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Sandbox

This is a sandbox question thread for testing purposes.

+ 2 like - 0 dislike
586 views

The Maximally Satirical Sandbox Model (MSSM)

The Maximally Satirical Sandbox Model is a model of an ideal sandbox, which can be stated as follows:- continued collaborative perturbations to the "Sandbox functional" results in \(\delta S=0\) where S is satire. This is equivalent through a simple R transformation (where R is reflection), to the Minimally Satirical Sandbox Model. The knowledge of the existence of knowledge of these two models can then be used to prove the PhysicsOverflow concavity theorem.

On the non-perturbative formulation of sock theory and the long standing missing sock problem

The Minimally Supersymmetric Sock Model (MSSM) is a supersymmetric \(\mathcal{N}=1\) matrix model that describes socks and sockinos. Socks are described as spin-2 massless particles while sockinos are described as spin 3/2 particles. Spontaneous supersymmetry breaking causes sockinos to gain mass, which explains the spontaneous disappearance of socks and their decoupling from the sockinos, besides when in a high-energy state, during which both sock disappear spontaneously and couple with their respective shoes and shoeinos (they are prevented from decoupling from their shoe because of Sock exclusion).

The Minimally Supersymmetric Sock Model exhibits S-duality with the Minimally Supersymmetric Shoe Model (MSSM). This means that a strongly-coupled sock is equivalent to a weakly-coupled shoe. This is unsurprising, because a very tight sock is the same as a very loose shoe, but the reverse is much more astonishing: a weakly-coupled sock is equivalent to a strongly-coupled shoe, i.e. a very loose sock is the same as a very tight shoe.

The Minimally Supersymmetric Sock Model also exhibits T-duality with the Minimally Supersymmetric Glove Model. This means that a very small sock is a very large glove, and vice versa.

This is an answer to the unsolved problem posted on Stack Exchange a few years ago, that was unfairly closed as being off-topic, despite the strong correlation observed by Shoesock Gloveson between the percentage of off-topic questions and the overall quality of a site.

april-fools

asked Mar 7, 2014 in Public Official Posts by dimension10 (1,950 points) [ revision history ]
edited Apr 3, 2015 by dimension10
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Testing testing (I am dimension10).      

@ArnoldNeumaier Fixed : )

please click again to confirm.

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comment - original revision.

If you see this comment, it means that there's a pretty annoying bug present in the autosave feature.

original revision of the comment.

31 Answers

+ 3 like - 0 dislike

$$\begin{align}I= \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} &= \frac12 \sum_{k=1}^8 \text{sgn}[\cos{(\arg{z_k})}] (\arg{z_k})^2 \\ &= \frac12 [2 (\arctan{\sqrt{\phi}})^2 + 2 (2 \pi - \arctan{\sqrt{\phi}})^2 \\ &- 2 (\pi - \arctan{\sqrt{\phi}})^2 - 2 (\pi + \arctan{\sqrt{\phi}})^2]\\ &= 2 \pi^2 -4 \pi \arctan{\sqrt{\phi}} \\ &= 4 \pi \, \text{arccot}{\sqrt{\phi}}\\\end{align}$$

\(\begin{align}I= \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} &= \frac12 \sum_{k=1}^8 \text{sgn}[\cos{(\arg{z_k})}] (\arg{z_k})^2 \\ &= \frac12 [2 (\arctan{\sqrt{\phi}})^2 + 2 (2 \pi - \arctan{\sqrt{\phi}})^2 \\ &- 2 (\pi - \arctan{\sqrt{\phi}})^2 - 2 (\pi + \arctan{\sqrt{\phi}})^2]\\ &= 2 \pi^2 -4 \pi \arctan{\sqrt{\phi}} \\ &= 4 \pi \, \text{arccot}{\sqrt{\phi}}\\\end{align} \)

answered Mar 26, 2014 by physicsnewbie (-20 points) [ revision history ]
edited Mar 26, 2014 by physicsnewbie

who's the joker upvoting these "answers"?

Hi physicsnewbie, I seriously like the look of these equations and it might well be that I steel them as a template some time in the future ... :-). Cheers

@dilaton oh lol :) They're not mine, I took them off mathstack exchange, because I wanted to test if align* worked under the Latex editor - and it does!

+ 3 like - 0 dislike
answered May 20, 2015 by polarkernel (0 points) [ no revision ]

@dimension10 test.

@dimension10: testing

@Test ok.

+ 2 like - 0 dislike
test\\ $a+b=c$ $$A+B=C$$
answered Mar 27, 2014 by Jia Yiyang (2,410 points) [ no revision ]

This is version 2 of my trial comment. This time I want to add it.

+ 3 like - 1 dislike

$R_{ab} - \frac{1}{2}R g_{ab} = 8\pi T_{ab}$

$$\begin{align}

-\frac{\hbar}{i}\frac{\partial \phi}{\partial t} &= -\frac{\hbar^{2}}{2m}\nabla^{2}\phi + V\phi\\

E\phi &= H\phi

\end{align}$$

answered Apr 1, 2014 by jerryschirmer (250 points) [ no revision ]
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hello (this is dimension10)

@dimension10 test

testing stuff regarding autosave. (1)

comment on answer - final revision.

comment on answer - eventual revision.

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original revision.

test comment.

+ 2 like - 1 dislike

Please use answers only to (at least partly) ask questions. To kill, murder, or assassinate for clarification, leave a knife instead. $\phi$  

      mmm     

answered Mar 18, 2014 by dimension10 (1,950 points) [ revision history ]
edited Apr 15, 2014 by dimension10

see, this got two up votes!

@VladimirKalitvianski A ping, as requested.

bla.

+ 1 like - 0 dislike

testing with Firefox.

Mozilla Firefox Microsoft Firefox

answered Mar 20, 2014 by physicsnewbie (-20 points) [ revision history ]
edited Feb 5, 2015 by dimension10
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@upvoters hello
self-pings clearly don't work. @dimension10.
@VladimirKalitvianski self-ping

testing test/..

another test

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testing tested @test, to test, please use the test where tests are tested as opposed to the test where tests are tested

@test tested tester test testably testing testably tested testing tests #test

+ 1 like - 0 dislike

\(\require{AMScd} \begin{CD} A @>a>> B\\@VVbV @VVcV\\ C @>d>> D \end{CD}\)

answered Mar 28, 2014 by dimension10 (1,950 points) [ revision history ]
edited Jun 2, 2014 by dimension10
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$\require{AMScd} \begin{CD} A @>a>> B\\@VVbV @VVcV\\ C @>d>> D \end{CD}$

@JiaYiyang: Is the above correct? If so, it works, at least for me.  

It looks good to me too, how did you fix it?

@JiaYiyang I don't know what went wrong for you. Did you not use the tex button?

I used dollar signs, shouldn't that also work?

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@JiaYiyang Oh, no I just meant "pre" and "p" by element type. SE's editor is basically a source editor (but takes markdown input unlike our source editor which only takes html input) so everything gets copy-pasted as plain text. There are three ways to avoid this problem:

  1. Paste in source mode.
  2. If you're using firefox 29.0+, you can install the add-on "Copy plain text" which allows you to ctrl+shift+C to get plain text copied.
  3. Paste it normally, but after pasting, select the pasted content and just change the value of the Levels menu from Pre-formatted text to Paragraph.

ok, thanks for the hints.

+ 1 like - 0 dislike

Heading Title

Heading Title

Heading Title

Heading Title

Heading Title
Heading Title 
Closed String spectrum
Sector Spectrum Massless Fields
RR \(\mathbf{8}_s\times\mathbf{8}_s\) \(C_0, C_1, C_2, C_3, C_4, C_5, C_6, C_7, C_8\)
RNS \(\mathbf{8}_s\times\mathbf{8}_v\) \(\Psi^\mu,\lambda'\)
NSR \(\mathbf{8}_v\times\mathbf{8}_s\) \(\Psi' ^\mu,\lambda \)
NSNS \(\mathbf{8}_v\times\mathbf{8}_v\) \(g_{\mu\nu}, F_{\mu\nu}, \Phi \)

Closed String spectrum
Sector Spectrum Massless Fields
RR \(\mathbf{8}_s\times\mathbf{8}_s\) \(C_0, C_1, C_2, C_3, C_4, C_5, C_6, C_7, C_8\)
RNS \(\mathbf{8}_s\times\mathbf{8}_v\) \(\Psi^\mu,\lambda'\)
NSR \(\mathbf{8}_v\times\mathbf{8}_s\) \(\Psi' ^\mu,\lambda \)
NSNS \(\mathbf{8}_v\times\mathbf{8}_v\) \(g_{\mu\nu}, F_{\mu\nu}, \Phi \)

answered Apr 6, 2014 by dimension10 (1,950 points) [ revision history ]
edited May 28, 2014 by dimension10
+ 1 like - 0 dislike

Rotating moderatorship:

answered Mar 24, 2015 by Vladimir Kalitvianski (-18 points) [ no revision ]

Hehe : )

+ 1 like - 0 dislike

testing green triangles

answered Nov 17 by Arnold Neumaier (11,395 points) [ no revision ]
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one green triangle seen

I found one, and deleted it.

another try

and one more

another attempt to get a green triangle

Most recent comments show all comments

Hm, I am not logged in as administrator (Dilaton.admin).

With respect to my own comments here, the new features work correctly too for me. What the heck ...?
 

Well, I get the right responses, too, but only since 15:11 Vienna time.

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