# This is a sandbox

You can use this question as a formatting sandbox (if you can edit CW questions), and you can post answers if you want to test out formatting there as well.

NB: It's hard not to like this meta post but please keep it downvoted to -8 or less (rather than upvoting it) so that it doesn't get bumped to the front meta page every time someone is experimenting in the sandbox, which usually is not interesting to anyone else.

• In line with meta.stackexchange.com/questions/3122, but math/formulas don't work there. Jan 31, 2011 at 19:26
• @ David Zaslavsky: thanks. Don't the formulas work here on meta? The original post is at physics.stackexchange.com/questions/2372/… Feb 1, 2011 at 9:01
• Oh wait, LaTeX doesn't work in MeTa Jun 28, 2013 at 15:26
• Links work in comments. Use the [text](url) syntax. Jul 2, 2013 at 19:33
• @Manishearth Can we have a periodic clean-up of this thread? It's a huge unsightly mess. A few comments are worth keeping, but it'll be a lot nicer to use this Sandbox if it gets a good plow-through now and then. (Edit: actually, no, let me make this a flag. stupid of me.) Sep 13, 2013 at 22:35
• @DImension10AbhimanyuPS That one was long, contained a lot of images (which made it hard to find the end of it/scroll), and was bothering others. Sep 14, 2013 at 4:12
• @DImension10AbhimanyuPS No need for it. FWIW, the post will still be bumped to meta front page anyway. Sep 14, 2013 at 4:23
• @Manishearth: No, it won't. I just edited this post with -10 votes and it doesn't bump the fIrst page. Sep 14, 2013 at 4:29
• @DImension10AbhimanyuPS This is meta. Sep 14, 2013 at 4:30
• @Manishearth: Even on meta. I retagged this post just now; it doesn't bump the first page. Sep 14, 2013 at 4:33
• @DImension10AbhimanyuPS Yet the formatting sandbox is at the top of the main page. There's some criteria other than votes which I forgot about. Sep 14, 2013 at 4:36
• Please don't rollback Sep 14, 2013 at 4:36
• @DImension10AbhimanyuPS IIRC the edits to the question don't bump, but edits to the answers and new answers do. The vote threshhold is -4. meta.stackexchange.com/a/48579/178438 Sep 14, 2013 at 4:38
• @EmilioPisanty: I have one thing to say to that: ♦.♦ Sep 14, 2013 at 15:18
• @Manishearth it's the weirdest thing, though. An edit to a deleted answer seems to have brought the page to the top. I don't know if that's by design or not; I'll ask the corresponding question/experiment in a bit. Sep 14, 2013 at 15:29

Some apparent redspace problems, 10 m/s (original post):

The period of a pendulum of length $h$ for small oscillations is $2\pi \sqrt{h/g}$, with $g$ the acceleration due to gravity, about $10 m/s$.

Sοurce:

The period of a pendulum of length $h$ for small oscillations
is $2\pi \sqrt{h/g}$, with $g$ the acceleration due to gravity,
about $10 m/s$.


Test post to test something (see comment)

gibberishwaE RscSA

gibberish .......qawr ASDCSaASFAF

gibberish

Mars Is Earth. , Manish Earth wrote this post.

• Identity theft! ‮ ♦ yksvalsaZ divaD May 8, 2012 at 13:56
• Just a note: The above comment is a test case for this: meta.stackexchange.com/q/131818/178438 . The comment was made by me, to display how usernames can be spoofed (doesn't work on MSO due to different styles for links). May 8, 2012 at 15:21
• I don't think I particularly needed to add the extra em dash, realized that now. May 9, 2012 at 4:51
• Identity theft! ‮♦ yksvalsaZ divaD May 9, 2012 at 6:17
• @dimension10 Oh, no problem, that comment is OK. ;-) Jul 6, 2013 at 8:26
• The original owner Sep 14, 2013 at 15:46
• Nope. Community wiki posts have different behavior when it comes to votes and edits, but otherwise the original poster is the owner Sep 14, 2013 at 15:50
• Big Bang ‮‮ Is that so? - Sep 14, 2013 at 20:23
• Test test test test test Nov 18, 2013 at 4:53

Given $\left\{ a_{n}\::\:n=1,\:2,\:3,\:\cdots\right\}$ is an infinite sequence in $\mathbb{R}$, and every term is positive. How to prove that the set

$$\left\{ \frac{2+a_{n}}{\sqrt{2+a_{n}^{2}}}\::\:n=1,\:2,\:3,\:\cdots\right\}$$ has a limit point?

Bolzano-Weierstrass theorem says that every bounded infinite subset of $\mathbb{R}$ has a limit point. But how to prove it is bounded? I tried this $$\left|\frac{2+a_{n}}{\sqrt{2+a_{n}^{2}}}\right|\leq\left|\frac{2+a_{n}}{\sqrt{a_{n}^{2}}}\right|=\left|\frac{2+a_{n}}{a_{n}}\right|$$ but does not seem work.

Or these is another way to show it has a limit point, without using Bolzano-Weierstrass theorem

Let me check if the sandbox at -8 goes to the front meta page if I add a new answer.

• Long message <html> <body><br> <br><br> </body></html>________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ end
– Qmechanic Mod
Dec 9, 2016 at 12:58
• Sandbox on Mother Meta (Warning: Slow to load.)
– Qmechanic Mod
Dec 9, 2016 at 13:29

I greet you, people from a decade ago, from the far future land of 2021.

I’m testing whether it’s still the case that activity on this highly-downvoted post is still hidden from the main page.

$$$$\partial_\mu \frac{\partial (g^{\alpha \beta} \partial_\alpha A_\beta)^2}{\partial (\partial_\mu A_\nu)} = \partial_\mu \Bigg(2 (g^{\alpha \beta} \partial_\alpha A_\beta) \frac{\partial(\partial_\alpha A_\beta)}{\partial(\partial_\mu A_\nu)} g^{\alpha \beta} \Bigg) = \partial_\mu (2(g^{\alpha \beta} \partial_\alpha A_\beta) \delta^{\mu}_{\ \alpha} \delta^{\nu}_{\ \beta} g^{\alpha \beta}) = 2 \partial^\nu (\partial_\alpha A^\alpha)$$$$

"However, I have seen a lot of books say that flux only depends on charge enclosed..."

Generally, when textbooks say that flux depends only on the enclosed charge, they are referring to the flux of a closed surface like a sphere or a cube. I assume your question is precisely about how the flux through a cube and a sphere is equivalent.

You can, by all means, do the integration to find flux enclosed by a cube, but the integration becomes a bit tedious given that the electric field isn't a constant and reduces as we move away from the centre towards the corners. You can go the extra mile by all means and ask " well if we put a charge very close to the surface wouldn't $$E$$ be changing for $$dA$$? How can their integral, ie flux, be equal???" Soon you realise that where $$E$$ is scarce on a portion of the closed surface, it is compensated by an excess of $$E$$ in other regions.

You can think of it this way. Take a spraypainter, the one you use for graffiti. Take it very close to the wall, about $$10 cm$$ and spray the paint. You notice that the region is painted thickly and darkly, but concentrated in only a small region. Now pull back the spraypainter by about half a meter and spray once again. This time you see that the area being painted increases. But at the same time, the painted region become lighter and thinner.

Image courtesy of Wikipedia

There's also another reason as to why Gauss' law is true. And that's because electric charges obey inverse square law

Had they obeyed inverse cube law, this may not have been possible. See this SE post