The h Bar - 2015-07-17 (page 1 of 2)

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user54412

1:11 AM

@Danu Here's my earlier statement about surface gravity, recast more in line with your notation. Consider a fixed charge Q and varying mass M. What do T and $(\partial T/\partial M)_Q$ look like?

user54412

user image

user54412

The region between $q = 1$ and $q = \sqrt{3}/2$ has $(\partial T/\partial M)_Q > 0$.

user54412

This derivative passing through 0 is equivalent to C diverging, as you can write C as something over this derivative with enough chain rule.

user54412

@Danu Another way of looking at this (probably related to all this phase transition stuff) is that normally we expect $T \to \infty$ as $M \to 0$. But as soon as there's any charge at all, $T \to 0^+$ as $M \to Q^+$ along a constant-$Q$ line.

user54412

user image

user54412

1:26 AM

That there is an extremum in T is in retrospect no more surprising than that $T \to 0$ as the hole becomes extremal (which is, admittedly, weird).

1:51 AM

@ChrisWhite I recorded a Let's Play of Morrowind

@0celo7 Link pls

Melee combat in this game is literally impossible

it's a proof of concept

not recorded in native quality

You're like a psychic

Also: my data science thing that I've spent the last 2 weeks on passed their expectations, so I'll be moseying on to the 2nd test

@0celo7 Lol, <30 sec in I see you ignored the Take All button & took dude's stuff one-by-one

@KyleKanos I wanted to see what he had...

And you're using a dagger

1:56 AM

i have nothing else...

@ChrisWhite how the hell do i hit people

Well I suppose you could always buy a sword

Or steal one (before or after killing someone, ofcourse)

That was like the first thing I did

Next step was getting a bow & arrow and some armor

This really is bringing back memories

I ought to get this on my Linux box

look at 8:15

fucking ridiculous

I'm at ~3

Wait...is this right at the beginning?

I have like one hour of game time

I've been fooling with mods this whole time

Okay

2:01 AM

I'll buy a sword.

I have 500+ gold

No man, don't buy one.

Find a schmuck with a sword & kill him for it

Or use some magic & blow up some daedric weaponry

There are also swords hidden all over the place

Lol you got rekt

3

Oh that made me laugh

I bought a sword and killed him with two hits

user54412

@0celo7 Practice on mudcrabs :)

2:34 AM

@ACuriousMind holy shit I randomly fired up Skyrim and my armor mod works now!!!!!

3:13 AM

This is hard to read: wikipeetia.org/Fluid

2

3:26 AM

We are not amused by your "amusing parody of Wikipedia."

Try this @KyleKanos

>8(

2 hours later…

5:10 AM

@ChrisWhite One can get it from rewriting the metric and comparing to Rindler (identifying something with the Rindler acceleration $a$) or, after analytic continuation and some more substitutions, to polar coordinates and force a periodicity in one of the coordinates. Both are directly coordinate related

@ChrisWhite This is kind of difficult (to interpret) because "phenomenologically" $q$ is what one should fix, not $Q$. A plot of just $\partial T/\partial M$ will look weird because at some point you'll go $|q|>1$ and stuff

So I didn't plot $T(M)$ or its derivative

@ChrisWhite I'm not sure your derivative is taken correctly, as $\mu$ of course also depends on $M$

(or did you take this into account, and just not write it out?)

user54412

5:28 AM

There's a lot of scratch work not written (though something still might be wrong)

user54412

I stand by my claim, though, that there should be a connection between $(\partial T/\partial M)_Q$ vanishing and C diverging

user54412

Besides, why should $q$ be fixed rather than $Q$? Certainly I can take a charged black hole and add neutral matter to it?

user54412

The reverse might be more difficult. Does Hawking radiation preferentially deplete charge imbalance somehow? On the one hand it seems like it should, on the other it feels like we should be constrained by the particles that actually exist.

5:44 AM

Send Professor Hawking an email and ask :P

Sorry, bad joke^

3 hours later…

8:32 AM

Hi @Hippalectryon

Did you read that article?

1 hour later…

9:42 AM

@skillpatrol which one ?

afk

10:10 AM

This one @Hippalectryon

Yeah I read it

What did you think?

It doesn't say anything new afaik

Doesn't it give you any ideas how to help yourself during exams?

Not really

10:15 AM

Automaticity could help with learning how to be careful, no?

Automaticity doesn't enable one to learn to be careful. It just enables one to do things a more automatic way

Indeed, automatically learn to be careful at the beginning.

Step 1 Read the problem carefully and think about its meaning.

That^ should be a "knee jerk" reaction

As I mentioned earlier :-)

@skillpatrol Btw do you have any idea on the problem I posted yesterday ?

13 hours ago, by Hippalectryon

Hi! I'm studying a usual linear accelerator for electrons and I'm trying to get the force that corresponds to the power radiated ($P=m\tau \ddot{x}^2$). I've tried calculating the associated work which gives me $W=m\tau\int_0^T\ddot{x}^2\mathrm{d}t=\int_0^T\vec{F_r}\cdot\vec{\dot{x}}\mathrm‌{d}t$, how do I continue ?

I would suggest posting it on the main site.

Just ask for a hint.

Sounds too homeworky for PSE. And it should be an easy question :/ I'm just somehow stuck

Which is why I'm asking in the chat

10:33 AM

Where have you used Work = Force * distance @Hippalectryon?

@skillpatrol $\int_0^T\vec{F_r}\cdot\vec{\dot{x}}\mathrm‌{d}t=\int_0^d\vec{F_r}\cdot\vec{ \mathrm‌{d}l}$

...and what about Power is the amount of Work done per unit time @Hippalectryon?

Isn't that what I used at the beginning ? $W=\int_0^T P(t)\mathrm{d}t$

That tells you what Work is equal to, right?

yes, in terms of an integral of $\ddot{x}^2$

10:43 AM

You are asked for the Force that corresponds to the power.

F = P(?)

ugh ? How can a force equal a power ?

I'm asked to find a force that can be associated with the radiation

Basically, when you add that force to the "radiation-less" model, you get the model with radiations

Had the acceleration being constant in the radiative mode I could indeed have used F=PT/d but it's not the case here

Can't you ask your teacher?

I don't see my teacher until next year

Do you have a textbook?

Nope, only notes

10:53 AM

How about a library?

What would I do there ?

The exercise doesn't come from a textbook or anything

There must be some substitution...

I thought integrating by parts could help but that square is troublesome

I basically want to transform that $\ddot{x(t)}^2$ into $f(t)\dot{x(t)}$

How about a change in variable?

:O $u=\dot{x}$

Wait no that doesn't work q_q (it does yield a result but no $\dot{x}$ appears)

10:59 AM

Maybe even F =ma is used somewhere

I gotta go :(

Ok

Sorry

Nah you tried to help me, thanks ;-)

Later pal

See you

11:17 AM

@KyleKanos I'll try my hand at streaming tonight so you can lol some more at my shittiness

11:51 AM

@xXx_based_dolan_xXx nice user name

1 hour later…

1:00 PM

@0celo7 Hmm. EVO is tonight (11a to 11p Eastern), so that might be tricky :/

1:12 PM

@KyleKanos you really think I can get a video and audio feed working in one sitting?

AFAIK, it's pretty simple on Windoze

Do you know if twitch streaming is free?

I believe it is free

Can't imagine too many people streaming if they had to pay for it

They also have a list of common broadcast software: twitch.tv/broadcast

Hello, are here any chat netiquette I should know, or can I just ask my question?

@Luke Just ask

If someone's interested, they'll respond. If not...well try again later when others are on?

1:21 PM

I am doing an exercise in basic quantum mechanics. I should show that Π := X - t/m P is a conserved quantity. First Idea was that it has to commute to the Hamiltonian if I recall correctly

but that led me to problems regarding the dimensions

I ended up at showing 0 = ΔX – XΔ

which does not make any sense to me in 3D

because the laplace operator maps to 1D, but X multiplies by the position vector yielding something 3D

Correct, commutation with Hamiltonian means conservation--the relation is dA/dt = [A, H] so [A, H]=0 means dA/dt=0

I forgot to mention it was about a free particle, so H = p²/2m

So you'll have [X - aP, H] = [X, H] - a[P, H]

(a=t/m, obviously)

that's how I started, yes

and since H = b P², [H,P] = 0

So your issue is just on the first term?

1:32 PM

indeed. ⇒ [X,H] = 1/2m(XP² - P²X)

which is, denying coefficients, the problem with Δ I explained.

Well don't forget that [A, BC] = [A, B]C + B[A, C]

Then recall that [x,p] = ihbar, no?

Kyle doing quantum mechanics?

⇒ [X,P²] = [X,P]P + P[X,P]

@0celo7 I've been known to do that sometimes.

hm, that seems rather trivial @KyleKanos, I wonder how I got to my conflict there.

but thank you

1:35 PM

Surely

@0celo7 I also have answered a few QFT questions too ;) (sadly, one is mislabeled (purely QM, no QFT) and one was really just a math question

@KyleKanos I don't have a dedicated USB mic

Will the mic on my phone earbuds work well enough?

@0celo7 You can't use built-in one?

@0celo7 Possibly.

@KyleKanos maybe

Not at home, can't check

I'll use whichever has the best quality I guess

I hope streaming doesn't bring me below 30 fps

I have no problem with 30

@KyleKanos huh

@KyleKanos I'll install a female nudity mod as an incentive

1:54 PM

@0celo7 Meh, just mod me & I'll be content

(i.e., make me moderator, not modify me)

There will be maybe two people in there

So?

It's going to be the official H Bar vidya stream

@KyleKanos I don't want you banning the other person

I probably wouldn't do much with moderator powers in a chat

then why do you want them?

be warned: now that I got all of my Skyrim mods to work, I'll probably stream that

2:09 PM

@0celo7 S&Gs

@KyleKanos pls define

Shits & Giggles

my precalc teacher said shiggles

"why do we have to square 234?" "shiggles."

holy shit I had no idea this chat was a thing

no cussing

there are children in here!

2:13 PM

eh about time children learned colorful vocabulary ahead of time

2:47 PM

@0celo7 lol :-P

@DavidZ it's no laughing matter

3:18 PM

@KyleKanos Do you have access to journals.aps.org/pr/abstract/10.1103/PhysRev.119.2082?

Yes

How long is it?

8 pages

Can you please see if he gives something like velocity = hyperbolic functions?

Yes he does

3:23 PM

There's something I've always been meaning to ask

Is there any possible way you could share how he comes up with that?

@0celo7 $U^\mu=(\cosh(\alpha\tau))L^\mu+(\sinh(\alpha\tau))M^\mu$

@FenderLesPaul "how does X troll have so much rep" cannot be answered

@KyleKanos I know that, I'm at a loss trying to derive it :/

It's the "obvious first integral" of $DA^\mu/d\tau=\alpha^2 U^\mu$

with $\alpha=\rm const$

@FenderLesPaul I don't think he's insane, he just holds rather non-standard positions on things

$\alpha^2=A^\mu A_\mu$?

3:25 PM

Are you going to write the Putnam? @0celo7

@skillpatrol no

I'm bad at math

@0celo7 section 6.2 of MTW

@FenderLesPaul is that in curved spacetime

@0celo7 Yes, he does say that

@0

oops

3:27 PM

Also, with his notation, $DA^\mu/d\tau\equiv {A^\mu}_{;\nu}$

@0celo7 ah no it's in flat space-time, sorry

@KyleKanos Not $U^\nu A^\mu{}_{;\nu}$?

does he say how to get that equality?

No, the line is $U^\mu=\frac{dx^\mu}{d\tau},\quad A^\mu=\frac{DU^\mu}{d\tau}=\frac{d^2x^\mu}{d\tau^2}+{\Gamma_{\nu\sigma}}^\mu\fra‌c{dx^\nu}{d\tau}\frac{dx^\sigma}{d\tau}$

that's all standard

I need the $DA/d\tau$ line

I already said it: $\frac{DA^\mu}{d\tau}=\alpha^2U^\mu$

I was just pointing out that the $D/d\tau$ was the covariant derivative; wasn't sure if you knew that

3:33 PM

@KyleKanos yes, that's standard

@KyleKanos no clue how to get that...

@0celo7 TIL...

unless

Oh shit, I lied to you. $\alpha^2=-A^\mu A_\mu$.

He says,

> Thus, the analog of a plane curve of constant curvature will be characterized by the diff. eq.'s
$\beta^\mu=\frac{D}{d\tau}\left(\frac{A^\mu}{\alpha}\right)-\alpha U^\mu=0,\quad-\alpha^2=g_{\mu\nu}A^\mu A^\nu$

3:50 PM

hmm

how the heck do you get that

$\beta^\mu=i\Omega B^\mu$ where $\Omega$ is the torsion (or second curvature) and $B^\mu$ the unit binormal (or second normal)

Does he say anything about calculating $DA/d\tau$?

Nope, just that it has an obvious first integral

fuck him too

ugh

any idea how to derive this?

Given the definition of $A^\mu$, he's saying $\frac{D^2U^\mu}{d\tau^2}=\alpha^2U^\mu$

Sans the 4-vector bit, that's really an $x''=\omega^2x$ equation with known solutions of hyperbolic functions

Isn't it?

3:56 PM

yes

I don't know how to get that equation to begin with

18 mins ago, by Kyle Kanos

> Thus, the analog of a plane curve of constant curvature will be characterized by the diff. eq.'s
$\beta^\mu=\frac{D}{d\tau}\left(\frac{A^\mu}{\alpha}\right)-\alpha U^\mu=0,\quad-\alpha^2=g_{\mu\nu}A^\mu A^\nu$

well I don't know how to get those either :P

Apparently there's an analogy to the Frenet-Serret equations

@KyleKanos do you see any quick solution to this?

I'm somewhat familiar with those

3:58 PM

6 hours ago, by Hippalectryon

13 hours ago, by Hippalectryon

Hi! I'm studying a usual linear accelerator for electrons and I'm trying to get the force that corresponds to the power radiated ($P=m\tau \ddot{x}^2$). I've tried calculating the associated work which gives me $W=m\tau\int_0^T\ddot{x}^2\mathrm{d}t=\int_0^T\vec{F_r}\cdot\vec{\dot{x}}\mathrm‌{d}t$, how do I continue ?

@KyleKanos would it fall under fair use to share the relevant pages?

@skillpatrol Quick, no. It's actually a kinda complicated thing

Thanks for replying

:-)

Well I guess not that complicated...this guy did it with all the units in it

wikipeetia.org/Streng%20thoery

reads like the typical PSE post

4:04 PM

@KyleKanos great thanks :D

dude streng theory is so dope

@0celo7 According to this site, whole pages probably would not fall under fair use. I can make whole quotes though

@KyleKanos I need enough to reconstruct the proof. As it stands I'm not very far.

@KyleKanos How does he define the binormal and the torsion?

and which one is zero so $\beta=0$?

and what is the random $i$ doing there?

@0celo7 He doesn't, he just states them

so how does he show $\beta=0$

4:12 PM

Via the claim

> The true generalization of the characteristics of straight lines and plane curves are obtained not by requiring the scalar curvature or torsion to vanish, respectively, but by requiring the curvature or torsion vector to vanish component-wise.

The torsion vector being $\Omega B^\mu$

and that's not defined anywhere?

It is: $$\Omega B^\mu=\frac{D}{d\tau}\left(\frac{A^\mu}{i\alpha}\right)+i\alpha U^\mu$$

wtf

how the hell does he get that??

By analogy to $$\mathbf{t}=d\mathbf{r}/ds,\quad\kappa\mathbf{n}=d\mathbf{t}/ds,\quad\omega\mat‌hbf{b}=d\mathbf{n}/ds+k\mathbf{t}$$

He references a book publishe in 1949 by Synge called "Tensor Calculus"

published*

I have no idea where the factors of i come from in that analogy though

4:19 PM

Those equations, btw, are standard formulas of the differential geometry of twisted curves in Euclidean 3-space

@KyleKanos what is k?

Sorry, that's a kappa at the end there

Which is curvature

wish I had my curve geometry book with me...

@0celo7 is this related to a project or just for fun?

@FenderLesPaul I fucked up an answer and I have to learn about accelerated motion in curved spacetime to fix it

one more month and I could go to the university library

4:22 PM

@0celo7 ah ok

Are you excited?

the chat notification sounds terrifying holy shit

yes

I turn it off

Dammit...Brandon's going to beat me to gold badge on LA :(

@0celo7 are you from the US, if you don't my asking?

4:24 PM

He's got a 250 review head-start

@FenderLesPaul read my bio

should say there

Click on the speaker in the upper right hand corner.

Well it's lunch time...I'll bbl

@ACuriousMind you're on the top GR users list, ahead of me now

nuclear engineering; what the fuck?

4:25 PM

@KyleKanos k

sorry :p

the chilluns!

you ruin their minds

ah Tennessee

cool

@FenderLesPaul read: applied nuclear physcis

I actually did a double take when I saw it said nuclear eng given your interest in GR haha

4:27 PM

I'm taking courses that allow me to branch out, should the need arise

but that's cool dude at least you get to study GR for fun and at the end of the day secure a job

I'm just not interested in a career in academia

that's fair

so you start this fall then?

yes

cool

4:29 PM

3

A: How long can a particle survive inside of the horizon of a black hole?

A particle cannot survive in region II indefinitely. Indeed, there does exist an upper bound on the survival time of the particle. As soon as the particle crosses the horizon into region II, its remaining proper time until it reaches $r=0$ is bounded by $$\tau_\mathrm{max}=\frac{\pi GM}{c^3}$$ W...

^^is incorrect

It is possible to strategically fire rockets and live marginally longer.

I want to write a mega answer explaining the paper John Rennie linked on the OP

@KyleKanos those are the 3-space equations, yes

however, $\vec b$ is defined using the cross product

Is that true? I'd look forward to reading it; the answer of $\pi M$ is one that was given to us in a HW problem at one point so I'd be curious as to when it can be circumvented

no clue how to define it in 4 dimensions

Presumably using the levi-civita symbol contracted with the 4-velocity of the comoving frame

He doesn't define it in the paper though

@FenderLesPaul the theorem from HE I applied (a very rigorous statement of the heuristic that geodesics are maximums of proper time) does not apply! It only applies curves connecting the same spacetime points, that means both the space and time coordinates of the points have to be the same. However, it is possible to accelerate in just the right way and land further along in coordinate time. Since you hit the singularity at a point in spacetime that cannot be reached by a geodesic, it's perfectly...

...fine that the geodesic is shorter

@FenderLesPaul Cornell?

I see; certainly I'd like to see your detailed answer if/when you write it :)

Indeed

4:36 PM

arxiv.org/abs/0705.1029

Oh cool, thanks!

the paper is understandable, but it relies on Rindler's work in eqs. (10), (11)

ah ok

HE = Hawking & Ellis

so many distractions!

:p

4:45 PM

@FenderLesPaul what are you interested in?

Other than Zeppelin :P

haha

physics wise you mean?

@FenderLesPaul @KyleKanos Does he give an explanation of why he thinks the binormal should vanish?

@FenderLesPaul yes

classical GR as a whole but specifically self-force problems and black hole perturbations, although currently my professor has me working on this quais-local rigid frames calculation in relation to these "super-rotations" in the BMS group

@0celo7 With $\Omega B^\mu$ vanishing, then we have a plane curve (of constant curvature)

4:54 PM

and also aspects of QG related to entanglement entropy, black hole information, scrambling and Ryu-Takyanagi-things of that nature

quasi*

@KyleKanos plane curve?

stupid question

@KyleKanos why is constant curvature necessary?

@0celo7 Does this mean you're retracting the above question?

@KyleKanos no

In mathematics, a plane curve is a curve in a Euclidean plane (compare with space curve). The most frequently studied cases are smooth plane curves (including piecewise smooth plane curves), and algebraic plane curves. A smooth plane curve is a curve in a real Euclidean plane R2 and is a one-dimensional smooth manifold. Equivalently, a smooth plane curve can be given locally by an equation f(x, y) = 0, where f : R2 → R is a smooth function, and the partial derivatives ∂f/∂x and ∂f/∂y are never both 0. In other words, a smooth plane curve is a plane curve which "locally looks like a line" with respect...

"curve in the Euclidean plane" does not compute

plane curve in curved spacetime?

Looks like a line along a set of coordinates?

what does that mean?

furthermore, why do constant acceleration worldlines have this property?

5:01 PM

> The vanishing of the "curvature vector $A^\mu$ is the well-known characteristic of a geodesic path.

for a geodesic $A\equiv 0$

literally the definition...

this might make a decent PSE question

@Jimself @ChrisWhite Unless you guys know anything about accelerated observers in curved spacetime

@0celo7 Asking why $\beta^\mu=0$ is needed?

@KyleKanos yes

And asking how valid this whole analogy is

@KyleKanos I can derive these

You want me to ask since I have the paper? Or do you want to ask since it's your interest?

Or wait for Chris and/or Jim to chime in?

5:16 PM

@KyleKanos yes

Okay

You think the paper is being unclear too?

I think it's more beyond my current level of knowledge than anything else

what does he say about the torsion?

That it vanishing along with the curvature means a plane curve

5:20 PM

I assume he doesn't explain why or how?

@KyleKanos wait, but the vanishing of the curvature means a geodesic

> In the differential geometry of twisted curves it; is well known that the vanishing of the curvature of a curve implies that the curve is a straight line, and the vanishing of the torsion implies that it is a plane curve

what does a constantly accelerating curve have to do with plane curves?

constant curvature -> constant acceleration, right?

Well he does also say,

> It turns out that any curve so characterized will be time- like along its entirety if the initial direction is chosen timelike.

5:37 PM

@KyleKanos this is true

1

Q: A space curve is planar if and only if its torsion is everywhere 0

Can someone please explain this proof to me. I know that a circle is planar and has nonzero constant curvature, so this must be an exception, but I am a little lost on the proof. Thanks!

differential-geometry planar-graph

6:21 PM

0

Q: Double-slit experiment with one electron per day

I am a mathematician and i am studying string theory. For this purpose i studied quantum theory. After reading Feynman's book in which he described the double-slit experiment (Young experiment) I was wondering if i send one electron per day or per month (even more), could i see the interference p...

quantum-mechanics experimental-physics double-slit-experiment

6:34 PM

@FenderLesPaul @KyleKanos Related: journals.aps.org/prd/abstract/10.1103/PhysRevD.3.1035

6:46 PM

oh that's a cool paper, thanks!

there's a related paper by Don Page if you're interested

arxiv.org/abs/gr-qc/9706029

also this paper by vishveshwara

researchgate.net/publication/…

I called the Library of Congress and ordered the issue of Physical Review that article is in.

I'm hardcore.

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Jul '1517

The h Bar

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