The h Bar

ACuriousMind

00:00

Chris is right, we're overcomplicating this

0celo7

@ACuriousMind Wolfram gets confused by \mathrm :(

ACuriousMind

@ChrisWhite That triangle looks good

user54412

@FenderLesPaul Just ask to visit some other time. The faculty are there anyway, as are the current students. Especially if you can still string the trips together so they're paying less anyway.

0celo7

@ACuriousMind I'm confused

what about surfaces that have mixed timelike and spacelike parts

user54412

For example, ask Berkeley if you can visit a bit earlier, and have them just worry about getting you out there. And then have UCSB get you down the coast and back home. Or anything like that.

m0nhawk

00:03

@0celo7 I'm not sure, but RHS looks like zero for me.

0celo7

@m0nhawk it is zero

it's zero because it's the integral of an odd function, I said that above

m0nhawk

Sorry, missed that.

0celo7

np

ACuriousMind

The normal vector to (0,0),(2,0) is $\pm (0,1)$, that to (2,0),(0,1) is $\pm (1,2)$, that to (0,1),(0,0) is $\pm (1,0)$

Now, the product of $(2,1)$ with those is: $\pm 1,\pm 0, \pm (-2)$

Integrating gives the lengths $2$ and $1$, so we get $\pm 2 + \pm (-2)$

0celo7

you've lost me

what are you doing

ACuriousMind

00:06

Hmm, now to figure out which choices of signs are inward/ outwards

0celo7

lol

ACuriousMind

@0celo7 Computing the boundary integral over @ChrisWhite's triangle

0celo7

I thought you didn't want to expend energy on this

not that I'm not appreciating it...

ACuriousMind

I didn't want to trace signs through theorems. This little computation is strangely fun ;)

0celo7

@ACuriousMind ok but the triangle does not have a uniform X-like boundary

so the theorem does not apply

ACuriousMind

00:09

Okay, outward choices would be - for the first sign and - for the second. So choosing both normals outward gives the correct result.

@0celo7 But they are "piecewise" either spacelike or timelike

0celo7

@ACuriousMind ok, and that's what I was saying earlier

ACuriousMind

And I've just found that choosing the normal vector outwards on all boundaries gives the correct result.

Wait

0celo7

Oh my god what if I found a mistake in Wald, Hawking-Ellis and Carroll

ACuriousMind

Oh god, signs

0celo7

lol

ACuriousMind

00:10

I relied too much on the geometric intuition

0celo7

@ACuriousMind me for the past week

@ACuriousMind hmm?

ACuriousMind

I determined "inward/outward" and "normal" with the Euclidean metric :/

Fortunately, the (0,1) and the (1,0) normals don't change

0celo7

Which book of Carrolls is this? — Mozibur Ullah 8 mins ago

Really?

If you don't know that appendix E of Carroll's GR book is around page 450, you're not a GR person

@ACuriousMind you're magic, I don't even know what you're calculating :P

are you just drawing a picture

on karo papier or something

ACuriousMind

In my head, actually :D

0celo7

:(

ACuriousMind

00:14

Okay, what I'm doing is this: I want to determine the normals to the Chris-White-triangle $(0,0),(2,0),(0,1)$ for metric signature $(-+)$.

0celo7

yes, how the hell do you do that

and how do you know which sides are timelike/spacelike

ACuriousMind

In Euclidean space, the normal to a vector $(t,x)$ would be $\pm (-x,t)$.

0celo7

yes

ACuriousMind

So in Minkowski space, it is $\pm(x,t)$

0celo7

oh :P

I thought you were a fucking magician

now note that you have to rescale each normal vector

they're supposedly unit.

so the part of the triangle $(0,0)(0,1)$ is the timelike side?

ACuriousMind

00:17

That means the normal to the edges are $\pm (0,1)$, $\pm\text{some factor}(-1,2)$ and $\pm (1,0)$

0celo7

MAGIC

draw a picture please

oh

I agree

carry on

mb draw a picture anyway :P

ACuriousMind

Now, we need to project the vector field onto these normals. Let's choose the vector field to be $(2,-1)$ so it is perpendicular to the hypotenuse in Minkowski space.

0celo7

@ACuriousMind Some factor = $1/\sqrt{5}$ btw

ACuriousMind

We end up with $\pm (-1) \cdot 2 + \pm (-2)\cdot 1$ for the integral over the boundary

0celo7

:( magic

wait a moment

you can totally use Euclidean intuition for the direction

or not, wtf

the normal to the hypotenuse is messed up

ACuriousMind

00:23

@0celo7 In Minkowski metric, $(2,-1)$ times $(0,1)$ gives the $-1$. The $2$ comes from the length of the edge. The integral over the second edge vanishes. $(2,-1)$ times $(1,0)$ gives $(-2)$

So we see we must choose the signs of the normals oppositely for this to give the correct answer, now what's left is to figure out which sign is inward and which is outward for Minkowski

0celo7

@ACuriousMind does that look right?

we SHOULD be able to pick the orientation by inspection.

the orientation is a topological thing/no metric dependence

ACuriousMind

Inward/outward are defined how again?

0celo7

Ugh

check the deleted answer

here:

ACuriousMind

The curve has to lie outside of the volume, right?

0celo7

No

it's a map into $M$

this is the standard def btw, even though I came up with it "on my own"

ACuriousMind

00:30

Huh? How can I then not find a curve $[0,\epsilon)$ for both choices of signs? Just take the geodesic starting at $p$ with $v$.

0celo7

@ACuriousMind you wanna Skype this :)

@ACuriousMind uhhh

because a manifold with boundary is not geodesically complete

so you need not have a solution of the geodesic equation with those initial conditions

ACuriousMind

Are you saying the definition fails if $M$ is geodesically complete?

0celo7

@ACuriousMind no I'm saying your counterexample fails

the geodesic equation doesn't need to have a solution right on the boundary

if it does, it won't be smooth

or $C^1$

ACuriousMind

I'm solving the geodesic equation in $M$, not on the boundary or the volume.

0celo7

@ACuriousMind Ohhhhh

bro just get your mic out, typing is hard

let's call the triangle $V$

then change $M$ to $V$ in the definition

I was viewing the triangle as a submanifold and calling it $M$

ACuriousMind

00:34

Yeah, that's what I was saying - you must restrict the curve to live either inside or outside of the volume

0celo7

Yes yes

the answer was about M itself, not V. but the results carry over

ACuriousMind

Yes, inside is the right choice.

0celo7

How do you know

ACuriousMind

Okay, so choosing both normals outward is both times minus in my thing above

0celo7

both what

ACuriousMind

00:36

But we get zero only if we choose them differently

So indeed, one of them has to point inward!

0celo7

:/

@ACuriousMind wait...ONE

shouldn't both have to point in

ACuriousMind

No, because one of the edges they belong to is timelike and the other spacelike

0celo7

dude please draw a picture

ACuriousMind

I wouldn't draw anything other than a triangle

0celo7

@ACuriousMind so there are two timelike normals, right

or two spacelike

there's two spacelike normals

ACuriousMind

00:43

Yes, but the hypotenuse is irrelevant

0celo7

huh?

ACuriousMind

We chose the vector field such that the contribution from the hypotenuse to the integral is zero.

0celo7

oh shit what was the vector field, anyway

ACuriousMind

$(2,-1)$

0celo7

right

so what's the issue again :P

sorry that I'm so confused

you then dotted this with what

ACuriousMind

00:47

With the normals

0celo7

$\pm(0,1)$ and $\pm(1,0)$?

ACuriousMind

Yep

0celo7

ok then what

ACuriousMind

Well, you get a surface integral by dotting the vector field with the normal vector and integrating the resulting function

0celo7

what did you integrate over

ACuriousMind

00:49

Since the vector field is constant, the integral over the edges reduces to just their length

@0celo7 The edge to which the respective normal vector belongs

0celo7

Minkowski length?

or Euclidean

ACuriousMind

@0celo7 No

At least, I used their Euclidean length

If you use Minkowski length, then one of them is negative, and you have to choose all normals outward

0celo7

@ACuriousMind so the conclusion is that you have to reverse one of them

does that prove the theorem

or just make it plausible

ACuriousMind

Plausible in the sense that it shows you need to do something with the normals, not just always choose them outward

0celo7

@ACuriousMind yes

Damn I wish I had some whiskey

ACuriousMind

00:56

We could determine whether to choose the timelike or the spacelike ones outward by choosing a vector field that's not orthogonal to the hypotensure

0celo7

@ACuriousMind yes.

ACuriousMind

Which still would not prove the theorem, but tell you whether to try to prove the Wald or the Carroll version

0celo7

@ACuriousMind Good point. I only tried to prove the Wald one.

And I...guess I recovered Carroll's version?

Part of it at least

I need to analyze timelike hypersurfaces more carefully.

ACuriousMind

@dmckee Not sure if someone is monitoring it, but a certain user that was recently suspended for low quality contributions seems determined to continue their former course of action.

0celo7

oooo link

ACuriousMind

01:00

nope, I'm certain dmckee will know what I'm talking about without one

user54412

@ACuriousMind I think that's a key point -- there's a subtle sign choice in deciding you want a positive volume element while integrating from a low coordinate to a high coordinate.

0celo7

@ChrisWhite ok so can you find the error in my proof?

user54412

I gave you a triangle, what more could you want?

0celo7

that's a counterexample, not a proof

ACuriousMind

@ChrisWhite wait

argh

I forgot that due to the orientation, we integrate from 1 to 0 on the left edge

user54412

01:03

53 mins ago, by ACuriousMind

Oh god, signs

0celo7

@ChrisWhite can you work your magic and find the original journal article with this result

@ACuriousMind lol

ACuriousMind

so the inward choice would follow for using the Minkowski length

0celo7

@ACuriousMind but why would you do that

the surface integral knows nothing about the metric, does it

And this, kids, is why Bajoran hates GR

And why you should only ever do Riemannian geometry

ACuriousMind

@0celo7 Dunno, maybe that's what Wald and all the other physicists consider the natural choice

0celo7

@ACuriousMind :/

ACuriousMind

01:06

And it might be. We can't see the difference here, but it would show up when curving the sides of the triangle

But I'm not computing that

0celo7

we could try the cube x R

but I fear with such a symmetric shape it won't work out anyway in a way that's obvious

ACuriousMind

Hmm...maybe one just needs to rotate the triangle

0celo7

or maybe one can prove this like a man

ACuriousMind

That would also lead to the Minkowski and Euclidean length differing by more than a sign

I think

@0celo7 I'm not trying to prove anything, just to help you determine whether the statement that one has to choose inward normals sometimes is actually correct

0celo7

@ACuriousMind yes, did I not say that

guess I only thought it

@ACuriousMind Yeah, the proof I delivered on the main site is circular and I was able to derive a contradiction only because of a sign error.

0celo7

01:26

@ACuriousMind this Mozibur Ullah guy is annoying...

0celo7

01:40

@ACuriousMind I think Carroll's really is a typo because he uses the "correct" result in his book.

HDE 226868

02:25

I answered this question, but after the editing and bounty, I no longer have any idea what it's asking.

If anyone does have an idea what it means, that would be great.

0celo7

02:39

@ACuriousMind Well...Gourgoulhon claims to have a proof

But it's not lucid

I'll devote time to this at some later point...I'm pretty burnt out on this.

Sir Cumference

03:15

-6

Q: Can there be life in black hole?

Can there be life inside a black hole is it possible that we are living inside a massive black hole right now? This is because i see no reason why we cant be inside a super massive black hole the only thing is that we cannot escape it? I HAVE ALSO HEARD THAT THERE CAN BE WHITE HOLES OF WHICH MATT...

supermassive-black-hole

Sigh...

0celo7

@SirCumference Yes, if it's large enough

Hari Prasad

03:36

Why Dirac monopole is a topological defect in a U(1) gauge theory?

Sir Cumference

03:53

@0celo7 What are you talking about?

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