.

Balarka Sen

19:01

@0celo7 Twice

He's on a roll

Semiclassical

eh, that wouldn't stop some parents @sid

0celo7

My advisor is on #3

Some people just have a talent

Semiclassical

my point is more that their economic/living situation is rather job-to-job and takes them across the country

0celo7

Understood @Semiclassical

They just need a super nanny

Sid

@Semiclassical Pakistan just fuels the fire by sending extremists to our side of the border. The fire that exists is hugely due to administrative fallacies by our former and current Governments.

Semiclassical

19:02

heh, maybe if tey struck acting gold

at this point my sister has to supplement her income by doing babysitting jobs

not when she's doing a show

Balarka Sen

@0celo7 We should just stop subscribing to the idea of marriage and be like we are in the 70's

Semiclassical

but, for instance, she was an understandy for a NYC show that they hoped would run a few months

but it's closing already within the first month

so there goes that

Anonymous

Acting is indeed one of the riskiest careers out there

0celo7

@BalarkaSen nuclear family or Gulag.

Semiclassical

(which is also important in that, had the job lasted longer, she'd have been able to get health insurance through the actors's guild (or whatever it's called) for about a year. as it is, it's more like six months)

Balarka Sen

19:05

That's also a good policy

Semiclassical

That's also a ~~good~~ policy.

Balarka Sen

That's also a ~~good policy~~

Semiclassical

~~That's also a good policy~~

Balarka Sen

.

GULAG

Semiclassical

damn

0celo7

19:06

Lmao

Semiclassical

well, grammar gulag > family gulag

probably

0celo7

The gulag was a pretty cool thing

Semiclassical

Siberia in general is pretty cool, yes.

0celo7

exactly :D

parvin

hello

there is a question I have

I have read that Energy is the effect of a force that has been transferred into an object

so when the object is still (when energy is rest energy) what would energy mean?

is there any force being transferred to it?

or does it have any other definition?

Balarka Sen

19:12

Gulag life was really rather painful, even though I use it as a meme. Read "One day in the life of Ivan Denisovich" by Solzhenistyn

Semiclassical

Well

Anonymous

@parvin Search "internal energy"

Balarka Sen

It's no concentration camp but straddles the line between human and inhuman

Semiclassical

it was painful as long as you were there

Emilio Pisanty

oh, btw folks, in case any hbar regulars are like me and really wanted a lego Saturn V but waited too long before ordering and didn't get one before they ran out

Balarka Sen

19:13

@Semiclassical Well, if you survive as long, that is

parvin

ok i'll search

Emilio Pisanty

the LEGO Saturn V is back in stock

Semiclassical

@BalarkaSen yup

Balarka Sen

"Maria Tchebotareva

Trying to feed her four hungry children during the massive 1932-1933 famine, the peasant mother allegedly stole three pounds of rye from her former field—confiscated by the state as part of collectivization. Soviet authorities sentenced her to ten years in the Gulag. When her sentence expired in 1943, it was arbitrarily extended until the end of the war in 1945. After her release, she was required to live in exile near her Gulag camp north of the Arctic Circle, and she was not able to return home until 1956, after the death of Stalin. Maria Tchebotareva never found her …

parvin

aha so the rest energy is caused by the energy of movement of body molecules

is that a force?

movement is not a force tho

where does the source of force come from?

Anonymous

19:15

Intermolecular forces...

parvin

what causes them?

Anonymous

I don't know any easy answer to that

Anonymous

Maybe some QFT person can explain

parvin

is kinetic energy caused by intermolecular forces too?

@Blue it's ok that's not a serious question

Anonymous

@parvin Kinetic energy of what?

parvin

19:18

of a moving object

Phase

@BalarkaSen jesus

I'm guessing Stalin wasn't a fan of moral relativism

Balarka Sen

Not. at. all.

Anonymous

@parvin No. There you have external force

parvin

kinetic energy is some how vague to me, I can't understand where the forces causing it have come from

I'll search more

Balarka Sen

and the story above is just nothing compared to the many many other fables out there

after all, the total death count in Gulag is about 30 million

lotta stories to tell

Semiclassical

19:21

and lots of stories that can't be told

parvin

@blue kinetic energy has some initial forces causing the motion, right?

Phase

was russia ever like, a leader of the developing world like the middle east and west europe have been before?

Anonymous

@parvin Pick up something and throw it. The KE that object develops will be caused due to the force you exerted on it

Anonymous

It's external force in that case

parvin

newton's law?

Balarka Sen

19:23

Depends on what you mean by development

parvin

no sorry

Semiclassical

looots of history here

Balarka Sen

It's certainly a leading figure in art and culture, mostly pre-USSR. It was also a pioneer in space exploration during cold war.

Semiclassical

but I think it is fair to say that Russia was not always a superpower, even compared to the relatively diminished capacity nowadays

0celo7

@BalarkaSen are you unironically a Soviet?

Balarka Sen

19:24

lmao

0celo7

Just curious. I can’t tell.

Phase

the important question

is how fucking old is @BalarkaSen

Anonymous

In his 60's

0celo7

He’s 16

Anonymous

Apparent age is 17

Sid

19:27

I think you mean, 70 and not 17. :P

Anonymous

I don't bet on that. He might actually be a Soviet mathematician pretending to be a school kid on the internet.

Balarka Sen

shit people knows I'm a KGB agent

0celo7

I think he’s Gromov.

Balarka Sen

Vat is Maanifolths?

Phase

if balarka is really 17, I'd be curious to know what got him into reading ahead

what was the first further ahead thing you read?

Balarka Sen

19:31

Communist Manifesto

Phase

fucks

sake

Balarka Sen

I think my mathematical interests were a bit of a clustercrap back in the days

Sir Cumference

19:49

@Mithrandir24601 I mean, the fact that SR is so unintuitive is why I chose physics in the first place. I always wanted to learn and understand the wacky, bizarre nature of our Universe. It was so interesting because you’d never expect it. But having to keep up with this speed is hell.

It’s kind of like being given a chocolate cake, but being forced to eat whole in under a minute. I should be loving this class, but instead I despise it.

Mithrandir24601

@SirCumference Oh, just you wait... ;)

Balarka Sen

gotta go faster

Mithrandir24601

For me, it just kept ramping up and up and up to frankly insane levels until MSc year, when everything calmed down to a nice and relatively sedate pace :)

Sir Cumference

@Mithrandir24601 MSc?!

Sigh...

Anonymous

@Mithrandir24601 I'm desperately waiting for those days

Anonymous

19:56

:P

Phase

Could I ask a SR question given the chat theme?

Sir Cumference

I have to go to class. Later

Anonymous

@SirCumference What's wrong with MSc?

Anonymous

@SirCumference See you

Phase

I've heard that proper acceleration is perpendicular to velocity, even for linear motion. Is that just for the particle frame, as position is just (t, 0, 0, 0)?

bolbteppa

20:00

@Phase four-velocity satisfies $u_{\mu} u^{\mu} = 1$ so it's derivative with respect to arc length $s$ is $2 w_{\mu} u^{\mu} = 2\frac{du_{\mu}}{ds} u^{\mu} = 0$ which shows acceleration $w_{\mu}$ is orthogonal to velocity $u^{\mu}$, regardless of the frame

(arc length = proper time)

Phase

Im still new to super and sub scripts. Like atm I'm trying to understand $\omega^\alpha = g^{\alpha \beta}\omega _\beta$

I can see that it looks like it should work but I don't know why

bolbteppa

@Phase why read Gourgoulhon when it's full of extra unnecessary notation complicating things

Phase

idk I've committed to it now

i'd feel like a coward if I ducked out now

bolbteppa

Landau does SR in 30 pages, that book does it in about a million

Phase

Does landau still talk about the metric etc?

Because I like the idea of doing SR in a tensorial way

0celo7

20:02

Read Gorgoulhon

bolbteppa

Yes, as if you only knew calculus and no linear algebra too

Physics Meta

0

Q: Typo on the tag description of $\text{aether}$

The name of Huygens in the aether tag description is spelled Huygen. BTW, where may one places this fix or improvment suggestion?

0celo7

@Phase if you learn tensor algebra now you don’t have to learn it later

Phase

Does the inverse of a metric map an element to its dual?

0celo7

Other way around.

Phase

20:08

Huh. What maps an element to it's dual then... The.. Metric...?

Balarka Sen

The metric/inner product is a map $V \times V \to \Bbb R$

Phase

Oh right

Balarka Sen

The map $V \to V^*$ is defined as $u \mapsto \langle -, u \rangle$

Phase

What is <-,u>

0celo7

@BalarkaSen pls teach him the musical shit

Balarka Sen

20:08

Where $\langle -, u\rangle$ is the linear map $V \to \Bbb R$ defined by $v \mapsto \langle v, u \rangle$

Slereah

Partial function application

0celo7

I’m trying to pay attention to algebra lecture

Balarka Sen

OK

Phase

oh i see so - is just a placeholder?

Balarka Sen

Exactly

Phase

20:10

So what's the deal with inverse non-injective functions

How do you define g^ab to be useful

:(

bolbteppa

@Phase the metric $g$ is a a certain function of two vectors $u, v$ mapping them to a number, $g(u,v)$, it linear in each argument so if you keep one vector fixed, say $u$, then it's a linear function of $v$, $f_u(v) = g(u,v)$, but the dual space is the space of linear functions on a vector space mapping the vectors to numbers, and so we now have this map from $u$ to $f_u = g(u, _ \ )$

0celo7

It’s a bijection

Phase

So $\omega^\alpha$ is a linear form?

also @0celo7 I thought that the metric applied to two vectors gives the inner product, is that really bijective?

im really confused by superscripts and subscripts. What's the difference between $g^{ab}$, $g_{ab}$ and $g^a_b$?

0celo7

@Slereah help him please

I can’t do this on my phone

bolbteppa

@Phase The point of even defining this stuff in SR all comes from looking at the line element, you have $ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 = 0$ for a light ray, and so $ds^2 \neq 0$ for non-light rays, so we can write it as $ds^2 = dx_0 dx^0 + dx_1 dx^1 + dx_2 dx^2 + dx_3 dx^3 = dx^0 dx^0 - dx^1 dx^1 - dx^2 dx^2 - dx^3 dx^3$ using the matrix $g_{\mu \nu} = (1,-1,-1,-1)$ (1,-1 on diagonals, 0 otherwise), if for some reason you want upper subscripts to be what you're working with, we set

Phase

20:16

small request: i know its not a big change but Im just used to -+++, :C

bolbteppa

lower subscripts to involve these minus signs, and so can define the convention that we sum along repeated subscripts without writing the $\sum$ notation

You can do all of this without even knowing linear algebra, and can use that way of looking at the subject as a means to motivate linear algebra

Phase

so $v_\alpha u^\alpha$ is equal to $g(v,u)$?

since its the sum over the repeated index?

But then what does changing the metric indices do :(

bolbteppa

If $u = u^{\alpha} \mathbf{e}_{\alpha}$ and $v = v^{\beta} \mathbf{e}_{\beta}$ then $g(u,v) = g(u^{\alpha} \mathbf{e}_{\alpha},v^{\beta} \mathbf{e}_{\beta}) = u^{\alpha}v^{\beta}g(\mathbf{e}_{\alpha},\mathbf{e}_{\beta}) = u^{\alpha}v^{\beta} g_{\alpha \beta} = u^{\alpha}v_{\alpha} = u_{\alpha}v^{\alpha}$

Phase

Oh I see.

Slereah

@Phase $g_{ab}$ is the metric tensor, giving you the norm of two vectors

$g^{ab}$ is the inverse of the metric tensor, which you can apply to two dual vectors

with the usual identity for inverse matrices, $g_{ab} g^{bc} = I^c_a$

$g^a_b$ will be the identity matrix because of this

bolbteppa

20:22

If $u = u^{\alpha} \mathbf{e}_{\alpha}$ and $v = v_{\beta} \mathbf{e}^{\beta}$ then $g(u,v) = g(u^{\alpha} \mathbf{e}_{\alpha},v_{\beta} \mathbf{e}^{\beta}) = u^{\alpha}v_{\beta}g(\mathbf{e}_{\alpha},\mathbf{e}^{\beta}) = u^{\alpha}v_{\beta} g_{\alpha} \ ^{\beta} = u_{\alpha}v^{\beta} g^{\alpha} \ _{\beta}$

Phase

Hang now I'm getting myself even more confused

Slereah

Can I write the generic solution to the wave equation as $$f(x) = \int_{k \in \mathscr C} f_k(g(k,x)) dk$$

Phase

The inner product is equivalent to putting an element of E into an element of E* right?

Slereah

With $k$ a null vector

@Phase if you have an inner product you can define such a map, yes

and vice versa

bolbteppa

@Phase en.wikipedia.org/wiki/Riesz_representation_theorem

Phase

20:25

tensors are really hard to dip my toes into

bolbteppa

@Phase this might help amazon.com/Explorations-Mathematical-Physics-Concepts-Language/… gives an overview of the language and explains it with nice examples etc

Slereah

Wait, does it make sense to write $g(k,x)$

If it's not Minkowski space

Phase

So the linear form $u(v) = g(u_\alpha e^\alpha, v^\beta e_\beta)$?

jesus christ what the hell happened there

$= u_\alpha v^\beta g^\alpha _\beta$?

bolbteppa

If you want to treat $u(v)$ as a linear functional then you expand it as a linear combination of functions $e^{\alpha}$ which act on basis vectors $\mathbf{e}_{\beta}$ to give $g^{\alpha} \ _{\beta}$
$u(v) = u_{\alpha} e^{\alpha}(v) = u_{\alpha} e^{\alpha}(v^{\beta} e_{\beta}) = u_{\alpha} v^{\beta}e^{\alpha}( e_{\beta}) = u_{\alpha} v^{\beta} g^{\alpha} \ _{\beta}$

Phase

OH

thats why it wasn't given a vector line in the book

I didnt even register that

bolbteppa

20:31

So $u$ is a vector in the dual space, but it is also a function acting on vectors in the original space too, so you write it first as a linear combination in the dual in terms of the dual basis, then let each dual basis vector function act on the vector in the original space, but THEN we define each dual basis function by how it acts on the original vector space basis to give $g^{\alpha} \ _{\beta}$

Phase

for minkowski space, ignoring signature, the metric is effectively just a kronecker delta?

Balarka Sen

@0celo7 lmao there's a new pewdiepie video

bolbteppa

The whole point of dual spaces is to let you change the components for a vector given an arbitrary basis, it's obvious to expand a given basis in terms of a new basis, but how do you represent the coordinates changing from one basis to another? This is a big issue, because old tensor analysis only works with the components of a vector completely ignoring bases

Infinite dimensions things get complicated, but so does the idea of components etc

Yes

Balarka Sen

@0celo7 this might potentially be racist to indians but i'm in tears

0celo7

Holy heck I’m in a talk where the person is reading from a piece of paper

Like literally everything

Phase

20:37

is $g^{ab}g_{ab} = 1$?

well

0celo7

@BalarkaSen like a new controversy?

Phase

$I$ i guess

0celo7

@Phase no, densjon

dimension of the space

dmckee

@SirCumference With practice you'll get better at drinking from the firehose.

Balarka Sen

@0celo7 no like literally watch it after you're out of the class

this is awesome

bolbteppa

20:38

$g^{ab} g_{bc} = \delta^a_c$, $\delta^a_a = n$.

Balarka Sen

(it's not actually offensive)

Phase

Im so lost. If it has one superscript and one subscript do you sum?

If they're the same

does it still count as repeated indices?

0celo7

What other reapeated indices would you sum 0.o

Balarka Sen

I might genuinely die from laughing too much

Phase

No i mean

I get that for something like v_a u_a you sum the indices but I wasn't sure if it's the same object with two indices the same if you sum them

Slereah

20:42

@0celo7 what does the phase term in wave solutions look like for non-flat spacetimes anyway

It can't be $g(k,x)$

$x$ ain't no vector

bolbteppa

@Phase You know the Pythagorean theorem, $r^2 = x^2 + y^2 + x^2$ right? Lets write $r = (x,y,z)$ as $r = (x,y,z) = (x^1,x^2,x^3)$ and write $r^2$ as $r^2 = x^2 + y^2 + z^2 = (x^1)^2 + (x^2)^2 + (x^3)^2 = \sum_{i=1}^3 (x^i)^2$. Now lets try to simplify notation by setting $x^i = x_i$ and saying that $\sum_{i=1}^3 (x^i)^2 = \sum_{i=1}^3 x_i x^i = x_i x^i$ where a repeated index up and down means to sum over them, while non-repeated indices means not to sum. Thus $x_1 x^2 = xy$.

Slereah

Do you use the separation vector or something

Emilio Pisanty

@Curio Yes

Phase

@bolbteppa but I thought x_i is the dual of x^i

Emilio Pisanty

ooooooooooooofff

Phase

20:44

or is that only for the bases

Emilio Pisanty

boy was this answer wrong

13

A: Can one force the electric quadrupole moments of a neutral charge distribution to vanish using a suitable translation?

Sometimes you can The obvious example is a purely dipolar charge which has been displaced from the origin, such as a dipolar gaussian $$ \rho(\mathbf r) =p_z(z-z_0)\frac{e^{-(\mathbf{r}-\mathbf{r}_0)^2/2\sigma^2}}{\sigma^5(2\pi)^{3/2}} . $$ This system is neutral, and it has a nonzero dipole m...

bolbteppa

@Phase ignore dual spaces for a moment, does all that make sense

Emilio Pisanty

anyone up for the octupole case?

@Slereah, maybe?

Phase

I guess but I don't understand the seemingly arbitrary decision to represent quantities that are effectively the same objects differently [one superscript and one subscript]

Slereah

Because they are not

A good example is test functions

Emilio Pisanty

20:46

@Phase they're almost the same objects, but they're not

Slereah

Test functions form a vector space

Balarka Sen

@0celo7 I am become dead

Slereah

But their dual vectors aren't test functions

Phase

idek what a test function is..

Slereah

They're distributions

Phase

20:46

:c

bolbteppa

@Phase This is Einstein summation notation, Einstein originally thought it up as a trick to simplify notation, just a rule to avoid clunky notation

Phase

so v_i and v^i ARE dual elements?

Slereah

He was too lazy to write the sum

Phase

well, one is dual of the other

Slereah

what a hack

Emilio Pisanty

20:47

@Slereah test functions might be a bit too complicated at this stage

@Phase yes

Phase

So it's like the hermitian conjugate?

raising or lowering an index i mean

Emilio Pisanty

@Phase not quite

0celo7

@Slereah ugh, yikes. See Straumann

Emilio Pisanty

at least, not like hermitian conjugate of full-fledged matrices

but it is like passing from a ket to a bra

Slereah

which part

It's a big book

Phase

20:48

At least I understand something on the page

Even if I dont know what $\Phi_g$ is

Emilio Pisanty

@Phase do you understand what $\underline u$ is?

Phase

Yeah it's the linear form right?

Emilio Pisanty

and how it depends on $\vec u$

?

Phase

Yeah

Emilio Pisanty

@Phase then you know what $\Phi_g$ is

;-)

Phase

20:50

Well

Maybe I dont then

Emilio Pisanty

it's the map that takes $\vec u$ and gives you $\underline u$

Phase

I understand the underlined u to be an element of E* that is a linear combination of basis vector functions of E*, such that underlined u(u) = inner product of u with u

but I dont get how to get $\Phi_g$ from that... Could you use something like <E* basis vectors|E> like in LA?

0celo7

@Slereah the part on optics in chapter 2

Emilio Pisanty

@Phase that's not quite right

Phase

but then you'd be taking the inner product of a function with a vector

Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

pain

Emilio Pisanty

20:53

$\underline u$ isn't determined by its action only on $u$

you need its action on all of $E$

Phase

Yeah

its action on all E would be the inner product of v in E with u in E

0celo7

You seem to get really triggered by the duality pairing notation

Phase

who me?

Emilio Pisanty

@Phase exactly

0celo7

Like, why? Same notation, different thing

@Phase yeah you

Slereah

20:54

Hm

Emilio Pisanty

i.e. $\underline u (v) = ⟨u,v⟩$ for all $v\in E$

Slereah

It seems to be some function $\psi$

Emilio Pisanty

is the difference clear?

Slereah

such that $\psi_{,\mu} = k_\mu$

Phase

Yeah I knew what underlined u was to a rough extent, the same as now. But I don't get what Phi is

D:

0celo7

20:55

@EmilioPisanty the book writes the LHS with angle brackets

Slereah

which makes sense I suppose

Emilio Pisanty

@0celo7 I'm just following @Phase's last notation

Semiclassical

@EmilioPisanty channelling every student from a grad EM class: NnnooooOoOoOo

Phase

Honestly I'd risk being dangerous in saying that I understand 'everything' in the book at least slightly, until I get when it writes "$\omega^\alpha = g^{\alpha\beta}\omega_\beta$"

Whoops

Emilio Pisanty

@Semiclassical ah, how bad can it be?

the quadrupole case only took me four years

(!)

scratch that

four and a half

bolbteppa

20:57

@Phase a vector $v$ in the vector space $V = \mathbb{R}^n$ with standard unit basis $\{ e_i \}$ (e.g. $e_1 = (1,0,\dots), e_3 = (0,0,1,0,\dots), \dots$) has components $v^i$, i.e. $v = v^i e_i = (v^1,v^2,\dots)$. In finite dimensions, a vector space $V$ is isomorphic to it's dual space $V^*$, so we can say that $v$ is an element of the dual $V^*$ too. We then say the dual space $V^*$ has the dual basis $\{ e^j \}$ where the dual basis $e^j$ is defined to be such that $e^j(e_i) = \delta^j_i$.

Thus the vector $v$ in the dual space can be written as $v = v_j e^j$. Notice the placing of indices tells us whether we are working in the vector space or it's dual. Thus a vector $w$ in the dual $V^*$ can be treated as a function of another vector $v$ in $V$ by forming $w(v)$, and then expanding the function $w$ in the basis $e^j$ as $w(v) = \sum_{j=1}^n w_j e^j(v) = w_j e^j(v) = w_j e^j (v^i e_i) = w_j v^i e^j(e_i) = w_j v^i \delta^j_i = w_i v^i$.

Semiclassical

Well why not do it for any 2^n-pole at that rate

It can’t be that bad /s

Phase

I get that @bolbteppa

but its like seeing the metric just fucks up everything in my head

Emilio Pisanty

@Semiclassical that's the end goal

but given how hard the finite cases have felt, I reckon another finite step is in order before making the full jump

Semiclassical

TBH I do wonder with some of those calculations hoe much can be done in the generic casr

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