The h Bar

12:07 AM

@ACuriousMind No physically natural notion of an inner product of two operators as in no physically natural way to combine them to get a scalar or no physically natural sense of 'overlap' between them / similarity in them or their behaviour?

(where the latter might require some object of greater dimension than a scalar to represent. These are the two senses in which I've learned (as still an undergraduate) to think about inner products and I'm curious which you meant, if either, because I think I'd agree in the first case and maybe disagree in the second.)

Charlie

12:56 AM

Have found the answer to my question ^

Rumplestillskin

5:46 AM

I was wondering can somebody provide me with a sanity check regarding the derivation of the GR equations of motion for a test particle subject to an arbitrary external force $F^\alpha$. I know the resulting system of equations should be the geodesic equation* = $F^\alpha$ ( technically, to use the term geodesic equation here is incorrect I guess). I want to do this using a variational principle but I have drawn a blank. I was thinking of defining a Lagrangian $L* \equiv L + D$ where $L = \frac{1}{2} g_{\mu\nu} u^\mu u^\nu$ and D is an arbitrary interaction function given by $D \equiv D(x^\…

7:41 AM

8:20 AM

Is Simulating Soft and Bouncy Jelly Possible? 🦑

►

ACuriousMind

9:30 AM

@perilousGourd I meant the first. More formally, I meant that the space of operators (or even of bounded operators) is not a Hilbert space.

There is a subset of operators - the Hilbert-Schmidt operators, those with "finite trace" - for which you have an inner product by summing the products of all matrix elements, but I couldn't tell you a physical relevance of that off the top of my head

skullpatrol

@EmilioPisanty thanks for sharing

Slereah

9:50 AM

@ACuriousMind What is the space of local functions

ie for a bundle $E$, $\mathrm{Loc}(E)$

Some paper talks about the horizontal complex being $$\mathrm{Loc}(E) \otimes (dx^\mu)$$

skullpatrol

Welcome sir @JohnRennie

Slereah

Was @JohnRennie knighted

John Rennie

@skullpatrol hi :-)

@Slereah benighted

Slereah

You know I looked it up once

If you're knighted in the UK

You don't get a sword

and you're not even allowed to carry a sword

Some knights

What will they even do if the realm is under attack

John Rennie

Self isolate

Slereah

9:54 AM

That may work for the black death, but what about the sarracens

or dragons

or even worse

The French

I don't approve of the British monarchy, but if you have to have one, at least give swords

John Rennie

There is an argument for 48% of us inviting a continental power to invade. It's worked before.

skullpatrol

Nobody has had an argument to stop Covid from invading.

Slereah

"A local function is a smooth functions on $J^k$ for some finite $k$."

Ah there we go

ACuriousMind

@Slereah A "local function" is essentially a Lagrangian, a functional that depends on the position and the value of functions and their derivatives at that position, i.e. it's a function on germs/jets

Slereah

Hm

He defined local functionals as $$F = H^n(d_H, \Omega)$$

With $\Omega$ the local functions and $d_H$ the horizontal differential $dx^\mu \partial_\mu$

What is $H^n$?

Apparently this is the space such that $L \sim L + d_H k$

ie local functions up to a total derivative

But what's the definition of $H^n$ here?

ACuriousMind

10:15 AM

It's a cohomology

Slereah

ah yes, that makes sense

$d_H k$ is an exact form?

Yuvraj Singh...

Cohomology is equal to homotopy

ACuriousMind

@Slereah yep

@YuvrajSingh... No, it isn't. The two are rather different concepts.

Slereah

Although... Why are we only considering closed forms

Is $d_H (L [+ d_H k]) = 0$ for the Lagrangian?

It's the horizontal differential, not the EL one

ACuriousMind

hm

can you link the thing you're reading?

Slereah

10:23 AM

arxiv.org/pdf/1010.0899.pdf

Section 2.4

ACuriousMind

Ah, $n$ is the top dimension

Slereah

yes

ACuriousMind

So everything is closed because there are no $n+1$ forms

Slereah

Ah yes, true

That's where remembering cochain diagrams would help

I guess it makes more sense that the BRST complex would be cohomology business if the Lagrangian is a cohomology

Yuvraj Singh...

@ACuriousMind I should have use '' dual to''

ACuriousMind

10:29 AM

That's also not true, cohomology is - as the name says - dual to homology

homology and homotopy are not the same

Yuvraj Singh...

10:42 AM

Maybe you can elaborate a bit?

@ACuriousMind

Luyw

11:25 AM

hello! as a motivation to give for studying the quantity $\vec{OM}(t)\times m\vec{v}(t)$ (angular momentum), would you think that mentioning Kepler's discovery that the area swept by the radius of a planet orbiting the sun sweeps equal areas in equal times, and if we analyse "infinitesimally" and see that for a time $\Delta t\to0$, the area is given by $\frac{1}{2m}||\vec{OM}(t)\times m\vec{v}||$ is a good one?

even though kepler's law is a consequence of the conservation of angular momentum in a central force field

Slereah

Sure

user434058

@Luyw Yeah, why not.

Luyw

the "even though" I mentioned is bugging me, as this all seems circular. :s