Tagging - 2018-04-02

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Martin Sleziak

05:05

Q: Deducing Newton's second law from Least Action Principle

I'm reading Simmons's Differential Equations and find that I have to obtain Euler-Lagrange equations of the following integrand (integrand of the action functional): $L = \frac 1 2 m \Big\lbrace \Big(\frac{dx}{dt}\Big)^2+\Big(\frac{dy}{dt}^2\Big)+\Big(\frac{dz}{dt}\Big)^2 \Big\rbrace - V(x,y,z)$...

Principle of least action

This article discusses the history of the principle of least action. For the application, please refer to action (physics). The principle of least action – or, more accurately, the principle of stationary action – is a variational principle that, when applied to the action of a mechanical system, can be used to obtain the equations of motion for that system. In relativity, a different action must be minimized or maximized. The principle can be used to derive Newtonian, Lagrangian and Hamiltonian equations of motion, and even general relativity (see Einstein–Hilbert action). The physicist Paul Dirac...

6 hours later…

Martin Sleziak

10:57

@amWhy: Do we really need an April Fools' tag on meta? — Asaf Karagila 1 hour ago

@AsafKaragila Maybe this could be a separate discussion, possibly in tagging chat room. As I have already mentioned there, the tag with the same name exists on Meta Stack Exchange. (And it is quite likely that we could get a bunch of questions where such tag could be suitable roughly at this time of the year.) — Martin Sleziak 29 secs ago

On meta.SE there is april-fools tag: meta.stackexchange.com/questions/tagged/april-fools They have even created quack-overflow - I'd say the latter would probably be too specific for our meta, but personally I'm ok with the former.

@Martin: True. But with maaaybe one question a year, the tag will fall into misuse and get deleted every year anew. So... what's the point? :) — Asaf Karagila 1 min ago

Some past questions about April Fools' topic:

Q: What is this game that appear in the right of all pages of MSE? (UNIKONG)

What is this game that appear in all pages of MSE from yesterday?! It happens only for me or it happens for all users?

support

A: What was that? Chat with an "expert"

The bot is here to fool people as 01 Apr is April Fool's Day. I chatted with the machine just now myself for about 5 min. This is just part of SE's way of having some fun with its users. Everyone, enjoy!

Q: What is this Unicoin business?

The question is in the title. What do the Unicoins do for us? Why are they on MSE? What is this mining business about?

support

This year there was also this question: Why does Quack Overflow start listening although I explicitly tell it I don't have a microphone? But the OP decided to delete it after being told in comments that this was already answered on Meta StackExchange.

@AsafKaragila I do not have strong feelings either way - whether the tag should stay or should be removed. In any case, to address your objections.

One of them was that the tag gets automatically removed until next year - since there is only one occurrence. I have added tag-excerpt and tag-wiki, so this is no longer an issue.

The other one is that the tag is likely to get misused. I am not sure, it's certainly possible. But probably from this viewpoint it is not worse than the already existing humor tag.

In any case, let's wait and see what the tag-creator (amWhy) has to say about this.

BTW the "What's the point" comment reminded me of MathOverflow.

in MathOverflow, Dec 3 '17 at 3:16, by Martin Sleziak

"it's not worth it", "let it all go down the drain", "not moving a finger any more"

in MathOverflow, Dec 3 '17 at 3:16, by Martin Sleziak

Reading some comments from MO users, one might get impression that they have rather pessimistic approach to the site. :-)

Martin Sleziak

12:12

The tag was edited away.

@AsafKaragila Not at all. It was only my mini April Fools contribution. — amWhy 8 mins ago

2 hours later…

Martin Sleziak

14:20

Am I the only one to think that the infinitelydivisible does not make much sense?

Q: Solution of $(F\ast F)(x)=F(\frac x2) \forall x $

Can the distribution of $X+Y$ be equal to the convolution of the distributions of $X$ and $Y$ when the random variables are not independent? In particular can this be true when $X=Y$?. This leads to $$(F\ast F)(x)=F(\frac x2) \forall x ...(1)$$ One example of such a distribution is the Cauchy di...