The h Bar

6:38 AM

@JingleBells I exactly asked this question to my teacher. President Obama had different parents from different races and he looks a bit mixed in between, so, the recessive or dominant feature thing doesn't seem to work out here.

@ACuriousMind You've a degree in Biology too!!? Wow!

Slereah

8:15 AM

Morning

Faded Giant

Morning

@Ankit Hi :-) Ping me when you're around and we can talk about this.

B. Brekke

8:41 AM

I am trying to understand crystallography and the space groups of crystals, but I have one major question bugging me. The book I am using adresses different lattice symmetries and applications of group theory, but I can't figure out how to handle crystals with a non-trivial basis. In which way is the formalism altered? Does it break the lattice symmetry? Where can I read about this? I guess this would be a very common issue in applications

It's over 40 years since I last did a crystallography course, but as I recall the point group is the symmetry of the lattice. When you consider the motif as well you need to use the space group.

B. Brekke

9:00 AM

Hmm, I am of the impression that the space group describes the lattice symmetry, and that this symmetry can be decomposed in some way into translations and the point group

10:18 AM

@JohnRennie I am here :)

@Ankit hi :-)

Have you learned about solving the Schrodinger equation for a free particle?

No . I am a high school junior student.

OK. Until you start learning quantum mechanics properly it's hard to describe things rigorously. Without a proper mathematical background it all sounds a bit like magic.

Probably black magic! :-)

But we can try and discuss it and see how far we get if you want.

Can you tell me the physical reason by avoiding mathematics.

Well particles in QM are weird things. We tend to think of particles being like little balls because we are all used to throwing balls around.

But once you get down to the length scale where quantum effects become important you find that the things we call particles are more like fuzzy clouds.

They don't have a clearly defined position and they don't have a clearly defined edge.

OK so far?

@Ankit hello?

10:29 AM

@JohnRennie I am reading your statements. Give me some time.

OK :-)

10:52 AM

@JohnRennie yes that's okay.

More Anonymous

I think the Von Neumann entropy increases if I time evolution of the sort if I measure $x$ I do $y$ else I do $z$ (conditional time evolution)

Is this already established?

@JohnRennie are you here?

@JohnRennie also how do we differentiate between two photons of same energy ?

More Anonymous

Is conditional time evolution allowed?

11:09 AM

@Ankit hi, sorry, I got called away.

Jack Rod

@JohnRennie hi

i am unhappy with the chemistry nobel prize award this

year i feel that it was previously worked by Japanese scientist in 1987 in previously in cpcir

skullpatrol

what's "cpcir" stand for? @JackRod

1:26 PM

In the FRW metric, why do we have to have $g_{0i} = 0$ and not just the same value irrespective of I

Shouldn't isotropy really imply that it should be some value that is the same for all directions as opposed to just 0?

@JakeRose FRW metric?

Yeah

What's it?

never heard that before

Freidman-Robertson-Walker metric

In classical metric $g_{ij} = 0,$ if $i \neq j$

1:28 PM

That's the Minkowski metric.

No, classical metric too

In classical physics you don't have a metric

Well, not one that takes into account time.

Anyway, I'm perfectly happy with that

because $ds^2 = -c \, dt^2 + dx^2 + dy^2 + dz^2$

Yes I know

I mean, classic metric tensor.

1:29 PM

This is different

FRW Metric seems something very different, I've never read about that...

@JakeRose If you want to know why Metric tensor has all it's non diagonal components = 0?

But, that doesn't extends to FRW...

I think this is something you're not aware of rn

so don't worry about it

FRW is the metric when you assume the universe is isotropic and homogenous

@JakeRose (Unrelated) Do you know Tissot's_indicatrix?

1:48 PM

nope sorry

a lil busy now with lectures so can't chat on none course stuff

Oh okay... :-)

enjoy

2:33 PM

@JakeRose Statements about $g_{0i} = 0$ are somewhat meaningless because they are coordinate-dependent (it's certainly possible to choose coordinates for a static spacetime in which the metric is not of that form!).

If you want to do this properly you need to define what "isotropic" or "homogeneous" means without referring to coordinates (the standard way is to talk about Killing vectors), and then one can show that for a static spacetime there always exist local coordinates with $g_{0i} = 0$.

Mhm so I'm not going crazy here

Lemme show you my lecture notes

Well, that's the way you cheat if you don't want to talk about Killing vectors :P

ahaha

Mhm. Can you offer any explanation without killing vectors

Specifically why you couldn't have terms like $\alpha dt(dx^1 +dx^2 +dx^3) $ where $\alpha$ is some constant

In my head that feels like that's not giving any direction a preference

but it at least should say that we can choose coordinates such that, even if it doesn't discuss why exactly "isotropy" means $g_{0i} = 0$.

we discussed isotropy qualitatively previously

Nothing too mathematical was put down tbf

Just essentially rotational invariance

So how can you choose a coordinate system such that αdt(dx1+dx2+dx3) is 0?

2:49 PM

@JakeRose it's not invariant under rotation - note that while $\sum_i (\mathrm{d}x^i)^2$ is invariant under rotations, $\sum_i \mathrm{d}x^i$ is not

it's the same principle as the Euclidean norm of a vector being rotationally invariant, but not the sum of its entries

Or another way to look at it: You could argue that the vector $(2,2,2)$ doesn't "give any direction a preference", but (apart from that being nonsense) it still has a direction, it's not a scalar like its norm.

3:02 PM

@ACuriousMind I do think I see this

So what type of cross terms would be permitted (if we choose a coordinate Fram that accepted them?)

Also, do you see what I'm doing wrong in this:

$dl^2 = \delta_{ij} dx^i dx^j$ under rotation $=\delta_{ij} R_{ik}R_{jl}dx^k dx^j$

Where $R_{ij}$ is the rotation matrix. I think this is actually slightly dodgy notation as it doesn't preserve up and down indices

Giving $R_{jk}R_{jl}dx^k dx^j$ which doesn't give back the delta because one of the R matrices has it's indices the wrong way around right|?

@JakeRose Well, in arbitrary coordinates you can have arbitrary cross-terms, that's why talking about the terms in specific coordinates is a hack :P

3:17 PM

MHmmm

I see

Any idea what I'm messing up with the rotation matrices

@JakeRose I don't know what you're trying to do there, but $\delta$ with both indices down is not a tensor, see physics.stackexchange.com/a/215388/50583

also, as you noted, your index positions are off

I was just trying to show explicitly that Euclidean space is rotationally invariant

yeah, you can't really reason that way about the $\mathrm{d}x^i$. What I said above was meant as intuition.

if you want formal statements it's Killing vectors I'm afraid :P

Surely you can show that the flat space line element is the same under rotation without resorting to killing vectors?

Well, what do you mean by the "line element" being "the same under rotation"? :P

In the end it's just a tensor, and tensors transform in certain ways e.g. under rotation. Anything you do to the line element itself isn't going to give you any insight beyond "it's a tensor"

3:25 PM

as in showing $dl^2 = \delta_{ij} dx^i dx^j$ is invariant when you apply a rotation

I'm not trying to gain any insight, just literally showing that rseult

*result

Do you mean $$\delta_{ij}\mathrm{d}x^i\mathrm{d}x^j \mapsto \delta_{ij}R^i_k R^j_l \mathrm{d}x^k\mathrm{d}x^l = {R^i}_k {R^i}_l \mathrm{d}x^k \mathrm{d}x^l = {R^i}_k {(R^{-1})_l}^i \mathrm{d}x^k\mathrm{d}x^l = \delta_{kl}\mathrm{d}x^k\mathrm{d}x^l?$$

ah yes

oh I was so stupid

well there's some issue with the index placement there too, but it doesn't matter since lowering/raising is free in Euclidean space :P

note that it's not that straightforward in GR since it is not clear what "rotation" means when we don't have fixed spatial coordinates

but at least you can see by the same logic why cross terms like $\mathrm{d}t\mathrm{d}x^i$ probably aren't going to be there in a frame where the isotropy is evident