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John Rennie

4:51 AM

Question of the day! :-)

-2

Q: Meteor Landings on Earth

Why do Meteors always land in craters? like seriously why not other places as well, Do y'all know?

gateprep

5:02 AM

need help@JohnRennie

Sending u a question from mechancs

John Rennie

@gateprep post it in the problem solving room:

Problem Solving Strategies

General chat for high school physics. For MathJax see [here](m...

@JohnRennie Ah Ha!!

@Sid :-)

@JohnRennie

pls go to that room

I cant find u there

Akash. B

7:17 AM

@Slereah I meant no slit

djsmiley2k

7:42 AM

UV isn't blocked by water is it?

I'm just having a crazy moment due to some 'new' research about sunscreens and being in water

John Rennie

@djsmiley2k far uv is blocked by water. UVA and UVB aren't.

djsmiley2k

8:43 AM

K.

ty

bolbteppa

11:37 AM

Consider $(-1)^{\dfrac{(p-1)(p-2)}{2}}$. For $p = 2n$ we have
\begin{align}
(-1)^{\dfrac{(p-1)(p-2)}{2}} &= (-1)^{\dfrac{(2n-1)(2n-2)}{2}} \\
&= (-1)^{(2n-1)(n-1)} \\
&= (-1)^{2n^2 - n + 2n + 1} \\
&= i^{2(2n^2 + n + 1)} \\
&= i^{4n^2 + 2n + 2} \\
&= - i^{4n^2 + 2n} \\
&= - i^{2n} \\
&= - (i^2)^{n} \\
&= - (-1)^{n} \\
&= (-1)^{n+1} \\
&= (-1)(-1)^{n} \\
&= (-1)(i^2)^{n} \\
&= - i^p
\end{align}
and for $p = 2n + 1$ we have
\begin{align}
(-1)^{\dfrac{(p-1)(p-2)}{2}} &= (-1)^{\dfrac{(2n+1-1)(2n+1-2)}{2}} \\

Slereah

12:04 PM

@bolbteppa Does that even work for $p = 1$?

Oh wait I guess it does, maybe?

Also are you using $\sqrt{AB} = \sqrt{A} \sqrt{B}$

like some kind of lunatic

bolbteppa

Wait

$(-1)^0 = 1 = - \mathrm{Re}[(1+i)i] = - \mathrm{Re}[- 1 + i] = - (-1) = 1$

Slereah

I guess the trick is that the exponent is never not an integer

hence no imaginary problems

bolbteppa

and for $p = 0$ we have $(-1)^1 = - 1 = - \mathrm{Re}[(1+i)] = - 1$

In fact, another proof of this insane identity is to note both sides change by the same factor under $p \to p+2$ so you only need to consider the $p=0,1$ cases

Yeah $p$ must be an integer

It's just the whole adding the $p$ even and $p$ odd cases together thing

I don't see any $\sqrt{AB} = \sqrt{A}\sqrt{B}$ kind of thing there

Slereah

All good then

bolbteppa

This is the price one pays for spinors in all dimensions

Slereah

12:11 PM

I guess the link would be... I guess that if we're doing $p$ even or odd, that would give $i^p = i^{2n} i^{\text{odd}}$

So $(-1)^{n} i^{\text{odd}}$

Hm, what's the behaviour of $\operatorname{Re}$ under multiplication

bolbteppa

Once you get the sum of two terms thing, it magically is just $\mathrm{Re}$, so the $\mathrm{Re}$ is not essential

Slereah

$$- \operatorname{Re}[(1 + i) (-1)^n i^{odd}] = - \operatorname{Re}[(-1)^n i^{odd} + (-1)^n (-1)^{odd}]$$

If $p$ is even, then that's $$- \operatorname{Re}[2 (-1)^n]$$

Wait that sounds wrong

Where does that $2$ come from

bolbteppa

\begin{align}
(-1)^{\dfrac{(p-1)(p-2)}{2}} = \begin{cases}
-i^p, & \text{if}\ p=2n \\
-ii^p, & \text{if}\ p=2n+1
\end{cases} =? - i^p - ii^p = -\frac{1}{2}[(1+i)i^p + (1-i)(-i)^p]
\end{align}

So obviously the $=?$ equality is wrong, but it's 'kind of' just adding them together then the rest

Slereah

Oh, there's a half

that would explain it

I don't want to besmirch the fine tradition of mathematical proof, but it's the kind of thing I think you can simply brute force

since there's only two alternatives

just prove that it is true for both cases

bolbteppa

Maybe something like $-i^p - ii^p = - i ^{2n} - ii^{2n+1} = - (i^{2n} + ii^{2n+1}) = - i^{2n}(i + ii)$

Slereah

12:25 PM

@bolbteppa

Wait

I think I have a Way

Use an indicator function of even and odd numbers

such as $\dfrac{(-1)^n+ 1}{2}$

That way you can add the two solutions together

bolbteppa

Slereah

$I_{\text{even}} (-i^p) + I_{\text{odd}} (-ii^p)$

bolbteppa

This might be what they're doing

Slereah

yeah it probably is

Captain Bohemian

are articles in Quanta Magazine generally believable? I think the authors of these articles aren't scientists in the fields of the articles they write, so do they accurately convey the ideas in the original papers which they base to write these articles?

Slereah

12:31 PM

I don't even know what quanta magazine is

Sounds pretty fishy

Sha Vuklia

Hey guys, could someone explain why the probability that an electron is found between $r$ and $r+dr$ in the ground state is $\vert R_{10}\vert^2r^2dr$? I am actually only confused about the $r^2$ factor; I was expecting just to work with $dr$. I guess it has something to do with the fact that we consider a whole shell, and not just a tiny ‘line’ element $dr$, but not sure how to find sources on this?

Slereah

@ShaVuklia Volume element

Volume element in spherical coordinates is $d^3x = r^2 \sin\theta dr d\theta d\phi$

Once you integrate over angular coordinates, you're left with $r^2 dr$

Sha Vuklia

is it correct that they left out some factors?

which are constant anyways

because I would think that if we integrate, that we get $4\pi r^2dr$

Slereah

The angular factors in the ground states are $4 \pi$

Which is the area of a sphere

Since in the ground state, the spherical harmonics are just $Y_{0,0} = 1$

Sha Vuklia

right, I got it, thanks!

DanielSank

2:10 PM

Hi, everybody.

Slereah

Hello

Captain Bohemian

I get eyestrain for reading on the computer screen.

Lozansky

2:39 PM

What is meant by a force acting on a magnetic moment?

Captain Bohemian

posting a comment on quanta magazine still requires waiting for approval!

Slereah

3:17 PM

You need to be OT 9 to publish in scientology

And yet

Sha Vuklia

Guys, why does the Hamiltonian commute with the displacement operator $Df(x)=f(x+a)$, if we are given a periodic potential $V(x+a)=V(x)$? I understand that $HDf(x)=Hf(x)$, but we need $DHf(x)=HDf(x)$, no?

Actually, I shouldn't say that $HDf(x)=Hf(x)$; it's just that $H$ works the same on $f(x+a)$ as on $f(x)$

Slereah

3:36 PM

I'm guessing the proof is similar to the momentum operator?

Since it's a discrete version of it

The Hamiltonian's kinetic term is already invariant under all translations

Physics Meta

0

Q: why I was banned?

As all community members know I am just a beginner of physics. Mistakes often happens from everyone .Why can't everyone give me a second chance to blot out my mistakes and this how I can improve my knowledge in physics .So I request community members to withdraw my ban.

support bug

Sha Vuklia

3:50 PM

@Slereah not sure what you mean by discrete version. but how can the kinetic term be invariant under translations? $\frac{d^2\psi(x+a)}{dx^2}$ surely isn't equal to $\frac{d^2\psi(x)}{dx^2}$?

Slereah

@ShaVuklia $e^{iqp} \psi(x) = \psi(x + q)$

Momentum generates translations, as you may recall

Sha Vuklia

I don't think I've ever heard of that

Slereah

$$e^{iqp} f(x) = \sum \frac{q^n}{n!} \frac{d^n}{dx^n} f(x) = f(x + q)$$

or something like that

probably some factors I forgot, but it's the idea of the proof

Sha Vuklia

right

Slereah

Basically the exponential of a derivative is the same series as the expansion around a point

bolbteppa

4:00 PM

How do you raise and lower indices on spinors for arbitrary dimensions, my god

Slereah

@bolbteppa $\varepsilon_{AB}$?

bolbteppa

Doesn't make sense right!

$\varepsilon_{13} = ?$

Slereah

True!

bolbteppa

You can only use $\varepsilon_{12\dots n}$ for $n$ spinors

Yikes

How do you do it for $D= 4$ even actually

These things are so tricky

Slereah

I dunno, you usually do it with Weyl spinors, so

it doesn't matter too much

I don't know how it works for Majorana spinors, though

Spinors are one of those physics things where they rarely talk about any dimension beyond 3

bolbteppa

4:06 PM

Just can't do that if you want to do superstrings :(

Slereah

just gotta use $\operatorname{Spin}(10)$

bolbteppa

Even the statement you can just take your spinor to be a Majorana spinor for simplicity for the 2-D superstring action is unbelievable, since you can only do that in $D = 2, 4 \mod 8$ for $D$ even...

Slereah

I'm guessing you need to find the metric which preserves the spinor inner product under $\mathrm{Spin}$ rotations

but who knows how that works

I don't even know how to prove it for $n = 3$

bolbteppa

I guess you just ignore it, I mean you can't even define $\mathrm{Spin}(1,D-1)$ as anything other than the set of operators $\exp (\frac{1}{8}[\gamma_m,\gamma_n])$ or something equivalent to this

Seems like you just define the Dirac conjugate with indices raised by definition

Slereah

It is indeed the matrix satisfying $\Lambda^T \varepsilon \Lambda = \varepsilon$

I suspect that the general case might be like

$\varepsilon$ has sub-matrices which are Levi-Civitta?

Maybe

bolbteppa

4:29 PM

In $D = 4$ the charge conjugation matrix reduces to a diagonal with two Levi-Civita's in some basis

pdf page 53 damtp.cam.ac.uk/user/examples/3P2Lb.pdf

Sha Vuklia

Guys, if we approximate a crystal potential by the Dirac comb, we need to do something about the fact that our solid doesn’t have infinite size. So Griffiths suggests to impose the following boundary condition: $\psi(x+Na)=\psi(x)$, where $N$ is the number of periods (of size $a$).

I don’t really understand why we have to do that; why can’t we just pretend that the crystal goes on forever? Or rather, why do we speficially impose these boundary conditions, and not something like $\psi(0)=\psi(Na)=0$?

bolbteppa

Because you want to describe real life crystals which don't go on forever

John Rennie

@bolbteppa well, real crystals are infinite to all intents and purposes i.e. their size is so much greater than the lattice spacing that it is effectively infinite.

@ShaVuklia these are wrap around boundary conditions - there's probably a posh name for them but i don't know it.

Sha Vuklia

yes, Griffiths describes it like that, but why did we choose those?

John Rennie

In effect when you leave the crystal at one face you re-enter it at the opposite face.

I think it makes the crystal a 3-torus

The part of the crystal you are considering is only a small part of an effectively infinite crystal, so you don't want to impose restrictions like $\psi = 0$ at the edges because that's not physically reasonable.

bolbteppa

4:41 PM

I think the reason you impose periodic boundary conditions is because you can effectively ignore the real boundary conditions because as John said you can basically treat them as infinite and you barely expect the boundaries to affect anything

John Rennie

Instead you just require the wavefunction to be continuous at the wraps

bolbteppa

@ShaVuklia chapter 5 of Griffiths says it's because he wants to apply Bloch's theorem

The first half of that chapter really confused me

John Rennie

@ShaVuklia if you imposed $\psi=0$ at the edges of your region you'd turn it into a 3D box and you couldn't have any plane waves passing through it - only standing waves.

Since Bloch waves are going to be used to model electron states, and since Bloch waves are essentially plane waves, this would be a big problem.

bolbteppa

I think it's a way to let you simply solve the equation in one interval, $0 < x < a$ for the Dirac comb and letting Bloch give you the solution for all intervals

Sha Vuklia

oh right, that makes sense. just one question then; is the potential outside of the crystal close to zero or rather "infinite", as an infinite well?

John Rennie

4:47 PM

The potential jump at the crystal surface is just the work function.

Definitely not infinite :-)

But for thermal excitations it is effectively infinite i.e. $\gg kT$

Sha Vuklia

right, I don't really understand what the work function is, but I'm imagining that if we have a crystal in empty space, then the potential outside of the crystal will rapidly decrease to zero

@JohnRennie in any case, this is the biggest reassurance for me that it's useful:p so thanks for the explanation

John Rennie

It takes energy to pull an electron out of the crystal, so the potential outside the crystal is higher than it is inside.

Sha Vuklia

oh right, that makes sense

John Rennie

I guess we'd say the potential outside is zero so the potential inside the crystal is negative.

In effect the crystal is a finite potential well

Sha Vuklia

right

enumaris

5:31 PM

@JohnRennie why's one of your comments duplicated in the sidebar?

unfair

ACuriousMind

@enumaris Look at the timestamps :P

enumaris

>.>

Sha Vuklia

6:14 PM

Guys, I don’t really understand why after $n=N-1$, we get no new solutions. Griffiths explains it as that the Bloch factor $e^{iKa}$ recycles, but I’m not sure why that means that there are only $N$ solutions possible. So say we have solutions $\psi$ and $\psi'$, belonging to $K$ and $K'$ respectively, and assume $e^{iKa}=e^{iK'a}$. Since it holds that $\psi(x+a)=e^{iKa}\psi(x)$, we know that $\psi(x+a)=e^{iK'a}\psi(x)$. But I should show that $\psi(x)=\psi'(a)$, right? How can I do this?

(if I haven't provided enough context, let me know)

typo; $\psi(x)=\psi'(x)$

enumaris

6:55 PM

weeee, achieved >1000 rep

ACuriousMind

Cute :)

enumaris

lol

Mr. 66k rep

Physics Meta

7:43 PM

0

Q: Seeking attention towards the unanswered questions

Is there a way to re-iterate the attention of the readers towards my questions which remains unanswered for a long time? Can I ask them again?

discussion questions

Lozansky

8:20 PM

If I know quantum numbers $j$ and $l$, I know $s$?

enumaris

question too vague

no way to answer

what quantum numbers are those?

Lozansky

$j$ comes from the total angular momentum which is the sum of the orbital angular momentum and the intrinsic angular momentum $\mathbf{J} = \mathbf{L}+\mathbf{S}$

And $s$ is the quantum numer associated with the spin and $l$ the quantum number associted with the orbital angular momentum

Given $l$ and $s$ I know that $j$ is restricted to the values $|l-s|, |l-s|+1,..., l+s-1,l+s$ but I don't know how I can determine $s$ from $j$ and $l$

enumaris

Angular momentum addition in quantum mechanics is a little more complicated. It's less straightforward than classical mech. See these notes: www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect15.pdf

Lozansky

I don't see a mention of the coupling of the quantum numbers :(

OK in the case of $j=1/2, l=1$ there is only one candidate $s=1/2$ since $s \geq 0$

Actually, that does not make sense unless $s$ is bounded somehow

enumaris

8:41 PM

o.o

usually s is just 1/2 or 1

depending on the type of particle

do you have a composite system?

of many different s?

Lozansky

Could have $j=1/2,l=1, s=3/2$ and then $1/2 = 1 + 3/2 - 2$, no?

It doesn't say anything of the particles

Just to find the angle between the orbital angular momentum and the intrinsic angular momentum given $j=1/2, l=1$

enumaris

o.O

hmmm...

I should probably go back and re-learn angular momentum addition in QM lol

haven't had to do it in like 7 years, super rusty

Lozansky

Yeah I only just read about it, seems a bit tricky :P

enumaris

8:59 PM

indeed

Physics Meta

9:16 PM

0

Q: Reopen request on a recently closed question

I would like to have this question reopened. In contrast to the users who expressed their apparent confusion, I found the question to be very clear. It also seems like one that would get a whole lot of Google hits. Both complained that one could not know the parameters, but there's only three a...

support reopen

enumaris

10:06 PM

I'm so bored -.-

bolbteppa

10:18 PM

Gauge covariant derivative

The gauge covariant derivative is a variation of the covariant derivative used in general relativity. If a theory has gauge transformations, it means that some physical properties of certain equations are preserved under those transformations. Likewise, the gauge covariant derivative is the ordinary derivative modified in such a way as to make it behave like a true vector operator, so that equations written using the covariant derivative preserve their physical properties under gauge transformations. == Overview == There are many ways in which to understand the gauge covariant derivative....

I don't fully get this beast yet

It's always bugged me how it differs from a normal covariant derivative you get from e.g. differentiating a vector field and it's basis

enumaris

11:43 PM

I think the gauge covariant derivative operates on some sort of fiber bundles that are different than the tangent bundle.....I think...

In other words the covariant derivative works on the tangent space to a manifold (space-time), but the gauge covariant derivative works on an arbitrary space that's attached to your manifold in a fiber bundle way.

...mmmmhm...hopefully I'm not just talking nonsense LOL

Problem Solving Strategies

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