The h Bar

0celo7

00:01

@yuggib I could take "Modern Algebra 1" instead of QM

yuggib

dunno, maybe...

Personally, I would prefer QM ;-)

0celo7

@yuggib the prof might say "come back when you've had EM"

yuggib

@0celo7 nah...they're not strictly related

0celo7

@yuggib I know quite a lot of QM

for a freshman that is

yuggib

it's up to you

0celo7

00:06

@yuggib but if I take grad level algebra I might be able to do some category shit

by the end of the year

wonder who's teaching that course

yuggib

@0celo7 maybe, but will it be useful?

0celo7

Eww, a number theorist

@yuggib probably not

I don't give a shit about algebra

I just want to understand ACM

yuggib

:-D

you can do some cat theory by yourself

ACuriousMind

@bolbteppa Nah, there's too many stars in there to do that nicely in the abstract. I'd remember that as "symmetric combination of directional derivative and curls".

0celo7

@yuggib I don't really care

bolbteppa

00:10

I know, I've posted like 2 questions about these nightmares, still can't remember them

0celo7

Dammit, no calculus of variations

ACuriousMind

Then don't remember it, look it up

0celo7

01:07

@ACuriousMind Gonna talk to Prof. Elf person tomorrow

Anthony

01:40

@DanielSank Oh, well please ping me if/when you do!

0celo7

04:16

@ACuriousMind I asked a question about integrating on a null hypersurface. We'll see what happens.

Secret

@0celo7 The answer to that might be useful to feed into that personal question of mine on calculating that minkowski metric question to explore it

0celo7

@Secret I take it you're the one who favorited it?

Secret

@0celo7 indeed, like one of the MSE users, I often use the favourite operation to act as a bookmarking for questions

You will notice I have quite a lot of favourite questions, because I will be constantly checking them in the future

11

Q: Would you ever favourite a question you down vote?

Today, I came across a user who posted lots of question in 1 area with no attempts on any of them. Many of these were very good questions and have answers. I would certainly benefit from learning from them, on the other hand, the user clearly made no slightest attempts on what appeared to me assi...

0celo7

@Secret so why no +1 >:(

Secret

I rarely vote or downvote a question, because I don't think it is necessary

I do, however, upvote answers I found useful

I am never knew to downvote any answers for reasons unknown

0celo7

04:32

@Secret what Minkowski metric question

Secret

you might say, the usual notion of favourite is also included

while I do use the favourite as a bookmarking system, in some cases I am more likely to favourite a question if I also found the question is a good one (you can easily see that as some of these have many uproots , or multiple favourites by other users)

As for that question, it is just a random curiosity, let me fetch the details...

Feb 24 at 15:14, by Secret

Let me rephrase my question

I am looking for a spacetime that is basically minkowski, but where points progressively became more degenerate the closer they are to the origin of (e.g. O)

Will the following metric can give what I want?

$$ds^2=(e^{x^2+t^2}-1)(-dt^2+dx^2)$$

?

Acuriousmind have answered this part, the next part is as follows

Feb 24 at 16:06, by Secret

@ Acuriousmind Hmm...make sense, interesting

I will try it out as an exercise later as I need the geodesics to investigate and to do so I need the first compute the christoffel symbols which I cannot do it within 5 mins

The gaussian should make this computation not very difficult as I knew gaussians tend to have nice properties when integrated

So this personal research is: Find anything interesting for this metric by calculating quantities that people calculate when given a metric

The issue is that I need heaps of concentration to get it started

0celo7

@Secret Does the Chinese language not distinguish have/has? One of my best friends in high school was Chinese and he always made that mistake.

"I am looking for a spacetime that is basically minkowski, but where points progressively became more degenerate the closer they are to the origin of (e.g. O)"

I'm not sure what that means

@Secret Ok, so that space is conformally flat. There's all sorts of wonderful theorems about conformally flat spacetimes.

Oh, nevermind, the origin is fucked up.

Secret

@0celo7 We chinese don't have the notion of tenses in our language, which is one reason why we tend to made that mistake

The basic idea of that spacetime is that as you move closer to the origin, the metric will become more and more like the zero metric, thus the degeneracy increases as you move to the origin

Acuriousmind said this thing might be non hasodoff

0celo7

The zero metric is not a metric

If your metric is undefined you basically just remove that point from the spacetime

And you end up with a space with a hole

Secret

ds^2=0(-dt^2+dx^2)
So this thing is not a metric?

0celo7

04:41

@Secret How are you defining a metric

A nondegenerate section of $S^2T^*\mathcal{M}$?

Secret

@0celo7 Not very familiar with that shape, is it related to a torus?

0celo7

@Secret uh it's the symmetric (0,2) tensors

Secret

yeah, I recall in our GR class all metric tensors we use are symmetric (0,2) ones

0celo7

the important point is that they're nondegenerate

that is, none of the eigenvalues of the metric are 0

Secret

I do recall reading about the acuberrie metric which has cross terms in the metric tensor, but even that one is nondegenerate

I am not sure if anyone actually worked with degenerate metrics, I recall Danu mentioned they are not very interesting

My system is kinda like a minkowski metric becoming progressively more degenerate as it approaches the origin

but from what you said, my system might actually have a hole in the origin instead

I am basically interested in how introducing degeneracy on a flat spacetime influence the description of events happening on that spacetime

But danu and Slereah said I cannot just work with a degenerate signature, because the direction where the spacetime is degenerate is (forgot exact terms) "not very different" from a spacetime without the degenerate direction, thus you essentially get the same results

This motivates me to introduce the degeneracy by using a gaussian dependence on the coordinates thus effectively creating a region of almost degeneracy near the origin

in order to investigate further

0celo7

04:54

@Secret that's really an oxymoron

a metric tensor is nondegenerate by definition

suppose we have a manifold $M$ and a "metric" $g$

and this "metric" is degenerate at $p\in M$

then everything works as usual on $M\setminus\{p\}$

kevin Tah N.

what do you guys even mean when you say it is "non-degenerate"

0celo7

@kevinTahN. it has an empty kernel

kevin Tah N.

ah,

0celo7

there is no $v$ such that $g(v,w)=0$ for all $w$

$v,w$ are vectors

kevin Tah N.

I see

Secret

04:58

https://en.wikipedia.org/wiki/Metric_signature

Ok, what I want is something that basically act like it has the signature r=/=0 at a point (or at least a small region) in a flat spacetime

I am not sure if there's a better way to describe it

0celo7

uh, what?

what is r?

Secret

r counts the number of zero eigenvalues of the metric g_ab (according to wikipedia)

0celo7

oh oh

I've never seen degenerate metrics used

Except for null hypersurfaces

which is annoying, but happens in physics

kevin Tah N.

I am sitting in on an elementry GR course and I don't even know about degenerate metrics lol

Secret

I don't knew about the existence of these either, until a discussion between Danu and Slereah on manifolds and tachyons some weeks ago caused me to stumbled into the metric signature wikipedia link

0celo7

05:07

@Secret you didn't talk about hypersurfaces in your GR course?

Secret

we talked about them (slices of xyz), but I don't recall they taught about null hyper surface, unless the light cone count as one

0celo7

@Secret the boundary of the light cone is a null hypersurface IIRC

Secret

but the minkowski metric is non degenerate, thus <please help me to complete this sentence>?

0celo7

@Secret huh?

@Secret are you ready for some math

I can give you some math

Secret

you mention you use degenerate metrics for null hyper surface

But the boundary of the light cone for flat spacetime came from a metric that is non degenerate, I am not sure how to get the rationale between these two points?

and sure, you can show me maths

0celo7

05:13

Let $S$ be a hypersurface of a spacetime $M$

that is, $S$ is a 3-dimensional submanifold of $M$

and we have the inclusion $\iota:S\hookrightarrow M$

now $M$ has a Lorentz metric $g$. the induced metric on $S$ is $\iota^*g$

if the normal vector to $S$ is a null vector, then $\iota^*g$ is degenerate

See Sect. 2.7 of Hawking-Ellis for a nice proof of this

@Secret you can think of it like this

let $p\in S$, then we have tangent spaces $T_pM$ and $T_pS$

now let $v\in T_pM$ and because $g$ is nondegenerate we have $g(v,w)\ne 0$ for all $w\in T_pM$

but maybe you can convince yourself that there is a subset of vectors for which $g(v,w)=0$

and that this set is $T_pS$

(proofs of everything can be found in Hawking-Ellis, of course)

Secret

Ah I see, so even though the metic of the manifold on M is non degenerate, the induced metric on the hyper surface (which must be a subset of the manifold) can be made degenerate

so that's why you said you use degenerate metrics to calculate things on null hyper surfaces

0celo7

yes

and what $\iota^*$ really does is restrict the inputs of $g$ from $T_pM$ to $T_pS$

Secret

I guess the reason I don't see that initially is I tend to approach problems from a global perspective, and not used to think of a subset of a problem

I am guessing, this $\iota^* g$ will have 3 zero eigenvalues? because the light cone boundary is 3 dimensional and must be spanned by three linearly independent vectors that are known to satisfy g(v,w)=0?

0celo7

05:32

@Secret ummm, I don't know off the top of my head, but that sounds right

check Hawking-Ellis ;)

I'm off to finish my algebra homework

Secret

ok

0celo7

05:44

@Secret umm, I can't actually find the theorem in HE which says the null cone is a null hypersurface

Is that wrong?

Hmm, I could have sworn it was a theorem...

Null hypersurface

In relativity, a null surface is a 3-surface whose normal vector is everywhere null (zero length with respect to the local Lorentz metric), but the vector is not identically zero. For example, light cones are null surfaces. An alternative characterization is that the tangent space at any point contains vectors that are all space-like except in one direction, in which vectors have a null "length". The metric applied to such a vector and any other vector in the tangent space (including the vector itself) is null. Another way of saying this is that the pullback of the metric onto the tangent space...

According to Wiki it is true

Secret

the stuff that is mentioned in the wiki is consistent to what you said above
As for hawking Elills, do you mean this book?
The Large Scale Structure of Space-Time

0celo7

yes

@Secret I know that I'm correct, I just want to know where the proofs are

I really should get Beem et al.

Secret

achronal=space like?

0celo7

@Secret No.

And I need to sleep, I'm mixing things up.

Goodnight.

Secret

night

Secret

05:55

http://download.springer.com/static/pdf/572/art%253A10.1023%252FA%253A1011390000002.pdf?originUrl=http%3A%2F%2Flink.springer.com%2Farticle%2F10.1023%2FA%3A1011390000002&token2=exp=1457503900~acl=%2Fstatic%2Fpdf%2F572%2Fart%25253A10.1023%25252FA%25253A1011390000002.pdf%3ForiginUrl%3Dhttp%253A%252F%252Flink.springer.com%252Farticle%252F10.1023%252FA%253A1011390000002

*~hmac=2f649b003035b8dc256355b982a1b0b504cfabcf4d1eb40fa0215aa981e85174

(the link cuts away the part at the bottom)

whatever, it can be googled easily

0celo7

@Secret I'll write a Q&A of the null cone thing once I have it proved.

But now I'll really go to bed!

Secret

ok

Secret

*notepad* Using the fact that the signatures p+q+r=dim(M) it means there isn't really much choice to play around

In order to have spacetime, I must have one - and at least one +

This means my possible options to investigate are (-,+,+,+) (minkowski) (-,+,+,0) (-,+,0,0) (-,0,0,0)

Let's see what happens for the (-,+,+,0) then

Now to formulate a metric tensor that has this signature and I am ready to go

DanielSank

08:50

AlphaGo beat Lee Se-dol, one of the two top Go players in the world, in the first of a five game series.

Slereah

11:35

"In order to clarify this situation, I decided to solve the classical equations of the Yang-Mills theory perturbatively in the strong coupling limit. Please, note that today I am the only one in the World able to perform such a computation having completely invented the techniques to do perturbation theory when a perturbation is taken to go to infinity (sorry, no AdS/CFT here but I can surely support it)."

Marco Frasca stop saying that shit

You make me lose trust in your papers

Emilio Pisanty

Folks, anyone up for bumping this past the line?

5

A: Community Promotion Ads - 2016

ACuriousMind

@EmilioPisanty Already upvoted it, can't do anything

Emilio Pisanty

@ACuriousMind cool

user116211

@EmilioPisanty yep :D

user116211

@ACuriousMind: o/

user116211

11:46

@EmilioPisanty when will they get featured in the site?

Emilio Pisanty

@user36790 They're live now

user116211

@EmilioPisanty yes! saw it.

0celo7

13:39

@ACuriousMind would you happen to know the proof of "the boundary of the light cone is a null hypersurface"

ACuriousMind

What is there to prove? Isn't the light cone basically the union of all null geodesics emitted from a point?

0celo7

@ACuriousMind Yes.

ACuriousMind

And a "null hypersurface" is a surface with null tangent vectors?

Hm, no, that can't be right

0celo7

@ACuriousMind The span of null vectors is two dimensional

ACuriousMind

That's why I said "Hm, no, that can't be right" :P

So what's a "null hypersurface"?

Hari Prasad

13:49

6

A: Why does matter exist in 3 states (liquids, solid, gas)?

Basically the existence of different states of matter has to do with Inter-molecular forces, Temperature of its surroundings and itself and the Density of the substance. This image below shows you how the transition between each states occur(called Phase transitions). These transitions occur ...

0celo7

@ACuriousMind The normal vector is null

Or the pullback metric is degenerate

Hari Prasad

@ACuriousMind Are there any non-liquefiable gases out there?

GPhys

@FenderLesPaul BNL was nice

@dmckee Has your work ever sent you out to BNL for something?

ACuriousMind

@HariPrasad Not that I would know

0celo7

@ACuriousMind By some theorem, we know that the light cone boundary is $3$-dimensional, so it has a normal vector.

@ACuriousMind It is also clear that some subset of the tangent spaces contain null vectors

@ACuriousMind Theorem: a timelike vector is never orthogonal to a null vector

So the normal vector cannot be timelike

ACuriousMind

13:58

I see, I don't know a proof.

0celo7

@ACuriousMind I know the proof of that.

But it could be spacelike

ACuriousMind

@0celo7 I meant of your original question.

0celo7

Oh.

It seems so obvious! I must be missing something.

ACuriousMind

@0celo7 Is the "normal" vector of the light cone perchance already its null tangent vector?

yuggib

General Relativity is so boring... :-P

0celo7

14:04

@ACuriousMind I don't know.

yuggib

Too much geometry

ACuriousMind

@0celo7 I mean, think Minkowski: The null tangent points in the mixed time/radial direction, and the other two tangents are angular. The null tangent is obviously orthogonal to itself, and also to the other two tangents because it has no angular component.

0celo7

@yuggib hush

@ACuriousMind ok, but how does that help

ACuriousMind

@0celo7 It shows that, in Minkowski space, the null tangent is also the normal vector of the light cone

Hence the light cone is a null surface because its normal vector is null.

John Duffield

32

Q: What is the proper way to explain the twin paradox?

The paradox in the twin paradox is that the situation appears symmetrical so each twin should think the other has aged less, which is of course impossible. There are a thousand explanations out there for why this doesn't happen, but they all end up saying something vague like it's because one tw...

special-relativity time reference-frames relativity

This has a new answer.

ACuriousMind

14:10

So I'm thinking one can perhaps generalize this idea that the light cone's null tangent (i.e. the tangent to the null geodesic through that point) is already its normal. Perhaps it's also enough to know it for Minkowski and one can do some local->global argument here, I don't know enough Lorentzian geometry to say

Bernard Meurer

14:22

@ACuriousMind Our universe is slightly De-Sitter right?

ACuriousMind

afaik, that's the currently favoured option, yes

Bernard Meurer

Sorry, I forgot to ask if I could ask that question, could I have asked that question I already asked?

ACuriousMind

No, had you asked to ask that question, you could not have asked it.

Bernard Meurer

:p

I see the day when one will have to ask to ask if one can ask a question

ACuriousMind

What a dystopian vision of the future

Bernard Meurer

14:30

I'm a doomsday prep

Hari Prasad

14:42

someone just gave me like 1100 reputation and the took it away!

no i'sm sorry i got +100 on all SE sites. What is happening?

it says we trust you in other sites

@ACuriousMind can you please tell me whats happening?

yuggib

SE gives you a bonus of points on the other sites if you have a certain rep on one (I think)

Hari Prasad

oh i see

ACuriousMind

@HariPrasad It's the association bonus. I think it should also take you to a site explaining that if you just click on the +100.

Hari Prasad

@ACuriousMind yes.

thanks

0celo7

14:57

@ACuriousMind Ok, maybe one can apply the Gauss lemma

@ACuriousMind But can't you also take the "other" null vector?

Your argument works for that case too

ACuriousMind

What's the "other" null vector here?

0celo7

@ACuriousMind well there's always two linearly independent null vectors at a point

The one that's tangent to the null geodesic and the "other" one

ACuriousMind

@0celo7 That's not what I was asking. You claimed my argument also works for that other one. So what is the other one? How can you say that my argument also works for it?

0celo7

@ACuriousMind I'll explain later when I'm not in a lecture

FenderLesPaul

15:16

Does anyone here have a mac and uses 'include graphics'?

sharaf zaman

consider a bomb, because of reactions it burst and all the walls if which bomb is covered just flies away. when the bomb bursts huge amount of energy is released

but only energy is released who provides the force to the bomb so it can break through the walls?

ACuriousMind

@FenderLesPaul Nope, but the people at TeX - LaTeX are usually very friendly and will be happy to help you ;)

FenderLesPaul

@ACuriousMind cool thanks!

ACuriousMind

@sharafzaman What do you mean "only energy is released"? There is the heated air rapidly expanding, there's shrapnel from the bomb (even if not built as such, it doesn't just disintegrate, does it?)...

0celo7

@FenderLesPaul yes.

sharaf zaman

15:23

@ACuriousMind yeah i mean how come shells are thrown because of some chemical reaction (because in newtonian mechanics there must be a force ) so what force is that?

0celo7

I'll help you if you tell me the proof for the null come boundary being a null hypersurface

ACuriousMind

@sharafzaman I do not understand the question.

sharaf zaman

@ACuriousMind wait a sec, let me think and explain my question!

yes, suppose i take a block from ground to some height and i keep it there!
the block's gravitational potential energy will increase (*agree?*)

@ACuriousMind ??

ACuriousMind

@sharafzaman What? Yes, of course I agree, get ot the point.

sharaf zaman

lol!!
now if i keep another block just next to it. i am sure that the block will still be there(i mean no force will act on block, so it doesn't move(neglect GMm/r^2 between blocks))

now in the same way because of some reaction in the bomb the particles get huge amount of internal energy and they just burst off, but why does they make particles next to it move
like in that block case even if i take the block to huge height (increasing its energy) it will not interfere in other blocks motion!

but then why does that bomb interfere in the motion of other particles if i increase the energy of a particular particle!!!
*hoping i am clear*

0celo7

15:36

Maybe one can use the Frobenius theorem

ACuriousMind

@sharafzaman Um, when you explode something it gets heated very fast. Heat is nothing but chaotic motion of the particle, more heat means more motion, and heated things expand. You don't "increase" the energy of a particle" when you explode a volatile substance. You turn its chemical energy into thermal energy very fast.

0celo7

@ACuriousMind Hmm, if the spacetime has holes, the boundary need not actually be generated by null geodesics...

ACuriousMind

@0celo7 This is another of these cases where I don't know why are you telling me that :P

0celo7

At least not geodesics that all emerge from the same point

@ACuriousMind I'm making mental notes while Dr. Finotti is rambling on about integers modulo m

ACuriousMind

This chat is not your notebook.

sharaf zaman

15:38

@ACuriousMind oh i see!! thank you...

0celo7

@ACuriousMind au contraire my friend

yuggib

15:55

shit...reviewing a three pages' russian functional analysis paper with: two theorems, and no proofs and being polite is rather hard

- is the paper clear? no
- are the results interesting? maybe, but who knows if they're true

probably for russian standards everything written there is trivial (and in fact the paper was accepted without problems, in a russian journal)

ACuriousMind

But if it's trivial, why publish it?

yuggib

trivial by russian standards

ACuriousMind

But you say it was accepted in a Russian journal?

yuggib

yes, and they accepted it without proofs for the proofs were trivial

I imagine

if else it is an announcement, and not a paper

i.e. not on the scope of the journal

sometimes those mathscinet reviews are tough to do...

Hari Prasad

@ACuriousMind Is it OK if i used a Question that i answered in my personal blog?

ACuriousMind

16:07

@HariPrasad If you link back to it, yes. All SE content is licensed under CC-BY-SA 3.0

Hari Prasad

@ACuriousMind Thanks

0celo7

@ACuriousMind Take the left line

There are null vectors which are tangent

but then take the Euclidean normal

this is also a null vector

ACuriousMind

But is it Minkowski-orthogonal to the tangent null vector?

(it's not)

0celo7

@ACuriousMind proof?

ACuriousMind

@0celo7 Compute it

0celo7

16:16

umm, it should be!

ACuriousMind

If you do not believe me and don't see it, do the calculation

I'm not doing that for you

0celo7

ugh...

@ACuriousMind why are you so smart

:/

Ok, new theorem: the integral curves of the normal are null geodesics

Proof: somewhere in Straumann

You cannot without other pieces of information. With a vector field $J$ (typically a timelike vector field) you can, defining the volume form $\omega = \star J$ and next integrating $f\omega$. — Valter Moretti 4 hours ago

@ACuriousMind So what would two Minkowski orthogonal null vectors look like on the above picture?

ACuriousMind

They would be parallel.

0celo7

sigh...

I had the wrong theorem memorized

> Two null vectors are orthogonal if and only if they are linearly dependent.

This is the correct result

Bernard Meurer

That one makes sense

ACuriousMind

16:21

@0celo7 that's "they would be parallel" in fancier speak :P

0celo7

@BernardMeurer do you even know what linear dependence is

@ACuriousMind No shit

Bernard Meurer

@0celo7 It's beginning to make sense to me

it's process

0celo7

@BernardMeurer a set of vectors $v_1,\dotsc,v_n$ is linearly dependent if there are scalars $c_i$ such that $v_1=c_2v_2+\cdots+c_nv_n$

ACuriousMind

@0celo7 That's why memorizing theorems without memorizing illustrative examples is not very useful

0celo7

@ACuriousMind probably, Euclidean intuition betrayed me

ACuriousMind

16:24

It has a tendency to do that

Bernard Meurer

So if I can define $v_a$ in terms of $v_b$ they are linearly dependent?

0celo7

@ACuriousMind you know, I probably just don't know the difference between space and spacetime

Bernard Meurer

and if there's no way to describe $v_a$ as the result of trivial operations on $v_b$ they are independent?

ACuriousMind

@BernardMeurer Think more geometrically. Two vectors are linearly dependent iff they are parallel. Three vectors are linearly dependent iff the third lies in the plane spanned by the two other ones. The formal statement is just the generalization of that to arbitrary dimensions.

0celo7

@ACuriousMind that seems like what I just said

ACuriousMind

16:27

@0celo7 Where did you say that? Do you expect @BernardMeurer to be able to translate the equation there into geometric intuition? That's something people comfortable with the subject do, but not beginners.

0celo7

@ACuriousMind But Shankar is a very good book and he makes that all clear!

Bernard Meurer

@ACuriousMind I see what you mean, did what I say make sense?

ACuriousMind

@BernardMeurer Yes, but it sounded a bit awkward ;)

Bernard Meurer

@ACuriousMind What have I said ever that didn't sound awkward?

ACuriousMind

@0celo7 Not all people learn as well from books as you do

0celo7

16:29

@ACuriousMind I don't learn well from them at all!

arxiv.org/pdf/gr-qc/0405108.pdf

0celo7

16:51

@ACuriousMind Ok, so if a null hypersurface has a normal vector $l$, then the integral curves of $l$ are null geodesics. Now, if $\Sigma$ is a union of a bunch of null geodesics, we can't conclude that the normal vector field we get from the tangents is normal, can we?

Since $l$ is both normal and tangent to $\Sigma$, it is clear that if $\Sigma$ is null, it is the union of null geodesics.

I can't turn that "if" into "iff".

ACuriousMind

Wait. doesn't "if a null hypersurface has a normal vector l, then the integral curves of l are null geodesics" imply that l is null, and that's all you wanted to show?

0celo7

@ACuriousMind Let's call the boundary of the null cone $C$. I don't know that $C$ has a null normal vector.

All I know is that it is constructed from null geodesics.

Ah, a hint! From HE: If $\mathscr S$ is a future set, $\partial\mathscr S$ is achronal.

So $C$ is achronal.

HE states the result without proof -.-

@ACuriousMind Ah, wait. We know that the normal vector cannot be timelike because a timelike vector is never orthogonal to a null vector. So let's show it cannot be spacelike. If it were spacelike, $\Sigma$ would be timelike. But an achronal boundary cannot be timelike (need to prove this). Thus the normal is null.

Ok, that should be easy to prove, I'm convinced.

@ACuriousMind If $\Sigma$ is a null hypersurface, I wonder if the integral $\int_\Sigma f\star\flat\operatorname{grad}f$ has interesting properties

ACuriousMind

17:14

@0celo7 Um...how do you suddenly know $C$ has a normal vector?

0celo7

@ACuriousMind It's an embedded 3-dimensional submanifold

ACuriousMind

@0celo7 Then they do you saY:

0celo7

Theorem in Hawking-Ellis

ACuriousMind

20 mins ago, by 0celo7

@ACuriousMind Let's call the boundary of the null cone $C$. I don't know that $C$ has a null normal vector.

0celo7

@ACuriousMind A null normal vector. I know it has a normal vector.

ACuriousMind

17:15

Uh...my point there was that from your "the integral curves of l are null geodesics" it follows that l is null.

0celo7

@ACuriousMind Yeah, so?

I don't know that there is such an $l$!

The $l$ there is the normal to $\Sigma$.

ACuriousMind

What? You said "If a null hypersuface has a normal vector l" and you just told me there's a HE theorem that says the surface has a normal vector.

0celo7

But I don't know that $C$ is null!

I don't know anything about $C$ other than it is the boundary of the future of a point.

ACuriousMind

Ohhhhhh

then I don't see the point of stating this at all, but fine :P

0celo7

Theorem in HE: $C$ is generated by null geodesics. Theorem in HE: $C$ is achronal.

These facts should be enough to guarantee it's a null hypersurface.

But what bothers me is that I need the second theorem.

It would seem that if something is a bunch of null geodesics it should be a null hypersurface.

@ACuriousMind $\Sigma$ is null and $l$ is the normal $\Rightarrow$ integral curves of $l$ are null geodesics that lie in $\Sigma$. I want to know if one can make that into a $\Leftrightarrow$

@ACuriousMind Does that make sense? Is it a reasonable conjecture?

ACuriousMind

17:24

@0celo7 That's pretty much what the "generalization" of my Minkowski example would have been, so is it not obvious that I think that's reasonable?

0celo7

@ACuriousMind Your mind is not obvious to me, no.

@ACuriousMind PSE question or is this too homework-y?

0celo7

17:40

Should be a nice exercise to verify that...

Hari Prasad

0

Q: Energy and Momentum: Impact

Boat A has just lost another race to Boat B. In a fit of rage, the driver of Boat A (mass = 500 kg) turns his boat towards Boat B (mass = 450 kg) which is at rest as the driver gloats over yet another victory. If Boat A has a speed of 25 m/s just before impact, what are the velocities of the t...

homework-and-exercises kinematics energy-conservation momentum collision

see the answer to this question

Transcript for