The h Bar

Emilio Pisanty

00:02

@tpg2114 yeah, I can see it now, but at the time I flagged it I got no inkling that that was the case

bolbteppa

01:07

@cows lol

@vzn stiffness of spacetime etc is the more general topic this stuff falls under, again it's really a misunderstanding to pretend one can go the other direction (i.e. GR from continuum mechanics and not the other way around), what are you using - either Newton's laws or the SR axioms, this simple question is something you should think about, i.e. the starting point for setting up a theory

vzn

05:04

@bolbteppa lol oh so now youre an expert on this bridge? its a fair question and the answer is again, theyre equivalent in a key way and therefore the mapping is bidirectional. my prediction, eventually it will be realized that its actually GR that is limited in the "limit cases" of extreme regimes ie very small (QM) and very large (big bang/ universe expansion), black holes, etc... just a matter of time™ :) the refs you found are already showing serious people are taking it seriously!

Captain Bohemian

07:44

@cows you are that cows who used to come here with a different avatar picture some time ago and said you are on a beach but need to go home for coding?

Slereah

08:55

Hello

skullpatrol

10:03

ello

1 hour earlier...

danielunderwood

11:37

llo

Secret

o

[]

Shing

I am working on a problem set about finding the ground state energy for $-1/\gamma = tan(ka/2)/ka$

I solved most of the parts, except one

"Calculate the ground state energy 𝐸1 , including corrections of order 1/𝛾 , but ignoring higher order ones."

that's so confusing me.

I don't understand what it means? especially "corrections of order 1/𝛾"

I tried $E \approx 2\hbar^2\pi^2/(m*a^2)(1+2/\gamma)$

by saying tan x ~ x - pi, since the gamma is very large

but that's what the correct answer, would anyone be kind enough to give me a helping hand there?

(it is a infinite potential well, with a Dirac delta well in between)

Shing

12:27

never mind, I think I have figure it out

Akash. B

14:22

@JohnRennie can you please help me to understand the term black body radiation?

Fermi paradox

14:34

Is there any python library for optics? I need some elementary formulas (Fresnel equations, etc.).

John Rennie

16:16

@DanielSank the Rindler metric doesn't describe a curved spacetime. If you compute the Riemann tensor from the Rindler metric you'll find it is zero. The Rindler metric is an example of a curved coordinate system on a flat spacetime.

Adirian

16:39

@JohnRennie The spacetime isn't flat, though, except maybe from the accelerating object's perspective, and that only if it is a pointmass. It's own gravitational field is Lorentz Contracted; Rindler Coordinates describe the situation where the spacetime forward of the accelerating object isn't uniformly contracted owing to the fact that an object's gravity also has to accelerate and lightspeed limitations mean that acceleration, and the resultant contraction, is a gradient.

If I got that right, how can the spacetime be flat?

John Rennie

@Adirian huh?

Adirian

If we consider a relativistic object's gravity, it must be Lorentz Contracted - is this part correct?

John Rennie

I don't know what that means ...

Adirian

The bowl-shaped well of gravity gets contracted along with everything else, such that gravity seems both stronger and to weaken with distance faster, from an outside perspective

John Rennie

OK, but what does that have to do with Rindler coordinates?

Adirian

16:47

You can arrive at Rindler Coordinates by considering that the contraction isn't uniform - lightspeed updates to the velocity of the entire bowl of gravity mean that the update isn't uniform; the parts of gravity closest to an accelerating object are more contracted than those further out, because they are moving faster.

John Rennie

No you can't. The Rindler coordinates are the coordinates of a uniformly accelerating observer in flat spacetime.

You derive them by requiring that the four-acceleration of the observer stationary at the origin be something like $(0, a, 0, 0)$.

Adirian

Additionally, the contraction occurs sooner forward of the acceleration, because you're increasingly further away from the part of the gravity well behind you that hasn't been updated yet and is still going slightly slower

John Rennie

There is no gravity well involved in the derivation of the Rindler coordinates.

Adirian

Is the effect I am describing accurate with respect to the way the physics should behave?

John Rennie

I wonder if you're mixing them up with some other coordinate system ...

@Adirian you're just describing the regular Schwarzschild metric in boosted coordinates.

I have seen that written down, but it wasn't very exciting and I have long ago forgotten what it looked like. A Google should be able to find it though.

Adirian

16:57

It should look exactly like the intersection of two gravitational fields, which is to say, it should look like gravity. AFAIK Schwarzchild metrics lack this?

John Rennie

Can we take a step back? How much GR do you know? Do you know how gravity, as experienced by us humans on Earth, is related to the curvature?

Adirian

Yes. I think in terms of spacetime density instead of curvature, which is related by the difference in surface area between curved space and flat space, but I have the concept down. Things orbit because the side closer to the earth have farther to travel than the far side does.

John Rennie

That is not a description of general relativity that would be widely accepted by physicists ...

Adirian

Gravity as we normally think of it is caused by the same effect, having further to travel, with the motion being through time as opposed to space.

John Rennie

I think you should learn the conventional treatments of GR before attempting to arrive at alternative explanations for it.

Adirian

17:05

Is it incorrect to say that the surface area of a curved space is greater than the surface area of flat space, implying greater internal volume of a curved-space sphere of equivalent Minkowski radius to a flat space sphere?

Sir Cumference

18:12

@RyanUnger you like diff geo right

Sir Cumference

18:26

so Spivak defines chains as such:

not sure what to make out of this addition. are we literally adding the parametric functions? or are we concatenating them or something?

Ryan Unger

19:32

@SirCumference like it? it's what I do...

@SirCumference he means exactly what he says...given a set $S$ you can create something called the free abelian group $F$ on $S$

do you know what an Abelian group is?

Sir Cumference

Yes, but how is addition defined here?

Ryan Unger

just take any set S

forget what's written above

Sir Cumference

This diagram would have me believe it's different from the usual addition defined for functions

Ryan Unger

ok so that's when you try to geometrically understand what's going on

but first you have to look at it formally

formally you have a set $S$ and it generates a free Abelian group $F$

Free abelian group

In abstract algebra, a free abelian group or free Z-module is an abelian group with a basis. Being an abelian group means that it is a set with an addition operation that is associative, commutative, and invertible. A basis is a subset such that every element of the group can be found by adding or subtracting basis elements, and such that every element's expression as a linear combination of basis elements is unique. For instance, the integers under addition form a free abelian group with basis {1}. Addition of integers is commutative, associative, and has subtraction as its inverse operation,...

Sir Cumference

Yeah, $S$ being the set of singular $n$ cubes?

Ryan Unger

19:36

what does he define as a singular cube

maps of the standard cube?

Sir Cumference

In the preceding paragraph

Looks like a plain old parametric function

Ryan Unger

it isn't a parametrization because it's not bijective onto its image

mapping all of $I^n$ to a single point is valid

Sir Cumference

A parameterization needs to be a bijection?

Ryan Unger

depending on your definition, yes

so you have the maps $I^n\to A$

and these form a set

Sir Cumference

@RyanUnger All right, so this is basically a generalization of the term

Ryan Unger

19:39

you now consider the free Abelian group generated by this set

which is literally just things of the form $4a_1+3a_2-a_3$ and with some natural associativity, distributative law, etc.

now the arrow is what we call an orientation

Sir Cumference

Yes, but how precisely is it defined? From the diagram, the sum appears to parameterize an oriented square

Ryan Unger

how precisely is what defined

Sir Cumference

the addition

Ryan Unger

abstractly

Sir Cumference

It's not just the usual addition of functions presumably

Ryan Unger

19:41

you can picture it in certain instances though

you can draw each $a_i$ in the sum right

now the picture of the sum is just: overlay all of the pictures and take into account the multiplicity in front

the sum is not defined in terms of the picture

Sir Cumference

So addition of n cubes is defined as the union of the images?

Ryan Unger

it's completely abstract

"the sum is not defined in terms of the picture"

Sir Cumference

@RyanUnger So basically Spivak is just saying "if you want, you can make the $n$ cubes into a vector space" without giving a useful example?

Ryan Unger

not a vector space, a free Abelian group (free Z-module)

Sir Cumference

We have scalar multiplication

Ryan Unger

19:44

basically yes

do you know what modules are?

Sir Cumference

oh crap now I'm remembering algebra

Ryan Unger

so the picture is saying that the boundary of the 2-cube is the sum of four 1-cubes

but four specific 1-cubes

not any old ones

and you need to make sure that the orientation is the right way

he'll talk about that in a bit

Sir Cumference

@RyanUnger Right, forgot that only integer multiplication was allowed here

@RyanUnger He only defines orientations for manifolds

Which is a whole other quagmire

@RyanUnger So is the orientation related to the order in which we sum the cubes?

Ryan Unger

the orientation here is just the sign in the sum

@SirCumference no, remember this is an Abelian group

Sir Cumference

@RyanUnger Right, I was about to ask

Ryan Unger

19:48

does he define the faces of a cube somewhere

let's see if I have a pdf

Sir Cumference

@RyanUnger Yep, he does

But let me just make sure about something. Is there a conventional definition for addition of $n$ cubes? Or is it irrelevant in this chapter?

Ryan Unger

he's saying everything I'm saying on page 97

@SirCumference the conventional defintion is completely abstractly as I said

you don't need pictures for any of this

Sir Cumference

If I have no clue how the addition is defined, how would I ever know when adding n cubes is relevant to a situation

Ryan Unger

I told you how it's defined

Sir Cumference

What is "completely abstractly"?

Ryan Unger

19:53

The free Abelian group generated by the cubes

There's only one up to isomorphism

Sir Cumference

@RyanUnger Oye, right, now I got it

Ryan Unger

Now in certain circumstances you can picture the sums

But you need to be very careful because most maps I^n -> A are unimaginable

Sir Cumference

@RyanUnger Unimaginable insofar as we can't picture dimensions higher than 3?

Ryan Unger

no like you can have a map $I^1\to I^2$ whose image is dense

you can have something low dimensional filling up something higher dimensional

who's to say geometrically what the boundary of such a cube is

the point is that this lets you do it algebraically

it's the formal (alternating) sum of the faces

replace "sum" with "formal sum" if you want for now. It's important to note that there's no summing of the images, or unions, or whatever

You're literally just defining chains to be elements of the FAG (free Abel gp, don't ban pls) generated by the cubes

Sir Cumference

So you're saying I shouldn't rely on a geometric intuition for what chains would look like, compared to their constituent cubes?

Ryan Unger

20:00

@SirCumference yeah except to convince yourself that it gives the right answer in the "standard" case

Like Figure 4-4

Sir Cumference

Ok, got it. Thanks a ton

Ryan Unger

@SirCumference work out $\partial I^3$ explicitly

so compute the faces and then try to understand the orientations

bolbteppa

20:12

and due to e.g. Stokes' theorem you want to then define the $\partial$ operator on the free abelian group sending an element to it's boundary and onward

@SirCumference right it's basically just abstract what you do when you integrate a function over a piecewise curve, $\int_{c_1 \cup c_2 \cup ...} ds = \int_{c_1} ds + \int_{c_2} ds + ...$ the right-hand side is why you might want to re-interpret those curves as an abelian group with integer coefficients... Generalizes to integrating over higher dimensional piecewise volumes,

enumaris

20:41

hmmm

super weird...running into an internalError at epoch 3...

I wonder if it's a memory error...

enumaris

21:09

looks likely to be a memory error...

memory leak....uh oh...

PM 2Ring

22:29

Q: What's purple & commutes? A: An abelian grape. — PM 2Ring Aug 22 '17 at 21:43

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