The h Bar

user54412

01:11

@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

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

01:26

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).

0celo7

01:51

@ChrisWhite I recorded a Let's Play of Morrowind

Kyle Kanos

@0celo7 Link pls

0celo7

Melee combat in this game is literally impossible

►

it's a proof of concept

not recorded in native quality

Kyle Kanos

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

0celo7

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

Kyle Kanos

And you're using a dagger

0celo7

01:56

i have nothing else...

@ChrisWhite how the hell do i hit people

Kyle Kanos

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

0celo7

look at 8:15

fucking ridiculous

Kyle Kanos

I'm at ~3

Wait...is this right at the beginning?

0celo7

I have like one hour of game time

I've been fooling with mods this whole time

Kyle Kanos

Okay

0celo7

02:01

I'll buy a sword.

I have 500+ gold

Kyle Kanos

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

0celo7

I bought a sword and killed him with two hits

user54412

@0celo7 Practice on mudcrabs :)

0celo7

02:34

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

Kyle Kanos

03:13

This is hard to read: wikipeetia.org/Fluid

2

skill patrol

03:26

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

Try this @KyleKanos

>8(

Danu

05:10

@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

05:28

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.

skill patrol

05:44

Send Professor Hawking an email and ask :P

Sorry, bad joke^

skill patrol

08:32

Hi @Hippalectryon

Did you read that article?

Hippalectryon

09:42

@skillpatrol which one ?

afk

skill patrol

10:10

This one @Hippalectryon

Hippalectryon

Yeah I read it

skill patrol

What did you think?

Hippalectryon

It doesn't say anything new afaik

skill patrol

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

Hippalectryon

Not really

skill patrol

10:15

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

Hippalectryon

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

skill patrol

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 :-)

Hippalectryon

@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 ?

skill patrol

I would suggest posting it on the main site.

Just ask for a hint.

Hippalectryon

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

skill patrol

10:33

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

Hippalectryon

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

skill patrol

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

Hippalectryon

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

skill patrol

That tells you what Work is equal to, right?

Hippalectryon

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

skill patrol

10:43

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

F = P(?)

Hippalectryon

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

skill patrol

Can't you ask your teacher?

Hippalectryon

I don't see my teacher until next year

skill patrol

Do you have a textbook?

Hippalectryon

Nope, only notes

skill patrol

10:53

How about a library?

Hippalectryon

What would I do there ?

The exercise doesn't come from a textbook or anything

skill patrol

There must be some substitution...

Hippalectryon

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

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

skill patrol

How about a change in variable?

Hippalectryon

:O $u=\dot{x}$

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

skill patrol

10:59

Maybe even F =ma is used somewhere

I gotta go :(

Hippalectryon

Ok

skill patrol

Sorry

Hippalectryon

Nah you tried to help me, thanks ;-)

skill patrol

Later pal

Hippalectryon

See you

0celo7

11:17

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

0celo7

11:51

@xXx_based_dolan_xXx nice user name

Kyle Kanos

13:00

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

0celo7

13:12

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

Kyle Kanos

AFAIK, it's pretty simple on Windoze

0celo7

Do you know if twitch streaming is free?

Kyle Kanos

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

Luke

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

Kyle Kanos

@Luke Just ask

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

Luke

13:21

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

Kyle Kanos

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

Luke

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

Kyle Kanos

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

(a=t/m, obviously)

Luke

that's how I started, yes

and since H = b P², [H,P] = 0

Kyle Kanos

So your issue is just on the first term?

Luke

13:32

indeed. ⇒ [X,H] = 1/2m(XP² - P²X)

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

Kyle Kanos

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

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

0celo7

Kyle doing quantum mechanics?

Luke

⇒ [X,P²] = [X,P]P + P[X,P]

Kyle Kanos

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

Luke

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

but thank you

Kyle Kanos

13:35

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

0celo7

@KyleKanos I don't have a dedicated USB mic

Will the mic on my phone earbuds work well enough?

Kyle Kanos

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

@0celo7 Possibly.

0celo7

@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

Kyle Kanos

13:54

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

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

0celo7

There will be maybe two people in there

Kyle Kanos

So?

0celo7

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

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

Kyle Kanos

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

0celo7

then why do you want them?

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

Kyle Kanos

14:09

@0celo7 S&Gs

0celo7

@KyleKanos pls define

Kyle Kanos

Shits & Giggles

0celo7

my precalc teacher said shiggles

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

FenderLesPaul

holy shit I had no idea this chat was a thing

0celo7

no cussing

there are children in here!

FenderLesPaul

14:13

eh about time children learned colorful vocabulary ahead of time

David Z

14:47

@0celo7 lol :-P

0celo7

@DavidZ it's no laughing matter

0celo7

15:18

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

Kyle Kanos

Yes

0celo7

How long is it?

Kyle Kanos

8 pages

0celo7

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

Kyle Kanos

Yes he does

FenderLesPaul

15:23

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

0celo7

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

Kyle Kanos

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

0celo7

@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 :/

Kyle Kanos

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

0celo7

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

skill patrol

15:25

Are you going to write the Putnam? @0celo7

0celo7

@skillpatrol no

I'm bad at math

FenderLesPaul

@0celo7 section 6.2 of MTW

0celo7

@FenderLesPaul is that in curved spacetime

Kyle Kanos

@0celo7 Yes, he does say that

FenderLesPaul

@0

oops

Kyle Kanos

15:27

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

FenderLesPaul

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

0celo7

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

does he say how to get that equality?

Kyle Kanos

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

0celo7

that's all standard

I need the $DA/d\tau$ line

Kyle Kanos

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

0celo7

15:33

@KyleKanos yes, that's standard

@KyleKanos no clue how to get that...

Kyle Kanos

@0celo7 TIL...

0celo7

unless

Kyle Kanos

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
$\beta^\mu=\frac{D}{d\tau}\left(\frac{A^\mu}{\alpha}\right)-\alpha U^\mu=0,\quad-\alpha^2=g_{\mu\nu}A^\mu A^\nu$

0celo7

15:50

hmm

how the heck do you get that

Kyle Kanos

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

0celo7

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

Kyle Kanos

Nope, just that it has an obvious first integral

0celo7

fuck him too

ugh

any idea how to derive this?

Kyle Kanos

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?

0celo7

15:56

yes

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

Kyle Kanos

18 mins ago, by Kyle Kanos

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

0celo7

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

Kyle Kanos

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

skill patrol

@KyleKanos do you see any quick solution to this?

0celo7

I'm somewhat familiar with those

skill patrol

15:58

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 ?

0celo7

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

Kyle Kanos

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

skill patrol

Thanks for replying

:-)

Kyle Kanos

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

0celo7

wikipeetia.org/Streng%20thoery

reads like the typical PSE post

skill patrol

16:04

@KyleKanos great thanks :D

FenderLesPaul

dude streng theory is so dope

Kyle Kanos

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

0celo7

@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?

Kyle Kanos

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

0celo7

so how does he show $\beta=0$

Kyle Kanos

16:12

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 $\Omega B^\mu$

0celo7

and that's not defined anywhere?

Kyle Kanos

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

0celo7

wtf

how the hell does he get that??

Kyle Kanos

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

FenderLesPaul

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

Kyle Kanos

16:19

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

0celo7

@KyleKanos what is k?

Kyle Kanos

Sorry, that's a kappa at the end there

Which is curvature

0celo7

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

FenderLesPaul

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

0celo7

@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

FenderLesPaul

16:22

@0celo7 ah ok

skill patrol

Are you excited?

FenderLesPaul

the chat notification sounds terrifying holy shit

0celo7

yes

I turn it off

Kyle Kanos

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

FenderLesPaul

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

Kyle Kanos

16:24

He's got a 250 review head-start

0celo7

@FenderLesPaul read my bio

should say there

skill patrol

Click on the speaker in the upper right hand corner.

Kyle Kanos

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

0celo7

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

FenderLesPaul

nuclear engineering; what the fuck?

0celo7

16:25

@KyleKanos k

FenderLesPaul

sorry :p

0celo7

the chilluns!

you ruin their minds

FenderLesPaul

ah Tennessee

cool

0celo7

@FenderLesPaul read: applied nuclear physcis

FenderLesPaul

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

0celo7

16:27

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

FenderLesPaul

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

0celo7

I'm just not interested in a career in academia

FenderLesPaul

that's fair

so you start this fall then?

0celo7

yes

FenderLesPaul

cool

0celo7

16:29

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

FenderLesPaul

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

0celo7

no clue how to define it in 4 dimensions

FenderLesPaul

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

He doesn't define it in the paper though

0celo7

@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?

FenderLesPaul

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

Indeed

0celo7

16:36

arxiv.org/abs/0705.1029

FenderLesPaul

Oh cool, thanks!

0celo7

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

FenderLesPaul

ah ok

0celo7

HE = Hawking & Ellis

FenderLesPaul

so many distractions!

:p

0celo7

16:45

@FenderLesPaul what are you interested in?

skill patrol

Other than Zeppelin :P

FenderLesPaul

haha

physics wise you mean?

0celo7

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

@FenderLesPaul yes

FenderLesPaul

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

Kyle Kanos

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

FenderLesPaul

16:54

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

quasi*

0celo7

@KyleKanos plane curve?

stupid question

@KyleKanos why is constant curvature necessary?

Kyle Kanos

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

0celo7

@KyleKanos no

Plane curve

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?

Kyle Kanos

Looks like a line along a set of coordinates?

0celo7

what does that mean?

furthermore, why do constant acceleration worldlines have this property?

Kyle Kanos

17:01

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

0celo7

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

Kyle Kanos

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

0celo7

@KyleKanos yes

And asking how valid this whole analogy is

@KyleKanos I can derive these

Kyle Kanos

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?

0celo7

17:16

@KyleKanos yes

Kyle Kanos

Okay

0celo7

You think the paper is being unclear too?

Kyle Kanos

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

0celo7

what does he say about the torsion?

Kyle Kanos

That it vanishing along with the curvature means a plane curve

0celo7

17:20

I assume he doesn't explain why or how?

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

Kyle Kanos

> 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

0celo7

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

constant curvature -> constant acceleration, right?

Kyle Kanos

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.

0celo7

17:37

@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!

0celo7

18:21

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...