The h Bar

Bml

00:55

Hi. Could someone have a look at this question of mine?

0

Q: Result from a paper of a potential, boundary value problem for a metallic disk with charge images

In this question of mine, @ErJio helped me to justify mathematically a result that in this paper was only asserted without derivation. Now, I am again in trouble with another result - only asserted, no derivation - of the same paper. Equation $(8)$ of the paper is a conformal map from the region ...

naturallyInconsistent

01:58

@imbAF Whatever ACM told you is definitely 100% correct, but I can add a bit of extra details. 1st of all, if you have any one of PDF, CDF, MGF, etc, then you have the probability distribution. Those are different representations of the same underlying thing. Now, in CM, a state is a point in phase space whereas in QM, any state, pure or mixed, is a (quasi-)PDF in phase space. You might link this to why we say that in QM, you need an ensemble just to define the state.

But more importantly, you can reconcile CM and QM by saying that in CM, the PDF is just a Dirac delta distribution. Then, you can say that both CM and QM use PDF in phase space, except that in QM case it is not a PDF but rather a quasi-PDF. It may well be Wigner's function, but you can also use any information-preserving integral transform of Wigner's function, for example, Huisimi's function.

@imbAF ACM is being calm and collected about your quip here. You had better keep the "realist" v.s. "deterministic" terminology correct here, because if you get them wrong, you are liable to bookshelves worth of angry arguments.

Silly Goose

04:41

Hartley transform

In mathematics, the Hartley transform (HT) is an integral transform closely related to the Fourier transform (FT), but which transforms real-valued functions to real-valued functions. It was proposed as an alternative to the Fourier transform by Ralph V. L. Hartley in 1942, and is one of many known Fourier-related transforms. Compared to the Fourier transform, the Hartley transform has the advantages of transforming real functions to real functions (as opposed to requiring complex numbers) and of being its own inverse. The discrete version of the transform, the discrete Hartley transform (DHT)...

naturallyInconsistent

@Bml Oh, conformal mappings are magic. One rarely knows what to try nor why. Instead, mathematicians simply compile a large library of random mappings they find, for fun and laughter, peace and joy, and people look through them to see which works, or approximately works. Sometimes even science itself works in mysterious ways.

@SillyGoose this is super beautiful

lemmeow star it

naturallyInconsistent

05:08

but apparently there isnt a lot of uses. FFTW's page on DHT asks people to tell them what it is useful for, if people used it.

Silly Goose

it sounded interesting, i found it from this paper: arxiv.org/abs/2406.19410

PM 2Ring

06:00

I made a new colouring for the Gyroid minimal surface unit cell which makes its hexagonal structure more obvious. It's essentially an embedding of a tessellation of the hyperbolic plane, with 6 hexagons meeting at each point.

sagecell.sagemath.org/…

Here's a Bookmarklet that opens the Sage 3D display iframe in a new tab or window, giving you an (almost) fullscreen view. javascript:(()=>open($('iframe').get(0).src))()

Bml

06:14

@naturallyInconsistent You are right about the conformal maps, even I have not been able to where they could come from or how they could be derived. But regarding my doubt about the potential with image charges, could you give me some more information? My difficulty is not with the conformal map (which I have now abandoned), but with the potential with image charges.

PM 2Ring

FWIW, the gyroid's {6, 6} tessellation looks pretty boring on the Poincaré disc (which gives a conformal mapping). Sometimes, you do want to preserve distances. And of course with this tessellation all the primary angles are identical anyway. And since the gyroid is a minimal surface it preserves the mean curvature. ;)

naturallyInconsistent

06:39

@Bml I understand the subject but I do not understand your question.

Oh, yet another pretty thing from PM 2Ring

Bml

@naturallyInconsistent What don't you understand about my question?

PM 2Ring

I also made a thing for plotting complex functions (of a single variable) in 3D. The norm maps to the height, the arg maps to a colour in a spectral cycle. So the (default) identity function maps to a rainbow cone. You can give arbitrary expressions & functions in the input box. Sage has a lot of built-in functions, and most of them should work. (Some don't work with symbolic variables, only numeric values). doc.sagemath.org/html/en/reference/functions/index.html

sagecell.sagemath.org/…

naturallyInconsistent

06:58

@Bml What am I supposed to answer to "regarding my doubt about the potential with image charges"? There is a lot of background that you should be doing properly, e.g. in Jackson's first 3 chapters that really really really show very clearly how these mathematical facts and physical facts lead to absolutely shocking result right at the start of the book. So, again, I am not sure where to begin helping you, or to aim to do what.

@PM2Ring Oh, this is for Bml? You even named the function as "make conformal mapping"?

PM 2Ring

I guess I should mention that Sage provides I as a constant for the imaginary unit, but it also accepts the usual Python notation for complex numbers, eg 2 +3j

Bml

In my question I posted, in points 1), 2), 3), my doubts about the result of the potential presented in the paper I referred to. What I would like to understand is where this result comes from, how it is obtained.
I also quoted Jackson, saying that the exercise in my opinion most similar to the derivation in the paper is however different.

PM 2Ring

@naturallyInconsistent No, that's just a coincidence. In that script, cmap is an abbreviation for colormap.

I actually wrote the core of that code a while ago. But while I was using it today I realised that I could make it more versatile. The old version didn't have the GUI input, you had to edit the expression into the script. I also added the new colormap, which I find less garish than the usual hsv map that's used for flat conplex plots.

Bml

07:25

@naturallyInconsistent Is there anything I could do to make my original question clearer?

Ryder Rude

08:51

there is a straightforward application of my yesterday's analysis in representation theory

theorem : if u apriori know that the Lie group is compact, then hermitian representations of the algebra are restricted to have rational eigenvalues, after re-scaling the generators

e.g. if we take the 1D lie groups, then $R$ is related to irrational numbers while $u(1)$ is related to rational numbers

i think these results can bring deep insights from number theory to group theory

i am trying to connect prime numbers to this line of thought

Bml

@naturallyInconsistent Is the expression for the potential in this paper (arxiv.org/pdf/1007.2175) trivial?

Allure

09:27

A question that might not be worth posting on the main site ...

Is it a coincidence that the electric potential energy between two charges (kQq/r) is also equal to the electric force * distance between them (= kQq/r^2 * r)?

Presumably the electric potential energy is equal to the force * distance, and since the force is constantly changing, we cannot just do F.S to get the potential energy - especially since the "distance" would be the integral from infinity to closest approach

The math checks out though. I am not sure how to interpret it, other than as a coincidence

Bml

@naturallyInconsistent OK, I properly re-read Jackson. The expression for the potential in the paper looks similar to one derived in Jackson, but it is not the same.

@naturallyInconsistent 1) In the paper there are logarithms, whereas in Jackson there are not. So, it seems they integrated these expressions. 2) In Jackson, q' = a/y q, whereas in the paper it seems q' = q and Q = 0.

Sahaj

Hi all; is this the right place to ask a question about the solution of a kinematics exercise in my book?

Allure

@Sahaj You could ask here, you could also ask on the main site (but see the policy on homework-like questions)

Sahaj

I've asked some questions on the main site earlier. All my questions get downvoted/closed there.

Allure

Might as well ask here then, with the caveat that you might not get a useful answer

Sahaj

10:15

This is the question; I have a doubt on how gravity's acceleration would act in my coordinate system (which is essentially just shifted 3D cartesian). I took new axes: a unit vector $\hat{m}$ along the incline, a unit vector $\hat{n}$ perpendicular to the incline and the $\hat{i}$ unit vector along the $x$ axis. Then I think that the acceleration due to gravity should be $\textbf{a} = -g\sin(37^{\circ})\hat{m} - g\cos(37^{\circ})\hat{n}$. Is that correct?

naturallyInconsistent

10:32

@Sahaj It is a bit fuzzy there; there are two angles, the wedge angle, and the projection angle. Are they both 37 degrees? Changing to the new directions is a good idea, but I am not sure it is a good idea to stick with it if you are stuck confused. Why not try to work with the usual x-y-z and see if you progress a bit?

Sahaj

Yes, both the angles are 37 degrees.

I think this is less tedious than using regular cartesian coordinates which is why I ask

naturallyInconsistent

11:27

@Bml I was just extremely busy at work just now. Yes, there are plenty you can improve. First of all, you should quote the equation numbers. I have a hard time searching for your snippets; citing the page numbers would really help us use less time.

@Bml e.g. this question---which potential? There are so many potentials there!

@Sahaj your suspicion is correct, but you might know how to proceed in the usual situation, whereas here you are stuck

@Bml Putting together what you are asking here and there, it is abundantly clear that you are trying to run before you can walk. That is not a sensible way to learn about anything. People who solve the problem you are asking about, first get themselves used to Jackson's level of mathematics over many different applications, before attempting the paper you are reading.

@Bml In particular, this (1) is not "they integrated these expressions". Instead, what is happening is that they are considering a 2D problem. The standard solution to Poisson's equation in 3D is the equation (2.8) that you see in Jackson, the $$\frac1{4\pi\varepsilon_0}\frac q{|\vec x-\vec y|}$$ thingy, whereas the corresponding standard solution in 2D is the logarithm. This is something every university course in physics will make sure students get to play around with.

And (2), Jackson was dealing with a sphere with radius $a$; that gives rise to the $\frac ay$ terms, whereas the equivalent for a mirror / flat plane is $q^\prime=q$ as is treated in every introductory book that is easier than Jackson (and Jackson does it by taking the sphere solution and taking appropriate limits to recover the flat plane, since it just means a sphere of infinite radius). Again, people get to have years of play time on these things.

(1) is just standard reasoning and mapping around the boundary conditions, just that it is in 2D rather than 3D.

(2) I cannot confirm nor deny your $\left(\frac R{\sqrt2}\right)^2$ assertion, too busy to try deriving it, but if $\bar v$ is not a typo (it doesn't seem to be; it appeared too many times to be a mistake) and also not introduced, then maybe it is the complex conjugate of $v$. That is one of the two possible standard notation for it.

(3) what does "image charging" mean?

Anyway, your problem seems to be too far a mismatch between the technical background you have and the paper.…

@Sanjana you might like youtube.com/watch?v=1uLi1I3G2N4

Sahaj

11:49

@naturallyInconsistent I've tried without changing any axes. I'm still more stuck. Can you point to any resources to confirm if my value of $\textbf{a}$ is correct or not?

Bml

@naturallyInconsistent Thank you for your feedback.

naturallyInconsistent

@Sahaj I dont think there is any resources like that. However, I computed it and I am in agreement with your result.

Bml

1. Where can I find a **detailed** reference on the treatment of a 2D image charge method that emphasises logarithmic derivation?
2. I know that the authors are considering a 2D problem, and I pointed this out in my question as well. Nevertheless, I still don't understand why they use the charge $q$ in 2D (writing $\frac1{4\pi\varepsilon_0}q$ rather than $\frac1{2\pi\varepsilon_0} \lambda$, where $\lambda$ is the linear charge density). That is why I used the expression "they seem to have integrated the expressions".

@naturallyInconsistent Where can I find an explicit and detailed derivation leading to the expression $q=-q'$ in a flat/mirror plane in a 2D case?

@naturallyInconsistent (2) Why would they use a complex conjugate of $v$? By taking the limit, subsequently this complex conjugate, or whatever symbol it is, disappears in the derivation. What is it supposed to represent?

@naturallyInconsistent (3) It means "method of charge images".

naturallyInconsistent

@Bml Jackson's coverage of the method of image charge is particularly good. Read chapters 2 and 3

@Bml I am not going to give you the answers to that. People learn about that from considering Gauß's law in the various number of dimensions, and it is a standard result. The are plenty of homework exercises on it. However, it is not usually written up as a spoonfed paper.

@Bml I will also not give you this, even if I have it.

@Bml I have not worked out that paper in full detail, so I do not know. I just know that it appears in the denominator, and it is not tooooooo outlandish to have something appearing in the denominator to be complex conjugated.

Sahaj

12:09

Thanks for your response. Then the initial velocity vector $\textbf{v}_0 = 8\hat{i}+6\hat{m} + 0\hat{n}$ and the accelaration vector $\textbf{a} = -6\hat{m} - 8\hat{n}$. Then the velocity vector at some point $t$, ie $\textbf{v(t)} = 8\hat{i} + (6-6t)\hat{m} - 8t \hat{n}$. Then in part C of the question one should be able to set $v(t) \dot \hat{i} = \frac45 || v(t) ||$ and find the values of $t$, is that right?

Thanks for your response. Then the initial velocity vector $\textbf{v}_0 = 8\hat{i}+6\hat{m} + 0\hat{n}$ and the accelaration vector $\textbf{a} = -6\hat{m} - 8\hat{n}$. Then the velocity vector at some point $t$, ie $\textbf{v(t)} = 8\hat{i} + (6-6t)\hat{m} - 8t \hat{n}$. Then in part C of the question one should be able to set $\textbf{v(t)} \cdot \hat{i} = \frac45 || \textbf{v(t)} ||$ and find the values of $t$, is that right?

Sorry for the messed mathjax in the earlier message

naturallyInconsistent

This is wrong.

Your general idea is correct. Your acceleration vector is wrong

It is much simpler than that

Sahaj

@naturallyInconsistent But, if one substitutes into $\textbf{a}=−g\sin(37^{\circ})\hat{m}−g\cos(37^{\circ})\hat{n}$ that $g=10$ one gets the mentioned acceleration vector

naturallyInconsistent

@Sahaj Let me put it this way. $\vec g=-6\hat{\vec m}-8\hat{\vec n}$ but that is not $\vec a$

Sahaj

Oh. But why is that? Isn't gravity the only acceleration on the body? Does one have to consider the normal force between the ball and the incline?

naturallyInconsistent

@Sahaj Obviously! Do you think the ball penetrates into the incline?

Sahaj

12:27

You're right: it makes sense now. My solution sheet kept on replacing $\vec{a}$ and $\vec{g}$ multiple times and I forgot to consider the normal. Thanks for your invaluable help!

naturallyInconsistent

meowth

Bml

@naturallyInconsistent I did not understand why you should not give me the answer. If you first say there is a lot of homework on this, and then you don't tell me where to get it, how should I ever learn?

naturallyInconsistent

@Bml I think you should follow a standard treatment of physics. I am not sure where nor how.

Allure

13:54

chat.stackexchange.com/transcript/message/65884665#65884665 Does anyone know about this?

Ryder Rude

15:10

@Allure hi we have $F=-\frac{dV}{dx}$. so if $V$ is a power of $x$, $F$ has one less power of $x$ because of the differentiation rule. but the differentiation also produces a constant which happens to be $-1$ by a co incidence here

Sanjana

16:01

@naturallyInconsistent yeah. I saw that too... Also another one by Prof. Dave

Bml

@naturallyInconsistent Why won't you tell me? Does Griffith's or Jackson's present this topic in detail?

naturallyInconsistent

@Bml Jackson brushes past it because it is aimed at people who already are good at maths and have seen E&M before. Griffiths might well have it, but it is a little too easy and focused upon the 3D case. You should try and find the relevant section there and modify it for yourself. However, most of all, you should step away from all these and focus upon learning the subject normally, in the 3D case, and not 2D, until you are famliar with the main theory.

Sanjana

What is the reason of people saying "string theory is background dependent"? Is it because the target space is a fixed background?

Bml

16:16

@naturallyInconsistent Is this post relevant?

2

Q: Electric field and potential of a point charge in a (strictly) 2D 'world'

I was trying to figure out how the electric potential and electric field are different in a 3D system versus in a 2D system (I take such a 2D 'world' to be the $xy$-plane, i.e. $z=0$, in a Cartesian coordinate system). The context/motivation has to do with the Coulomb interactions that electrons/...

electrostatics gauss-law dimensional-analysis coulombs-law dirac-delta-distributions

naturallyInconsistent

@Bml yes, mostly giving the full answer.

not sure if you can intuit the full understanding from that, though

Bml

@naturallyInconsistent Yes, it is super-clear!

naturallyInconsistent

great!

Bml

16:34

@naturallyInconsistent But I still don't understand why, instead of $\varepsilon_0/2$, many (the paper itself, too) use $1/8pi$. Could you explain why?

naturallyInconsistent

where? Which page? which equation?

I know why. Even though SI units have clearly won, the Americans still like their Gaußian units. In Gaußian units the $\frac{\varepsilon_0}2$ turns into $\frac1{8\pi}$

Bml

@naturallyInconsistent But is it correct? Eq. 2, page 3. I know this equation with $\frac{\varepsilon_0}2$, but are we sure that $\frac{\varepsilon_0}2 \ \mapsto \frac1{8\pi}$ in Gaußian units?

naturallyInconsistent

I am absolutely certain. They are just being silly, attempting to set $\frac1{4\pi\varepsilon_0}=1$, which just ends up sending the $4\pi$ around. See

Gaussian units

Gaussian units constitute a metric system of physical units. This system is the most common of the several electromagnetic unit systems based on cgs (centimetre–gram–second) units. It is also called the Gaussian unit system, Gaussian-cgs units, or often just cgs units. The term "cgs units" is ambiguous and therefore to be avoided if possible: there are several variants of cgs with conflicting definitions of electromagnetic quantities and units. SI units predominate in most fields, and continue to increase in popularity at the expense of Gaussian units. Alternative unit systems also exist. Conversions...

Arjun

16:56

Hello people! Has anyone here read the book by Melvin shwartz,his book on electrodynamics.I've done first 8 chapters from Griffiths electrodynamics and am planning to study Melvin schwartz now and also simultaneously do Griffiths 9-12.You guys think it would be a good idea to do that? Plus eventually I'm definitely going to do Jackson/zangwill

To give some context I'm eventually going to have a course in electromagnetism ,the course is called special relativity and electromagnetism..it tries to teach str and give a relativistic picture of classical electromagnetism

naturallyInconsistent

The contents page seems to suggest that it is quite short

Arjun

17:11

@naturallyInconsistent the course content is essentially the first 9 chapters of landau vol 2,I'm thinking of supplementing landau vol 2 with more comprehensive books

naturallyInconsistent

17:35

meowth

Bml

19:28

@naturallyInconsistent I don't like them...

@naturallyInconsistent So, what is the result of Eq. 13, pag. 4 of the paper (arxiv.org/pdf/1007.2175)? In the denominator there is $2$, so...? Would it be a $\frac{1}{8\pi\varepsilon_0}$, or a $\frac{1}{4\pi\varepsilon_0}$?

@naturallyInconsistent My confusion stems from the fact that on pages 2 and 3 they talk about 3D geometries, so $\frac{1}{4\pi\varepsilon_0} = 1$, but on page 4 they talk about a 2D geometry, so the electric field (and the potential) should have a factor $\frac{1}{2\pi\varepsilon_0}$ in the denominator, and by placing $\frac{1}{2\pi\varepsilon_0} = 1$, the denominator of Eq.13, pag. 4, becomes $\frac{1}{4\pi\varepsilon_0}$. What do you think?

imbAF

22:37

Regarding coherence, in wikipedia the following is said:
In physics, coherence expresses the potential for two waves to interfere.
Even if one has a destructive interference, that's an interference pattern. So how exactly non-interferance looks like ?

Allure

You mean what do two waves look like if there's no interference?

They just look like one wave, with the other not existing

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