@ACuriousMind : I have knowledge of it. And the moot point is this: Can you step outside and look up to the clear night sky and point out a Lorentzian manifold? Or a light cone? Or a world line? Or a closed timelike curve? How about a sphere on two-dimensional Minkowski spacetime? Or an inward vector stuck into it like an arrow? The answer is no. Because the map is not the territory.

@ACuriousMind eye twitch

odd number

I've just recovered my original proof that the direction should not matter

I really wonder where these books got the statement about the direction from

@ACuriousMind huh?

@0celo7 I thought we were posting irrelevant pictures.

@ACuriousMind ugh, are you interested in figuring this out with me

I'm gonna delete it

@Danu Yes. But can you Quess why. -I can.

@0celo7 I meant I wonder if they showed it themselves and decided not to include the proof, if they copied it from someone else or if they...just made it up

I'm back to my earlier idea that it's because GR books can't do anything right and their indices are fucked up

I already said I don't care enough about this issue to spend time and energy on it

@ACuriousMind I know what you meant

And get another non-answer like from 't Hooft :D

@BernardMeurer They are not Mad. They just seem like "mad" to the people who don't understand them. Ie. Wittgenstein was surely not a recluse. I can guarantee you that ie. 98% of people thinks I am such. But I am not. I have my friends. More than I need. 80% of the so called "mankind" is pure shit to me. Witgenstein put it word perfectly; "Wo von man nicht sprechen kann, darüber muss mann schweigen." in English, "Whereof one cannot speak, thereof one must be silent." ,,,, and that your "recluse"??

@ACuriousMind lol

Here you have a Finnish "recluse" youtube.com/watch?v=57PWqFowq-4

@ACuriousMind Should I delete the answer until I can fix it?

Well, if you no longer think it's correct then I'd do that, yes.

@ACuriousMind : 't Hooft will doubtless tell you gravity is holographic and the universe is two-dimensional, and furnish the mathematical proof, QED. Then before you know it, it will be in the media, then textbooks, then 0celo7 will be writing answers about it.

@0celo7 : don't delete it, just put a note on it for the benefit of the OP.

@ACuriousMind I see where the issue is. My argument is circular. I need to use a different approach, perhaps by aligning the coordinates with the normal vector and doing everything in that basis shudders

@ACuriousMind actually, flipping the normal vector inwards would be a contradiction

so I must have messed up somewhere above, or the theorem is just wrong

@ACuriousMind I want to provide the OP with the TeX code so he can look at it, how should I do that

Probably leave it undeleted, then

Ugh...stupid signs

@ACuriousMind Yup, I'm back to that theorem in Lee that states for any manifold, regardless of metric, contracting the volume form with the OUTWARD normal gives a consistently oriented volume form on the boundary

so maybe these books are implicitly reversing the orientation on the boundary

I wouldn't put it past the physicists to do something like that

Yeah, that is possible

Sigh...and then you apply the theorem which says on Riemannian manifolds, the positive oriented volume form is the one with $\sqrt{g}$

now I'll try to review the proof of this again

the only thing that changes in the Lorentzian case is that you get a $-$ in the square root

You could always puzzle about whether the flow of time points up or down.

@JohnDuffield huh?

I like muffins and ponies too!

Are we just saying random things now

There is no direction to the flow of time, because there is no flow of time.

@JokelaTurbine Chill.

@JohnDuffield Dunno why he would tell me that

Most people don't answer question X with something completely unrelated to X

That's a you thing

@JokelaTurbine I'm an admirer of Wittgenstein's works and I myself face people thinking I'm somewhat recluse. I'm not calling them mad, I wouldn't read any of them if I thought so, but their persona is of the "mad genius"

They're absolutely sane, it's just how they appear to the average individual

In similar vein there are no inward or outward vectors.

Okay, I wouldn't say absolutely

OK, gotta go.

@JohnDuffield What?

pastebin is broken :(

@0celo7 Banana! Yes.

It's all abstraction. What's not is that when space is homogeneous light goes straight and your pencil doesn't fall down. When it isn't light curves and your pencil does fall down. That's it. We model it as curved spacetime, but the map is not the territory. There are no world lines, or light cones, or spacelike boundaries, or volumes or spherical surfaces with positive or negative arrows sticking in or out.

Ergo the question merely concerns definitions and consistency within mathematical abstraction.

@JohnDuffield lol

what are you talking about?

Physics.

I'm not

so why are you talking about physics?

Because this is the physics stack exchange, and I'm explaining why I gave you that downvote.

@JohnDuffield it was a mathematical physics question

not physics

Now I really have to go. Mañana.

Bye

@ACuriousMind I could answer this in one line by saying physical systems are indistinguishable if they're diffeomorphic...but he's reading Hartle

I think it's even worse than Schutz D:

@ACuriousMind wow what a terrible answer on that integration question

It's pretty ridiculous

@ACuriousMind Note that at present I am not convinced one should flip the normal inside. But it requires a hell of a lot more proving than what he gave.

@0celo7 the answer doesn't even say "don't flip the normal" (then it would be an answer). It says "use the form where you don't need a normal", which doesn't answer the question.

@ACuriousMind I know

Ok, that statement does not make sense.

@ACuriousMind Straumann handy?

Silly question: Have you taken some simple function on Minkowski space, and computed both sides of Stokes' theorem once with the flipped and once with the usual normal to see which is correct?

Nope.

I don't actually know how to do that.

I need some closed surface

Well, that would at least decide whether you should try to prove the flipping or disprove it, no?

Wait a moment

@ACuriousMind there's something amiss here

the books are saying that the orientation only matters if the surface is closed

why would that matter

lol, what?

How would you get a non-closed surface as the boundary of a volume, anyway?

$\partial^2 = 0$.

no no

uh

ignore ^

at least until I figure out what the physics books are saying

@ACuriousMind Ok, so I need a vector field on Minkowski space and I need a compact region that has an everywhere spacelike boundary

any suggestions?

@Danu after giving it more thought

I'm now pretty much split between UCSB and Berkeley

having a hard time deciding :/

(compact so the integrals are finite, spacelike so we can test the flip)

@0celo7 Vector field something like $(1/t^4,1/x^4,1/y^4,1/z^4)$ so compactness doesn't matter, region of integration something like $\mathbb{R}\times [0,1]^3\subset \mathbb{R}^{1,3}$?

make it to the fifth power to be sure :)

Yeah, choose it large enough the silly thing converges

ok then

Wait

Waiting.

Ugh I have to calculate the normal vector to that region

No, carry on; I'm now wondering if there is a compact subset in Minkwoski with everywhere spacelike boundary

I'm bad at math

oh boy, time to calculate $\int_R\partial_\mu X^\mu\,\mathrm{d}x^0\cdots\mathrm{d}x^3$

@ChrisWhite can you help here

First off, $$\partial_\mu X^\mu=oh boy calculus$$

Hmmm, maybe take an $D^3$ instead of the $[0,1]^3$ as the base so that the boundary is smooth, then you don't need to split the integral into an integral over the various faces

@ACuriousMind that makes the normal WAY harder to calculate, no?

@0celo7 In polar coordinates the normal to the sphere is just the radial vector

But they, it might get ugly

@ACuriousMind true, but what about that vector field

and then you need the metric in polars

Well, the vector field was just a random guess, I'm not sure if that actually makes for nice computation

Actually, something like $(1/\sqrt{t^2+x^2+y^2+z^2}^n,0,0,0)$ might be better

Ok so $$\partial_\mu X^\mu=4/t^3-4/x^3-4/y^3-4/z^3$$

I think!

I can't actually calculate stuff

@ACuriousMind what

why even give it $x,y,z$ dependence

Because a constant vector field will not have finite integral

make it decay in time bruh

For that, you'd need a compact region

@0celo7 Oh, right

Then just scaling a constant vector field with $\mathrm{e}^{-t^2}$ or something might be best

I dunno, I've never had to come up with "exercises" before :D

lol

^ this guy

@BernardMeurer I know exactly what goes on their mind. But it's not only that they appear "recluse" to average. After certain level You are "recluse" simply through the fact that if the gaussian has given you ie. the 1/1000 000 cards, it means that there is only 500 people like you in Europe. And as they are all also different age, language etc, you just are poor f-king "recluse" though you don't want to be and though your life might even be really enjoyable etc.

@ChrisWhite How has the cube everywhere spacelike boundary?

::facepalm::

@ChrisWhite we're integrating over a region with boundary a spacelike hypersurface

so the normal vector is timelike

@ChrisWhite If I understand correctly, the claim is that you have to flip the normal vector only for spacelike boundaries.

@ChrisWhite unfortunately I'll only be able to visit one or the other

which is why I'm trying to decide ahead of time

also their visit date is a week before the deadline to accept an offer

which sucks

@ACuriousMind Ok, $X^\mu=(\mathrm{e}^{-t^2},0,0,0)$ is good. I like this. Then $\partial_\mu X^\mu=2t\mathrm{e}^{-t^2}$.

@BernardMeurer ...and notice that it's simply more fun to be alone than debate stupid dialogues with others. That pretty much what Perelman did.

the negative gets cancelled by the negative in the metric

And $R:=\mathbb{R}\times[0,1]^3$.

$R$ for "region" (I'm so original)

Random thought about "Carroll has the "outward" and "inward" directions switched. " - does he use a different sign convention for the metric?

@ACuriousMind No, I checked.

'kay :)

Of course, maybe wald is wrong...

But let's check this.

Let $\mu=dx^0\wedge\cdots \wedge dx^3$ be the metric volume form on Minwowski

@ChrisWhite yeah

they directly overlap

Berkeley is 6th-7th and UCSB is 7th-8th

Then $$\int_R\partial_\mu X^\mu\,\mu=(\int_0^1\mathrm{d}x)^3\int_{-\infty}^\infty 2t\mathrm{e}^{-t^2}\,\mathrm{d}t$$

@ChrisWhite and 7th is the day I get to meet Berkeley faculty so I can't jet early

which sucks because there's a professor at Berkeley I really want to meet

Raphael Bousso

can someone do those integrals above pls

Ask Wolfram ;)

@ACuriousMind LOL

the integral vanishes, it's an odd function