The h Bar

0celo7

12:51 AM

(removed)

skill patrol

1:56 AM

(removed)

Bernard Meurer

3:05 AM

The hell's going on?

0celo7

@BernardMeurer Abuse.

Bernard Meurer

@0celo7 Ah, usual

vzn

4:09 AM

hi all, there is a large contingent in here who are aggressively pursuing higher edu & advanced readings in physics, has anyone looked into coursera courses for physics? surprised it hasnt been mentioned. its very good for cs stuff, an initial focus, but lately expanded into general areas eg physics. was looking for GR related stuff which is a big topic in here, found some. (may be better for SR.) some have complained about book prices in here & heres some alternative. coursera.org

John Rennie

5:51 AM

"multiple dementions" - a Freudian slip? :-)

-4

Q: is this possible

I'm not a physicist and I only have a basic understanding of science but I was wondering if gravity can move between universe in string theroey wouldn't it be possible to prove multiple dementions buy waiting untell gravitational waves to be found that has no point of origin and therefor must be ...

string-theory

skill patrol

:D

combined with a well thought out and informative title :P

2physics

6:37 AM

@vzn thanks for mentioning it. coursera is very helpful and also www.khanacademy.org and www.ocw.mit.edu/courses are of the same kind.

khanacademy.org and ocw.MIT.edu

user_na

@JohnRennie How does one put links to questions into the chat like this?

John Rennie

Write something like:
[this is the link text](http://www.etc.com/)

2physics

72

Q: Adding links to Comments

While giving comments many times I wanted to add links to my comment. But instead of using some short hand like Read "this" (this is a link in this case) I had to write Read "http:\\www.somelink.com". I tried ctrl+L to insert a link to my comment but it was of no use. I have seen people inserti...

support comments hyperlinks

user_na

6:52 AM

@2physics thumbs up. This is the correct way to show my lack of own research in the meta pages, I guess.

2physics

@user_na lol. that's alright :)

skill patrol

It's all apart of the learning process :-)

agree

@JohnRennie

@Bhavesh Hi.

6:59 AM

hi

John Rennie

The IXS Enterprise uses a conventional Alcubierre drive.

Bhavesh

so there isn't any difference?

John Rennie

No

However there is lots of tweaking of the Aclubierre metric that can be done to improve efficiency.

Bhavesh

k

John Rennie

The IXS Enterprise design is more about fine tuning to use a reasonable amount of exotic matter.

The initial designs requirted a mass of exotic matter greater than the mass of Jupiter! :-)

Bhavesh

7:01 AM

do we have exotic matter?

John Rennie

@Bhavesh No, as far as we know it does not exist.

And there are good theoretical reasons for suspecting it cannot exist

Bhavesh

so everything is hypothetical?

John Rennie

Yes, the Alcubierre drive is just an exercise in having fun with mathematics.

Bhavesh

we don't have concrete proof whether this warp drive can work practical

John Rennie

But that's OK, as these sorts of studies can help us learn more about the way GR works.

Bhavesh

7:03 AM

Gamma Rays

?

John Rennie

@Bhavesh not exotic matter. In fact not matter at all since gamma rays are just high energy photons.

Bhavesh

yeah i know that

we need negative energy or matter?

John Rennie

The terms exotic matter and negative matter/energy mean the same thing. And yes, any form of FTL drive requires exotic matter. Hawking proved that in his paper on chronology protection.

Well, he proved that closed timelike curves require exotic matter, but since any FTL travel implies the existance of CTCs his theorem covers FTL travel as well.

Bhavesh

CTC?

John Rennie

Closed Timelike Curve

Bhavesh

7:08 AM

oh

ok, but we can get negative energy in small amounts, right? (I mean the casimir effect)

John Rennie

It isn't clear whether the Alcubierre drive would work even if exotic matter were available. There are arguments based on some ideas in quantum gravity that it could not work.

I have to disappear for a while now (work!)

Bhavesh

ok thanks

@JohnRennie thanks for lcearing my questions :D

t*clearing

Slereah

8:32 AM

Whaaaat

I must disagree with several things

Chronology protection and FTL are unrelated

Although IIRC Krasnikov did prove that FTL requires energy condition violation

Also we can already produce exotic energy

At least in two forms

John Rennie

Any form of FTL allows CTCs. Yes?

Slereah

It's complicated

All metrics for FTLs can be deformed into CTC metrics (so far), but chronology protection implies that you can't actually perform that deformation

But this doesn't imply that FTL violate the NEC

The argument for that is more complicated

arxiv.org/pdf/gr-qc/0204038.pdf

When I say that all of them can be deformed into it

It may just be a coincidence

there's no theorem for it

We only have like two such metrics

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

8:49 AM

@yuggib o/

yuggib

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ \o

how are you?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Fine thanks, how are you doing? @yuggib

yuggib

fine! I'm in Japan B-)

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

cool

yuggib

yep

Slereah

8:52 AM

Will u go to Yukawa's grave

yuggib

no, I'm not in tokyo/kyoto (I don't know where he's resting exactly)

;-P

Slereah

Will you ask the government for apologies for Nanking

yuggib

?

I'm not chinese

actually, in that very period my country was ally of the japanese...T__T

Slereah

No need to ask for apologies

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

you ask some of the most tangential political questions of anyone I know :P

Slereah

9:00 AM

My AMA is gonna be all about the truth of 9/11

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

cool, can't wait

A = anything

Slereah

Wake up sheeples!

John Rennie

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ you're a Pink Floyd fan?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Yep @JohnRennie

John Rennie

Cool :-) I was a huuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuge Floyd fan in my youth though I've grown old and boring since.

user116211

9:03 AM

@Slereah Just send an e-mail to Ron Maimon; he has a great theory on it.

user116211

@yuggib: o/

John Rennie

These days the only group I support big time is Hawkwind and a spin-off group called Hawklords.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

They are second to the raiders @JohnRennie

John Rennie

Raiders?

user116211

@JohnRennie: Do you follow cricket?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

9:06 AM

Oakland Raiders @JohnRennie nfl football

John Rennie

@MAFIA36790 not really. If there's a test match on I'll make a point of checking the scores regularly, but I can't honestly say I'm a real fan.

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I thought their latest album was rubbish

user116211

@JohnRennie oh. so, you must follow Ashes; that is one of the great test series in a season.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Not even rugby? @JohnRennie

John Rennie

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ again, when the Six Nations is on I'll check the results. But I only rarely watch a match, and even then only on the TV never at the ground.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Pink Floyd's music will never grow old for me.

Slereah

9:10 AM

Last known photo of Pink Floyd

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:D

They are timeless.

John Rennie

I think Dark Side of the Moon is genuinely a seminal album that really changed the music world. But Wish You Were Here and Animals mainly consisted of tracks written at the same time as DSOTM. So basically they did nothing new until The Wall.

And by the time they did The Wall the band was disfunctional.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Agreed.

John Rennie

The Wall was in effect a Waters solo album, though the best track on it (Comfortably Numb) was cowritten with Gilmour.

The albums following The Wall were, well, just OK.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Have you seen them live?

John Rennie

9:15 AM

Yes.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

In England?

John Rennie

I don't think they're that special live. When I see a live band I like to really get into the spirit i.e. dance about like a loon and scream and shout. The Floyd are more a band you sit down and listen to.

The very, very best live show I've been to was Ozzy Osbourne's first solo tour in (I think) 1979 or 80.

The atmosphere was just incredible

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Ozzy? Now that is interesting

John Rennie

I've seen Black Sabbath, but by the time I was old enough to go to their shows they were already well past their best.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Heavy metal at its prime.

John Rennie

9:19 AM

@Slereah: if I put the stress energy tensor for a point mass into the Einstein equation is the solution the Schwarzschild metric?

Slereah

Yes

But then

You have to use generalized functions

Things become complicated

John Rennie

@Slereah because it's a point mass i.e. a distribution?

Slereah

Pretty much

John Rennie

I was thinking about the perennial question of whether a moving mass generates an increased gravitational field.

Slereah

The sET is gonna be like $\approx v^a v^b \delta(x)$

John Rennie

9:22 AM

If you write $G = T$, where $T$ is the stress-energy tensor for a point object then you can get a moving object just by applying the same Lorentz boost to both sides. Yes?

Slereah

You can apply the boost directly to the metric, too

John Rennie

Well we know the metric is Lorentz covariant

I was trying to develop an argument that starts with a moving particle and shows that the metric is the same as for a stationary particle by Lorentz boosting into the rest frame of the particle.

Slereah

Probably a lot of work for not much

I think it's the same as boosting the metric directly

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

9:51 AM

Muhammad Ali public memorial service information

►

Ali is back home :'(

John Rennie

10:15 AM

Reincarnation? Really?

-4

Q: What are your views on this explanation of reincarnation using physics?

https://www.youtube.com/watch?v=GT8iVYfg9eM "Skeptical scientists keep dismissing the concept of reincarnation based on the fact that there is no need for a soul. However, one of the most skeptical and smartest scientists ever, Carl Sagan, the author of the best book on skepticism 'The Demon Hau...

entanglement

While it's true I'm happy to answer questions from non-physicists I expect those questions to at least make sense.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

"entanglement" indeed ;-)

Secret

10:34 AM

My two cents on that question:
There are many issues postulating that entanglement as the essence of a soul
1. Entanglement, as far we knew, is way too fragile to exist for more than brief periods of time
2. Saying that some atoms in one neuron is entangled with another in explaining a soulmate seemed to be a gross simplification of the complexity of the human brain. How can you be sure that basically such a massive dynamical system will not drown out any possible quantum effects that can manifest. Also with so many entangled pairs (in the billions) we expect them to last less than an alto…

John Rennie

@Secret I can't of course prevent you from taking the question seriously, but it does make me wonder if you really haven't got anything better to do :-)

Secret

My way of handling cases when scientific concepts is being mystified to explain esoteric concepts is not just to treat it a nonsense (and as it turns out, they often are), but to explain why it is nonsense and how their proposals will fail by highlighting the key areas that will fail.
(well most of the time, such analysis mostly benefits myself in convincing myself why they are nonsense/not testable anyway, as ever since I become an agnostic (although still mostly in science), I are aware myself becoming quite suceptible to esoteric concepts thus I have to stay alert)

Especially if you actually look closely at art, science, esoterics, religion, business, culture, the way they use their jargon shares a very simialr structural pattern.

It is as if they are using completely different terms to describe the same thing, or that this similarity is basically highlighting how we humans use language

> but it does make me wonder if you really haven't got anything better to do :-)

The good thing is that typing the above wall of text does not take me much time, thus it is a reasonable investment

@0celo7 Actually, screw what I and Acuriousmind talked about yesterday, $b$ is ultimately found to be $0$ because
$$b=b+0=0^2+0=0(0+1)=0*1=0$$

Now I am suspecting that as long both multiplicative and additive identities are present along with distributive law, it will guarentee there exists a unique multiplicative absorbing element 0. Wikipedia on absorbing element also said that if 0 ever absorb itself, then that algebraic structure can only have 0 as th absorbing element

Slereah

10:52 AM

Hm

I think I found a paper with a solution for bipolar coordinates, but...

jstage.jst.go.jp/article/ieejfms1972/98/12/98_12_607/_pdf

Secret

I cannot read japanese

Slereah

Yeah me either

innisfree

Has the AMA happened already?

Secret

not yet, the next one up is Slereah's

16

Q: Physics Stack Exchange "Ask me anything" (AMA)

Purpose The user base of Physics Stack Exchange is rather diverse in terms of cultural background, career progress, field of expertise, and personality. While the discussion in hbar is lively, it is mostly focused on physics, with some side conversations on a variety of topics from cooking to fa...

discussion chat-events

(***)

innisfree

We've had one already?

@Slereah do you want to do a PhD?

Secret

11:03 AM

Attempt to break the proof
Let elements a,b,c,d,e,f,g,h,i,0
Then: $$0^2=b=c+d=0(e+f)=0g=h$$
$$bg=0^2g=0(0g)=0h=i$$
Now to determine what constraints do a to h need to obey to ensure $b\neq 0$

(Above showed my engineer attitude to maths problem (try to rig it until the desirable outcome is acheived, or die trying))

Slereah

@innisfree that I do

innisfree

What's stopping you? You seem more capable than heaps of first year PhD students I encountered.

Slereah

I tried

In France tho, you need to be paid to do a phd

So people arequite picky

And not too keen on theory

user507974

hey slereah, i saw @DanielSank and his question on stochastic processes

Secret

$$b\neq 0$$ implies
$$c+d \neq 0$$
$$0(e+f)=0e+0f\neq 0$$
$$0g \neq 0$$
$$h \neq 0$$

Line 2 suggest there cannot be 1) additive inverses, 2) c,d are not zero sum elements
Line 3 suggest that 0e, 0f are not zero sum elements

user507974

11:08 AM

what do you need to know before you can read up on stochastic processes and stochastic calculus

innisfree

Did you want to work on a very specific topic?

Secret

Hmm
https://en.wikipedia.org/wiki/Zerosumfree_monoid

books.google.com.au/…

0

Q: What if the speed of light was not contant (Luminiferous aether is real)

How would the laws of physics change if the speed of light was not constant and the Michaelson-Morley experiment was instead proven to be true? Obviously relativity would not be true, but on a small scale, what would the differences be? Would Maxwell's equations hold up properly? Would electric...

science-based hard-science scientific-development

0celo7

11:46 AM

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Oh shit we were discovered

Secret

Line 4 suggests multiplicative identity cannot exist
$$0g\neq 0$$
As otherwise we have at least one case where 0 is absorbing, and we will be screwed because the proof will survive

Slereah

The last two characters look identical

0celo7

they are not

one is psi, the other phi

get glasses

Slereah

Still

Bad typesetting

Proper psi is $\psi$

I don't want to sound racist but those coordinates look alike to me

0celo7

@Slereah : I grabbed Visser from my storage unit

where is that proof of the toroidal spacetime energy divergence thing

Slereah

11:59 AM

The proof is the divergence for a simplified wormhole spacetime

0celo7

where is it

Slereah

Chapter 22

And 23

Eq. 23.5 for the important result

0celo7

@Slereah what the heck am I looking at here

Slereah

Read chapter 20 through 23 for details

Secret

So it does seems, all it takes is additive identity and distributive law (and that the binary operation is only defined as the juxaposition of the two elements that form its argument) to guarentee the existence of an absorbing element for the smallest size semiring

The existence of even just one neutral element (an element a such that a+b=b where b is some elements in the proper subset of the algebraic structure in question) is enough to lock up nearly half of the entries in the + Cayley table. The existence of an additive identity constrains all elements in the + cayley table.

Also $0+0$, because of distributive law, will determine the value of $b+b$

Most importantly, any such structure must not also have multiplicative identity elements (or at least, cannot exist multplicative neutral elements e for 0 such that $0*e=0$) else it will immediately collapse into the trivial ring

typo: all elements involving 0 in the + cayley table

Secret

12:31 PM

Since a multiplicative identity cannot exist in a distributive structure without implying the existence of the unique absorber 0, division by zero algebric structures MUST be non distributive, or has binary operations defined nontrivially (not as an element which is the juxaposition of its two arguments)

Nontrivial semirings without absorber can exist with distributivity and identities only if the binary operation is nontrivial, such as this one:

4

Q: Examples for almost-semirings without absorbing zero

What is an instructive example of a set $X$ equipped with two monoid structures $(X,+,0)$, $(X,\cdot,1)$, such that $+$ is commutative, the distributive laws hold, but $0 \cdot x = 0$ or $x \cdot 0 = 0$ do not hold? Notice that in case these two absorbing laws hold, one calls $(X,+,0,\cdot,1)$ a...

ring-theory examples-counterexamples

(proofs will be avaiale later once I figure out how to prove them)

FrancescoS

Hi! I am studying graph theory applied to feynman diagrams and amplitues. There is a chapter in the book relating to find divergent subgraphs. I call $G$ the graph and $l(G)$ it's internal lines. At a certain point, the author says "we can always restrict ourselves to consider subgraphs which are regular (no isolated vertices) and which furthermore correspond to subsets of internal lines of G. Hence there are exactly $2^l(G)$ such subgraphs in G. "

My question is: why $2^l(G)$? I tried to compute all possible subgraphs of the graph with two vertices and three legs for each one contracted one-by-one . And, I see there are 7 subgraphs (including the graph itself)

0celo7

1:12 PM

@ACuriousMind I'm still confused. We have our unit vector field $v$ with its zero at the origin, and we take two spheres $S_\epsilon$ and $S_\delta$ that work in the definition of the index. Sure $S_\epsilon$ and $S_\delta$ are diffeomorphic, but how does that induce a homotopy on the maps $v:S_\epsilon\to S^n$ and $v:S_\delta\to S^n$?

In particular, these maps have a different domain!

Let $f:S_\epsilon\to S_\delta$ be an orientation preserving diffeomorphism.

Let the map $v:S_\epsilon \to S^n$ be denoted $v_\epsilon$ and let the map $v:S_\delta\to S^n$ be denoted $v_\delta$. Then I'm willing to believe that $v_\delta \circ f$ and $v_\epsilon$ are homotopic and have the same degree.

But I'm still not sure why $v_\delta$ and $v_\epsilon$ have the same degree.

@ACuriousMind Oh wait, $\mathrm{deg}(f)=+1$, so if $v_\delta\circ f\sim v_\epsilon$ then $\mathrm{deg}(v_\epsilon)=\mathrm{deg}(v_\delta \circ f)=\mathrm{deg}(v_\delta)\mathrm{deg}(f)=\mathrm{deg}(v_\delta)$?

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Now don't ever tell me there isn't someone out to get us.

I've done literally nothing wrong.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

It's not about right or wrong anymore @0celo7

Big Bother is watching

QED

0celo7

I wonder if there's a special mod taskforce for us @Sᴋᴜʟʟᴘᴇᴛʀᴏʟ

God they must hate us

Isn't it tiring?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

nah, just lay back and be calm and cool

0celo7

Calm, cool and collected?

I can do that.

It's called not throwing a fit after being banned for a month for no reason.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

1:27 PM

Been there, done that.

Got a T-shirt :P

0celo7

You've gotten a month before?

What the f**k

What did you do

Why did you remove that

The mods can still see it

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Good **

Now you're learning

cya

0celo7

bye

@ACuriousMind Ah! Assume WLOG $\epsilon<\delta$ and let $\eta:=\delta-\epsilon$. Consider the map $g:S_\epsilon\times[0,1]\to\Bbb R^{n+1}$ which for fixed $t$ is the diffeomorphism $g_t:S_\epsilon\to S_{\epsilon+t\eta}$.

Thus it it smooth in $t$ and $g_1=f$ from above.

Then we have a homotopy $F:S_\epsilon\times[0,1]\to S^n$ between $v_\delta\circ f$ and $v_\epsilon$ given by $F(x,t)=(v\circ g_t)(x)$.

@ACuriousMind I would appreciate you letting me know if this is alright, I came up with this myself (I'm impressed with myself if it's correct)

I guess I should say $g_t$ is the diffeomorphism obtained by scaling the vectors pointing from the origin and thus it's smooth in $t$.

user116211

1:58 PM

A $r$-neighbourhood is always a uniform neighbourhood.

user116211

Hmm, why? I'm not getting that ;/

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