The h Bar

school is always rather tedious

@0celo7 Hope to talk to you soon.

heather

@Secret i often don't talk about it because i've tried to move on (school switch, not due to what happened, separate issue) but I got bullied a not insignificant amount, especially in sixth grade. but i've never found that talking about it has helped. just sort of letting go of it has done more than anything else.

Letting go is hard for me due to the way my memories work. They are like incredibly stubborn. They will not give up until whatever their goal is fulfilled, which frankly, I am not very sure what kind of "closure" they are seeking for given how the bullies are no longer relevant after grade 10 and every physical damage is resolved, and given how twisted the dark side had became. It's as if they are chasing for something that is... no longer existed

I sometimes thought I have let go due to how in my daily life and stuff I seemed to be fine without actually actting (I then to be myself in face of most people), but as many of those dream logs have reminded me, they are like those peripheral sensory information I refer to earlier, seemed to be held in the mind for reasons not so certain

Math question: I know that if you have $n>1$ point charges with the same sign (and let's also specify to equal magnitude of charge) arranged in any non-coincident positions, any gaussian sphere that contains all of the point charges must also contain a point where the magnitude of the electric field vanishes. I know this must be true. However, I had a difficult time proving this to someone mathematically the other day. I'm sure there's a math theorem covering this, anyone know which?

1:14 PM

Jim: Is that some kind of newton shell theorem?

It'd be nice to just say a name instead of go through the rigor

@Secret you tell me

For any arrangement of charges on the surface of a sphere (assuming it is not a conductor nor dielectric), it must corresponds to some effective charge within the sphere when you superimpose all the electrostatic fields they produce. So I guess an argument might be somehow the gaussian sphere never enclose that effective charge?

ACuriousMind

@Jim if the charges are all equal, shouldn't that place be at the center of mess?

but I don't think that is possible because then gauss theorem in maxwell equation will be violated as the effective charge is always the total enclosed charge within the gaussian sphere?

@ACuriousMind no. Consider if you had a few closely packed charges and then one far off elsewhere. The center of mass is closer to the clump whereas the point of zero E field is closer to the loner

ACuriousMind

1:35 PM

@Jim Right. How's this: Since the field falls of at infinity, we can view it as a function on $S^3$. It's not defined at the locations of the point charges, so it's actually on an $n$-punctured 3-sphere, which is homotopic to the $n-1$-bouquet of $S^2$s, which has Euler characteristic $2(n-1)$ and therefore for $n \geq 2$ we have to have a zero of the vector field by the Poincaré-Hopf theorem.

@ACuriousMind This does not sound right. I don't think P-H works for noncompact fellows. I mean just take a nowhere zero vector field on S^3, then puncture it n times - that's a nowhere zero vector field on the n-punctured S^3.

Irrelevant, but in fact any noncompact manifold should admit nowhere zero vector fields: that's just the self-intersection number of the zero section in the tangent bundle and because everything is noncompact it should just be 0. Intuitively one should be able to "push the zeroes off towards infinity"

I think an ideal P-H for noncompact manifold ought to track information about the behavior of the vector field near infinity (end of the manifold). I dunno how to do it.

1:51 PM

you all are making this more complex than it should be. I just want know the name of whatever theorem would more succinctly prove that for any arbitrary arrangement of a finite number of similar charges, there exists a set of finite coordinates where the electric field vanishes. Use 2D if it makes it easier.

Maybe I should have gone to math.se

Consider two point charges on a sphere of radius $\rho$ located on the xy plane as follows ($\theta$ is the angle of the position vector of the 2nd charge from the y axis):
$$E_1=k_e\frac{q}{(\vec{r}-\rho\vec{x})^2},E_2=k_e\frac{q}{(\vec{r}-\rho \cos \theta\vec{x}-\rho \sin \theta\vec{y})^2}$$
The total electric field at any point $\vec{r}$ is given as:
$$E_1+E_2=k_eq\left(\frac{1}{(\vec{r}-\rho\vec{x})^2}+\frac{1}{(\vec{r}-\rho \cos \theta\vec{x}-\rho \sin \theta\vec{y})^2}\right)$$
Expanding:

I am however currently too integration symmetry occupied (one of simpleart's projects) to do the n charge case

@Jim I wasn't really answering your question, but I don't know the answer. I think It's more likely that a physics person would be able to answer this better than a math person.

(sorry the stuff above hasa typo, fixing it now...)

@Secret please note, I did not say the charges were on a sphere. I said that it is true for any sphere that contains the charges. They can be anywhere in the sphere, but all of them have to be in the sphere for it to also contain the zero point. Also, that seems incorrect. For two charges of equal magnitude and sign, there will always be a point of zero field exactly in the middle of them, which is inside the sphere if they are on the surface

good old secants

Seems that's a sign I have been leaving behind electrostatics for wayy tooo long, sigh...

2:28 PM

...
How would one proof that the following series
$$\sum_{i=1}^{\infty}\frac{1}{\vec{r}_i^2}$$
subjected to $x^2+y^2+z^2=\rho^2$
must contains a zero...?

Slereah

Does $\delta S = 0$ make sense, mathematically

I'm not sure it does

The well defined operation is $$\frac{\delta S}{\delta \phi} = 0$$

@BalarkaSen He's probably not coming back

Perhaps.

Well, depends on what you mean by $\delta S$. $dS = 0$ makes sense in terms of differential forms.

Mike Miller

You can do anything you want towards infinity. Pick your nonzero vector field, make it unit length, then multiply by your favorite smooth function to make it arbitrary length.

There's a version with boundary whenever your vector field is outward pointing.

Sorry, that's only true in even dimensions.

@Secret one wouldn't. You need a $\hat r_i$ inside the sum. Assuming, of course, that $\vec r_i$ is the displacement from the hypothetical zero point, and not simply the coordinates of the i-th point

Mike Miller

@Jim B and I are from math.SE. Your question says "charges", so I object to calling it a math question. Make it about manifolds and vector fields and i can help.

2:43 PM

Jim: I am currently thinking about picking an origin, and define all my fields by the ith charge with position vector $\vec{r}_i$. Then the field is zero only when they sum to the zero vector, which is always possible since the zero vector is always linearly dependent. Then it is easy to see that the charges will be distributed in a way that there exists a basis formed by the unit vectors such that it spans either $\mathbb{R}^3$ or its subspaces.

Since the picking of an origin is arbitrary and does not affect the physics, the above arugment should still hold and thus there is always a zero…

@Secret yes, I know. I want to know the name of the theorem that applies to this

typo: "charge will be distributed in a way" should be "given any charge distribution made of point charges"

Slereah

2:56 PM

@BalarkaSen The usual variational calculus

Mike: I think Jim's question is basically asking given any sums of vector fields of this form defined in the 3-ball $$\frac{\hat{r}_i}{||\vec{r}_i||^2}$$ with $||\vec{r}_i||^2 \leq R^2$ for all $i$, the name of the theorem that guarantees the sum always give zero somewhere in the 3-ball of radius $R$

@Slereah Oh, then I don't know what it's about.

3:11 PM

@Secret the inequality you give is not really necessary. $\vec r_i$ is the displacement vector between the zero point and a charge, which could have a magnitude larger than R, if the zero point were at one side of the "3-ball" and the charge at the other

Kenshin

sup peeps

who's ready for TRUMP

I think that is the same as if I set the origin at the centre of the sphere and the zero point will then be some linear combination of the unit vector of the above vector fields that give zero

Kelthar

@DanielSank I'm not saying "requirements" in the strict sense, I'm saying subjects that may be useful, as I have a bunch to choose from, for my master's. Just want to make sure I choose the most useful

@BernardoMeurer just saw that you're from Lisbon, me too! xD Don't tell me you're in the same uni, Instituto Superior Técnico (IST) as well? xD that would make the xilinx issue much more relatable :P

Kelthar

3:33 PM

@BernardoMeurer lol, it was displayed right below your name... feeling stupid now... anyway cool :)