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cOnnectOrTR 12

7:25 AM

Hi @JohnRennie

John Rennie

@cOnnectOrTR12 Hi :-)

cOnnectOrTR 12

I was doing some numericals I got confused where to use E=sigma/e0 and where to use E=sigma/2e0. I concluded that when it’s thin plate ir sheet we use E=sigma/2e0 and when it’s a thick conductor we use E=sigma/e0

I doesn’t matter if it’s a conductor or non conductor. If it’s thin we use E=sigma/2e0

And if it’s thick plate Conductor we use E=sigma/e0.

Right?

John Rennie

A plate has two surfaces. Yes?

cOnnectOrTR 12

Yeah

John Rennie

So the total area charge density of the plate is the sum of the charge densities on both surfaces.

cOnnectOrTR 12

7:36 AM

Ok you told that to me.

John Rennie

i.e. if the total charge density is σ then the charge densities on the two surfaces are both ¹⁄₂σ.

So when we are doing a calculation, we just need to think about whether we are including the field from both surfaces or just the field from one surface.

cOnnectOrTR 12

Yes

John Rennie

It isn't to do with how thick the plate is, it's to do with how we do the calculation.

Do you have an example problem where you are finding it unclear which charge density to use?

cOnnectOrTR 12

Let me tell. When we have a thick conductor we have total charge as 2sigmaA. Sigma being the surface charge density. We take the Gaussian surface through the plate and out on both sides or we could have taken one side of Gaussian surface inside the conductor. End result is E=sigma/e0.

But when we have thin conductor or non conductor sheet we have total charge as sigmaA. And we get E=sigma/2e0.

Right?

So doesn’t it matter that whether it’s a very thin sheet or thin conductor or if it’s a normal conductor, not a thin one.

Because amount of charges change.

From 2sigmaA to sigmaA

John Rennie

6 mins ago, by cOnnectOrTR 12

But when we have thin conductor or non conductor sheet we have total charge as sigmaA. And we get E=sigma/2e0.

I don't understand why you say this.

Can you give an example?

cOnnectOrTR 12

7:51 AM

I mean a thin sheet of conductor or non conductor. If it’s thin we have only sigmaA as total charge

John Rennie

Are you asking about the definition of the symbol σ i.e. whether it means the total charge or just the charge on one of the two surfaces?

cOnnectOrTR 12

No! When we need to find E we use gauss law. There we need to find total q. That is sigmaA.

So if it’s a thin sheet of conductor or non conductor we will only have sigmaA as total charge in place of total q in gauss law

John Rennie

OK, so you are taking σ to be the total charge density when we add both surfaces together. Yes?

cOnnectOrTR 12

No . sigma as surface charge density.

John Rennie

So the total charge density when we add both surfaces is 2σ?

cOnnectOrTR 12

8:00 AM

Yes

In resnik it’s surface charge density!

John Rennie

The point is that σ is just a symbol and we can define it any way we want.

cOnnectOrTR 12

Yeah

John Rennie

You can define σ to mean the surface charge density or you can define it to mean the total charge density including both surfaces.

Some authors will do one and some will do the other.

It makes no difference to the end result as long as it is clear which definition is being used.

cOnnectOrTR 12

Back to OP…

If it’s thin conductor or non conductor we have E=sigma/e0

Sorry!

E=sigma/2e0

John Rennie

I don't understand why the thickness matters. That is what is not clear to me.

cOnnectOrTR 12

8:05 AM

Because if it’s thick conductor we will have charge’s on both the surfaces of conductor where as if it’s very thin conductor we will have only 1 layer of charges producing field on both sides

John Rennie

You mean for the thin plate we approximate it as having zero thickness so the two surfaces are in the same place?

cOnnectOrTR 12

Yeah something like that

Because in the solution they have used E=sigma/2e0 for plates

John Rennie

It's just down to how you define σ.

Suppose we start with a thick plate where σ is the charge density on one surface so the total charge density is 2σ

Now imagine shrinking down the plate to make it thnner and thinner.

When we have reduced the thickness to zero the two surfaces occupy the same plane, and since the charge hasn't changed this means the charge density in this zero thickness plane has to be 2σ i.e. the sum of the two surfaces that we started with.

cOnnectOrTR 12

So that will yield E=sigma/e0 even for a very thin plate?

John Rennie

If we define σ as the charge distribution on a single surface of a thick plate that implies that for a zero thickness plate we need to define the charge density as 2σ to be consistent. That means the field would be ¹⁄₂(2σ/ε) and that is just σ/ε.

But we could also define σ differently for the zero thickness plate i.e. we could define σ by saying that it is the surface charge density for a thick plate and the total charge density for a zero thickness plate.

It's up to you what definitions you want to use.

cOnnectOrTR 12

8:19 AM

See this solution: quizlet.com/explanations/textbook-solutions/…

John Rennie

That needs me to log in to see the page and I don't want to do that.

cOnnectOrTR 12

Wait

share.icloud.com/photos/0jk8hiWapXj9EjLfm-znkqn4A

John Rennie

There is extra complexity here because in this case we have two plates that are parallel, and the plates have equal and opposite charge densities.

cOnnectOrTR 12

share.icloud.com/photos/0x-eORKokYuc3UftDZizAptsA

John Rennie

In this situation all the charge goes to the inner surfaces of the two plates. So the charge density on the outer surfaces is zero.

cOnnectOrTR 12

8:30 AM

share.icloud.com/photos/0LXJo7ZuiAJHh220sFAaKCITg

share.icloud.com/photos/0jCP4kBwdd15WkSguK6EL2c1g

John Rennie

The charge distribution looks like this.

cOnnectOrTR 12

yeah. ok

@JohnRennie ?

John Rennie

Yes?

cOnnectOrTR 12

you didnt respond

John Rennie

It isn't clear to me what you are asking.

cOnnectOrTR 12

8:45 AM

In this question they have used E=sigma/2e0. is it correct

I attached the solution too. you can see

John Rennie

Since all the charge lies on one surface the charge density on the inner surface and the total charge density are the same. This differs from a single plate where the surface charge densities are half the total. OK so far?

cOnnectOrTR 12

ok

John Rennie

Oh wait, no that isn't how they did the solution.

cOnnectOrTR 12

They are using E=sigma/2e0

John Rennie

I can explain how they did the solution but I'll need to draw a diagram and I don't have time to do that right now. I'll have to come back to this later.

cOnnectOrTR 12

8:50 AM

Ok! anything short you want to add up

John Rennie

Not right now.

cOnnectOrTR 12

bye

Mad Spaces

9:48 AM

Why when i integrate the moment of inertia of a cuboid a b c with the integration boundries from 0 to a, 0 to b, 0 to c.. i do not get the same result when i do the same integration but with the boundries.. -a/2 to a/2 .. and so on?

fqq

@Semiclassical it's not ambiguous. Some manipulations involving real exponentials don't extend to complex ones

geocalc33

11:44 AM

@bolbteppa alright thanks so much - I'll check out that book

geocalc33

11:58 AM

0

Q: decompose real coordinate space into hyperbolae. Infintesimal mapping onto themselves?

Decompose the real coordinate space of $\Bbb R^2 \setminus\{0\}$ into leaves of the form $xy=s^2,$ where $s \in \Bbb R.$ How do you express the continuous mapping from the leaves onto themselves? Can this be expressed as a flow? The reason I am stuck is because the infinite space of hyperbola...

also i know this sort of a math problem, but can it be viewed in terms of SR?

maybe change the space to $\Bbb R^{1,1}$ with the lorentz metric. I still have no idea how to rigorously define this however

it's obvious to me that 1/x can be "mapped" to 2/x by multiplying by a scalar qty

it's very unclear how this could happen when the "size" of the set of functions is on par with the cardinality of the real numbers. That is when you have infinitely many leaves foliating $\Bbb R^{1,1}$ (maybe in the light cone region only).

due to Hilbert's hotel problem, you can indeed "translate"/"shift" infinities -but I can't figure out the rigorosities

SnoopyKid

1:12 PM

Regarding the image here images.newscientist.com/wp-content/uploads/2013/05/… what is that at the center? Is that electron or proton? According to the original paper, Hydrogen Atoms under Magnification: Direct Observation of the Nodal Structure of Stark States, those red, blue colours are electron densities.

I thought at the center, there are protons

Semiclassical

1:53 PM

@SnoopyKid electrons. see for instance the (2,0,0), (3,0,0), and (4,0,0) states here: commons.wikimedia.org/wiki/File:Hydrogen_Density_Plots.png

(the ground state isn't shown but would just be a single blob)

the point is that the s-orbitals of the hydrogen atom are spherically symmetric

and iirc the electron density would be highest at the center. but the size of the nucleus is very much taken into account for that

which is reasonable, since the radius of the electron cloud is on the order of the Bohr radius (so ~10^-10 m)

by contrast, the size of the nucleus is about 10^-5. so 100000 times smaller than the electron cloud

which is way way too small to see on that image.

Semiclassical

3:03 PM

Should have been “electron orbitals do not account for size of nucleus” b/c it’s so small as to usually be negligible

Typically dealt with by perturbation theory instead

Tim Crosby

4:11 PM

I just did the derivation of the wavefunction and energy eigenvalues for the delta potential. How is it possible that the wavefunction is non-zero at x=0,i'e where the potential spikes up to infinity. Does QM alow Total energy of a particle to be infinity?

Mad Spaces

Good evening ! i have been struggiling for hours with this question, i really hope someone here can help me!
given is a hamillton function.
$H(p,q)=p2/2−a/q$ and the transformation $q \rightarrow Q^2$
Find a function F such that the transformation is canonical.

I know the relationships of $\frac{df}{dq}= p$ and such on.. but i am not reaching anywhere with these equations..........

Mad Spaces

6:04 PM

No one can help? welp. thats disappointing

ACuriousMind

6:35 PM

@MadSpaces looks like a case of a type 3 generating function to me - with $G_3 = -pQ^2$ you get $q = Q^2$.

Mad Spaces

How did you reach this result?

ACuriousMind

I looked at the different generating functions and chose the one that allowed me to specify what $q$ is in terms of $Q$ (in this case via $q = -\frac{\partial G_3}{\partial p}$)

and the simplest way to get $Q^2$ as the $p$-derivative is just to take $Q^2p$

Mad Spaces

I do not understand these types... we only learned that a generating function produces a canonical transformation and the relationship $ q = - $\delta F \ \delta p $ is true.. along side with other relationships

but we did not learn about types..

ACuriousMind

6:51 PM

well, in that case ignore the types, since your $F$ has the same relationship to $q$ as my $G_3$

Mad Spaces

Yes.. but like you literally.. guessed a function..

i did your same approach but i integrated according to p and got qp and so on.. and it went on and on...

How do you know that this guessed function fits?

ACuriousMind

@MadSpaces well, the only constraint I have is that the $p$-derivative is $Q^2$, right?

Mad Spaces

Right

ACuriousMind

and the simplest way to get some non-$p$-dependent term as the $p$-derivative is to use it as a prefactor to a linear $p$ term, i.e $Q^2p$

because $\frac{\partial(cx)}{\partial x} = c$ for any $c$ that does not depend on $x$

Mad Spaces

Yea.. i got that part..

Okay lets suppose this is the right function then is $P= 2pQ $ and $q = Q^2 $
If we insert this into our hammilton function we get
$P^2 / 8Q^4 - a/Q^2 $

But we know that because it is canonical then it must be true that $ H - H´ = \frac{\delta G}{\delta t } $

However this is not the case...?

ACuriousMind

6:59 PM

why wouldn't $H-H'$ be zero?

you got the "new" Hamiltonian by just replacing the old $q,p$ by the new $Q,P$, so as functions they are equal

Mad Spaces

Hmm i do not know.. i am used to .. for example when doing a transformation with a lagrangian, you get that $ \Delta L = \frac{dF}{dt} $ so i am just carrying the principles

So.. i guess its that simple huh?.. just guess a function.. lol

Its obviously a lack of understanding from my side of these transformations that i do not really understand at an intutive level.. i am just learning for my exams .

But thanks for your help.

Semiclassical

9:43 PM

Here’s a question that came up in conversation today. I feel pretty strongly about what I think the answer is, but I’ll refrain from saying which one to avoid bias

Suppose Alice and Bob share a Bell state |00>+|11> ( Z-basis for each). One can check that this is equivalent to |++> + |—> in the X basis.

Alice and Bob will each choose to measure either X or Z

Now, two statements I’m confident about;

1) if Alice communicates nothing to Bob, then Bob knows nothing about his subsystem and will describe it with reduced density matrix $I/2$ (maximally mixed)

2) if Alice measures in the X direction and obtains +1, then Bob knows with certainty that he’ll obtain X=+1 as well. So now his reduced density matrix is the pure state |+><+|.

Now: suppose that all that Alice communicates is that she made a measurement in the X direction but not the result. Does Bob’s description of his subsystem change?

(Or to put it differently: The result of a measurement certainly counts as “information.” What about the choice of measurement itself?)

glS

@Semiclassical mh, interesting question. I'd say the point is essentially how Bob should incorporate the "prior information" Alice is giving him by telling him what would happen if he were to perform a specific type of measurement

Semiclassical

Tbh, you’re the exact person whose opinion I wanted on this :D

glS

9:58 PM

if she were to say both direction and measurement result (say, $|+\rangle$), Bob could confidently describe his state as $|+\rangle$ I think. Of course this also assumes he has the prior knowledge that the initial shared state was a maximally entangled one.

now, if she doesn't tell the result.. I'd say Bob should describe his state as a mixture of the two possibilities? So $\frac12(\mathbb P_++\mathbb P_-)$ in this case?

Semiclassical

Yeah, I’m prepared to grant that they both know the initial shared stated with certainty

@glS that’s my line of thinking as well

glS

I admit I never quite thought about this. It's interesting: the generalisation would be that given a bipartite state $\rho$, what information does Bob gain from knowing Alice performed a POVM $\{\mu(a)\}_a$, without knowing the outcome? Analogous thinking would lead to say that his state should be $\sum_a {\rm Tr}_1((\mu(a)\otimes I)\rho)$... I think

Semiclassical

To lay my cards on the table, my own conclusion is that knowing Alice’s choice of what to measure does not change Bob’s situation at all

Any more than knowing that Alice obtained +1 tells Bob nothing if he doesn’t know if it’s X or Z that was measured

The prof I’m grading for said in lecture today that knowing the choice of measurement did change things for Bob, on the grounds that he now knows which ensemble to use

But I couldn’t see any operational sense in which that matters. Knowing that you’ve either got |+> or |-> with equal likelihood is just another way to say you have the maximally-mixed state

@glS I’ll have to think on that

glS

10:39 PM

@Semiclassical well, I'm tempted to agree with you here, simply because $\frac{1}{2}(\mathbb P_++\mathbb P_-)=\frac12 I$. Or more generally, $|u\rangle\!\langle u|+|v\rangle\!\langle v|=|0\rangle\!\langle 0|+|1\rangle\!\langle 1|$ for any pair of orthonormal vectors $|u\rangle,|v\rangle$. So in this case the state Bob gets is identical to the one obtained by just partial tracing

glS

10:58 PM

it might not be true in general though, if the initial state is not maximally entangled or something like that

actually, scratch that. You are right, it's always true. Let $\rho$ be an arbitrary bipartite state, and $\{\mu(a)\}$ an arbitrary POVM for Alice. The probability of getting the outcome $a$ is $p(a)= {\rm Tr}((\mu(a)\otimes I)\rho)$. The post-measurement result for Bob is $\rho(B|A=a)=\frac{{\rm Tr}_1((\mu(a)\otimes I)\rho)}{{\rm Tr}((\mu(a)\otimes I)\rho)}$. Thus Bob's state knowing Alice performed the POVM $\mu$, but not knowing the outcome, is
$$\sum_a p(a) \rho(B|A=a) = \sum_a {\rm Tr}_1((\mu(a)\otimes I)\rho) = \rho_B$$

in other words, knowing what measurement was performed but not the outcome provides exactly zero information to Bob. That's not what I'd have expected intuitively. I'll think about it

this PRL seems to be stating the exact opposite btw: journals.aps.org/prl/abstract/10.1103/PhysRevLett.110.260502. I haven't read past the abstract though. Idk, I might be missing something now. I'll think about it more tomorrow

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