The h Bar

Abcd

6:00 AM

Can you explain without going into the details of maths?

Sid

He he he..

Nothing easier? Or more instantaneous?

John Rennie

@Abcd I'm not sure what the difference between those three is.

Aren't they all the same thing?

Abcd

@JohnRennie Just a minute. Let me upload an example picture.

anna v

@DavidZ I no longer see links to META messages in the main page . Also my reputation is not highligheted when an answer is checked. I just got an uptick physics.stackexchange.com/questions/342011/… and it did give the checks on the answer but it does not appear on the achievements (next to the inbox on top).

Abcd

@JohnRennie Here's an example picture^^ . how do I ascertain the particular orbital from the graphs?

John Rennie

6:06 AM

@Abcd OK. The hydrogen wave function can be written as a radial part and an angular part: $$ \Psi(r,\theta,\phi) = R(r)Y(\theta, \phi) $$

Abcd

@JohnRennie Yes,

John Rennie

The top graph shows the radial function $R(r)$

Abcd

Alright.

John Rennie

These functions are the Laguerre polynomials

Abcd

ok

John Rennie

6:09 AM

Was there anything more you wanted to ask about $R(r)$, or shall we move on to the probability graphs?

Abcd

@JohnRennie let's move to the graphs.

John Rennie

OK. The function: $$P(r) = \Psi^*(r) \Psi(r)$$ is the probability density.

Abcd

@JohnRennie ok.

John Rennie

If you consider some small volume $dV$ then $P(r) = \Psi^*(r) \Psi(r) dV$ gives the probability of finding the electron in the small volume element $dV$.

Abcd

@JohnRennie Ok.

John Rennie

6:12 AM

That's why it's called a density, because it is a probability per unit volume i.e. multiply it by a volume $dV$ and the get the probability the electron is in that volume.

So the second set of graphs is just the top set of graphs squared.

OK so far?

Abcd

yes

0celo7

John Rennie

The last set of graphs is the one that tends to confuse people. Beause it is showing something slightly different.

The last set shows the probability that the electron is at a distance between $r$ and $r+dr$ from the nucleus.

Abcd

@JohnRennie ok. How to interpret which subshell/orbital we are referring to from any given graph?

John Rennie

@Abcd Look at the top row i.e. the graphs of $R(r)$.

Abcd

6:15 AM

@JohnRennie ok. Next?

John Rennie

If you look at the $s$ orbitals the $1s$ has no zeros, the $2s$ has one zero, the $3s$ has two zeros and so on.

So if it's an $s$ orbital the number of zeros tells us which $s$ orbital it is.

And we can identify the $s$ orbitals because they have a non-zero value at $r=0$.

So you can tell immediately from the graph if it's an $s$ orbital and if so which $s$ orbital it is.

Abcd

@JohnRennie can you tell briefly what radial wave function gives us (don;t give the mathematical definition please..)

@JohnRennie Understood.

John Rennie

@Abcd the radial wave function doesn't have a simple physical interpretation because it's not something an experiment can ever observe directly.

The square of the wave function has a physical interpretation as a probability density.

And various other functions of the wavefunction have physical interpretations, but the wave function does not.

Abcd

What is it?

John Rennie

It is just a function that contains all the information about the system and from which we can calculate observable quantites.

Abcd

6:19 AM

Just tell me what R on the Y axis in the first graph indicates.

John Rennie

@Abcd there isn't an answer to that. $R$ is just a part (the radial part) of the wave function.

Abcd

@JohnRennie Ok. How do I interpret the p orbitals from this graph?

John Rennie

@Abcd what do you mean by interpret the p orbitals?

Abcd

@JohnRennie How do I identify which p orbital is represented by a given graph?

John Rennie

Remember what I said about the $s$ orbitals and the number of nodes?

Abcd

6:23 AM

Okay. So 2p has zero nodes @JohnRennie?

Therefore answer is A ?

John Rennie

@Abcd Correct. The $3p$ has one node, the $4p$ two nodes and so on

And all $p$ orbitals have a zero at the origin i.e. at $r=0$

Abcd

@JohnRennie Another thing: but p orbitals have one nodal plane :(

John Rennie

Remember that at the moment we are only considering the radial part of the wavefunction. But there is also the angular part $Y(\theta,\phi)$.

The $s$ orbitals are spherically symmetric so we can just ignore the angular part of the wave function. However the $p$ orbitals are not spherically symmetric.

Abcd

@JohnRennie Okay. Understood :)

David Z

@annav That post is community wiki, you don't get any reputation for upvotes on it

John Rennie

6:26 AM

For the $p$ orbitals $Y(\theta,\phi)$ has a nodal plane.

Abcd

@JohnRennie Ok. Let's move to the next type of graph.

John Rennie

And that applies to all the $p$ orbitals. So we can nodes where $R(r)=0$ and also nodes where $Y(\theta,\phi)=0$.

@Abcd the middle row?

Abcd

@JohnRennie Yes

John Rennie

That's just $R^2(r)$ i.e. just the top graphs squared.

Abcd

@JohnRennie Can R also be written as $\psi$

John Rennie

6:29 AM

No, $\Psi = R(r)Y(\theta\phi)$. The function $R$ is just part of $\Psi$.

Abcd

@JohnRennie I am talking about the second set, not the third

anna v

@DavidZ fine, what about the link to meta on the main page?

John Rennie

@Abcd Yes.

Abcd

@JohnRennie But you said "top two graphs"

John Rennie

@Abcd no I didn't. Where did I say top two graphs?

David Z

6:31 AM

@annav No idea. Perhaps you could post a bug report on meta with a screenshot?

Abcd

@JohnRennie I am extremely sorry. I misread this as top two.

@JohnRennie Then how do I solve this question using the method we discussed because here we have psi instead of R?

John Rennie

That's a misleading graph. The $y$ axis has been labelled $\Psi$ when it's really $R(r)$.

Abcd

@JohnRennie Ok.

@JohnRennie Next one then

@JohnRennie There are three zeroes here :(

How to identify the correct graph?

John Rennie

When we use the term node we normally ignore $r=0$ and $r=\infty$. That is we only consider zeros at some non-zero finite value of $r$.

So only one of those graphs has one node.

anna v

@DavidZ OK . Had trouble finding meta link, that is why I called you on the h bar. found meta in my profile.

Abcd

6:35 AM

@JohnRennie a,d have one node

John Rennie

The graphs are poorly drawn. For example (a) shows a $1s$ orbital and that only goes to zero at infinity. The graph makes it look as if it goes to zero at a finite value or $r$, but it doesn't.

That applies to all the graphs. That's a really poorly written question.

Abcd

@JohnRennie Alright. let's move to the next set of graphs (of R^2)

John Rennie

Before we move on, can you now answer that question?

Abcd

@JohnRennie I can't.

John Rennie

Well, ignore any zeros at $r=0$, and the graphs aren't really zero at large $r$. So there is only one graph that has a single point where $r=0$.

Abcd

6:40 AM

@JohnRennie Is it considered zero when the graph line and the X axis simply touch as in the case of a and d

John Rennie

Remember that graphs appear to touch zero at the right end of the $r$ axis because they are badly drawn. They should tend to zero asymptotically as $r \rightarrow \infty$.

Abcd

@JohnRennie So b is the answer. c has two nodes, am I right?

John Rennie

So (a) really has no zeros and (d) has only the one zero at $r=0$.

Yes, (b) is the answer. And you should be able to tell me what orbital (b) shows.

Abcd

@JohnRennie It can be 2s or 4d

John Rennie

Why could it be $4d$?

$d$ orbitals are like $p$ orbitals in that they always have $R(0) = 0$.

Abcd

6:44 AM

@JohnRennie 4d has (n-l-1) = (4-2-1) = 1 node. Therefore it can be 4d

John Rennie

You're quite correct that $4d$ has one node, just like $3p$ and $2s$. But $d$ orbitals are zero at the origin.

Abcd

@JohnRennie and s orbitals don't have R(0) = 0 , right?

@JohnRennie Understood again.

John Rennie

Yes $s$ orbitals are the only ones that are non-zero at the origin

So (b) is a $2s$ orbital.

Abcd

@JohnRennie Correct. Let's move to the next set

John Rennie

OK

Abcd

6:48 AM

@JohnRennie It's simply the square of the first. Thus, all the negative stuff would come above the X axis and we are done. Am I right?

0celo7

@JohnRennie if I just turn off the monitor while the downloads are going everything should be fine, right?

John Rennie

@0celo7 Yes.

@Abcd Yes.

0celo7

cool

night

Abcd

@JohnRennie Thanks. What about the third set of graphs? how do I identify them?

John Rennie

The third set are a bit more complicated.

Abcd

6:52 AM

oh

John Rennie

Remember that I said $\Psi^2 dV$ is the probability of finding the electron in the small volume $dV$?

Abcd

@JohnRennie what is d in your formulas?

John Rennie

$dV$ means an infinitesimal volume element. The prefix $d$ basically means infinitely small.

Abcd

@JohnRennie ok

John Rennie

So $d$ doesn't mean anything on its own

Abcd

6:54 AM

@JohnRennie Understood

John Rennie

$dV$ means an infinitely small volume, and as we'll see in a moment $dr$ means an infinitely small distance along the $r$ axis.

Abcd

ok

John Rennie

Anyway, are you happy that $\Psi^2 dV$ is the probability of finding the electron in the small volume $dV$

Abcd

@JohnRennie Yes

John Rennie

OK, but the third graph doesn't show the probability of finding the electron in a small volume. It shows the probability of finding the electron at a distance $r$ from the nucleus.

Abcd

6:57 AM

@JohnRennie ok

John Rennie

To explain how this works I'll have to draw a diagram ...

Abcd

Ok

@JohnRennie Wait. I have one.

let me upload

@JohnRennie have a look at the top right corner

John Rennie

Ah, OK, yes that was the diagram I was going to draw. Here's my version:

Abcd

@JohnRennie Ok. Next?

John Rennie

The point is that the volume of the spherical shell from $r$ to $r+dr$ is $dV = 4\pi r^2 dr$

Is that OK so far? This is a key point so you need to be confident you understand it.

Abcd

7:04 AM

@JohnRennie Your diagram has r1,r2, and only dr instead of dr1, dr2. That's little confusing.

John Rennie

I was only part way through drawing the diagram when you uploaded yours. The point I was going to make is that the volume of the shell at $r_2$ is greater than the volume of the shell at $r_1$. Both shells have thickness $dr$.

Abcd

@JohnRennie Ok.

John Rennie

I was going to lead up to saying $dV = 4\pi r^2 dr$ but if you're happy with that we don't need to repeat it.

Abcd

ok

John Rennie

The probability is still $\Psi^2 dV$, but now we have $dV = 4\pi r^2 dr$. If we substitute for $dV$ we get the expression for the probability: $$P(r) = \Psi^2 4\pi r^2 $$

So the bottom row is just the middle row multiplied by $4 \pi r^2$.

Abcd

7:10 AM

@JohnRennie The probability is psi^2dV. Do I need to understand this or simply accept it?

John Rennie

That's a fundamental principle in quantum mechanics. It's called the Born rule.

So yes you need to just accept it because it's one of the fundamental building blocks of quantum mechanics.

Abcd

@JohnRennie Alright. Accepted.

John Rennie

Born rule

The Born rule (also called the Born law, Born's rule, or Born's law) formulated by German physicist Max Born in 1926, is a law of quantum mechanics giving the probability that a measurement on a quantum system will yield a given result. In its simplest form it states that the probability density of finding the particle at a given point is proportional to the square of the magnitude of the particle's wavefunction at that point. The Born rule is one of the key principles of quantum mechanics. There have been many attempts to derive the Born rule from the other assumptions of quantum mechanics, with...

Back to that bottom row of graphs.

Abcd

@JohnRennie yes

John Rennie

Because $P(r) = \Psi^2 4\pi r^2$ that means when $r=0$ the probability $P(r)=0$ (because anything multiplied by zero is zero).

So the main difference between the second and third row is that the graphs are always zero at $r=0$ even for the $s$ orbitals.

Abcd

7:14 AM

@JohnRennie is psi^2 same as R^2? Notations are confusing me

John Rennie

Good point, I should have written $R^2$ not $\Psi^2$.

Abcd

Yes.

John Rennie

Anyhow, there isn't a lot of difference between the shapes of the graphs in the second and third row except that they are always zero at $r=0$.

Abcd

@JohnRennie How is that possible? The middle graph indicates that there's probability of finding electron near the nucleus but the bottom graph denies that.

John Rennie

This is what usually confuses people.

Abcd

7:17 AM

oh

John Rennie

It's because the third graph shows the probability of finding the electron at a distance between $r$ and $r+dr$ from the origin.

And the volume of the shell from $r$ to $r+dr$ goes to zero as $r$ goes to zero.

So $R^2$ is indeed non-zero at the origin but $dV$ is zero.

And therefore the product $R^2 dV$ is zero at the origin.

Abcd

@JohnRennie Why is R^2 non zero at origin, if I may ask?

John Rennie

$R^2$ is non-zero for the $s$ orbitals.

Which is where the confusion arises.

Abcd

Why?

John Rennie

Do you mean why does the Schrodinger equation produce solutions with non-zero $R^2$ at the origin?

i.e. why is $R(0)$ non-zero for the $s$ orbitals?

Abcd

7:22 AM

@JohnRennie Umm... leave it. I will just accept it again.

Let's continue.

John Rennie

I was going to say you probably just have to accept it at your level :-)

Abcd

I understood the zero part of the third graph.

What next?

John Rennie

The third row graphs are pretty much like the second row excpet for the behaviour at $r=0$. The overall shape is a bit different because of that factor of $r^2$, but they have the same number of nodes.

Abcd

@JohnRennie Acceptes. Now here's my little summary:

The first set of graphs gives radial wave function R which (for time being) is meaningless and simply tells about the nodes in the orbital. The second graph which is The Plot of Radial Probability Density gives us the probability of finding an electron at a distance r from the origin. The third graph gives the probability of finding it in a spherical shell of thickness dr. And now I know how to interpret the graphs.

@JohnRennie ^^^^. Am I right?

John Rennie

You need to be a bit more precise about the second row.

Abcd

7:29 AM

How?

John Rennie

The second row gives the probability of finding an electron in a small volume dV at a distance r from the origin.

Abcd

@JohnRennie Okay.

@JohnRennie TYSM. You are really helpful and kind!!

John Rennie

Contrast this with the third row that gives us the probability of finding an electron in a spherical shell at a distance r from the origin.

Abcd

@JohnRennie Right.

John Rennie

That's the key difference that confuses people between the second and third rows.

Abcd

7:30 AM

ok

user029857

7:48 AM

I want to get in to qm. Which linear algebra and calculus books would you advise me to study?

Physics Meta

8:14 AM

0

Q: not seeing a meta link on main page

The physics Meta panel on the right is a chat , not active since june 24 bug? or it only appears when an interesting/popular subject is being discussed?

bug

Slereah

9:48 AM

Oh boy

I got a package from CERN

4

You know what that means!

The particle data group book is here!

oh no the chinamen have gotten to it

Aw yis

there's a big GR section

that's the good stuff

user685272

12:58 PM

why would you want to learn GR from the Chinese?

user685272

1:15 PM

how do you like your new computer so far? @0celo7

0celo7

1:33 PM

@user685272 I'm unimpressed by how easily I can use 80% of my GPU

I expected more from a 1080

but it works great

Slereah

1:48 PM

@0celo7 Jevon's paradox

0celo7

@Slereah pretty much

I'm using like 20% of my GPU for hair physics

Slereah

yeah

I'd rather see more ressources being used for the physics engine these days for like

Destructible environment

That would be fun

ACuriousMind

2:54 PM

@Slereah I think the problem there is not the engine but that you'd have to have graphics for everything in various states of destruction. That's a lotta work

Slereah

I'm not asking for anything too fancy

I just want to shoot at walls with my rockets

doesn't have to look great, it must just be satisfying

0celo7

@ACuriousMind the combat in TW3 takes some getting used to

ACuriousMind

Are you playing with a controller?

0celo7

I tried to kill a bunch of Drowners but had to give up

@ACuriousMind no

ACuriousMind

It's much better with a controller

0celo7

3:05 PM

Ugh. I don't want to buy a controller

The Dark Side

@Slereah :: Looks Amused:: You have a physical copy of Particle Data Booklet.

Just why?

Slereah

@TheDarkSide it's freeeee

The Dark Side

Is it?

How?

Who pays for printed pages?

I mean, do they recover the cost spent in printing the pages, or are they running for charity?

DanielSank

3:21 PM

Hi, everybody.

The Dark Side

Hello Dr. Sank.

DanielSank

What's all this video game nonsense? You guys should just play Smash Bros.

The Dark Side

What a nice name.

Smash whose bros?

DanielSank

Super Smash Bros. Melee

Super Smash Bros. Melee is a crossover fighting game developed by HAL Laboratory and published by Nintendo for the GameCube video game console. It is the second game in the Super Smash Bros. series, following the 1999 release of the original game. It was released in Japan and North America in 2001, and in Europe and Australia in 2002. The game features characters from Nintendo video game franchises such as Mario, The Legend of Zelda, Star Fox, and Pokémon. Melee includes all playable characters from the first game in the series on the Nintendo 64 and also adds new characters from franchises such...

Best multiplayer game ever.

The Dark Side

22 mins ago, by 0celo7

I tried to kill a bunch of Drowners but had to give up

Drowners ...

wink, wink @DanielSank :P

0celo7

3:30 PM

@JohnRennie gyazo.com/44c8d626f35192846628e61ca1648745

I'm the cool uncle :)

Slereah

3:42 PM

@TheDarkSide The TAX PAYERS

The Dark Side

@Slereah How does one get it delivered (to themselves) for free?

Direct answer please.

Slereah

You just order it on the official website

It takes a while, though

I waited 5 months for it

The Dark Side

You mean the order link here?

Slereah

yes

The Dark Side

OK. Let me try.

heather

4:40 PM

it's quiet today

dmckee

4:57 PM

Yeah. Too quiet.

::looks around for the action sequence::

Hmmm ... must be an art house film.

Sid

Aren't weekends supposed to be quiet?

Emilio Pisanty

5:31 PM

@dmckee tvtropes.org/pmwiki/pmwiki.php/Main/ItsQuietTooQuiet

you brought it on yourself

0celo7

5:50 PM

@TheDarkSide what?

@ACuriousMind I'm getting the hang of it. Killed the Griffin without trouble

Sid

you killed someone? :o

Abcd

Two bullets are fired simultaneously, horizontally and with different speed from the same place. Which bullet will hit the ground first?

Please explain.

Answer is both will hit simultaneously

Run like hell

Hey guys I have a question which is not really quantitative so I thought about asking here. My teacher was explaining time ago the basics of how the LIGO experiment worked detecting that famous merging of black holes. In my notes at a certain point I've written that since there were only two interferometers working, we could figure the distance of the event, but not the position, its direction. Why is it like that? And how many interferometers would we need to figure out the direction?

Sid

@Abcd yeah, they will

Abcd

@Sid Why???

Sid

5:54 PM

Though we are assuming that there is no resistive force

Abcd

Yeah that's fine

Sid

and horizontal=pure horizontal. There is no vertical component of velocity

Abcd

Right

Sid

From same place=Same height.

Abcd

yes

Sid

5:55 PM

Now, a ball is going to drop only if it experiences some force downwards.

What force?

Abcd

@Sid Gravitational acceleration

Sid

Correct!

Abcd

Then?

Sid

Now, use equation of motion. S=ut+1/2at^2

u=0.

Since vertical component of velocity=0

S=same. a=g.

Hence, time after which they will drop= same.

0celo7

@Sid why wouldn't I?

Sid

5:57 PM

However do remember that they will not drop at the same place.

Abcd

@Sid But in projectile motion we consider that no gravitational a acts on horizontal component..

I know we ain't talking about projectile

but I am referring to horizontal component of velocity

Sid

It's sort of projectile with Angle of velocity=0 degrees

Abcd

Ok.

Sid

I don't understand what you are asking

Abcd

@Sid No acceleration acts on horizontal velocity right?

Sid

5:59 PM

Gravity always acts on a body irrespective of whether it has any horizontal velocity or not

Emilio Pisanty

1

Q: LIGO Gravitational wave discovery - how did they know the cause of spike?

I understand how unbelievably lucky the discoverers were to catch the wave produced billions of years ago by an event that happens so rarely one hour into a test run of their equipment. But one thing is still not clear to me – how did they know what exactly caused the spike? Was that merely a con...

black-holes gravitational-waves ligo

Sid

Yes, along the horizontal direction there is no acceleration.

But, acceleration due to gravity exists

and that is the only cause of the dropping of the ball

horizontal velocity is just to confuse you..

Emilio Pisanty

Having three interferometers would significantly narrow down the probable origin region

Abcd

@Sid How S = same?

Sid

S= height from ground

Emilio Pisanty

6:00 PM

If the detectors were perfect, they would be enough to triangulate to a unique direction

Sid

Do remember that we are using equation of motion on vertical direction only

Emilio Pisanty

Real detectors are imperfect, of course

Abcd

@Sid oh, I see

Sid

Hence, u=0

Emilio Pisanty

Having two detectors restricts the signal to a cone in the sky

the finite width in the cone comes from uncertainty in the timing

Sid

6:02 PM

I had a dream last night that nuclear war will start soon..

Abcd

@Sid Yes. Thanks.

Emilio Pisanty

the uneven weight in along the cone comes from the fact that the waves are polarized

Sid

(I really hope I am wrong)

Emilio Pisanty

having a third detector would narrow it down to a square-ish region of the same width as the swathes in the image

Run like hell

@EmilioPisanty Thank you for explaining, I'll read your answer to the question, it's a bit clearer now anyway

Anonymous

6:08 PM

@Sid Not unbelievable. ¯\_(ツ)_/¯

Sid

lol..

Secret

6:34 PM

@Sid Doomsday dream are very common for me. I once had a dream where in the year 2019, the air on earth became so thin to breath that a lot died. In that dream one of the person who died is my mum

and my siblings and I have to figure our survival on our own

Sid

well... considering climate-change is a hoax, I wouldn't be surprised if that happened..

Secret

Interestingly, back in the dark ages of 2016 (2016 is often reputed as the worst year ever due to many baby boomer celebrities died) I secretly wished the world will simultaneously continue and end at the same time, for some agenda

Abcd

7:23 PM

I have differentiated and got v = u +2a(t-2s). What next?

Emilio Pisanty

7:38 PM

pffffffffffff, for goodness' sake, this made HNQ?

5

Q: Zee's use of Kronecker Product in "QFT in a Nutshell" to represent Dirac matrices

In his book Quantum Theory in a Nutshell (2nd edition, p. 94), Zee describes the Dirac gamma matrices and lists a representation using Pauli matrices and the identity matrix. For example he writes $$ \gamma^0 = \begin{pmatrix}I&0\\0&-I\end{pmatrix}= I\otimes \tau_3,$$ where (I assume that) $\ta...