The h Bar

For instance, I'm still getting the hang of list comprehensions. It's not that they are all that mysterious, it's just that they aren't part of a language I already know well, so I have to develop the habits that go with them.

0celo7

@dmckee How pretentious is it to write Hodge duals in vector calculus stuff because I don't like $\mathrm{d}\mathbf{A}$

dmckee

It depends entirely on how the person who is going to grade that work feels about it.

0celo7

@dmckee PSE answer.

dmckee

If the grade is a TA who doesn't know that notation it is not likely to end well.

Well, then you'll only get votes from people who feel comfortable with that notation. For instance, I know I've heard of them (Hodge duals) and seen them written once or twice, but I couldn't write one down without looking stuff up. I'm unlikely to vote for an answer relying on them.

0celo7

@dmckee See, here's the problem

The answer is part Newtonian, part GR

the GR part needs the Hodge dual

like, there's no other way to write it

but the newtonian part could do without it

dmckee

3:17 AM

Well, then. The audience for the answer knows them. So you're fine.

0celo7

@dmckee Ok, that may be, but is $\int\star g$ just as understandable than $\int \mathbf{g}\cdot\mathrm{d}\mathbf{A}$

even for a GR audience

dmckee

How would I know?

0celo7

@dmckee you seem to be a pretty smart guy

dmckee

Note what I said above about Hodge Duals. They are not part of my mathematical vocabulary and I can decode them only with considerable looking up of references and cognition.

user54412

4:15 AM

@Bill- Not to be rude but I no longer reply to comments questioning my reason. — The Void 16 mins ago

user54412

I think that wasn't quite what the person meant to say, but it came out even... better... this way.

Secret

5:31 AM

journals.aps.org/prd/abstract/10.1103/PhysRevD.93.011501

Gluon contribute more to spin in the proton that quarks do.

0celo7 might ask how is this related to GDP

The next step is to wait for an electron ion collider, they say

0celo7

5:58 AM

@ChrisWhite What's the canonical reference on ADM, preferably with a heavy math flavoring?

user54412

other than the ADM paper?

0celo7

@ChrisWhite Yes.

user54412

dunno

user54412

what do you want to do with it?

0celo7

@ChrisWhite Read.

user54412

6:09 AM

like, the quantum gravity literature is separate from the numerical relativity literature

user54412

and both are separated from the causal structure of spacetime literature (which is basically just Wald)

0celo7

@ChrisWhite And Hawking-Ellis, Beem et al., Geroch, Penrose, ...

There's a whole book on asymptotic structures

Are you sure you're a GR person?

user54412

I thought I was a code monkey

3

0celo7

Then there's another book on topology + Cauchy problem

@ChrisWhite It's turning out to be the case...

> Thistlethwaite

Hahah

What the fuck

We have a prof with that name

@ChrisWhite some mathematician has to have written a thing on ADM, right?

user54412

probably

user54412

6:23 AM

at least 339 citations to the original article, which was then made into a chapter of (a different) Witten's book, which was then republished into a paper with at least 738 more citations

user54412

feel free to look for anything interesting :P

user54412

actually those 3 published about a dozen closely related articles over a span of 3 years

user54412

guess they knew how to milk a good idea

DanielSank

Yo @0celo7, thanks for reminding me that asking other people in the field questions is a good idea.

It's hilarious how often we forget that.

DanielSank

8:39 AM

@Danu, since you were making fun of my insisting on defining dimensionless (or conveniently dimensioned) quantities, I offer you a challenge:

Go find an E&M book which uses "cgs units" and another using "SI units" and explain to me how to get equations written in one of those systems into the other.

If you can do this in a clear way that is not doing what I insist, I will mail you a box of snacks of your choice.

user116211

8:53 AM

@DanielSank I'm sure you can get Purcell's E&M original Berkeley text in the archives. Then you can check the recent books modified by Morin. The former is on CGS and the latter is in SI.

Danu

10:24 AM

@DanielSank Lmao, that's funny

I might take you up on that

However, I am currently in Greece for holidays so it might not be soon ;)

@ChrisWhite That's actually the father of "current Witten" (if you're talking about Louis)

DanielSank

10:51 AM

@Danu Well, I am happy that what I recall was not clear to be a long lived adventure is still bringing you to Greece.

Go to Syphnos.

I guess it's actually probably better written Sufnos.

Whatever.

Danu

11:16 AM

@DanielSank Συφνός; :)

@DanielSank Still holding on ;)

Slereah

12:09 PM

I was about to answer a question and then bam

it is closed

v. rude

Danu

You can still answer it.

ACuriousMind

@Slereah Was it some crazy time travel question that you wanted to answer with crazy GR? ;)

Slereah

IT WAS

I mean it's either that or answering some gravitational waves, lately :p

ACuriousMind

Haha

Danu

Hey, I got mad points for some algebra on David Z's question :)

Easiest points I got in a while

Slereah

12:21 PM

The most point I've ever done was basically saying that chemical bonds have p. weak mass

SE doesn't have that good return on investment :p

ACuriousMind

12:41 PM

That is true

0celo7

@DanielSank I figured you guys would have a private PSE thing.

"you guys" = quantum computer nerds

0

Q: Minkowski geometry definition

The general relativity is based on Minkowski geometry definition with its special properties. The general relativity cant be approved wihout Minkowski geometry definition. Why Minkowski geometry is "True"? I can understand that the general relativity has been approved using measuring of light dev...

Maybe this person is coming at this from a Cartan geometry angle

then the question makes sense

ACuriousMind

@0celo7 Such a person would ask "Why is the GR connection the Levi-Civita connection?" or somesuch (and we have those questions, too!), not ask about "Minkowski geometry".

Slereah

Well

IS IT???

0celo7

@Slereah no, it's all a lie

ACuriousMind

@Slereah is what what?

Slereah

12:48 PM

in almost all cases the Levi Civitta connection and the Cartan connection are the same in GR

Hm

0celo7

@Slereah since when do you know about Cartan connections

Slereah

I wonder if any classical source would generate non-metricity, with a general connection

Since I've read about Lucazevitz or whatever his name is

dude who wrote a big book on GR torsion and such

0celo7

oh, from a physics book :P

ACuriousMind

The reason the GR connection is the LC connection is that the E-H action gives the LC connection as the solution to the equations of motion for the connection if you treat the metric and the connection as a priori independent

Slereah

well duh

0celo7

12:50 PM

@ACuriousMind yes, we've all read Qmechanic's post

Slereah

@ACuriousMind Not quite

I mean in a vacuum, sure

But with matter terms they can be different

0celo7

this is actually true

get #rekt @ACuriousMind

ACuriousMind

@0celo7 How can you claim a bad memory and remember that some random post you read sometime in the part on this site is written by Qmechanic? :P

@0celo7 It's true, doesn't make what I said wrong - if you add matter, you don't have the E-H action anymore :P

0celo7

@ACuriousMind that is the post you're referring to, right?

Slereah

Well sure but that doesn't mean much :V

"A flat connection is L-C"

Well duh

ACuriousMind

12:52 PM

@0celo7 I had to look it up to confirm the author, but yes

Slereah

Tho I guess empty space can be non-flat

No torsion waves :p

0celo7

@ACuriousMind oh, please link it

I never favorited it

ACuriousMind

No, practice your Google-fu

0celo7

but my memory is too crap for me to find it

ACuriousMind

(I haven't favourited it either :P)

0celo7

12:53 PM

@ACuriousMind oh you're so evil

Cartan geometry:

Slereah

I wonder what a source term looks like in metric affine gravity

0celo7

12

Q: Why is it so coincident that Palatini variation of Einstein-Hilbert action will obtain an equation that connection is Levi-Civita connection?

There are two ways to do the variation of Einstein-Hilbert action. First one is Einstein formalism which takes only metric independent. After variation of action, we get the Einstein field equation. Second one is Palatini formalism which takes metric and connection are independent. After variati...

general-relativity gravity lagrangian-formalism differential-geometry variational-principle

Secret

http://sciencedemonstrations.fas.harvard.edu/presentations/yo-yo

guys I need some help to understand some rotational mechanics stuff again

I am trying to convince myself that the yoyo in figure b is rotating clockwise (instead of counterclockwise) but I don't know how to move from the frame where point p is not rotating to a frame where point c is not rotating

Any ideas on how to think about it?

Slereah

inspirehep.net/record/1246899/files/arXiv%3A1308.1642.pdf

what the fuck is equation 7

ACuriousMind

@Slereah Well, how do matter terms look that generate torsion?

0celo7

12:56 PM

@Slereah fucking physicists

watching physicists do math is like monkeys in an opera

skull petrol

Captain subtle^

:P

0celo7

@ACuriousMind German time

is the "g" in "Wagner" hard in Pfalzisch?

or does one kind of soften it

Slereah

@ACuriousMind $T \approx \Theta$

Or something

0celo7

what?

Slereah

Not sure what's the term for non-metricity, though

Torsion is proportional to the spin tensor

0celo7

1:00 PM

ah

Slereah

$\Theta \approx \frac{\delta \mathcal{L}}{\delta \omega}$

Well I guess technically $\frac{\delta \mathcal{L}}{\delta \omega}$ will be the same with non-metricity

ACuriousMind

@Slereah I'm not sure you can generate non-metricity. A non-metric connection would not be a connection on the orthonormal frame bundle anymore, i.e. the structure reduction to get the spin connection would fail.

Slereah

Still part of the connection

Well I know that some theories use non-metric connections

Generated by the hypermomentum

not sure what that is tho

0celo7

it's super super momentum

Slereah

super duper

Isn't supermomentum a SUSY thing

Momentum, supermomentum, hypermomentum, gigamomentum, ultramomentum

yuggib

1:04 PM

@Danu to motivate the "big deal" about the operators proof, see here:

20 hours ago, by ACuriousMind

@yuggib Okay, thinking about it, it's not as simple as I intuitively assumed. For self-adjoint bounded operators, you get that the "expectation value" is a norm. Since you can uniquely reconstruct the scalar product from a norm obeying the parallelgram identity (and that's not trivial!), self-adjoint operators who agree in their expectation value are already the same. For general operator, split into self-adjoint and anti-self-adjoint parts.

and the following discussion

I'm not saying it's impossible, but also trickier than many people may think

Slereah

"From the physical point of view, it would be interesting to study some concrete examples of matter fields coupled to connection, namely those fields for which the hypermomentum does not vanish. A (phenomenological) candidate to source connection dynamics could be a semiclassical spinning dust matter distribution, alias a generalization of the perfect fluid in the case of non-vanishing spin, a fluid otherwise dubbed Weyssenhoff fluid "

what

Oh wait

I think "hypermomentum" is a generic term

For a source of both torsion and non-metricity

So hypermomentum is still written as $\frac{\delta \mathcal{L}}{\delta \omega}$

Sidarth

Hey guys...I cannot find a question as "what is space time"....can anyone help me with a good link?

@ACuriousMind?

@Slereah

@DavidZ?

ACuriousMind

Pseudo-Riemannian manifold

In differential geometry, a pseudo-Riemannian manifold (also called a semi-Riemannian manifold) is a generalization of a Riemannian manifold in which the metric tensor need not be positive-definite. Instead a weaker condition of nondegeneracy is imposed on the metric tensor. Every tangent space of a pseudo-Riemannian manifold is a pseudo-Euclidean space described by an isotropic quadratic form. A special case of great importance to general relativity is a Lorentzian manifold, in which one dimension has a sign opposite to that of the rest. This allows tangent vectors to be classified into timelike...

Sidarth

thanks dude..

:)

so "space time fabric" actually refers to this?

@ACuriousMind

Slereah

Spacetime is a triplet of a second countable, Hausdorff, paracompact manifold, a metric of signature n - 2 and a Levi-Civitta connection

There you go

Sidarth

1:11 PM

@Slereah of course it is.

ACuriousMind

@Sidarth "Spacetime fabric" is a word without technical meaning.

Sidarth

a bad jargon?

ACuriousMind

It's something people use who also think the rubber sheet analogy explains GR.

Secret

http://sciencedemonstrations.fas.harvard.edu/presentations/yo-yo

So when it slides, is it slide to the right because the horizontal component of the tension is pointing to the right?

Sidarth

so the actual thing to be learnt is that manifold huh?

ACuriousMind

1:13 PM

If you want to say something that is not handwaving, yes

Of course, many people seem completely fine waving their hands :P

Sidarth

hmm....actually it was a serious question...

:P

yuggib

@ACuriousMind or if you want to handwave at a higher level :-þ

ACuriousMind

@Sidarth I answered it seriously, I just mentioned there are people who might disagree with me

@yuggib Heh, well....true

Sidarth

@ACuriousMind I never questioned your seriousness...;)

Slereah

of course in a fancy spacetime you are also expected to define a time orientation vector field and a volume form :p

ACuriousMind

1:19 PM

@Slereah The existence of a volume form is no restriction.

Slereah

Orientable one

ACuriousMind

Ah, but that has nothing to do with the volume form, just with the orientability of the manifold

Slereah

Hm

I recall a paper defining a thing for the orientation of the spacetime

What was it

ACuriousMind

Oh, sorry

The correct statement is: Existence of a volume form is equivalent to orientability

Slereah

Well there :p

Although in this case, it is a 3-form

Because IIRC time orientability + space orientability is equivalent to spacetime orientability

According to Visser and that weird ass paper on De Sitter spaces identifications

Secret

1:28 PM

I am also currently (out of pure curiosity) trying to do some simple hand calculations on a spacetime based on the Minkowski spacetime

The spacetime I am trying to get a metric to describe is like the usual spacetime diagram, except that for points lying closer to the origin, their 0 character of the metric increases in a gaussian like fashion

i.e. I am trying to investigate a spacetime that looks roughly like this:

$$ds^2=-dt^2+dx^2+?(e^{x^2-t^2})du^2$$

where $?(e^{x^2-t^2})du^2$ is a term that take care of the fact that points closer to the origin are more degenerate

0celo7

@ACuriousMind Salsa and eggs

@ACuriousMind Do you know the proof? I feel like this is one of those things everyone knows but the proof is in some topology book from the 60s

ACuriousMind

1:50 PM

@0celo7 Wikipedia has the proof.

0celo7

@ACuriousMind Huh? Where?

ACuriousMind

@0celo7 In the last paragraph of the section is linked.

0celo7

@ACuriousMind It's not very clear.

ACuriousMind

Well, you can of course also get out the sheaves ;)

0celo7

@ACuriousMind They don't explain what SL has to do with anything or why the top form bundle has to be trivial.

Once one shows that, I agree that the theorem follows.

But they simply claim the most important part of the proof :V

ACuriousMind

2:07 PM

@0celo7 1. An $\mathrm{SL}(n)$-structure is equivalent to a volume form by the preceding paragraph. 2. A volume form is a nowhere vanishing section of the top exterior bundle, and vector bundles with global nowhere vanishing sections are trivial.

ACuriousMind

"right now we are struggling to maintain the quality of the site under an onslaught of ill thought out, poorly phrased or even utterly barking questions" - JohnRennie

The last few times I had that feeling, someone dug up statistics that showed not more questions are closed than usual. Is it different this time? If not, could that mean we are operating at maximum closure capacity, not that we don't get more bad questions than usual?

The Dark Side

@ACuriousMind Not just low quality questions. Data point 1

Data point 2

ACuriousMind

That's one data point because it's the same user :P

The Dark Side

Yes :P

And second answer to the same question.

ACuriousMind

also, not-an-answer answers aren't really concerning me. There are always people who inexplicably feel the need to write garbage as answers, and they invariably get flagged, downvoted and deleted.

The VLQ queue is not overly full

However, the close queue does seem unchararistically full to me, and I suspect it's only not fuller because the frequent close voters use up most of their votes in review and can't send more questions into it after that

ACuriousMind

2:27 PM

This is an interesting sequence of edits

It suggests OP is perfectly capable of writing an intellegible grammatically correct question, they just didn't care to do so until they didn't get an answer and got downvoted.

The Dark Side

@ACuriousMind You are right. Close vote queue isn't accessible to me, but I see 39 items in there. And obviously you are done for the day I guess.

21

A: A guide to moderating crypto.stackexchange yourself - close voting

Rationale: why do we close questions? There are really two ways to do moderation. If you've been on any of the .moderated newsgroups you'll be well aware that to get a post on there, it needs to undergo a review process first and be accepted. The aim is to keep problem discussions from arising. ...

^ I think these guys have put together a nice FAQ there. So, guys like me can contribute to the process by flagging, & that does send VLQ stuff into that queue.

John Duffield

2:44 PM

All: this is not a low quality post, John Rennie asked for comments in meta, and then suggested that commenters give their views in an answer. That's what I did.

The Dark Side

@ACuriousMind So, considering that flagging begins at 15 rep, I don't think the close voters are to be blamed. They have their place in the story, but items can be sent there by others too.

So, it is a question of participation.

0celo7

@ACuriousMind What? Circular proof is circular.

The Dark Side

3:00 PM

"Comedy is tragedy that happens to other people":

22

Q: Multimeter exploded during ACV measurment - What did I do wrong?

simulate this circuit – Schematic created using CircuitLab A DT-830B meter. I bought a new transformer, and I was trying to measure the output voltage. I plugged the probes on 'VΩmA' (not 10 A) and COM. and set it to 750(not 100% sure if i put it on 750 or 200) ACV. Then I put my probe...

safety multimeter voltage-measurement

ACuriousMind

@0celo7 There is no circularity: $M$ is orientable $\iff$ $M$ has a $\mathrm{GL}^+(n)$ structure $\iff$ $M$ has a $\mathrm{SL}(n)$-structure $\iff$ $M$ has a volume form $\iff$ $\Lambda^n(T^\ast M)$ is trivial.

The first $\iff$ is because such a structure is precisely given by positive-determinant transition functions, the second because GL+ deformation retracts onto SL, the third by the preceding paragraph in the Wiki article, the fourth because global non-vanishing sections are equivalent to triviality.

0celo7

@ACuriousMind I will examine your wizardry later.

ACuriousMind

@TheDarkSide Yes, it undoubtedly is

I didn't want to "blame" the close voters, but I don't think many people except the regular reviewers flag questions for closure at all

0celo7

@ACuriousMind I don't understand the third one.

ACuriousMind

@0celo7 Well, that doesn't mean the proof is circular :P

What do you not understand?

user116211

3:09 PM

PSE is stalled!

The Dark Side

@ACuriousMind So, maybe publicizing this post is one way forward.

ACuriousMind

@TheDarkSide It's already upvoted enough to appear as a community ad. What more could we do?

The Dark Side

@ACuriousMind Certainly nothing more. Wait for more people to see it, maybe?

And hope that they pick up the message.

ACuriousMind

Also, the review link is right there at the top of the page. The issue is not that people don't know they can review, I think, at least not for >3k people

It's that they don't do it. Because they don't have the time, energy or inclination, and I don't see what we could possibly do about that

user116211

@ACuriousMind 100% true.

Secret

3:14 PM

Let me rephrase my question

I am looking for a spacetime that is basically minkowski, but where points progressively became more degenerate the closer they are to the origin of (e.g. O)

Will the following metric can give what I want?

$$ds^2=(e^{x^2+t^2}-1)(-dt^2+dx^2)$$

?

The Dark Side

@ACuriousMind No, but VLQ queue can be populated even by those having rep score > mere 15. (By flagging). And the solution may lie there.

ACuriousMind

@TheDarkSide How? The VLQ queue has no issues. It's fast moving, and almost always decided unanimously

And "recommending closure" for a question in the VLQ queue does nothing but send it into the close queue

@Secret What do you mean by "degenerate"?

Secret

If I recall correctly form my discussion with Slereah and Danu, a degenerate point on a manifold is one where the distance between it and any other point vanishes

I am trying to introduce a single dengenerate point into the minkowski spacetime model just to play around and see what interesting things will pop up

Feb 9 at 13:10, by Slereah

For a metric with a degenerate point, every distance between that specific point and any point is 0

ACuriousMind

@Secret 1. A manifold where that happens cannot be Hausdorff. That means it's terribly pathological (although I know people are considering them). 2. What is stopping you from just calculating the distance for the metric you give?

The Dark Side

@ACuriousMind Oh, OK, so I missed it then. But, if eventually everything is settled only in the close queue, why do we need a lower level queue? There is no division of labor then. The tier-2 queue only forwards it to the tier-1 queue!

ACuriousMind

3:20 PM

@TheDarkSide Answers are dealt with within the queue, and most of the VLQ flags are on answers

I actually don't think it's possble to manually cast a VLQ flag on a question

The only questions that appear there are those the system automatically detects as spam/LQ

The Dark Side

@ACuriousMind Are you sure?

So, this VLQ flag doesn't send it to the VLQ queue?

Danu

@yuggib After applying the polarization identity, you have that all matrix elements are teh same

The Dark Side

Or did I miss something somewhere?

Danu

I want to say that, under reasonable conditions, this completely determines the operator---at least generally enough to assume it in physics

Secret

@ACuriousMind
2. I am not sure if the metric I wrote (by inspection) is the one that has the properties I want. Thus I want to make sure my guess is sensible before I start using it to calculate distance

1. So even if the function that introduce the degeneracy is continuous, as long it has a degenerate point, it will still not be haudoff. Is it because there exists one point (the degenerate point) whoose neighbourhood with other points are not disjoint?

ACuriousMind

3:26 PM

@Secret Um, actually, the definition of a degenerate point is wrong. A point is degenerate if the determinant of the metric vanishes there (which it does at 0 by inspection for yours). You can't write down a metric that's not identically zero that would have a point whose distance to every other point is zero.

Danu

@ACuriousMind what do you think?

ACuriousMind

@Danu "Applying the polarization identity" is exactly what my "norm that fulfills the parallelogram identity induces unique scalar product" does.

You're claiming the argument is trivial by hiding it behind a fancy name :P

Danu

But that's not fancy at all AFAIK

ACuriousMind

@Danu That always true. Equal matrix elements always in the same basis mean the operators are equal, that's the definition of a basis.

Danu

It's just saying that $2\text{Re} \langle \psi \mid A \mid \phi \rangle = \langle \psi + \phi \mid A \mid \psi + \phi\rangle - \langle \psi \mid A \mid \psi \rangle - \langle \phi \mid A \mid \phi \rangle $

@ACuriousMind Okay, so how is this not easy, then?

something like that

Yeah okay I see how one can get some pendantic trouble ;)

ACuriousMind

3:32 PM

It's not the hardest proof in the world, but it's not the one-liner I thought it would be at first glance, either

Danu

Hmkay.

No, actually, come on

Polarization also gives the imaginary part

@ACuriousMind I think it really is a one-liner, then.

Secret

>You can't write down a metric that's not identically zero that would have a point whose distance to every other point is zero.

This point does not sound quite straightforward to me. Isn't that when the equation of a metric has functions of the coordinates will mean that it vary with the coordinates (c.f. Schwarzschild metric where it has a 1/r dependence on the dt dr and d$\Omega$ term,

Thus if I wrote that metric of mine above, I can have its determinant vanishes only at the origin but not elsewhere? thus you can have a metric that contains a degenerate point without it being the zero m…

ACuriousMind

@TheDarkSide oO

I can't flag a question a VLQ. The only options under flag for me are "spam", "rude or abusive" and "in need of mod intervention"

I have no idea what the VLQ option there does. How can a question VLQ in the sense that it should go into a queue, but not be either closeworthy or spam/rude/in need of mod intervention?

@Secret The distance being zero means that $\int_\gamma \sqrt{g(\dot{\gamma}(t),\dot{\gamma}(t))}\mathrm{d}t = 0$ for every geodesic $\gamma$ (at least locally). $\gamma$ stays a geodesic if you shorten it, meaning that the function $g(\dot(\gamma),\dot(\gamma))$ is zero when integrated over arbitrary intervals as long as they are not too large. There is no $f$ with $\int_a^b f = 0$ for a continuous range of $b$ that's not zero (except on a null set)

One has to do a bit more to conclude from this fully that $g$ vanishes identically everywhere, but I am pretty confident it does

The Dark Side

@ACuriousMind Haha, you earned a lot of rep very quickly :D

I'm sure these flags were a respectable chunk of my 868 helpful flags (to date)!

0celo7

@Danu What's up with the huge spacing there?

@ACuriousMind sorry for the delay, had to take an exam

And now I have to do homework

ACuriousMind

3:47 PM

@TheDarkSide Oh, I flagged questions before getting 3k, too. I just never chose the VLQ option, always a reason for closure.

The Dark Side

:)

ACuriousMind

So I didn't remember that there's a VLQ flag on questions for <3k users

And now I have no idea what that flag is supposed to do

I'll ask meta about it later, I think

0celo7

@ACuriousMind Is the curve timelike?

The Dark Side

@ACuriousMind :) C'ya.

0celo7

Wait, the metric cannot vanish.

What are you talking about?

@ACuriousMind I get it.

0

A: What is a metric?

It is a way to quantify the behavior of the space, whether it is flat or curvilinear. $ ds = g_{\mu \nu} \triangle x_{\mu} \triangle x_{\nu} $ Where ds is the interval, g is the metric and the x's are the co-ordinates. As it has been said the x's can be drawn analogous to pythagoras in euclidea...

What's up with those YUGE triangles?

@ACuriousMind Do you have any PSE posts with a commutative diagram?

5

A: Can radioactivity be slowed through time dilation?

Yes. The classic example is that this is the only reason muons produced by cosmic radiation high up in the atmosphere live long enough to reach the ground.