The h Bar

@0celo7 I don't see a reason why momentum conservation can't be used. CM momentum stays zero/has zero time derivative and is the sum of external forces. So yes! $Mg$. Even though you do actually have to do work by dragging the rope more and more quickly.

Kyle Kanos

You guys and not letting me get my purely-fictional victory

Stan Shunpike

3:07 AM

I'm proud to say that I started paying closer attention to how I tag and QMechanic hasn't had to edit my posts as much.

I noticed he always cleaned up after me, so I tried to do a better job myself.

ACuriousMind

BTW, I've got a good suspicion who he is, I think. I won't tell anyone, though :P

Kyle Kanos

Like IRL?

ACuriousMind

Yep

Kyle Kanos

There are a few people who it's pretty obvious who they are

Like Chris White.

user54412

>.>

Kyle Kanos

3:10 AM

He's a Lacrosse player

user54412

LOL

Stan Shunpike

MLL

ACuriousMind

xDD

Stan Shunpike

(Major League Lacrosse)

Lacrosse is effin dangerous man

I went to a soccer camp once

And a teammate of mine there had a collapsed lung because someone had hit him between the blades with a lacrosse stick during a lacrosse game

0celo7

@ACuriousMind Is he a professional?

I bet you saw an article with $\approx$.

ACuriousMind

3:11 AM

As I once said, I mainly just dont want to torture you peple with my umlauts. My real name is Björn Jüliger, you don't want to try and type that every time :D

Stan Shunpike

we could just call you BJ

lmao

0celo7

Does ü work?

ACuriousMind

@0celo7 That was a hint, yes

@StanShunpike You're not the first to "invent" that :D

Stan Shunpike

But I like ACuriousMind better than BJ

I won't be the last. It's not very original.

0celo7

I would not have guessed Björn.

Kyle Kanos

3:13 AM

Is Bjoern okay for those w/o the umlauts?

Stan Shunpike

The Bjorn Identity

Kyle Kanos

Or would you prefer Urs?

Since they apparently mean the same?

ACuriousMind

@KyleKanos It...hurts, actually.

0celo7

He would love Urs.

Stan Shunpike

(Bourne Identity...anybody seen it? lol)

0celo7

3:14 AM

Correction--he loves Urs.

Kyle Kanos

Hmm. Not having any umlauts in my name, I cannot feel the pain

ACuriousMind

And yes, Björn is Scandinavian (Norwegian I think) for bear, which Urs is Latin for.

Kyle Kanos

Actually, @ACuriousMind is easiest because if I need to talk to you, it pings you.

0celo7

@ACuriousMind Hence Ursa Major/Minor

user54412

@KyleKanos at that point you should just go with Beorn

Kyle Kanos

3:15 AM

@ChrisWhite I was thinking that as I was typing that too

Stan Shunpike

@ChrisWhite So yesterday we were discussing the definition of a black hole. And someone mentioned some definition basically involving subtracting points from a spacetime manifold $M$. Do you know of any books that discuss blackholes in these terms but that also give them an information theory treatment?

user54412

@StanShunpike umm, well, the information theory aspect isn't quite settled, so...

0celo7

@StanShunpike You'd have to read journals and articles for that.

ACuriousMind

Call me whatever you like :) But it'll still only ping me if you use the alias.

0celo7

@ACuriousMind We gonna have a BJ symmetry one day?

Stan Shunpike

3:17 AM

@ChrisWhite oh, Susskind was very ocal about that....

vocal*

I thought it was certain. That's confusing then.

ACuriousMind

@0celo7 Since it's usually the surname, it'll be the Jüliger symmetry :P

0celo7

@ACuriousMind I realized.

@ACuriousMind You need to make SUSY even more symmetric.

ACuriousMind

But a BJ symmetry would be good :D

0celo7

HYSY. Hypersymmetry

user54412

ubersymmetry?

0celo7

3:19 AM

ÜBSY

ACuriousMind

I don't think there's much symmetry undiscovered by SUSY and BRST, at least in the usual sense.

0celo7

@ACuriousMind There are more dualities probably.

Überduality

Stan Shunpike

@ACuriousMind You explained the tensor product very nicely. You should me both how to define it and that it could be used to define $k$-tensors out of $1$-tensors. This was helpful because I had both a definition and an application. Can you now give me an application of the wedge product? Something I might use it for. I'm having trouble understanding what the definition you gave really signifies...

0celo7

@ACuriousMind Speaking of tensor products, what the hell is the direct product?

ACuriousMind

Product (category theory)

In category theory, the product of two (or more) objects in a category is a notion designed to capture the essence behind constructions in other areas of mathematics such as the cartesian product of sets, the direct product of groups, the direct product of rings and the product of topological spaces. Essentially, the product of a family of objects is the "most general" object which admits a morphism to each of the given objects. == Definition == Let be a category with some objects and . An object is a product of and , denoted , iff it satisfies this universal property: there exist morphisms...

0celo7

3:28 AM

Urs?

I thought you said it's the same as the direct sum?

ACuriousMind

@0celo7 Perhaps :D The direct product is an object with projections onto its components such that every map into it is completely determined by maps into its components

@StanShunpike The most "natural" setting is probably the idea of seeing the exterior algebra in degree $k$ as the space of (sums of) $k$-hyperplanes. The plane spanned by two vectors $v,w$ is represented by $v\wedge w$ - if $v$ and $w$ are collinear, they do not span a plane at all, and their wedge is zero.

0celo7

@ACuriousMind So when I write $A\oplus B$ for something, $A\operatorname{(insert direct product here)}B$ also works?

ACuriousMind

@0celo7 Yes. You call an operation the "direct sum" when it is the categorial product as well as the categorial coproduct.

0celo7

@ACuriousMind The Urs in you is too powerful. I bow to you.

I gots to sleep.

ACuriousMind

@0celo7 Then sleep well :) (Oh god, it is 4:30 am. My sleep cycle is totally out of sync)

user54412

4:17 AM

This is awful. I'm pretty sure now that there are two different sign conventions for the electromagnetic field tensor running around (even given the same metric sign convention).

Stan Shunpike

Yeah, I think I saw that too. Its very confusing.

user54412

Hmm, I think my convention is the same as Wikipedia's, but the opposite of MTW's.

user54412

But Carroll agrees with MTW

ACuriousMind

@ChrisWhite Is it perhaps resolved by the way the Lorentz force equation is written?

user54412

and Wald agrees with the other books

user54412

4:24 AM

@ACuriousMind Now there's an equation I haven't written down in years. There's an ambiguity?

ACuriousMind

@ChrisWhite I'm not sure what exactly you mean by sign convention, but the Lorentz force is usually wriiten as $\frac{\mathrm{d}p}{\mathrm{d}t} = qF(u,-)$. I imagine the sign could be due to switching the slot the four-velocity $u$ is inserted in

(Since $F$, as a two-form, is antisymmetric)

user54412

In that case there must be disagreement on which index is divergenced to get $J$ in Maxwell's equations

ACuriousMind

@ChrisWhite I think people also disagree about whether the current is naturally a one-form or a three-form.

user54412

Let's see: Wiki says $F^{\alpha\beta}_{;\alpha} = J^\beta$

user54412

@ACuriousMind You're right. This really sucks.

user54412

4:35 AM

$2^n$ different sign conventions, for large n

ACuriousMind

Heh, yeah

user54412

Well, Carroll says $F^{\alpha\beta}{}_{;\beta} = J^\alpha$

bolbteppa

The signs are given here en.wikipedia.org/wiki/… been meaning to figure it out tbh

user54412

@ACuriousMind I think that when forced to be component-free, I go with $\mathrm{d}(\star F) = \star J$, so $J$ is the 1-form associated with the charge current density

ACuriousMind

@ChrisWhite But is he considering curved spacetime properly? You don't see the difference between being "naturally a three-form" and "naturally a one-from" easily in fixed metrics, but Wikipedia calls $J^\alpha$ a "tensor density", which, for me, seems just to mean that it is the Hodge dual of a true three-tensor $J_{\beta\gamma\delta}$, and the odd transformation by the determinant of the metric is induced by the dependence of the Hodge dual of that three-form on the metric.

bolbteppa

4:48 AM

@StanShunpike I'm just a student doing projects at the moment lol

ACuriousMind

@ChrisWhite The issue is that the four current is a density, and that makes its transformation differ from that of an usual one-form by the determinant of the metric.

So it would be more "natural" to consider the Hodge dual, which is a three-form, and transforms then indeed as an usual three-form instead of as a "density".

user54412

@ACuriousMind Um, well I mean "density" as in "per unit volume" not "tensor density."

user54412

At least all the wiki articles I'm looking at self-consistently have $J$ as a 3-form

ACuriousMind

@ChrisWhite Because the usual one-form would be a density.

user54412

also, where did wiki get its convention?

user54412

4:52 AM

I see lots of Griffiths in the references

ACuriousMind

@ChrisWhite That is a good question. I bet just some editor thought about this and decided all things should be proper forms, and so Hodge-dualized all densities to get equations where no unexpected determinants show up

But not everyone does that, so you get sign issues even in the Minkowskian case, since $\star^2 = -1$ there.

or someone just copied Griffiths :D

user54412

@ACuriousMind $\star^2 = (-1)^{s+p(n-p)}$ for p-forms in n dimensions with s negative signs in the metric signature?

Stan Shunpike

@bolbteppa what kind of projects? Solving the 3D ising problem? :p

user54412

so it sign flips 2-forms but returns 1- and 3-forms unchanged?

bolbteppa

@StanShunpike general CFT stuff :)

ACuriousMind

4:56 AM

@ChrisWhite It...would seem that way. I don't want to say with certainty though

Stan Shunpike

@bolbteppa haha sounds fun

user54412

What's most frustrating is that no one comments on the sign ambiguity with F -- everyone just writes down their version.

user54412

oh well. at least I think my adviser and I are using the same convention

ACuriousMind

@ChrisWhite Chasing such signs is just so annoying that I think everyone just says: Choose my version and it works out with my equatons

I'm currently reading a text where $[-,-]$ denotes a commutator as well as an anticommutator, depending on what kind of things you insert into it.

It's convenient for conciseness, but hell if you actually want to calculate stuff

user54412

[-,-] definitely an emoticon of some sort

ACuriousMind

5:04 AM

@ChrisWhite Oh no, one more meaning! :D

Stan Shunpike

@ACuriousMind So you said a natural application of the wedge product is the exterior algebra?

ACuriousMind

@StanShunpike It's, again, kinda the definition of it.

Stan Shunpike

What do hyperplanes have to do with $s\wedge t := s \otimes t - t \otimes s$?

user54412

@StanShunpike You should take a look at the cross product formula with that equation in mind ;)

Stan Shunpike

@ChrisWhite hmmm :: runs off to wikipedia ::

ACuriousMind

5:11 AM

@StanShunpike Idea: Take $s,t$ as ordinary vectors. Then, the plane spanned as $s\wedge t$ is the plane whose orientation (the orientation of planes is an oriented circle within them) is from $s$ towards $t$. The plane spanned as $t\wedge s$ is the same plane with reverse orientation. If $s,t$ are collinear, i..e $t = as$ for some number $a$, then the two vectors do not span a plane at all, it's only a line, and $s\wedge t = a (t\wedge t) = 0$.

This should motivate that the wedge is the right thing to describe oriented planes.

Stan Shunpike

Are we talking about regular planes or hyperplanes?

ACuriousMind

@StanShunpike In that example, I'm talking about usual planes.

It extends to hyperplanes, but there our intuition forsakes us ;)

Stan Shunpike

Are you considering the case where the vectors are 2D?

ACuriousMind

@StanShunpike I'm thinking of the case where the vectors are 3D, but their dimensionality is formally arbitrary until now.

Stan Shunpike

Your first statement was "Take $s,t$ as ordinary vectors". So these are $n$ dimensional

ACuriousMind

5:19 AM

@StanShunpike Yep

Stan Shunpike

Doesn't the word plane imply 2D?

ACuriousMind

The plane they span is 2D - two vectors span a 2D object in any dimension.

Stan Shunpike

Oh, damn. you're right of course.

Alright, so then are we saying $s\wedge t$ spans this plane too?

ACuriousMind

@StanShunpike Not really - I am saying we can use $s\wedge t$ to uniquely denote that plane together with it's orientation.

It's not "spanning" it because it is not a vector in the same vector space as $s$ and $t$ are.

Stan Shunpike

We didn't discuss orientation when we talked about tensor products. Why does it show up here?

What do we mean by oriented?

ACuriousMind

5:25 AM

Perhaps the case of three dimensions is illustrative - there, every oriented plane has unique normal vector, and that normal vector is $s\times t$, with $\times$ the usual cross product. $s\wedge t$ lives in a different space, but it is isomorphic to the description by $s\times t$, that's the Hodge dual I talk about in the answer you asked earlier about

Stan Shunpike

I've got a definition in one of my books, but I don't recall it off hand

Hmmm...interesting

ACuriousMind

@StanShunpike Oh, I discuss it because I think it is the most "intuitve" way to see in a basic application what the wedge is

The "wedge" is actually much more deeply entangled with such geometric structures like planes and hyperplanes inside the larger space, but that would require a full-blown course on (co-)homology to make precise.

Stan Shunpike

Okay, let's suppose I accept what you have said about using $s \wedge t$ to denoate the plane together with it's orientation. How does it do that exactly? How does $s \otimes t - t \otimes s$ give us information about the plane or it's orientation?

Like the only information I could see is that

If $s = t$, then we know it must be zero.

Here's another source of confusion. Yesterday, we were working with $1$-tensors not vectors. Now we are using vectors. So it doesn't make sense because how can I apply the tensor product to vectors? I get the idea that...if we take $s \otimes t$ to be a vector, then the subtraction makes sense.

ACuriousMind

@StanShunpike Okay, let's say you have a metric on the space. Then, you have that $s\otimes t$ can be fed two other vectors (by raising indices) $v,w$, and since $(s\otimes t)(v,w) = s(v)t(w)$, you have that this is zero whenever $v$ is orthogonal to $s$, and $w$ is orthogonal to $t$.

Stan Shunpike

Okay, that's enough for me.

Like I accept that

ACuriousMind

5:31 AM

The antisymmetric combination is now such that $s\wedge t$ is zero whenever the two vectors fed into it are orthogonal to both $s$ and $t$.

So, $s\wedge t$ defines the plane by telling you all the vectors that are orthogonal to it - since being orthogonal to a plane means being orthogonal to the vectors that span it

Stan Shunpike

If both $v$ and $w$ were orthogonal to both $s$ and $t$, would they be collinear (I think that's the term you used before)?

ACuriousMind

@StanShunpike In three dimensions, yes - that's why you can represent a plane by it's unique normal vector. In higher dimensions, no

also, collinear simply means that the vectors are multiples of each other (equiv.: are parallel), sorry for using unnecessary terminology.

Stan Shunpike

$(s\otimes t)(v,w)$ is a $2$-tensor right?

ACuriousMind

@StanShunpike No, that's a number :P $s\otimes t$ is the tensor.

Stan Shunpike

oh right!

duh sorry you're right

But is $(s\otimes t)$ the metric in this case? So we can raise or lower indices then? Is that what you are saying? Just in this example obviously.

ACuriousMind

5:40 AM

@StanShunpike No, that's not the metric. I've been just supposing that there is some metric with which we can raise or lower indices (you can just think of the Euclidean case, where raising/lowering is a do-nothing operation)

I guess if I could draw you some pictures that would be really helpful :D

Stan Shunpike

How can $s$ be orthogonal to $v$? Can a tensor be orthogonal to a vector?

ACuriousMind

@StanShunpike Ah, I said you should take $s,t$ to be usual vectors. By lowering their index, they are also covectors, which are $1$-tensors.

I'm sorry, I think I am presupposing too much on your part here

Stan Shunpike

Why?

I'm just slow. Let me try lol

ACuriousMind

@StanShunpike "Lowering an index" means usuing the equivalence of vectors $v$ and covectors $g(v,-)$, where $g$ is the metric. (Recall that covectors are members of the dual space, which in turn are linear functions on the original space)

Also, I should really get to sleep now, the sun has risen :D

Stan Shunpike

That's fine.

ttyl

to be continued

:D

ACuriousMind

5:45 AM

I'll be back ;)

Cya

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