The h Bar

danielunderwood

00:01

Sounds like you need a software consultant that's paid a very large sum of money nudge nudge wink wink

enumaris

lol

I don't get to make any spending decisions

I'm a contractor myself

danielunderwood

Ahh welcome to the club!

enumaris

haha

Secret

01:55

Last night dream: A Mario game where Mario and his friends are in some Mario Party like board game. Initially there are only two boards unlocked and others will be unlocked as the game progress (including Doom Mountain in Board 2, which requires completing anything else in Board 2). In one of the levels in space, called Causality of Black Holes, Mario and his friends get sucked into a black hole, which is actually a one way transversible wormhole, where space (bluish) and time (yellow) were not

only swapped inside the neck of the black-white hole, but is twisted so much that the time axis is rotated by 180 degrees while the space axis remains the same. The result is that after they enter the black hole, they remerge from it shortly after, but in reality, they are now in the past of the universe.

(pics later)

Secret

02:30

In an apartment, I locked the door to my room to prepare a jump into the past by hopping into a blue portal. My family eventually got in, just as I stepped out of the balcony. Due to the time travel, my present (which is still outside the portal) changed discontinuously to that where the apartment is worn out, My parents still standing in the same position, but now they looked like they were being hit by something, and a figure that look like me passed by

Comments:

Causality Game Full Walkthrough (All Levels)

►

The time travel basically follows the Causality model shown here, and it is so sensitive that as soon something foreign to the initial past end up in the past, the future (present) is already changed. The apparent discontinuous looking events can be explained by the transition between these two versions of the timeline, which are both self consistent, connected causally by the time traveler's action to enter the portal, which is what creates the discontinuous events

and things seemly popping out of nowhere

This dream provide an interesting framework to explore the Grandfather Paradox if the time traveler is present at both ends of a time portal since only part of their body stepped through. If the very act of stepping through eventually undoes the act of stepping though, how will the present reflect that change

Perhaps we will end up with an interesting scenario, similar to an elaborate mirror, that if you step through, the timeline changes into the configuration that supports an ontological loop of your stepping through, with the moment when you stepped through is both the beginning and end of the cycle.

And if you then decided to pull your leg away, then the timeline reconfigure into a linear one where you never stepped through

So the result is that, similar to walking past a mirror where your reflection is only visible if you are at the mirror, you have two versions of timelines which the timeline can take depending on whether you step through the portal, like a mirror

ChoMedit

Then timelines are superposed but be observed only when you are in?

Secret

Thus if you repeatedly step through and pull you leg back, and assuming the two timelines have a red and blue wall respectively, then you will be greeted with an interesting display that the wall becomes red or blue at the frequency you put your leg into the portal and pull it out

@ChoMedit Well, pretty much, as long something that sustain that version is present, then that version is actualised, and if not, then the other version is actualised

I think now the more interesting question is, assuming the two version of the timelines, A and B, one has a vase and one does not. What happens if you step your leg through the portal and thus the timeline takes the image of A, take the vase, and then pull your leg out again. Will the timeline become B

(assuming taking away the vase will not change anything else in the timeline other than missing the vase)

Actually not, you will end up with the timeline taking the version of C, because the vase was not there before

But to anyone else other than you, the vase seemed to came from nowhere, since events are discontinous at the time portal, thus nobody knew you have taken a vase from the portal unless in the immediate area

Or more accurately, since only one version of timeline can exist at each instance, the vase seemed to spontaneously emerge from the portal

So yeah, if time travel works in the way where history is revised, it will be analogous to pulling a duplicated object from its reflection in a mirror

Secret

03:17

(To rob if he saw this, the above ping is a misping because my phone kept the message number when loading the page on mobile)

Secret

Actually no, the above analysis is only correct if there can only be two versions of timeline possible. In more realistic scenarios, the timeline cannot possibly compute what future will be caused when the leg is stepped through the portal, so only the timeline with the least rearrangement, which means most resemble the original, will be visible at the present side of the portal since it has the highest entropy

invadingdingo

I had a quick question

rob

@Secret Pinging someone and then deleting the message makes the "new chat reply" indicator behave strangely. Thanks for the explanation.

invadingdingo

In the figure a nonconducting rod of length L = 8.34 cm has charge -q = -4.47 fC uniformly distributed along its length.

That's the beginning to my problem. Do you think there's a typo with -q? I just don't think it makes sense to say -q = -charge..

rob

@invadingdingo Hmmm, interesting.

Semiclassical

03:23

are there any other charges in the problem?

rob

That's a very roundabout way to write about a positive charge.

What's the rest of the problem? It might be more obvious with the rest of the context.

invadingdingo

In the figure a nonconducting rod of length L = 8.34 cm has charge -q = -4.47 fC uniformly distributed along its length. (a) What is the linear charge density of the rod? What are the (b) magnitude and (c) direction (positive angle relative to the positive direction of the x axis) of the electric field produced at point P, at distance a = 14.7 cm from the rod? What is the electric field magnitude produced at distance a = 83 m by

(d) the rod and (e) a particle of charge -q = -4.47 fC that replaces the rod?

It reoccurs in part e.

Semiclassical

weird

invadingdingo

edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c22/… There's the image

rob

@invadingdingo Deep linking seems not to work ... I get a login page.

invadingdingo

03:27

Try that

Seems to work on my end

Semiclassical

they might be doing it so that you can have $q=|q|$

but...why bother

rob

@invadingdingo The illustration looks like they are trying to emphasize that it's a negative charge.

Like, the charges in the rod are little minuses (rather than little pluses).

invadingdingo

Ahh..

That's a really dumb way to go about that haha. It totally looks like the negatives should cancel

Semiclassical

well, in some sense they do cancel. $q$ is a positive quantity

but the charge is $-q$, not $q$

rob

@invadingdingo Well in English a double negative doesn't not cancel.

invadingdingo

03:30

HAHA

rob

Which I guess can mean whatever you want.

With the figure and the text, I'd interpret as negative charge.

But I agree that text alone is a little unusual.

invadingdingo

So use the quantity -4.47fC

not +4.47mC?

fC *

Semiclassical

what I'd do to disambiguate it is to write the charge of the rod as Q=-q = -4.47 fC

invadingdingo

Ah, so just associate it with a new variable altogether?

Semiclassical

yeah. the point is that the charge of the rod is $-q$, not $q$

rob

03:34

If I were handing the problem in to a person, I would write an explanatory sentence about why the charge labeling is ambiguous and give both solutions for the direction of the field. But if you're submitted an answer to an illiterate robo-grader, too bad.

invadingdingo

Charge density along a line is lambda = q/L, right?

Yeah, I hate the freaking software that this school uses.

Semiclassical

it is, if q is the charge of the object

in this case, the charge is -q

(in their notation)

invadingdingo

And we only get 5 different attempts at a submission, which defeats the purpose of physics problem solving altogether. I shouldn't be scared to try new methods to get an answer..

Ok, so that's what I was going to ask

Chegg has it listed as -q/L for this equation

Semiclassical

yeah

invadingdingo

which really confused me

Semiclassical

03:36

i.e. Q/L

rob

@invadingdingo Linear charge density is a differential, $\mathrm dq /\mathrm dL$.

Semiclassical

eh. if it's uniform density, then it's also total charge / total length

rob

For a constant charge density along a line, both integrals trivial, and $q/L$ is the same value.

invadingdingo

It does say uniformly distributed

rob

But I've had students who memorized $\lambda = Q/L$ as a "fact" and then did really creatively boneheaded things when I gave them a nonuniform charge distribution.

Semiclassical

03:38

yeah

fair point

invadingdingo

I thought integration wasn't involved until working with surfaces and volumes

Semiclassical

it's involved in the same way as $x=x_0+\int_0^t v\,dt$

rob

But the sign of a charge density follows the sign of the charges. If I take an empty region of space and smear around some excess negative charges, the charge density in that region is negative.

Semiclassical

i.e. if $v$ is constant then $x=x_0+v t$ i.e. $v=\Delta x/\Delta t$

invadingdingo

Wait, I'm a little confused on what that statement means.

Semiclassical

03:41

remember that velocity is defined as $v=\frac{dx}{dt}$

rob

@invadingdingo Which statement?

invadingdingo

@Semiclassical's

Semiclassical

And therefore $x(t)-x(0) = \int_0^t \frac{dx}{dt}\,dt = \int_0^t v\,dt$

invadingdingo

Ooh, I get it now

Semiclassical

plus, any time you have something non-uniform you should expect at least some calculus

and the concept of linear charge density had better apply to non-uniform distributions as much as uniform ones

invadingdingo

03:47

I got you. So we can think of formulas that don't use calculus as an "easier" method that only works sometimes, i.e, when it is talking about the entirety of a charge?

And anytime it isn't the entire charge, we have to use calculus.

Semiclassical

pretty much.

invadingdingo

Alright, I don't think we covered that in class, but it's good to know.

Avantgarde

I was surprised when this had -2 points some time ago!

0

Q: How do the two different force formulae ($F = ma$ and $F = G\frac{m_1m_2}{r^2}$) relate to each other?

$F = ma$ and $F = G\frac{m_1m_2}{r^2}$ are two formulae. Aren't they measuring the same thing? How do they relate to each other?

invadingdingo

So I worked part a of the problem and am getting -5.38e-14C/M. Does that look right? I want to make sure before going forward with the problem.

Semiclassical

$F_{net}$ vs. $F_g$

Avantgarde

03:49

It's way deeper than that.

The asker seems to have captured a very deep aspect.

Semiclassical

Unless you're referring to the equivalence principle, I dunno what you mean

I mean, the simple fact that $F=ma$ applies to non-gravitational forces tells you that they can't be measuring the same thing

Avantgarde

Equiv principle, yes.

I quite enjoyed writing an answer to that haha.

Semiclassical

@invadingdingo i get -5.36e-14 C/m, but that's close enough

I think you're giving them too much credit, frankly.

There's nothing in that question that at all indicates that they've got the equivalence principle in mind as compared to the garden variety confusion between a resultant force and a particular (in this case gravitational) force.

Avantgarde

Sure. I realize that. Equivalence principle itself was devised when the universality of free fall was generalized by Einstein.

The 'principle' is a postulate, anyway.

invadingdingo

I'm a little confused on part b.

I don't know how to go about finding the magnitude

Semiclassical

04:01

It's not a simple equation like E=k q1 q2/r^2 here, no: the negative charges at the left end of the rod are farther from P than those on the right

and indeed the distance r to point P varies continuously as you move along the rod

so you won't be able to apply Coulomb's law to the entire rod at once.

invadingdingo

Ok, so integrate and "add" all the pieces together?

Semiclassical

yeah.

invadingdingo

So I'd be integrating the E equation, kQ/r^2?

Semiclassical

you'll end up doing something like that, yes. but I'd start by considering an infinitesimal portion of the rod dx (at position x, say) and consider the field this bit of the rod will produce

once you've computed that infinitesimal contribution dE, you can sum that up via integration

invadingdingo

how does $-dq$ = lambda($dx$)?

Semiclassical

04:12

for now, let's take $Q=-q$

so $dQ=\lambda \,dx$

(more properly, $\lambda = \frac{dQ}{dx}$. but $dQ=\lambda\,dx$ is typical notation)

invadingdingo

But wouldn't we need to take the derivative of λ as well?

I don't understand how it doesn't change

Semiclassical

no. $\lambda= dQ/dx$ is a definition, for one

suppose you take a finite length of the rod $\Delta x$. It'll carry some charge $\Delta Q$.

Hence the average charge per unit length of that portion is $\overline{\lambda} = \Delta Q/\Delta x$

if you now take that $\Delta x$ to be infinitesimally small, then that converges to $\lambda = dQ/dx$

Tanuj

05:05

@JohnRennie Hi :)

John Rennie

@Tanuj morning :-)

Tanuj

@JohnRennie How have you been ?

John Rennie

@Tanuj much as usual - life goes on. How are things at your end?

Tanuj

@JohnRennie I've finally landed up in NIT - UK :)

3

John Rennie

NIT?

Tanuj

05:09

National Institute of Technology

John Rennie

@Tanuj in the UK?

Abcd

@JohnRennie I think he means Uttarakhand

John Rennie

Aha!

National Institute of Technology Uttarakhand

...

Abcd

Uttarakhand

Uttarakhand (English: ), officially the State of Uttarakhand (Uttarākhaṇḍ Rājya), formerly known as Uttaranchal, is a state in the northern part of India. It is often referred to as the Devbhumi (literally "Land of the Gods") due to a large number of Hindu temples and pilgrimage centres found throughout the state. Uttarakhand is known for the natural environment of the Himalayas, the Bhabhar and the Terai. On 9 November 2000, Uttarakhand became the 27th state of the Republic of India, being created from the Himalayan and adjoining northwestern districts of Uttar Pradesh. It borders Tibet to the...

John Rennie

@Tanuj anyhow, congratulations :-) Doing engineering?

Tanuj

05:12

@JohnRennie Yes ! :)

Avnish Kabaj

@Tanuj C's?

Tanuj

@AvnishKabaj Computer Science and Engineering , yes .

John Rennie

Ah, something's up at work. I need to work for a few minutes ...

John Rennie

08:35

Damn and blast. I'm out of milk and I don't like black coffee.

user1732

08:54

have a cup of tea?

with lemon and honey

John Rennie

I've never been keen on tea. I don't like the tannin aftertaste.

Physics Meta

11:48

0

Q: Policy regarding posting the same question in different SEs

In my thought, posting the same question in different SEs can help getting answers from different perspectives and points of view, which is great. Is it against the SE policy to post the same question in to different forums, for example in Physics SE and in Mathematics SE? If there is a policy re...

discussion

Anonymous

12:35

@Tanuj Congrats :)

user 2646

12:54

we have two Mithrandirs now?

Mithrandir24601

@user2646 There have been 2 of us for a while now :) :P

user 2646

not in here?

:-)

Mithrandir24601

@JohnRennie Then make some decent loose tea instead of those terrible tasting things they sell as 'tea' in the supermarkets

user 2646

welcome @Mithrandir

Mithrandir

I just don't usually hang out in this room.

Mithrandir24601

12:58

@user2646 Possibly not - I haven't been around as much as I'd normally be in here in the past month or so as I've been away quite a lot, so I don't really know so much about who's been around :/

user 2646

yeah, i've noticed

i thought you changed your avatar :P

like blue

Mithrandir24601

Speaking of which, @Mith - it's hard to see with the small images, but that's violet, right?

@user2646 I did become a mod of quantum computing SE, if you missed that :)

Mithrandir

@Mithrandir24601 correct

Mithrandir24601

Incredibles 2 was sooo good

Mithrandir

Haven't watched it yet. Missed it in the theater.

user 2646

13:10

is that a cartoon?

Mithrandir24601

@Mithrandir it was absolutely fantastic. When I saw it, the entire audience (most if not all adults) was just constantly laughing throughout

Definitely better than the first one

@user2646 There are 2 animated films - The Incredibles and The Incredibles 2

Mithrandir

Most people seem to think it's the other way around...

Mithrandir24601

@Mithrandir As much as I loved the first one, I've got to disagree with those people

user 2646

yeah, animated movies have come a long way...since the lion king days

which was my last encounter with them :-)

omg! 24 years ago?

Mithrandir24601

@user2646 Lion King is really good though - if you ever get the chance to see it live, it's well worth it

user 2646

13:18

will do

Mithrandir24601

But I've got to admit, my favourite animated film is unquestionably Spirited Away

user 2646

hmm, thanks i'll check it out

Nobody recognizeable

13:32

In a bungee jump how is the gravity restoring force ? It acts away from the centre not towards . So how do you count that as a restoring force?

nitsua60

@Nobodyrecognizeable (only half the time, though)

Nobody recognizeable

13:53

@nitsua60 what do you mean ?

Semiclassical

if you have an object oscillating on a hanging spring, then half the time it'll be above the center of oscillation and half the time it'll be below

in both cases, the net force (gravitational plus spring) is always towards the center of oscillation

(though this assumes that there's always some minimum stretch in the spring, which I"m not certain is true in the bungee case)

nitsua60

14:32

@Nobodyrecognizeable Sometimes the object bungeeing is above the equilibrium position, so gravity works toward the center. Other times the object bungeeing is below the equilibrium position, so gravity's working away from the center as you say.

(And I'm just saying "half" colloquially--I haven't done the math but I wouldn't be surprised if there's an asymmetry.)

Semiclassical

@nitsua60 there's probably asymmetry inasmuch as, as one nears the top of a bungee jump, the bungee will go slack. so you'd spend some of the time only being subject to the gravitational force

nitsua60

@Semiclassical Yeah, that slack-length turns something into a piecewise, and I don't feel like Laplacing it in my head. Classes start back up on Monday so I'm on my last day of being able to not math for a while =)

Semiclassical

yup

Abcd

15:12

What? There are 2 Mithrandirs!

Nobody recognizeable

@Semiclassical @nitsua60 thanks.

danielunderwood

And they're both mods!

Jack Clerk

Good morning/evening/night to everyone!

I would like to know some (undergraduate/introductory) references about Special Relativity in non-inertial frames.

John Rennie

@JackClerk weren't you asking about this before? Or was that someone else?

Jack Clerk

@Jonh Rennie Acctually not (I guess)

John Rennie

15:25

The definitive book is by some guy whose name I can't remember (not much help, I know) but it starts "Gorg..."

Jack Clerk

Gourgulhon?

It's a little bit advanced.

John Rennie

@JackClerk that's the one.

I don't think you'll find a treatment that's much simpler. You kind of have to know differential geometry to handle non-inertial frames.

ACuriousMind

@JohnRennie Ah, the famous book by Gorgonzola

John Rennie

You could start with the Rindler geometry I suppose. Any decent SR book will cover than.

@ACuriousMind :-)

@ACuriousMind "Gourgulhon" isn't a name I find easy to remember.

In fact it's "Gourgoulhon"

Jack Clerk

Thank you.

Emilio Pisanty

15:35

@JohnRennie just saw this and thought you'd find it at least chuckle-worthy: nature.com/articles/40456

~~dunno if~~ it's paywalled

presumably available on elbakyan-net

John Rennie

@EmilioPisanty yes, scihub to the rescue.

Though the article doesn't say much.

Semiclassical

hmm. I know $|jm\rangle = C(j,m) (J^-)^{j-m}|jj\rangle$ for some $C(j,m)$. But I can't seem to work out what $C(j,m)$ would have to be

or at least I can't convince myself I know the answer :/

Lozansky

Suppose I got a charged sphere with total charge $Q$, radius $2a$ and then hollow out a smaller sphere with radius $a$ as in the picture: imgur.com/a/P5RmATh

Now I'm interested in finding the potential at $P$

Semiclassical

I think it works out to be $C(j,m)=\dfrac{(j+m)!}{(2j)!(j-m)!}$

Lozansky

So I guess one approach would be using superposition as follows: compute the potential from the larger sphere and subtract the potential from the smaller sphere

Semiclassical

15:47

but that seems wrong somehow

@Lozansky that'll work best if you work in terms of charge density rather than total charge

Lozansky

But what is a suitable place to put my origin? I was thinking the origin of the larger sphere, but I think that makes calculations for the potential of the smaller sphere a bit messy?

@Semiclassical But the charge density is just $3Q/4\pi (2a)^3$?

Semiclassical

Where $Q$ is the charge of the larger sphere, yes

Lozansky

Homogenous charge btw...

Semiclassical

the point is that the charge of the sphere you remove will not be $Q$

Lozansky

No but it's proportional

As in $Q (a/(2a))^3 = Q/8$?

Semiclassical

15:50

Sure. My point is that while the charge isn't the same, the charge density of the sphere you remove will be

so it's more convenient to work in terms of $\rho$ throughout

Lozansky

Yes, but with opposite sign and then I add them together?

Semiclassical

(And if you do that, the problem ends up becoming a lot simpler to understand.)

Yes.

Lozansky

That was what I was thinking of doing

Semiclassical

Okay. I'll also suggest that it'll be easier to think in terms of the electric field.

Lozansky

Let's start with this approach :P

Anonymous

15:52

@Lozansky Involves some boring integration tho!

Semiclassical

Yeah, if you don't do it via the electric field then I think it ends up being a tedious integral

Lozansky

@Blue That's what my formula book is for :P

Semiclassical

I may be wrong on that tho

Anonymous

What's the electric field method?

Lozansky

Path integral?

Semiclassical

15:53

Well, what's the field inside a sphere with uniform charge density $\rho$?

enumaris

hint: use the shell theorem

Lozansky

@Semiclassical $\dfrac{1}{\epsilon_0} (r/R)^3 Q \hat{r}$

Semiclassical

not quite.

you pick up a factor of $r^2$ from the electric flux integral

Lozansky

Ah right

Divide by $4 \pi r^2$ :P

Semiclassical

I forgot you were doing it in terms of $Q$

At this point, it's convenient to work in terms of $\rho = Q/(4\pi R^3/3)$

Anonymous

15:56

$$\vec{E} = \frac{\rho \vec{r}}{3\epsilon_0}$$...looks like a nice form

Semiclassical

Yep.

Lozansky

Got it

Semiclassical

Note that it's really $\vec{r}-\vec{r}_0$ if you assume a sphere centered at $\vec{r}_0$

Now do superposition :3

@JohnRennie sounds like the name of one of the alien races in A Hitchhiker's Guide

Anonymous

So the net electric field points along the common axis of the two spheres I guess

Semiclassical

Yeah. More to the point, the electric field inside is constant.

Anonymous

16:03

Cool. Ah, so this doesn't require integration. However, if it were a hollowed out disc instead of a hollowed out sphere, integration would be necessary!

Anonymous

(I think)

Semiclassical

Yeah.

That said...

John Rennie

@Semiclassical He's French I think, though it doesn't sound like a French surname.

Semiclassical

this isn't as useful as I imagined it would be. It does tell you that the change in electric potential between two points in the hollowed out region will be $$\Delta V= V(\vec{r}_2)-V(\vec{r}_1) = -\vec{E}\cdot(\vec{r}_2-\vec{r}_1)=-\frac{\rho}{3\epsilon_0}\vec{a}\cdot \Delta \vec{r}$$ where $\vec{a}$ is the position vector of the center of the hollow sphere

Lozansky

@Semiclassical If the total charge of the sphere is $Q$ then the large sphere should have the charge $8Q/7$ and the small sphere the charge $-Q/7$

Semiclassical

16:06

Looks plausible.

Lozansky

And the point $P$ is on the surface of the sphere

Semiclassical

So once you know the potential at a single point in the hollow, you know it everywhere inside the hollow.

But: You still need to know the potential at a single point in the hollow first :/

In which case you might as well pick the point of interest to be $P$ from the start and calculate it directly

Lozansky

So the potential at that point from the large sphere is the same as that of a point charge in the center, i.e. $V_{large} = \dfrac{1}{4\pi \epsilon_0} \dfrac{8Q}{7} \dfrac{1}{2a}$ and likewise for the small sphere $V_{small} = -\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{7} \dfrac{1}{a}$

$V_{tot} = \dfrac{3Q}{28 \pi \epsilon_0 a}$

Semiclassical

Looks right.

Though, come to think of it, there is one weird thing. Suppose you wanted to compute the potential at the center of the inner sphere.

Never mind, I see my error

Yeah, that makes sense. (In terms of $\rho$, that's $V_{tot} = \rho /7\epsilon_0 a$.)

The advantage of the electric field computation is that you could now compute the potential at, say, the center of the the hollow sphere without doing any integration

(blah, ignore my previous statement about $V_{tot}$, I can't do algebra right now)

the change in position would be $-\vec{a}$ where $\vec{a}$ is the center of the sphere. Hence the change in potential would be $\Delta V =( \rho /3 \epsilon_0)a^2$

So that's kinda handy.

John Rennie

@EmilioPisanty have a look at this. The usual caveats apply: it's one person's opinion. However that person is a physicist who has worked in fusion research.

Semiclassical

16:19

the old canard springs to mind as ever: "Fusion is the energy of the future, and always will be."

vzn

as JR mentioned earlier there seems to be new inroads on stellarators, have been seeing some scattered "signs of life" last few yrs, we discussed this some on tues biwkly mtg

Emilio Pisanty

16:38

@JohnRennie a bunch of red herrings for sure. but some nontrivial points buried in there.

still tho, tl;orh

(only read half)

tl;guht ?

user351417

@EmilioPisanty guht?

Emilio Pisanty

gave up halfway through

John Rennie

@EmilioPisanty over and over we've found that what seems impossibly hard seems routine a generation later. The tech industry is a prime example of this. And what the article says is that ITER seems impossibly hard.

user351417

Ahh of course. Story of my last 3 popsci books!!

John Rennie

The author seems to think it will remain impossibly hard. Maybe so and maybe not.

But if we don't try we won't find out.

Emilio Pisanty

16:46

@JohnRennie pretty much.

Semiclassical

Eh, the overall vibe I get isn't just "ITER seems impossibly hard" but "it seems both impossibly hard and not capable of giving you what you were promised"

(plus, stuff like "perpetual motion machines" also seemed impossibly hard...and turned out to be impossibly hard.)

2

John Rennie

@Semiclassical that's a bit unfair. ITER was never designed to actually generate electricity. It was designed to prove that a burning plasma could be maintained long enough to be used in a reactor that does produce electricity.

Emilio Pisanty

@JohnRennie still, the bits about how it's represented are on point

Semiclassical

yeah, "designed as" != "sold as"

John Rennie

@EmilioPisanty yes, but the same can be said of e.g. string theory.

Emilio Pisanty

16:51

"it generates X MW of power" is indeed pretty misleading if there's no underlying model for how much of that you can capture

Semiclassical

except that string theory doesn't capture nearly as much money as ITER

enumaris

it's all about the Benjamins

John Rennie

I don't want to find myself the champion of ITER, because I'm rather sceptical of fusion power myself. But if we're going to criticise it let's keep the criticism to technical issues.

This isn't the Daily Mail. We are all physicists here.

4

Emilio Pisanty

@JohnRennie what if I'm a physicist who sends mail daily?

2

enumaris

@JohnRennie I am not a physicist anymore unfortunately :(

John Rennie

16:53

As it happens I'm a physicist who is daily male.

enumaris

I got my membership revoked last week

John Rennie

(let's not go into what happens at night :-)

enumaris

the correct etiquette is to put a closing parenthesis

Even if the smiley consists of a parenthesis

John Rennie

Aaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrrrrrrrrrrrrrrrrrggggggggggggggghhhhhhhhhhhh‌hhhhhhh!!!!!!!!!!!!!!!!!!!

4

Semiclassical

I feel like that's a dodge. To the extent that I care about ITER, it's as a highly visible example of public science funding. If it's been oversold, that damages the credibility of such.

enumaris

16:55

Like so (:))

ACuriousMind

@enumaris That's a malformed Lisp expression, not a smiley :P

2

Semiclassical

I mean: "As of 2016, the total price of constructing the experiment is expected to be in excess of €20 billion." (wikipedia)

enumaris

lol

Let's get fusing!!

Semiclassical

(I don't have a good sense of the total EU science budget tho)

Loong

@Semiclassical Hm, our project costs here are probably already above ten billion Euro.

Semiclassical

17:02

wouldn't surprise me

Lozansky

$|\dfrac{z+h+\sqrt{s^2+(z+h)^2}}{z-h+\sqrt{s^2+(z-h)^2}}| = |\dfrac{h-z+\sqrt{s^2+(z-h)^2}}{-(h+z)+\sqrt{s^2+(z+h)^2}}|, s>0$?

Loong

Last official estimate in 2012 was 8.5 billion.

Lozansky

$z,h \in \mathbb{R}$

If you really need something to be true, does that make it so?

Semiclassical

The thing to do is probably move the denominators to the other sides, and then multiply out to get differences of squares

In which case it looks right?

Lozansky

Yeah, that gives $|(z-h)^2| - |-(z-h)^2| =0 :)$

Abcd

17:20

@ACuriousMind Wanted to ask you something related to flag for moderator option.

ACuriousMind

@Abcd Shoot

Abcd

@ACuriousMind Is it OK to request the moderator to move something to trash using "flag for moderator" option because you think its unnecessary noise.

coniferous_smellerULPBG-W8ZgjR

John Rennie made an edit today which I think broke the sentence into grammatically incorrect. anyone over 2k points want to modify 1 word?

ACuriousMind

@Abcd Not really. You should reserve flags for serious issues.

coniferous_smellerULPBG-W8ZgjR

or is such an edit impossible?

Abcd

17:28

@ACuriousMind Okay.

ACuriousMind

You could, however, simply ping the owner of the room and ask them to move the messages

Abcd

They arent online.

coniferous_smellerULPBG-W8ZgjR

can someone add back the "is" ? physics.stackexchange.com/questions/427290/… so the sentence becomes Boiling is a nonequilibrium process.

coz currently the grammar is broken due to an edit ^^

Mithrandir24601

@Abcd They'll still get the notification when they do come online

Abcd

@Mithrandir24601 Can you do it for me right now? You have the same powers

Mithrandir24601

17:32

@Abcd If you seriously feel that you want it removed, OK, I'm happy to do that, although I'll also remind you that it's just chat and having a harmless message is fine

enumaris

done

Abcd

Thanks.

Mithrandir24601

@Abcd no problem :)

Anonymous

@coniferous_smellerULPBG-W8ZgjR JR just edited the tags

coniferous_smellerULPBG-W8ZgjR

oh

ah right the asker modified it on his own

Anonymous

17:42

What's the "ULPBG-W8ZgjR" in your name?

coniferous_smellerULPBG-W8ZgjR

but thanks enumaris for fixing it!

random chars

Anonymous

Hurts my eye

coniferous_smellerULPBG-W8ZgjR

remember when my username was an entire random string of chars

well, that's how i finished the last part of it now

first part is obviously not random

Anonymous

@coniferous_smellerULPBG-W8ZgjR I do. I had put you on Ignore for that alone ;)

coniferous_smellerULPBG-W8ZgjR

oh really. ouch!

Anonymous

17:44

'JD's poodle' was a nice name tho

Anonymous

Sad that it isn't allowed XD

coniferous_smellerULPBG-W8ZgjR

uh why not?

Anonymous

I dunno. Why did you change it?

Anonymous

I assumed that it would classify as "impersonation"

coniferous_smellerULPBG-W8ZgjR

18:22

@Blue I just got what you mean. i didnt know what a poodle was

i used to be john's duffield dog yeah

not really impersonification. they renamed my nick stating i dishonored JD or something like that

Anonymous

18:42

@coniferous_smellerULPBG-W8ZgjR lol

Anonymous

Hmm, I really need to learn bash scripting

Anonymous

Tired of doing these things manually

danielunderwood

Just stick a bunch of commands in a file with #!/bin/bash at the top!

Anonymous

heh :P

danielunderwood

And scratch your head for years wondering why case had to end with esac

And the syntax of case in general

something something ;;

Anonymous

18:47

@coniferous_smellerULPBG-W8ZgjR Ah, your location was "Poole". My subconscious mashed "Poole" and "dog" together.

Anonymous

To produce "Poodle".

Anonymous

Well, that subconscious mashing of stuff is why I have weird dreams...

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