The h Bar

Emilio Pisanty

12:01 AM

@DanielSank you got anything special coming up this cinco de mayo?

DanielSank

12:29 AM

@EmilioPisanty Might go snorkeling :-D

You?

heather

4:53 AM

bit quiet in here

Sid

Everyone sleeping, I presume

heather

true

Sid

I survived a supercyclone. I am feeling pretty lively out of that experience.

Physics Meta

0

Q: About stack exchange problem solving website

My reputation is 61. I can't upload photo in problem solving website. When i able to upload photo in this website ?

feature-request

Sir Cumference

@heather Oh hey you're back!

heather

5:19 AM

@SirCumference i've been around but haven't popped in here much =)

Abhas Kumar Sinha

6:01 AM

@Sid you live in Odisha? I too survived that

Sid

@AbhasKumarSinha Yes

Abhas Kumar Sinha

@Sid ah okay

Novalium Company

8:40 AM

How exactly does an electron get the energy of a photon to get to a higher energy state? Maybe a part of the photon's momentum is transferred to the electron? If that's the case, why don't we see the reflected photon with the left momentum? Also when the electron goes down to a lower energy state and releases a photon, how exactly does that happen? Where does the photon pop up from?

John Rennie

@NovaliumCompany when there is no photon the electron is in an energy eigenstate i.e. a solution to the Schrodinger equation for the atom. OK so far?

Novalium Company

Nope, didn't understand that sorry. Do you mean when no photon is hitting the electron, the electron is like in a default energy state?

John Rennie

For a hydrogen atom (to keep this simple) the Schrodinger equation is:

$$ \frac{-\hbar^2}{2m}\nabla^2\psi + V\psi = 0$$

Where $V$ is the potential energy, which is just the usual electrostatic potential:

$$ V = -\frac{ke^2}{r} $$

Novalium Company

But when the electron is at it's default state and "can't go more in to the nucleus" does that mean that it has no V?

John Rennie

When we solve this equation we get the energy states of the hydrogen atom i.e. the $1s$, $2s$, $2p$, etc.

@NovaliumCompany default state?

Novalium Company

8:49 AM

@JohnRennie The energy level where the electron can't go no lower.

John Rennie

That's the $1s$ state. Are you thinking the lowest energy level should be when the electron is sitting directly on the proton?

Novalium Company

Also, the nucleous has protons, which should attract the electrons, why don't the electrons collapse into the nucleus?

John Rennie

5

Q: What prevents an atom's electrons from "collapsing" onto its protons?

Forgive me if the answer to this is obvious. I have no formal physics training, and I remember that when I asked my physics teacher this, she just frowned and said "Good question." An electron is negatively charged. A proton is positively charged. Based on basic principles, it seems like it w...

quantum-mechanics electromagnetism nuclear-physics electrons atomic-physics

Novalium Company

Simple answer, please? :p

John Rennie

That's about as simple as it gets ...

@NovaliumCompany If the electron is sitting stationary on the atom then we know precisely where it is, so the uncertainty in its position is zero. Yes?

Novalium Company

8:52 AM

Yes but we don't know it's momentum then?

due to $\Delta x\Delta p>=\frac{h}{4\pi}$

John Rennie

Correct. The uncertainty principle tells us that $\Delta x \Delta p \approx \hbar$ so if $\Delta x = 0$ that means $\Delta p = \infty$

Novalium Company

Yep

John Rennie

But energy and momentum are related by $E = p^2/2m$, so if the uncertainty in the momentum is infinite the uncertainty in the energy must also be infinite.

Novalium Company

Got it, what's that formula called for the energy (I'll google)

John Rennie

The equation just comes from the kinetic energy $E = \tfrac{1}{2}mv^2 = \tfrac{1}{2}m^2v^2/m = (mv)^2/2m = p^2/2m$

Novalium Company

8:56 AM

Got it

John Rennie

Anyhow, the only way the uncertainty in the energy can be infinite is is the energy is infinite. So having the electron stationary on the proton would be a state that has infinite energy.

Novalium Company

Ok? But what's that distance between the nucleus and $1s$?

David Z

@JohnRennie I've seen enough people get confused by the finite version of that statement that it bothers me somewhat

John Rennie

@DavidZ white lies :-) We can apologise for them when the students start their first proper QM course.

Novalium Company

I think I'll settle down on just "that's how the universe works" answer. What about the original question, or it's also complicated?

David Z

9:02 AM

@JohnRennie That's rather a large white lie ;-)

John Rennie

@NovaliumCompany where I came in was that with no photons present the electrons have well defined energy levels i.e. the $1s$, etc. And if you put the electron in one of those states it just stays there.

Novalium Company

Doesn't it collapse to a lower state?

I thought the universe likes to get rid of it's energy :D

John Rennie

Where the photon changes things is that a photon has an electric field and that electric field exerts a force on the electron. In our original Schrodinger equation we had the potential $V = ke^2/r$, but the electric field of the photon changes this potential.

Novalium Company

So photons are charged?

John Rennie

The $1s$ etc states are the states for the original Schrodinger equation, and when the presence of a photon changes the Schrodinger equation our original $1s$ states are no longer solutions to the new Schrodinger equation modified by the presence of the photon.

@NovaliumCompany see:

1

Q: Where do photon gets its electric field without carrying charge?

Electric field is determined by the electric force per unit charge, but photon clearly doesn't carry charge so where do the electric field came from?

electromagnetic-radiation photons electric-fields

I'm afraid this is getting to be a bit of a rabbit hole :-)

Novalium Company

9:08 AM

Yep :p But I understood what you are saying. The incoming photon changes the electron's V due to the photon's electric field.

John Rennie

OK, the solutions to the Schrodinger equation depend on the potential. Without the photon the solutions are the $1s$ etc, but once the photon changes the potential the solutions are no longer the $1s$ etc. Instead they will be some new wavefunction $\psi$.

OK so far?

Novalium Company

Ok

John Rennie

We don't know $\psi$, but what we can do is express it as a sum of the original solutions i.e.

$$ \psi = a_1 1s + a_2 2s + a_3 2p + etc $$

where we adjust the coefficients $a_1$, $a_2$, etc to make this equation correct.

So our new wavefunction is a mixture of the original $1s$ etc states

Novalium Company

Yep

John Rennie

Now suppose we make an observation of the atom state. Our observation collapses the superposition of the electron states, so we find the atom is back in a $1s$, etc state but not necessarily in the original state. There is a probability $a_2^2$ that we will find it in the $2s$ state, a probability $a_3^2$ that we'll find it in the $2p$ state etc.

Novalium Company

9:18 AM

What is $a_2^2$ and $a_3^2$ and $2p$?

(my mind is about to blow, literally)

John Rennie

5 mins ago, by John Rennie

$$ \psi = a_1 1s + a_2 2s + a_3 2p + etc $$

Novalium Company

Why do you square it? To make it a probability density of position?

John Rennie

@NovaliumCompany I think the problem is that without a basic knowledge of quantum mechanics everything I say sounds like voodoo magic :-)

Novalium Company

I'm currenly reading a book, so things are about to get better :P

I forgot which question you are answering :D

Bcs as u said, a rabbit hole... and it get's deeper.

John Rennie

Basically if you start with a 1s state and a photon then the interaction between them mixes them up. When you observe the new state it could collaspe back into a 1s state and a photon, i.e. nothing happened, or a 2p state and no photon, i.e. the photn got absorbed and excited the atom from the 1s to 2p state.

Novalium Company

9:23 AM

What is that 2p state? what is the p, I though it was s?

Ohh, 2p for the uncertainty of the new wavefunction on the new uncertain second level or something :P?

John Rennie

For reasons that will take us even deeper down the rabbit hole a photon cannot excite an electron from a 1s state to a 2s state.

The 1s state can only go to a p state i.e. 2p, 3p, 4p and so on.

Novalium Company

Is there like a diagram of those somwhere?

Like, where does the p state go to?

I feel like physicists will always find a way to make things even more complicated...

John Rennie

If you google hydrogen atom orbitals or something like that you'll find lots of articles describing the states.

Novalium Company

Got it, shall we continue?

John Rennie

@NovaliumCompany actually physicsts spend their time trying to make things simpler. This all looks complicated because you haven't learned the basic principles that underlie it. Those principles are surprisingly simple.

Novalium Company

9:28 AM

Alright, let's just get out of the rabbit hole and I'll make sure I do my homework before I ask any more questions. :)

John Rennie

:-)

Novalium Company

:--)

@JohnRennie By getting out of the rabbit hole I mean continuing with the explaining, because I'll be unsatisfied the whole day otherwise :D If don't have the nevers to continue, please say so.

John Rennie

I've lost track of where we got to now.

Novalium Company

I forgot even what question you were answering... xD

The incoming photon transfers part of it's momentum to the electron, correct?

John Rennie

@NovaliumCompany the incoming photon becomes entangled with the electron so now they aren't separate objects. Instead the form a single system where the two particle are mixed up.

Novalium Company

9:37 AM

pholectron?

xD

John Rennie

This new state can then evolve in time. It can evolve back to the original state i.e. nothing has changed - the photon wasn't absorbed.

Or it can evolve to a new electron state and no photon i.e. the photon got absorbed and transferred its energy to the electron.

Novalium Company

@JohnRennie The photon gets reflecated as it hasn't hit the electron at all?

John Rennie

@NovaliumCompany when you get down to length scales where quantum effects dominate the particles are not like little balls that collide. They are more like fuzzy objects with only vaguely defined positions.

Novalium Company

@JohnRennie If you observe them, they have positions, if you don't observe them, properties such as position, momentum don't even exist, right?

John Rennie

The photon overlaps the electron and starts interacting with it, and instead of two separate fuzzy clouds you get a single merged fuzzy cloud that contains both the electron and the photon.

@NovaliumCompany correct. More precisely the particle exists in a superposition of many different positions and many different momenta

Novalium Company

9:42 AM

@JohnRennie Ahh, I love QM

So what happens in the cenario where the electron and photon have become "one" and are now at a higher energy state? (meaning, the photon's frequency was high enough to not dismiss the electron which will bring us to the first cenario)?

John Rennie

@NovaliumCompany it sounds like voodoo magic, but the mathematical description of this is very simple once you've learned it.

@NovaliumCompany this new state is unstable and evolves in time.

Novalium Company

@JohnRennie Does it at some point collapse in a way that the electron goes back to it's original s1 and the photon is emitted with energy equal to the potential energy lost due to the electron's "falling"?

John Rennie

@NovaliumCompany The collapse can take place in various ways and we can calculate the probabilities for these different collapse routes.

What we find is that unless the photon energy matches an energy difference between two energy levels of the atom then the merged system evolves back into the original state and the original photon with a very high probability i.e. nothing happened. The photon wasn't absorbed and heads off leaving behind the atom in its original state.

Novalium Company

If the photon's energy is not enough to take the electron-photon merged system to the next energy state, the electron will stay in it's current state while the photon just dissapears?

John Rennie

> the photon just dissapears

The photon carries on travelling away from the atom in the same direction that it came in. It doesn't disappear in the sense of cease to exist.

Novalium Company

9:58 AM

So in simple sense, if the incoming photon's energy is not enough to get the electron to the next energy state, the photon and electron will just pass through each other?

John Rennie

@NovaliumCompany yes, exactly

Novalium Company

Alright @JohnRennie. I can't tell you how thankful I am for your existance. Here's my plan: For now I'll continue to learn the basic quantum mechanics stuff (read my book) and hopefully the answer to the more complicated questions will unravel in the meanwhile.

John Rennie

@NovaliumCompany My number one guideline is to read what you enjoy reading. If you're enjoying reading QM books then go for it, and you can always ask here if you get stuck.

Novalium Company

@JohnRennie Alright :) And again, thank you!!!

pss 1

Why does the atom lose energy when an electron is added to it....

Novalium Company

10:03 AM

@pss1 Another rabbit hole incoming xD

pss 1

Then is it true that electron while moving towards a positive charge (isolated from everything ...(atom too)....loses energy (I mean physically giving energy as some radiation or.......

Novalium Company

@pss1 As far as I know, the electron has negative potential energy, which means that it's "bound". But don't listen to me!

pss 1

Yeah....and when it moves Towards the positive charge it’s change in potential energy is -be meaning it lost energy so we might think it gains kinetic energy but the potential energy is negative..how does that make sense ... so what my notion is that the electron is said to have some pseudo energy meaning, the energy it has to gain to move to infinity...

pss 1

10:39 AM

So accordingly how to proceed

Abhas Kumar Sinha

12:53 PM

@JohnRennie This guy has been a blessing to the life of many high school students here. Whenever they get lost, he always helps them. No wonder that someday, he becomes an inspiration for high school students to become a great Physicist. :)

Novalium Company

1:57 PM

@AbhasKumarSinha I couldnt agree more. @JohnRennie just know that there are people in the world that admire you and are thankful for all your help. You are really helping me, and others, in every problem we have. So THANK YOU!!!

You can go to bed every night, knowing that you have helped someone and made the world a better place. :)

@JohnRennie I will remember you and hopefully I can meet you in real life some day.

Novalium Company

2:14 PM

Also, how can empty space have zero-point energy when there are no particles to have that minimum amount of energy?

Abhas Kumar Sinha

2:39 PM

@NovaliumCompany Copy that Bro :)

Manvendra Somvanshi

@NovaliumCompany empty space isn't really empty.

John Rennie

@NovaliumCompany that's a good question, but a surprisingly complicated one

@NovaliumCompany if you dare (because it's complicated :-) have a look at my answer here:

16

A: Are vacuum fluctuations really happening all the time?

I think it’s possible to give a beginners guide to what is meant by vacuum fluctuations, but it necessarily involves taking a few liberties so bear that in mind in what follows. Before we start let’s remind ourselves of the following key point about superpositions. Suppose we have an operator $\...

Abhas Kumar Sinha

@JohnRennie Should I learn QM or GR first?

John Rennie

@AbhasKumarSinha are you learning for fun on your own?

Abhas Kumar Sinha

2:54 PM

@JohnRennie Yes, also I know a bit of Lagrangian and Hamiltonian Mechanics and for GR, I don't have anything such, except a very bit of knowledge about Rel. Mechanics (Lorentz Transformation and stuff)

John Rennie

I would start on QM as GR uses a branch of maths that you'd need to learn (differential geometry)

Abhas Kumar Sinha

@JohnRennie What are mathematical background required for QR and GM?

John Rennie

i'm not the best person to advise on learning QM. @DanielSank or @dmckee would be good people to ask.

Abhas Kumar Sinha

@JohnRennie For GR? What are the requirements?

John Rennie

@AbhasKumarSinha just basic calculus really

Abhas Kumar Sinha

3:01 PM

@JohnRennie For GR, basic Calculus!? But, isn't that hard?

John Rennie

GR is not as hard as you think.

Abhas Kumar Sinha

@JohnRennie People claim GR is harder than QM and is very hard....

John Rennie

I would have said QM is harder than GR. QFT is certainly harder than GR.

Abhas Kumar Sinha

@JohnRennie Oh okay, that's another level, so, QFT doesn't contradict GR?

pss 1

May I know why my question is not being answered.....(sorry If I interrupt)...but please....doubt which is actually confusing (though basic if you feel.....after all I’m high school)

Abhas Kumar Sinha

3:05 PM

@pss1 What's your question, I've not seen it, If you can link please?

John Rennie

What you've probably heard is that quantum gravity doesn't seem to work, so that extent there is a problem applying quantum mechanics to GR. But I wouldn't say they contradict each other.

@pss1 this?

5 hours ago, by pss 1

Why does the atom lose energy when an electron is added to it....

pss 1

Yeah

Emilio Pisanty

110

Q: A list of inconveniences between quantum mechanics and (general) relativity?

It is well known that quantum mechanics and (general) relativity do not fit well. I am wondering whether it is possible to make a list of contradictions or problems between them? E.g. relativity theory uses a space-time continuum, while quantum theory uses discrete states. I am not merely looki...

quantum-mechanics general-relativity special-relativity quantum-gravity faq

John Rennie

5 hours ago, by pss 1

Yeah....and when it moves Towards the positive charge it’s change in potential energy is -be meaning it lost energy so we might think it gains kinetic energy but the potential energy is negative..how does that make sense ... so what my notion is that the electron is said to have some pseudo energy meaning, the energy it has to gain to move to infinity...

Emilio Pisanty

@JohnRennie @pss1 Not quite. The system itself, which goes from the configuration (positive ion + electron) to the configuration (neutral ion) loses energy in that transformation.

this can be by emission of a photon (known as photorecombination), if there is nothing else around to carry the energy

John Rennie

3:10 PM

@pss1 one of the weird things about potential energy it that its absolute value is not defined. That is, PE is never negative or positive or even zero. All we can ever define is a difference in PE.

Emilio Pisanty

it can also be by shoving that energy off into some other system (say, another nearby atom) if those channels are available.

@rob it sounds like that took a pretty good dive ;-)

Re your last paragraph (v1, "climb the hill"): you have to be careful whether the masses you find are nuclear masses, whose measurement would require a completely-stripped nucleus in a mass spectrometer, or the masses of neutral atoms, including all the electrons and the electronic binding energies. The "mass excesses" I use in my answer are for neutral atoms, and those data make it clear that the reaction $$\rm ^{124}Xe + 54e^- \not\to {}^{124}I + 53e^- + \nu_e$$ is energetically forbidden. — rob ♦ 20 hours ago

I'm not sure I understand the details, though

there is a common term of $53 m_e$ which obviously doesn't change anything

you're saying that the electronic binding energies make a difference?

I guess 200keV isn't that crazy if there are tightly-bound electrons involved

but it still sounds weird to me

pss 1

4:05 PM

@EmilioPisanty yeah what about the neutral to ion case

John Rennie

@pss1 suppose you start with the ion and the electron widely separated i.e. effectively infinitely separated.

We tend to define the potential energy as zero for this case, but that isn't an absolute value for the PE, it just means we'll be measuring changes in the PE from this initial state.

@pss1 OK so far?

pss 1

Yup

John Rennie

@pss1 if we now let the electron fall towards the ion then the PE decreases and the KE increases. The total energy stays the same because no energy is being added or removed from the system.

So when the electron is near the ion the PE has decreased but the KE has increased by the same amount.

The trouble is that in this situation the electron has too much KE to bind to the ion. Basically it's moving too fast so it flies past the ion and speeds away to infinity again.

It isn't possible to form the neutral atom unless there is some way to take energy out of that system i.e. to reduce the electron KE enough for it to bind to the ion.

pss 1

@JohnRennie Yeah doubt with the 2nd statement!

John Rennie

Yes?

pss 1

4:13 PM

Is it not like the potential energy is going to -ve terms(becoz it decreased means that it goes to less than zero right....) just to increase the kinetic energy.......

I mean it’s physicslly had

John Rennie

> becoz it decreased means that it goes to less than zero right....

The PE isn't really zero at infinity. The PE doesn't have an absolute value. It's just convenient for us to define it as zero there.

So the fact that the value of the PE is a negative number doesn't signify anything deep. It just means the PE is less that it was at infinity.

pss 1

so say we started with something huge as potential energy then it kind of feels good right? Imeam we could have some potential energy left such that kinetic energy is also present

@JohnRennie Yeah though continuing as you said ...the 4th statemtn

John Rennie

This one?

7 mins ago, by John Rennie

It isn't possible to form the neutral atom unless there is some way to take energy out of that system i.e. to reduce the electron KE enough for it to bind to the ion.

pss 1

Ya

How do we extract that energy now

@JohnRennie Will the system try to lose itself (by photrecomb.....as mentioned )...so to form a neutral and stable compound....

John Rennie

There are various ways the excess energy can be lost. For example it could be emitted as a photon, or the ion/electron could collide with something, e.g. another atom, and transfer energy to that other atom as kinetic energy.

pss 1

4:22 PM

Why it should lose energy ....I mean the electron could go far from atom (to infinity)....is it to form a stable neutral atom?

DanielSank

@JohnRennie Hidee ho there.

@AbhasKumarSinha What the heck are QR and GM?

5

John Rennie

@pss1 yes. When the electron reaches the ion it has gained too much speed to stick to the ion and form a neutral atom. It can only form a neutral ion if the kinetic energy of the electron can be removed in some way.

@DanielSank ah yes, I hadn't noticed. An acronym spoonerism :-)

pss 1

Is there also possibility that electron could escape to infinity again instead sticking to atom

John Rennie

@pss1 yes indeed. That may well happen.

DanielSank

General mechanics and quantum relativity?

2

John Rennie

4:26 PM

@pss1 in that case the electron will keep going till it meets the next ion and tries again.

pss 1

Thank you very much sir!

John Rennie

Chewing versus sex in duck-billed dinosaurs. Now there's a headline you don't see very often.

user8718165

@JohnRennie Good evening :-) Will you be there tomorrow morning?

John Rennie

@user8718165 yes, I'll be around from about 5 a.m. UK time as usual.

user8718165

@JohnRennie Ok thanks :-)

John Rennie

4:33 PM

@user8718165 or if it's a general question you can ask now

user8718165

@JohnRennie I'm having trouble with circuits...solving equivalent resistance:-(

pss 1

Any idea ? physics.stackexchange.com/questions/477846/…

John Rennie

@user8718165 you mean using things like star-delta transformations?

@pss1 what do you mean by:

> radiation is a volumetric phenomena

pss 1

Like the emission of radiant energy is by volume not by surface

Imean the photoelectric effect is surface phenomena....

Like that radiation is volumetric ...

My physics teacher said so actually....

user8718165

@JohnRennie this one sir...not getting it imgur.com/lrmQwpo

John Rennie

4:41 PM

How about this analogy: water is volumetric in the sense that it occupies a volume and the volume occupied is proportional to the amount of water present. OK so far?

pss 1

Ah yeah

user8718165

@JohnRennie I know the equations of star-delta transforms...though I haven't used them much till now:-)

John Rennie

@pss1 But suppose have water flowing at speed v out of a pipe with area A. The flowrate is equal to $vA$ i.e. the flow rate is proportional to the area of the pipe.

It's basically the same with radiation except that the speed is always the speed of light.

If you imagine the radiation as flowing out of the surface then it's obvious the total amount of radiation is going to be proportional to the surface.

Sir Cumference

Election ends in two days

John Rennie

@user8718165 tomorrow. It's actually not that hard, but it's too late in the day for me to want to start doing calcultaions.

Sir Cumference

4:46 PM

Wonder who'll win

user8718165

@JohnRennie I'm also too tired...just need your help...anytime:-) Good night sir

John Rennie

See you tomorrow then. I'm off now as well.

pss 1

Nice one

pss 1

5:13 PM

Ya now conclusion:fluid flow is also kinda volumetric phenomena which depends on surface area....thus so is the radiation

PM 2Ring

5:33 PM

@SirCumference We're all waiting with bated breath. :rolls eyes:

rob

@EmilioPisanty You made me look: the K-shell binding energies are about 30 keV. The problem is stoichiometry. I left you another comment.

Sir Cumference

5:51 PM

@PM2Ring Here's hoping for NewAlexandria

Down with the aristocracy

The top 1% of users own 50% of the rep

PM 2Ring

@NovaliumCompany SR might turn out to be false Oh, we knew SR was false the day it was published, in the sense that it's a simplified model that assumes spacetime is flat, i.e., it totally ignores gravity. But it's still very useful, because there are plenty of situations involving high speed where we can ignore gravity.

Also, GR says that if you zoom in close enough to curved spacetime it looks flat, and you can then use SR on that flat section. That's just like how we make a useful atlas of flat maps of the curved Earth.

@SirCumference I wish them well, but I'll be extremely surprised if they get many votes... but you never know.

Sir Cumference

@PM2Ring If they win, it'll say something about our method of electing moderators

It's a long shot but who knows lol

PM 2Ring

@rob I was always taught that nuclear processes happened at such a high energy level that they were virtually impervious to anything that affected the electrons. So it came as a bit of a shock to learn that ionization & chemical bonds can have a big impact on electron capture decays, but of course it makes perfect sense.

rob

@PM2Ring Well, most "chemistry" is eV-scale interactions among valence electrons, so that advice is totally right.

PM 2Ring

@pss 1 I think you'll find John Baez's article about torsors helpful.

rob

6:05 PM

There is a lightweight nucleus which is unstable against electron capture, but stable against positron emission. So it doesn't occur in chemical sample from Earth, where atoms are neutral, but the bare nucleus is a component of cosmic rays.

But the cases where that happens are pretty rare.

PM 2Ring

@rob True, and even with EC nuclei electron stuff normally only changes the half life by 1% or so. Unless EC is the only decade mode, and you totally ionise the atom to a bare nucleus. ;)

rob

@PM2Ring Just so.

PM 2Ring

I remember reading something a few months back about a fairly recent experiment where that total ionization effect was demonstrated, but I'm not having any luck finding it again. But Wikipedia does list a bunch of nuclides that do EC but the energy is too low to do positron emission, eg rubidium-83.

I think it's cool that K-40 does both negative & positive beta decay, although the positive is usually through EC (and gamma emission) rather than positron emission.

heather

6:25 PM

morning

ACuriousMind

evening :)

PM 2Ring

6:36 PM

Hi @heather We haven't seen in you in the Python room for a while.

ACuriousMind

Dammit, the firefox extension signing thing has robbed me of my auto-comments

danielunderwood

@ACuriousMind isn't that fixed now? My extensions are all back now it seems

ACuriousMind

@danielunderwood On Debian, I get only the ESR release for which no fix has been provided yet

rob

6:56 PM

@ACuriousMind Man, I love a lot of things about debian, but Firefox ESR is not one of them.

ACuriousMind

aye

danielunderwood

Debian users?!

ACuriousMind

@danielunderwood ...is there something wrong with that?

Novalium Company

youtube.com/watch?v=mPbtR4vorgY At 5:10, does this guy use anything else besides the hydraulic cylinder and a solenoid to make the cylinder work? But he sure must have a pump right? Also, how does he reverses the flow of high pressure gas (I suppose it's gas) that's coming into the two channels of the press?

danielunderwood

@ACuriousMind very!

really I'd probably be much happier with debian than my current broken/addon-filled ubuntu nonsense

heather

7:14 PM

@PM2Ring i was just in yesterday =)

Novalium Company

Where's @Blue? :P

heather

I think he's going by @SanchayanDutta now

Novalium Company

@SanchayanDutta Where are you buddy?

heather

they've been doing a lot on quantum computing, so.

Novalium Company

@heather youtube.com/watch?v=mPbtR4vorgY at 5:10, this guy uses a hydraulic cylinder, a solenoid, a pump and a valve to make the cylinder work?

heather

7:27 PM

@NovaliumCompany that was...out of context

why do you ask?

Novalium Company

Because I don't know.

I'm just learning about hydraulic systems and it looks to me like he is hiding a pump and a relief valve?

heather

i honestly probably know less than you do.

i just know what it says in the video.

Novalium Company

oh ok

how old r u?

heather

mid-teens

not going to post my exact age on the internet, but i'm in highschool

Novalium Company

me 2

that's nice

heather

7:30 PM

are you studying hydraulics for fun, or for a specific purpose?

Novalium Company

fun

I'm 16

soon 17 :P

heather

ah, nice.

Novalium Company

I'm just having a hard time finding friends with my interests, so I study stuff

although I do have a few friends with not my interests but still satisfy my social needs which are a must-have in order to not go crazy.

@heather Why are you here in general, just to ask questions when you have problems?

Captain Bohemian

can friends not sharing similar mental interests as yours satisfy your social need? I doubt.

Novalium Company

Well, they don't... but it's still better than being alone (which I used to be). Just 2 years and I'll finally be able to travel and meet with people which actually talk interesting stuff. My current friends mainly talk about where to go and get drunk, what cigarets they smoke...

I live in Bulgaria and you hardly find friends (my age) with my interests...

But I'll deal with it the next 2 years and that's it. :P

Captain Bohemian

7:44 PM

yea, I used to be too lonely because I had no colleague at all in my research institute so I talked with the security guard of that institute.

but the security guard can't understand a lot of things I want to talk about. He even doesn't know much English, so I often got frustrated.

Mithrandir24601

@heather I don't think so in chat

Novalium Company

I've come to the point where the only criteria for someone becoming my friend is mutual interest... at least one... please ;D

But again, friends which you have nothing in common (or very less) is still better than no friends.

PM 2Ring

@heather Ah. I have to catch up on the Python room transcript. :)

heather

7:59 PM

@NovaliumCompany studying is interesting, but i'm sorry you're having a hard time.

@NovaliumCompany well, partly because i enjoy the community, partly because physics is cool. partly to hang out.

and yeah, partly to ask questions =)

Novalium Company

8:54 PM

@heather Oh, dont be sorry. Life is tough sometimes and we just have to deal with it. Thats what makes us stronger.

Captain Bohemian

@NovaliumCompany actually I don't treat all people talking with me as my friends. When I am too lonely but can't find people to talk to meaningfully, I may quarrel with people who I treat as my enemies.

Sir Cumference

9:54 PM

Spent 4 hours trying to solve a proof in Spivak before I realized the claim is incorrect

This book is the worst

heather

@SirCumference which one?

his new one?

Sir Cumference

@heather Calculus on Manifolds

heather

@SirCumference ah. funny, i've heard good things about that book.

Sir Cumference