The h Bar

Shing

2:11 AM

8

Q: What is the smallest length scale ever measured?

And, by the way, what is, or are, the measured values?

It has been a few years, I wonder if the scale has ever deceased?

John Rennie

5:39 AM

@Shing As far as I know the LHC still holds the record.

Shing

@JohnRennie I see, thanks Rennie!

BannedUser

5:59 AM

Charge distribution doesn't makes sense to me by only charge quantity unless there is electron or proton present

Think of charge of less than electron does that even exists ?

To me the question needs to say how much electron or proton it holds.

What is definition of charge to you guys

?

really confusing

let's say I got 1 electron on one end of wire and another with 0 charge that electron won't move a bit if it is in rest

so I am really confused why do you guys say q/2 on one end and q/2 for another end. Unless thee are like 2 proton I think it is valid. If for 4 then 2 of them will be stuckedon wire

Manas Dogra

If we start in the first excited state of a harmonic oscillator and calculate the probability of finding it in the ground state after some time, then the probability may be 1 for specific values of time, right? So there's an energy difference due to the transition, where did that extra energy go?

BannedUser

how do you know charge distribution if you don't have proton and electron?

Manas Dogra

I could say the extra energy is carried away by the photons, but when I write the hamiltonian of the harmonic oscillator there is no coupling to the EM field, so there is no interaction, then how will the energy be carried away by them...and if the photons don't carry the energy off, how do we explain the energy difference?

BannedUser

6:55 AM

The number of electrons that surround the nucleus will determine whether or not an atom is electrically charged or electrically neutral. The amount of charge on a single proton is equal to the amount of charge possessed by a single electron. A proton and an electron have an equal amount but an opposite type of charge.

123

7:18 AM

Hello all...

Hi @JohnRennie Sir

Manas Dogra

7:31 AM

Kindly ignore my silly question :)

ACuriousMind

8:29 AM

@BannedUser the charges you're supposed to think about in classical electrostatics are so large they're composed of millions of electrons/lack of electrons/whatever else carries the charge

you're right that if you only have one or two charge carriers, then the charge can't divide infinitely, but that's not the situation you're considering here

John Rennie

@ManasDogra Yes, as you've spotted we need to introduce an interaction term for the probability of the transition to be non-zero. Typically we do this using Fermi's golden rule.

ACuriousMind

concrete physical theories are always just a model of the world appropriate for certain situations. When thinking about how charges will distribute in a conductor, the model of classical electrostatics is not appropriate to think about very small charges, but it is appropriate for situations where the charge is effectively infinitely divisible

Manas Dogra

@JohnRennie I was thinking mistakenly that the probability of finding the system in a ground state after I let it evolve from say the first excited state may not be zero, but it is always zero.

John Rennie

The states are the eigenstates of the time independent SE, so they are (not surprisingly) time independent.

If you have an SHO in the first excited state then it will stay in that state forever unless you change the Hamiltonian so the states are no longer its eigenfunctions.

Manas Dogra

Yes, but my mistake was that I evaluated (somehow idk how :) ) <1|e^(-iE_0t/hbar)|0> to be nonzero...it is zero right?

John Rennie

8:36 AM

Yes :-)

Obviously because e^(-iE_0t/hbar) is just a number so you get e^(-iE_0t/hbar)<1|0> and the eigenfunctions are orthogonal so <1|0> = 0.

Manas Dogra

Thing is I evaluated something like <0|U(t)| ((|0> + |1>)/sqrt(2)) to be non-zero many months ago, and many months later in my head the info became "If we let any "state" evolve in time, it may go to another state, because there is time dependence" which is clearly wrong.

It works for a superposition of states in general but not for an eigenstate

And you know I went on arguing with the professor saying in Griffiths it is written in perturbation theory chapter that there should be a time dependence(completely missing the fact there should be time dependence on the potential not evolution of states) :)

and now I feel ashamed...thinking what the professor thought of dumb me

John Rennie

Yes, correct, a superposition of states is a solution to the time dependent SE but not a solution to the time independent SE.

So superpositions evolve with time.

Manas Dogra

Hmm :) I shouldn't have relied on my memories of some calculation done months ago...I should have thought of that at the moment and answered

John Rennie

If you have a state (|0> + |1>)/sqrt(2) you'll find it oscillates with time, and the frequency of the oscillation is exactly hf = E₁ - E₀.

That's why it emits a photon with a frequency given by hf = E₁ - E₀.

Manas Dogra

Hmm Thanks I have a quick question, these annihilation operators have eigenfunctions(coherent states) but the creation operators don't (I have seen a short math proof of it), but intuitively why it should be so?

John Rennie

8:52 AM

Don't know, sorry. @ACuriousMind can probably answer that.

@ManasDogra Check this:

11

A: Eigenvalue for the creation operator for a coherent state

To add to Innisfree's correct answer, I'd like to emphasize something that the OP does not seem to know and that is that the creation operator has no eigenvectors (nor, therefore, eigenvalues). It is easy to see this: write a general state as a row vector $(\psi_0,\,\psi_1,\,\cdots)$ of superposi...

That seems like a nice simple explanation.

ACuriousMind

9:14 AM

couldn't have said it better

Manas Dogra

9:36 AM

@JohnRennie really nice...

John Rennie

:-)

Manas Dogra

You are here...let me annoy you with another question then? :)

John Rennie

OK ... ?

Manas Dogra

If I am given frequency distribution of some radiation can I comment whether it is from magnetic dipole radiation or electric dipole radiation or some other higher order multipole radiation?

I was wondering how people are saying that pulsar radio waves originate from magnetic dipole radiation(cz we consider that only after electric dipole radiation is zero)..

John Rennie

You mean EM waves with a distribution of frequencies?

Manas Dogra

9:41 AM

Yes

Intensity vs frequency plot say

John Rennie

I don't know I'm afraid. Googling took me to a Wikipedia article that suggests it can be done:

Multipole radiation

Multipole radiation is a theoretical framework for the description of electromagnetic or gravitational radiation from time-dependent distributions of distant sources. These tools are applied to physical phenomena which occur at a variety of length scales - from gravitational waves due to galaxy collisions to gamma radiation resulting from nuclear decay. Multipole radiation is analyzed using similar multipole expansion techniques that describe fields from static sources, however there are important differences in the details of the analysis because multipole radiation fields behave quite differently...

Manas Dogra

I saw that, most of it is from Jackson's electrodynamics but it's an inverse problem...
Trying to guess the source nature from the fields..I see no reason why it should be unique..

Actually I am more interested in finding out about how people found out that pulsar radio waves arise from magnetic and not electric dipole or electric quadrupole moments...

I am reading Pranab Ghosh's book on pulsars where he just assumes that the rotating neutron star will have a magnetic dipole moment. Upto this it's okay, but then he says that this magnetic dipole moment will cause the radiation. This means he must have assumed electric dipole moment=0..So the question is Why would a generic neutron star have no electric dipole moment?

John Rennie

10:22 AM

@ManasDogra For any large object to have an electric dipole moment requires there to be a large scale charge separation, and it's hard to see how any such charge separation could be caused.

By contrast a magnetic dipole moment just requires a current.

So for example the Earth has a magnetic dipole due to currents in the core, but no electric dipole moment because, well, what could possibly cause a charge separation in the Earth?

Manas Dogra

10:35 AM

But dipole moments can be created without charge separation too, no?
If we put a point charge at some point, then wrt the origin it has a dipole moment. no equal and opposite charges and seperation of sort required...

It depends on the coordinate system chosen though

ACuriousMind

@ManasDogra but a dipole moment as such does not generate radiation

John Rennie

It has to be an oscillating dipole

ACuriousMind

and some charge moving around far from your chosen origin is very far from being an oscillating dipole (even though it may radiate simply as an accelerating charge)

Manas Dogra

Why can't it behave like an oscillatory dipole? I can't think of a configuration in which the fields will be such that the Poynting vector doesn't vanish, but surely there can't be any configuration???

ACuriousMind

@ManasDogra "oscillating dipole" means that you have an actual dipole (with charge separation!) that reverses its polarity periodically

just having a dipole moment doesn't magically create radiation like that of such an oscillating dipole

Manas Dogra

10:45 AM

@ACuriousMind Oh yess....Dipole moment has to have non-vanishing second order time derivatives.

Thanks thanks

Slereah

2:00 PM

"A thin homotopy between paths $f,g: I \to X$ in a topological space $X$ (with $I = [0,1]$ the standard interval) is a homotopy $I\times I \to X$ which, roughly speaking, has zero area."

How do you have a zero area homotopy except for the identity

ACuriousMind

@Slereah why are you asking that when it explains what it means in the very next sentence

Slereah

I would question the validity of saying "explain" here

Could use an example

ACuriousMind

if you're in the smooth case, it's a homotopy whose derivative never has full rank

I think that's a perfectly clear explanation

as for examples: Any reparametrization is an example, i.e. cases where the image of $f$ and $g$ is identical.

Slereah

I mean fair enough, but also basically what I said!

ACuriousMind

how?

reparametrizations are very different from the identity

Slereah

2:11 PM

Not for unparametrized curves!

ACuriousMind

just because you probably intuitively identify paths "up to reparametrizations" doesn't mean mathematicians don't have to write down that notion

and if you try to write down what it means for a homotopy to be a reparametrization, you essentially end up with the notion of "thin homotopy"

Slereah

are there any thin homotopies that aren't just identity up to reparametrization?

ACuriousMind

well the thing is that the people who make that definition want to use it in contexts where you don't have access to "reparametrization"

Slereah

I see

ACuriousMind

It also depends on whether or not you consider reversing a path a reparameterization, I guess

Slereah

2:15 PM

Well it is in the two connected element reparametrization group

ACuriousMind

traditionally a reparametrization would keep $\gamma(0)$ and $\gamma(1)$ fixed but the homotopy that turns a straight line "around" by contracting it to a point and expanding it again would also be thin

@Slereah Ah! Of course there are thin homotopies that aren't reparametrizations: E.g. the homotopy contracting the line to a point

Slereah

ah yes, true

ACuriousMind

unless you unnecessarily wobble around with the line while contracting it, that'll sweep out zero area, too

Slereah

I guess I was thinking only of loops and not lines

Slereah

2:34 PM

I guess the rank of the matrix being less than two means "the homotopy goes at best along the curve"

ACuriousMind

@Slereah the idea is that if it had full rank anywhere, you could locally invert it there by the inverse function theorem and then there'd be a patch in the image that's diffeomorphic to a part of $\mathbb{R}^2$, so not "zero area"

this thin thing doesn't really deform anything, it just reparametrizes/contracts/stretches

note that stretching is included: Taking a short line and stretching it linearly into a long line is also a thin homotopy

you're just not allowed to "move" the path

Slereah

2:55 PM

I know all about homotopies from here : youtube.com/watch?v=6iw_rdx11V0

123

4:51 PM

What is the condition of pure translation of rigid body under multiple non concurrent forces?