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psa
7:25 AM
@JohnRennie hey, where does this inequality come from?
 
@psa hi :-)
 
psa
Is the energy of this particle just $\frac{1}{2}mv^2$?
 
Yes, and that can also be written as $p^2/2m$.
 
psa
are we just ignoring potentials $V(x)$ here?
 
The potential is zero everywhere inside te box.
 
psa
7:28 AM
ohh right
woops, thanks
 
:-)
 
7:47 AM
@JohnRennie Hello sir :)
 
@Knight hi :-)
 
@JohnRennie I want to understand what happened here $$ \int \mathbf J (\vec {r’}) \delta ^3 (\vec{r}- \vec {r’})dV’ = \mathbf J (\vec r)$$
Or let me write it without the arrows $$ \int \mathbf J (\mathbf {r’}) ~ \delta ^3 ( \mathbf r - \mathbf{r’} ) dV’ = \mathbf J (\mathbf r) $$
 
The expression:
$$ \mathbf J (\vec {r’}) \delta ^3 (\vec{r}- \vec {r’}) $$
is zero everywhere except at $r' - r$ where it is equal to $\mathbf J(\mathbf r)$. Yes?
 
I think it is zero everywhere except when $\vec{r} - \vec{r’} =0$. Am I missing something?
 
Oops, that's a typo. It should have been $r' = r$
 
7:58 AM
Yes
 
So the integral is just over the (infinitesimal) region where $\delta(r'-r)$ is non-zero, and over this region we can take J constant since the region is infinitesimal. That means we can take J outside the integral to get:
$$ \mathbf J (\mathbf{r}) \int \delta ^3 (\vec{r}- \vec {r’})dV’ $$
 
Yeah! :)
 
And the integral of $\delta$ is one.
 
But sir my problem is that $\mathbf J$ is the volume current density and it is a function of $\vec r$’ so what does it mean when we write $\mathbf J (\mathbf r)$
$\mathbf J$ is a function of primed variables
 
The point is that the integrand is zero everywhere except at $\mathbf r' = \mathbf r$.
We can split our integral into parts:
1. at $\mathbf r' \ne \mathbf r$: in this region $\mathbf J(\mathbf r') \ne \mathbf J(\mathbf r)$ but we don't care because the integrand is zero anyway due to the $\delta$.
2. at $\mathbf r' = \mathbf r$: everywhere in this region $\mathbf J = \mathbf J(\mathbf r)$
 
8:07 AM
Okay
I think I should present more context here.
 
So in region 2 $\mathbf J$ is effectively just a constant and equal to $\mathbf J(\mathbf r)$.
 
Thank you so much sir for calling me son.
 
Oops :-)
 
WHAT!
That’s cheating
Please revert it :)
 
I can't now. You only get two minutes to edit posts.
 
8:12 AM
Hahahahaha Oh my God
Sir my main problem is what does $\mathbf J (\mathbf r)$ means when J is a function of r prime.
 
$\mathbf r$ is just a constant in this context. Lets take a simple example. Consider the function $y = x$.
And suppose we want to do:
$$ \int y(x) \delta(x-a) dx $$
where $a$ is a constant e.g. $a = 2$
 
Yeah we would get y(2)
 
The answer is $y(a)$
@Knight yes.
Well your integral of $\mathbf J$ works in just the same way.
 
So what does $\mathbf J (\mathbf r)$ mean? $\mathbf r$ is a constant, how? $\mathbf r = \langle x, y , z \rangle$
 
Take my example and write it as:
$$ \int \mathbf J(x) \delta(x-a) dx $$
The $x$ is just the variable used for integration, and we can call it $x'$ if we wish, so we can write the equation as:
$$ \int \mathbf J(x') \delta(x'-a) dx' $$
And $a$ is just a constant so we can call this constant $x$ if we wish:
$$ \int \mathbf J(x') \delta(x'-x) dx' $$
And now the result is $\mathbf J(x)$
But in this calculation $x$ is just some constant, just like $a$.
> but we cannot measure J at x as J is a function of x’
 
8:23 AM
Okay means x is just designation to a constant
?
 
Yes, exactly.
In your original integral $\mathbf r$ is a constant vector.
 
So, what does $$ \nabla \times \mathbf B (\mathbf r) = \mu_0 \mathbf J (\mathbf r) $$
means?
 
In that expression $\mathbf r$ is a variable. The notation $\mathbf r$ means different things in the two equations.
There's no law saying we can't use $\mathbf r$ as either a constant or a variable, we just can't do both in the same equation.
 
But the thing that we did above was a proof of this $$\nabla \times \mathbf B (\mathbf r) = \mu_0 \mathbf J (\mathbf r)$$
 
This method is used frequently in derivations. I think you have a deeper issue than just this one derivation. You need to go back and consider the mathematical methods used.
e.g. the Fourier transform of a functon $f(x)$ is $\int f(x)e^{ixp}dx$
 
8:33 AM
Missing $ ?
@JohnRennie Okay, then?
 
And this gives us a new function $g(p)$
So in the integral we take $p$ constant, but the result gives us a new function $g(p)$ dependent on the variable $p$.
 
Yeah! It'a ll very doubtful.
 
The trouble is that I'm so used to doing this that I genuinely can't see why anyone would have a problem with it.
2
 
HAHAHHAHAHAHA
And you know these problems stop me from moving forward and even if they are solved I get nothing useful.
 
@JohnRennie hi
 
8:41 AM
Suppose you have a function of two variables $f(x,y)$ then you can compute a new function $g(y) = \int f(x,y) dx$
 
Yeah
 
i.e. you're integrating to remove the dependence on $x$.
Well that is all that is being done here. You have a function $f(r,r')$ and you're computing $\int f(r,r') dr'$ to remove the dependence on $r'$.
 
Okay !
 
@Nobodyrecognizeable hi :-)
 
Why come intensity of first polarizer is halved? @JohnRennie its not mole law.
@JohnRennie ^^
 
8:46 AM
Suppose we have light with some electric vector $E \hat j$, and we pass it through a polariser at an angle $\theta$ to the electric vector.
 
@JohnRennie yeah.
 
Do you know the expression for the amplitude of the light transmitted through the polariser?
 
@JohnRennie yeah $i*cos^2 {\theta} $
 
That's the intensity, i.e. $E^2$. If were dealing with the amplitude we get $E' = E\cos\theta$. Yes?
 
@JohnRennie but why in first polarizer intensity is halved?
@JohnRennie of course.
 
psa
8:49 AM
Because the light was initially unpolarized.
 
@Nobodyrecognizeable ah, OK, sorry I misunderstood what you were asking. You're asking why when unpolarised light passes through a polariser its intensity is halved?
 
@psa so moles law does not hold for polarizer?
@JohnRennie yeah.
 
psa
There's a result that for unpolarized light passing through a polarizer, $I = \frac{1}{2}I_0$. John can tell you about the justification for why.
 
What is moles law?
 
8:51 AM
@JohnRennie that cos law which I wrote.
 
psa
It's basically because unpolarized means the light is "randomly" polarized, so the overall result is that half of the light is absorbed.
 
3 mins ago, by Nobody recognizeable
@JohnRennie yeah $i*cos^2 {\theta} $
This one
 
I must admit I've never heard that name used for the result.
 
psa
It's Malus' law, not moles.
 
But anyhow, as psa says that tells you what happens to polarised light when it passes through a polariser.
 
8:52 AM
@JohnRennie yep professionals do only business. :) -
@psa yeah sorry.
 
@psa of course! :-)
 
@JohnRennie sorry too.
 
You can derive the transmission for unpolarised light if you want to. Ut's a relatively straightforward integral.
 
@JohnRennie so for unpolarized light it'll get halved.
 
Yes
 
8:54 AM
@JohnRennie all right. Another one?
 
OK ... ?
 
psa
You could also think of it in terms of breaking things up into orthogonal components. If the light is randomly polarized, half of the electric field is oriented orthogonal to the polarizer, and half of the electric field is oriented parallel to it. So half goes through, half gets absorbed.
 
@psa that would mean $E/E_0 = 1/2$ so $I/I_0 = 1/4$ ! :-)
 
psa
I was just thinking about that... you're right
damn
welp, the integral works
 
8:57 AM
You do it by writing the unpolarised light as an integral $\int \mathbf E(\theta) d\theta$
 
psa
I meant the integral John was referring to, not that one. :)
 
@JohnRennie please refer to the question above.
@JohnRennie it should be valid in + and - $\pi$ right?
 
psa
OK, hand-wavey but more precise, you can think of it intuitively like this: without a preference for polarization, perfectly depolarized light must dump half its energy into a polarizer: you can take this as a kind of "definition" of depolarized light if you like.
 
You're asking the wrong person about maths, though I think the result is $\cos(\pi) + \cos(-\pi)$
But that doesn't match any of the options.
 
9:02 AM
@JohnRennie no problem.
@JohnRennie Thanks for the help. Have a nice day professor. Goodbye
@psa bye.
 
psa
bye!
@JohnRennie I just used a very precise calculator to find an answer for this, but is there a nicer way to find an answer to this?
In order for $Nln(N)-N$ to be an approximation to $ln(N!)$ accurate to $10^{−10}$, how large does N have to be?
 
No idea.
 
psa
haha OK
It's in regards to the Stirling approximation
 
I don't think there is an analytic expression of it.
 
psa
yeah I kind of thought so
@JohnRennie Something else. In the process of gas expansion to vacuum, according to the first law of thermodynamics, dU= 0. However, if dU=−PdV+TdS holds, because the expansion process doesn't do work and dS > 0, dU must be greater than 0. This seems to cause a contradiction? dV for the system is zero if we treat the box and not the gas as our system, yet the change in entropy is positive and the change in internal energy is zero, which doesn't sit quite right with me.
 
9:14 AM
Isothermal expansion?
Or a Joule expansion?
 
psa
Irreversible, so a Joule expansion.
 
I give up
 
psa
no worries
 
@psa I am unconvinced that dU=−PdV+TdS can be used for irreversible processes.
 
psa
@JohnRennie I'm not sure either. My prof said so long as pressure and temperature are well defined at the starting point and endpoint of the process (i.e. there is equilibrium at the start and end), the identity should hold.
 
psa
9:41 AM
@JohnRennie I guess as long as it's quasistatic it holds. So maybe we could consider free expansion but quasistatically, and then sum up all of those processes to get the overall free expansion. That still doesn't explain my confusion, but yeah.
 
10:37 AM
@JohnRennie Hello !
 
@Jasmine hi :-)
 
I had a question from nuclear physics
 
OK ... ?
 
An electron of an Hydrogen atom is in its ground state and having energy equal to that of an excited electron of He+. Find the ratio of magnetic field created by electron in Hydrogen atom's ground state upon nucleus to that of electron in He+ in excited state upon nucleus.
What I did - qvB=mv^2/r
the electron is in 1st excited state in He+
But I got a wrong answer
 
I don't know how you calculate the magnetic moment for a hydrogenic atom.
Presumably the question intends you to use the Bohr model, then treat the problem classically.
 
10:47 AM
@JohnRennie Ok, then
 
So you're comparing the $n=1$ level in hydrogen with the $n=2$ level in He+?
 
@JohnRennie yes
@JohnRennie Whats wrong here
11 mins ago, by Jasmine
What I did - qvB=mv^2/r
 
I think the He+ n=2 Bohr orbital will have the same radius as the H n=1. I have a recollection that for a nuclear charge Z the Bohr radii are $a_0/Z$.
So you'll get a velocity for the Bohr electron $v^2 \propto Z$
 
@JohnRennie the radius is proportional to n^2 as well
 
And the magnetic moment will be proportional to $v$, so the magnetic moment in the He+ ion is a factor of $\sqrt{2}$ greater.
 
10:57 AM
@JohnRennie we need to find magnetic field and there is only one electron
Anyways answer given 32
 
I have no idea how this calculation is done.
I would guess you're supposed to take an electron moving in a circle so it is equivalent to a circular current.
 
@JohnRennie yes
 
Then the moment is $\mu = IA = I\pi r^2$, where $r$ is the orbital radius.
 
@JohnRennie why are we supposed to calculate magnetic moment
 
Isn't that what the question is asking?
 
11:00 AM
@JohnRennie the question asks about magnetic field
 
As in $B = \mu_0 I/2R$ ?
 
@JohnRennie yes
@JohnRennie If I do that way I get a different answer
 
And the answer is 32?
 
@JohnRennie yes
32:1
 
I have no idea. Sorry.
 
11:05 AM
@JohnRennie nvm ok :-)
 

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