1:27 AM
School canceled for tomorrow on account of winter weather. Someone up there has something against my laboratory section.
And I had a really good lab scheduled: measuring the field inside and outside of solenoids. Each group dose it as a function of current and by sharing results at a fixed current they get it as a function of turn density, too.

2:06 AM
Despite aggressive efforts to rapidly close homework questions, I don't think it's working. For example, user41607 has asked 5 homework questions. All 5 have been put on hold yet all 5 received answers user41607 deemed worthy of accepting.
So if I were user41607 I'd just keep asking homework questions all day. Sure we're not a homework help site in name, but we are one in practice.

@BrandonEnright Yeah, we may need an extra mod to help out :/

@KyleKanos the mods are in a terrible place on this. Unilateral closes draw really negative attention from a vocal few. I think the mods need to draw up some rough guidelines for themselves (maybe solicit feedback in a meta post) for closing homework questions. Then, any homework question that meets the guidelines gets closed as soon as a mod sees it. That way we reduce / deter the asking of so many homework questions.
And if that doesn't work to close the bad ones fast enough, then maybe we could get another mod or two to help reduce the load.

A vocal few who don't do anything except complain
And soliciting meta for feedback is going to suck because those vocal few will complain about this being wrong and we shouldn't do that
I wish they would give a solid answer backed with examples of "What is a homework question to you"

2:47 AM
@KyleKanos We've been intentionally letting the community cover the homework issue recently. And I agree, that it's not enough despite quite credible efforts on the part of a lot of the community.

@dmckee I thought that experiment was over?

The problem is that a single earnestly helpful or rep hungry user can beat the close.
May be formally over, but I suffer from inertia on these things.
I've always been reluctant to delete such answers outright, because we don't have a policy calling for that and it seems so much more destructive than closing---which at least leave the post visible to all.

I can see that being a problem
The last item, I mean
I suppose someone could propose such a policy in Meta.

Well, from time to time I do get the urge to rule with an iron fist, but my turban-with-a-point-in-the-middle is out at the cleaners.

Actually, I'd probably be a terrible mod because all I'd do is use the powers to rule with an iron fist

2:54 AM
How's your mad cackling? I've always felt that a really good mad cackle makes up for at least a little bit of iron fist.

Sounds like that

Nice.

3:41 AM
I have that system
How would I find the equation of the moment?
It can be left in terms of variables of course.
I have the equations of forces already but this eludes me
Any idea of how I may get it?
I know it is a static system so the moment is zero

user54412
@Gigi10012 It will only work for those who have something like chatjax manually installed

user54412
but most people who frequent this room either have that or are fluent enough in tex that you don't have to worry

user54412
5:37 AM
anyone out there want to explain Gauss's Theorem to me?

Sure:

user54412
lol

user54412

But now that the site can shorten it through bit.ly, it would probably work again
So what are you having a hard time understanding? Maybe I can help

user54412
my problem is that no one talks about curved spacetimes

5:41 AM
Well, applying it to that will be beyond me I think. If you just needed some descriptions of the theorem I could probably help

user54412
I've read half a dozen papers that make an enormous leap from a differential flux-conservative equation to an integral one, without any citation or hint as to what they are doing

Okay -- so I might be able to help there

user54412
and MTW doesn't do it for curved spacetime - I used to think that book had everything

What Gauss's theorem says is that the volume integral of the variation of a field is equal to the integral of the flux through the surface of the volume.
So if you have a differential form of a governing equation, \Nabla \cdot f = s for example
You integrate both sides in X, Y, Z (over the volume) and the left side becomes the surface integral of f \cdot \hat{n}
While the source term would still be a volume integral of the source within the domain
This by the way is how you go from finite difference to finite volume forms of governing equations

user54412
that all sounds well and good, but is \nabla the covariant derivative?

5:47 AM
I suppose for 4D spacetime, you would integrate the differential form of the equation over (t, X, Y, Z) and then it will become the integral of your variable dotted with the normal of your spacetime surface...

user54412
also, in GR there's pesky appearances of \sqrt{-\det(g)} everywhere

Okay, now it's over my math-head... But maybe this will help?
On a differentiable manifold, the exterior derivative extends the concept of the differential of a function to differential forms of higher degree. The exterior derivative was first described in its current form by Élie Cartan; it allows for a natural, metric-independent generalization of Stokes' theorem, Gauss's theorem, and Green's theorem from vector calculus. If a k-form is thought of as measuring the flux through an infinitesimal k-parallelepiped, then its exterior derivative can be thought of as measuring the net flux through the boundary of a (k+1)-parallelepiped. Definition T...
It says that's what allows the extension of Gauss's theorem, but it doesn't show how

user54412
everyone seems reluctant to write out those details

Hanging out around here just makes me appreciate how lucky I am that fluid dynamics doesn't get more complicated than vector or 2nd order tensor analysis
@ChrisWhite I bet there is a Russian paper/book with the details somewhere. And now might be a good time to learn Russian since we might all be speaking it in our lifetime ;)
2

user54412
6:07 AM
indeed

6:24 AM
-1

The system starts at rest, and those two masses are connected with massless rod. Question : When v1=v2, what is the speed at that point? There is no friction, and the hint given was that normal force does no work. The bottom half is what I have tried myself. I though the velocity will equal...

Swiftly shut that one down please

user54412
7:15 AM
Will I get suspended when one day I crack and start a long personal rant about how some people on this site are useless to society and are not welcome on the internet?

user54412
Because every time I venture onto meta, I get one step closer to that point.

@ChrisWhite Mod messaged, yes. Suspended, depends :/
Want to rant? My email is open :p

user54412
maybe I'll just do something physical

user54412
I suppose I can shovel snow right now

user54412
7:29 AM
-1

I think His question is: "does anyone understand why magnetism and electricity are so fundamentally related?". It's a "phenomenon" because we don't know "why" electrical current through a conductor produces a magnetic field, we just know that it does and how to play with it.

user54412
^ the sad thing is I was "taught" E&M from a professor who believed this

9:47 AM
is it possible to solve Shrodinger equation for a photon to get wave function for double slit experiment?
For photon potential is zero
$$V=0$$
so Shrodinger equation will look like this:
$$\nabla^2\Psi + \frac{2mE}{\hbar^2}\Psi =0$$
$$\frac{\partial^2 \Psi}{\partial x^2} + \frac{\partial^2 \Psi}{\partial y^2} + \frac{\partial ^2 \Psi}{\partial z^2} + \frac{2mE}{\hbar^2}\Psi=0$$
then separate Wave function:
$$\Psi (x, y, z) = X(x)\cdot Y(y, z)$$
is this correct? what should i do next

10:04 AM
after separating i will get
$$\frac{\partial}{\partial x}\left(\frac{\partial}{\partial x} (XY)\right) + \frac{\partial}{\partial y}\left(\frac{\partial}{\partial y} (XY)\right) + \frac{\partial}{\partial z}\left(\frac{\partial}{\partial z} (XY)\right) + \frac{2mE}{\hbar^2}XY=0$$
then i will get:
$$Y\frac{\partial^2 X}{\partial x^2} + X\frac{\partial^2 Y}{\partial y^2} + X\frac{\partial^2 Y}{\partial z^2} + \frac{2mE}{\hbar^2}XY = 0$$
then?

1 hour later…
11:30 AM
is it possible to solve this?

1 hour later…
Anonymous
12:58 PM
@dmckee Fine, I will no longer participate in the review queues, I will try to completely stop visiting this site, just stop calling me rep-hungry.
2

2:19 PM
0

\begin{align} a = b \label{my equation} \end{align} Equation \ref{my equation} is my equation.

2:55 PM
Just as a heads up, this answer from Martin Beckett is at the top of the hot list. It does not answer the question, it is factually wrong, and it has twice the upvotes of the (very reasonable) question. It does not make the site look good, so you may want to take a look at it.

@DIMension10 Do you have a recent history of rushing to give complete answers to homework questions in the scant hours before they are closed? No? Then why imagine that comment was directed at you?
2
Context, man, context.
@tpg2114 A safe bet on many, many topics.
@ChrisWhite It might have been nice if someone had mentioned the relativistic origin in a comment before it was deleted. But not nice enough that I'm going to undelete it, make the comment and re-delete it. Of course, it would probably turn out that this user doesn't believe in relativity.

1 hour later…
Anonymous
4:21 PM
@dmckee Ok, ok, it was because I saw the words "vocal few", etc., near the "rep hungry", so that immediately made me think it was directed at me and others.

4:43 PM
How exactly we got this? $\Delta(xy) = x\Delta y + y\Delta x$ ?
And more importantly, what exactly does it mean?

4:55 PM
@ShuklaSannidhya Seems a straight-forward application of the product rule:
In calculus, the product rule is a formula used to find the derivatives of products of two or more functions. It may be stated thus: :(f\cdot g)'=f'\cdot g+f\cdot g' \,\! . or in the Leibniz notation thus: :\dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{dv}{dx}+v\cdot \dfrac{du}{dx}. In the notation of differentials this can be written as follows: : d(uv)=u\,dv+v\,du. The derivative of the product of three functions is: :\dfrac{d}{dx}(u\cdot v \cdot w)=\dfrac{du}{dx} \cdot v \cdot w + u \cdot \dfrac{dv}{dx} \cdot w + u\cdot v\cdot \dfrac{dw}{dx}. Discovery Discovery of this rule is credi...

5 hours later…
10:07 PM
"You do realize that we are all just your normal everyday people, and not scientists?"
@KyleKanos I guess you aren't a scientist.
I guess that goes for the rest of you too.
-2

You do realize that we are all just your normal everyday people, and not scientists?

@BrandonEnright I got a good laugh out of that one too

It's very unfortunate that some how "scientists" have become some abstract concept of "people out there" and not any sort of real profession that the average person sees happen.
I feel like the average public understands better what it is like to be a movie star or professional athlete than they understand what it is to be a "scientist".
5

10:41 PM
@BrandonEnright If possible, make a point to be "A Scientist" to the kids in your neighborhood. Build pin-hole cameras to observe eclipses. Make your jack-o-lanterns glow evil blue. Show them the conservation of angular momentum the next time you have a wheel off your bike (or they have one off theirs).
2
Also there are a lot of "sciencey" toy you can give away cheep. A couple of neighbor kids got enormous use out of a pair of US\$7 binoculars I bought. Fool things had crappy plastic lenses, but they did the basic trick: they made things look bigger and closer.
Ohhh...I almost forgot film canister rockets!
Magnifying glasses, rubber-band powered model planes (ones that actually generate lift, please), toy sized parachutes, the difference between balancing a pencil on your palm and doing the same with a broomstick, Newton's third law and conservation of angular momentum can be done on skates (ice or roller).
And on and on. You get a reputation as goofy but fun.
And kids go forth and learn, and then they know that (1) they can learn and (2) that there are systematic ways to investigate the world.
2

11:29 PM
@dmckee I couldn't agree more with the sentiment. I show off science-y things as much as I can. Unfortunately I don't really live in the typical American neighborhood with two car garages and families with kids. Families have been completely priced out of various silicon valley neighborhoods. I mostly only influence friends and coworkers.