morning all

Is thermal oscillation in dense systems like solids and liquids literally mechanical displacement (eg. phonons?), which then produces the charge oscillation in the electron density to create black body radiation?

The electrons and nuclei do not oscillate in synch with each other.

ohh thanks @JohnRennie what is the outcome of that?

Ahh @JohnRennie , so this form of radiation is mostly due to oscillation in charge density between the electrons & nuclei, rather than oscillations in charge density between whole atoms/molecules?

or another way, the 'dipole' is formed between electrons & their nuclei not between eg. two atoms (or molecules) of different charge?

I'm not sure to be honest ...

No worries, you already helped me alot :) Thank you!

sometimes i wish to just study the phenomena i want to understand, but quickly it seems many topics are related to eachother

why didn't i do a double degree in undergrad like my friends?? oh well

anybody up to any interesting physics these days?

also here's a page which I bet not many people have seen: physics.stackexchange.com/questions/greatest-hits

Here's something, why is the amplitude for n particle scattering in a first quantized theory given in path integral form as $A(p_1,..,p_n) = \sum_{topologies} \int Dx e^{i S + i \sum_k p_k x_k}$ e.g. for a relativistic point particle we usually start from $Z = \int Dx e^{iS} = \int Dx(\tau) e^{i\int d \tau L(\tau)}$, there are some real jumps from this to the other expression that are not like second quantization and just adding a $J \phi$ term

@NiharKarve eh, "The current algorithm divides the number of page views with the total amount of question and answer feedback received (adding a bonus for high view counts), excluding questions with less views than the median"

so compared to "most views" and "most votes" this is a "most views" list that emphasizes questions with many answers (since this increases the potential total feedback per view) and topics many people will vote on

not sure that's really a useful metric

a lot of them are stuffed with high-school level physics keywords and SE has good SEO, so that is to be expected

it's just a nice easter egg - I don't think the page is linked anywhere else on the site

@bolbteppa what's wrong with Qmechanics answer to the question you just bountied

$= \neq \geq$, showing it's exactly half rather than just the inequality is non-trivial, unless I'm over-complicating something simple, maybe it's supposed to be as maximal as possible because you're using $p$'s but my comment about the light-cone gauge means you can't really even use that thinking

hmm

is there something you can do with like changing $\Gamma\cdot p$ to $\Gamma\cdot p+m\Gamma_{11}$ and then taking the massless limit

obviously kappa symmetry confirms that exactly half the components are fake

That just squares to give $p^2 + m^2 = 0$ which is the same problem, Becker etc... state it using it in that form and also just state it as a fact that that squaring to zero implies half of the components go away

This is the motivation for even thinking to look for kappa symmetry in the first place so whether it's half or more than half which are trivial is important

the real motivation for looking for kappa symmetry is to convince oneself that GS is as good as ✨RNS✨

Yeah it's amazing how complicated the basic setup of this really is

unless I'm missing something this argument wasn't really all that complicated, it's just that the "the matrix squares to zero" is a red herring

(I posted an answer :P)

(For the final matrix, if $A$ is nilpotent then $A v = \lambda v \to 0 = A^2 v = \lambda^2 v$ implies $\lambda = 0$, is the word eigenvalue bad and word rank better or...)

we should probably talk about singular values and not eigenvalues, yeah, and in the end it's indeed only the rank that matters

I've edited the answer to make that clearer, this is just physicists being sloppy with the math again (and it is a bit concerning that the "squares to zero so half the eigenvalues are zero" argument seems to have entered the physics folklore :P)

greetings

I want to ask something about capacitance

Can anyone help me

I have asked the question in problem solving room

I want to ask about E between parallel plates

I have asked a question and I was told to come here. Anyone

> Don't ask about asking, just ask.

if anyone wants to answer your question, they will, but just asking if people are around will generally produce far worse results that actually asking the question - in that case even people that come along later might be able to answer it

I was solving a problem with parallel plates. There E was there from positive plate to negative plate. I guess they are just representing the direction of E and not a vector at a point

I want to ask what will be the E vector at a point. Is it going to be the integration of all the charge elements from the positive plate.

@cOnnectOrTR12 the electric field vector at a point is tangent to the field line at that point

what we usually draw are the field lines, not the vectors

But how do we find vector at a point

Due to parallel plates I mean in betweeen them

Is it going to be the integration of all the charge elements from the positive plate

why would you only look at the positive plate and not the other plate?

Gauß' law is usually an easier way to figure out such electric fields than using superposition directly

I don’t know what will be the E vector at a point between the plates

Ok! I read the E value will be same everywhere in the field between the plates

I guess there is some mathematical representation for this I don’t know.

I read this is different from the E from a charged wire which changes with positions. At equal distances it will be same but if we move closer and further the y component will not cancel and the E will change

physics.stackexchange.com/q/110480/50583 discusses various ways to think about the field between two infinite plates

I have not read gauss law. So I can’t understand

I was going through a problem where there are two plates. Positive and negative. A proton leaves from positive and electron from negative plate at same time

@cOnnectOrTR12 you don't need to know the actual magnitude of the field to answer that

you just need to know it's constant and perpendicular to the plates

I know . I was confused where this vector E starts. If a particle releases from the plate then it will as soon be under fields effect. But I couldn’t imagine a vector E from the surface of the plate.

How do you calculate a field that starts at the surface of the plate. There is no distance after all

From the charge to that point

@cOnnectOrTR12 Regarding your original problem, you would need to setup a differential equation for their separation

The two forces you need to include are the electric field between the plates and their mutual attraction

There's also radiation resistance to consider

It would not be linear so the best way would probably be using a software such as Mathematica

@cOnnectOrTR12 I don't understand what you are trying to ask

What do you mean by "a field that starts at the surface of the plate"?

fields don't start at surfaces of plates

@cOnnectOrTR12 the field can be calculated using Gauss's theorem.

fields exist for all space

@VincentThacker I think this is supposed to be an oversimplified high school textbook exercise where you ignore that the particles repel each other and just figure out that the ratio of their accelerations is the ratio of their masses

@cOnnectOrTR12 you are quite correct that right at the surface the field changes discontinuously so it is undefined.

@cOnnectOrTR12 if your problem is about $E$ directly on the surface just imagine the particles starting a very small distance above their respective surface

Though this doesn't happen in real life as the finite size of atoms means the surface of the conductor is not sharply defined.

fields right on boundaries are an a bit more subtle issue that's completely irrelevant here

@JohnRennie can you please elaborate that field changes dis…

And by the way it’s from resnik . Yes here the force between particles is ignored.

In a charged conductor the charge always resides at the surface of the conductor. You can easily show this using Gauss's law. So in an ideal case the charge forms an infinitely thin layer at the surface.

In real life the charge does go to the surface of a conductor but it doesn't form and infinitely thin layer - it forms a layer with a finite thickness that depends on various properties of the conductor.

You said it changes. Why will it change

Anyhow, most introductory texts will do their calculations assuming the ideal case, because for most purposes this is a pretty good approximation and of course it simplifies the calculations.

@cOnnectOrTR12 The electric field across a surface charge $\sigma$ changes by $\sigma/\epsilon_0$ across the surface charge, in the normal direction

I am confused 😕

Do you mean why does the electric field change at the surface?

Yes

Suppose we have a negatively charged plate - just a single plate for now. Give me a moment and I'll draw a quick diagram.

I think it’s getting out of hand now

I have drawn a plate with some positive charge. We are looking at the plate edge on.

We know the charge goes to the surface of the plate, so the charge will form a thin sheet at the upper and lower surfaces as shown by the red lines. OK so far?

Ok

I am still studying electric field and will next jump to gauss theorem. I can’t absorb too much at the moment. What is your opinion when solving such problems at the moment

@cOnnectOrTR12 Learning Gauss's theorem is a really good way to understand how the field is related to the charge that produces it.

So that should be your next step.

@cOnnectOrTR12 Consider just an infinite plane with surface charge density $\sigma$. The field is $\sigma/2\epsilon_0$ (in the normal direction) on each side, yes?

Go on

@VincentThacker It's optimistic to expect a beginner in EM to understand that.

@cOnnectOrTR12 Now place this plane in some region already with an electric field

Electric fields simply add by vector addition

@cOnnectOrTR12 Suppose the electric field already there is $E$. When the plane is placed there, the field on one side is $E + \sigma/2\epsilon_0$ and the field on the other side is $E - \sigma/2\epsilon_0$, because the field $\sigma/2\epsilon_0$ points away from the plane

Can you now see why it changes by $\sigma/\epsilon_0$ when we go from one side of the plane to the other?

@JohnRennie Well you only need that and the principle of superposition