# 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

# Please don't use computer-generated text for questions or answers on Physics.

In the past couple of weeks, a new generation of computer language-generating tools has become available to the public. The main bit of news is about a product called "ChatGPT," but that's just the most recent iteration of a class of software "chat bots." (Whether it's appropriate to refer to these systems as "artificial intelligence" is a philosophical question.)

Within a few days of ChatGPT's release, Stack Overflow issued a temporary don't-use-this policy, stating

Overall, because the average rate of getting correct answers from ChatGPT is too low, the posting of answers created by ChatGPT is substantially harmful to the site and to users who are asking or looking for correct answers.

The primary problem is that while the answers which ChatGPT produces have a high rate of being incorrect, they typically look like they might be good and the answers are very easy to produce. There are also many people trying out ChatGPT to create answers, without the expertise or willingness to verify that the answer is correct prior to posting.

On Stack Overflow, the blanket ban has mostly been a volume problem. Physics is a much smaller community, and we have so far detected only a smattering of such posts. However, the ones we have found have been pretty terrible, ranging from low-information word salad to obvious physical errors. For example, the sentence

In the case you described, the Drapher's point [sic] corresponds to a temperature of approximately 3,631 K and a wavelength of approximately 3631 nanometers.

should raise the eyebrows of anyone whose physics education has gotten as far as Wien's Law. (It may not, however, raise the eyebrows of anyone who has tried to teach Wien's Law to reluctant intro-astronomy students.)

In another post, the asker ended their question with "I asked an AI, but it didn't help me," followed by a properly-quoted paragraph which hadn't helped them because it didn't make any sense. I had a little flashback to when my children were small, and would sometimes run excitedly up to me, saying, "this thing! i found it on the floor! it tastes so gross! you have to try it!"

Some posts have even crossed the line from well-intentioned to deceptive. On one now-deleted post, a commenter asked the user who posted the answer to include references, and the post was edited to include

Some references for spin fluctuation are:

• "Pairing in Type-II Superconductors Induced by Spin Fluctuations" by D. J. Scalapino, E. Loh, Jr., and J. E. Hirsch, Physical Review Letters, Vol. 50, No. 4 (1983)
• "Spin Fluctuation-Mediated Pairing in Type-II Superconductors" by D. J. Scalapino, E. Loh, Jr., and J. E. Hirsch, Physical Review B, Vol. 34, No. 6 (1986)
• "Spin Fluctuation-Mediated Superconductivity: A Review" by D. J. Scalapino, Physics Reports, Vol. 250, No. 3 (1995)

It's instructive to compare these "references" to a search of this time period at the Physical Review, which should turn up the first two. Scalapino and Hirsch coauthored a number of papers on superconductivity in the early 1980s, including one in PRL v50 (1983) and another in PRB v34 (1986). However, Loh doesn't seem to have joined the group until 1986, and none of the team's coauthored titles includes the phrase "spin fluctuations." Likewise, the best candidate for the third reference has a different issue number and title. Is it a good use of anyone's time to pursue this detective work into thirty- and forty-year-old literature to see whether these rhymes-with-correct citations address the question at hand? Almost certainly not.

These are just examples which have been posted here on Physics by actual users. (That's why they're not linked: this is a policy discussion, not a name-and-shame.)

Note that my "please don't do this" isn't a new fancy policy tailored to existence of an exciting new chatbot which superficially appears intelligent. Our community has a number of established posting standards which are violated by these low-quality contributions:

• Originality. User contributions on this site are expected to be primarily the poster's own original work. If properly cited, including a small passage from a third party is fine, but complete answers are not.

• Attribution. Content which originally appeared elsewhere, including your own content, must be posted with attribution. Plagiarized content may be hidden until appropriate attributions are added, or may be removed altogether. It isn't common, but some serial plagiarists have found their site-use privileges suspended.

• Respect for others. If a user posts a question or an answer, our community needs to be able to expect that the post is a good-faith effort to learn things, or to help other people to learn things. Note that the network-wide policy is that "abuse of the system or the community," including cat-on-keyboard gibberish posts, can reasonably be flagged using the "rude or abusive" option, where enough flags will automatically delete the post and apply a reputation penalty. Surreptitiously involving Physics users in your tests of some chatbot software is rude. Generating "citations" without any idea whether they refer to real documents or not, much less whether the cited documents are relevant, is an abuse of other people's time.

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

"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

I'm going to use that $$\nabla$$ is the Levi-Civita derivative and the Riemann tensor is given by

$$R_{abc}^{\quad d} \omega_d = [\nabla_a, \nabla_b] \omega_c,$$

where $$\omega_c$$ is a 1-form.

The Lie derivative with respect to $$X$$ of $$\nabla_a \nabla_b \phi$$ is given by

$$\mathscr{L}_X \nabla_a \nabla_b \phi = X^c \nabla_c \nabla_a \nabla_b \phi + (\nabla_a \nabla_c \phi) \nabla_b X^c + (\nabla_c \nabla_b \phi) \nabla_a X^c.$$

On the other hand

\begin{align} \nabla_a \mathscr{L}_X \nabla_b \phi &= \nabla_a (X^c \nabla_c \nabla_b \phi + (\nabla_c \phi) \nabla_b X^c)\\ &= (\nabla_aX^c) \nabla_c \nabla_b \phi + X^c \nabla_a \nabla_c \nabla_b \phi + (\nabla_a \nabla_c \phi) \nabla_b X^c + (\nabla_c \phi) \nabla_a \nabla_b X^c\\ &= \underline{(\nabla_a X^c) \nabla_c \nabla_b \phi} + R_{acb}^{\quad d} X^c \nabla_d \phi + \underline{X^c \nabla_c \nabla_a \nabla_b \phi} + \underline{(\nabla_a \nabla_c \phi) \nabla_b X^c} + (\nabla_c \phi) \nabla_a \nabla_b X^c, \end{align}

where the underlined components are present in $$\mathscr{L}_X \nabla_a \nabla_b \phi$$. Using the symmetries of the Riemann tensor, we get

$$R_{acbd} X^c \nabla^d \phi = R_{bdac} X^c \nabla^d \phi = R_{bda}^{\quad c} X_c \nabla^d \phi = ([\nabla_b, \nabla_c] X_a) \nabla^c \phi.$$

Notice also that

\begin{align} \nabla^c \phi ([\nabla_b, \nabla_c] X_a + \nabla_a \nabla_b X_c) &= \nabla^c \phi (\nabla_b \nabla_c X_a - \nabla_c \nabla_b X_a + \nabla_a \nabla_b X_c)\\ &= \nabla^c \phi (\nabla_b \nabla_c X_a + \nabla_b \nabla_a X_c - \nabla_b \nabla_a X_c\\ &\quad \quad \ \ + \nabla_a \nabla_b X_c + \nabla_a \nabla_c X_b - \nabla_a \nabla_c X_b\\ &\quad \quad \ \ - \nabla_c \nabla_b X_a - \nabla_c \nabla_a X_b + \nabla_c \nabla_a X_b)\\ &= \nabla^c \phi (\nabla_a \mathscr{L}_X g_{bc} + \nabla_b \mathscr{L}_X g_{ac} - \nabla_c \mathscr{L}_X g_{ab}\\ &\quad \quad \ \ + \nabla_c \nabla_a X_b - \nabla_a \nabla_c X_b - \nabla_b \nabla_a X_c). \end{align}

Plugging the former in $$\nabla_a \mathscr{L}_X \nabla_b \phi$$ gives

\begin{align} \nabla_a \mathscr{L}_X \nabla_b \phi - \nabla^c \phi (\nabla_a \mathscr{L}_X g_{bc} + \nabla_b \mathscr{L}_X g_{ac} - \nabla_c \mathscr{L}_X g_{ab}) &= \mathscr{L}_X \nabla_a \nabla_b \phi\\ & - \nabla^c \phi (R_{acb}^{\quad d} X_d + \nabla_b \nabla_a X_c) \end{align}