The h Bar

2:49 AM

SCP

Secret's criteria of abnormality: A phenomenon that displays nonlocal classical causation without any involvement of rituals or other guiding context

A phenomenon is classically nonlocal if 1) there is no quantum correlation (duh), and 2) there are no causal sequence bounded by the speed of light that connects from the source to the occurrence of the phenomenon. In other words "it just is"

Thus a phenomenon displays nonlocal classical causation if every time event A happened, event B happened some time later but there is no causal chains that can connects the influence of A to B below the speed of light.

Timelike separated events that are causal, but not positioned next to each other in some phase space (including locally in spacetime) nor mediated by any fields is an anomally

A cup that change the color of the sky when rotated is an example of an anomalous object because no physical models can fully reproduce the causation it demonstrates

Ryan Unger

3:56 AM

@Slereah whom did you email here about that thesis anyway

Simplex

6:06 AM

I'm trying to understand what is meant by "fractional timing precision" in the following: "Over the past year, the precision of optical lattice clocks has advanced dramatically, to a fractional timing precision of (∆t/t)∼10^−18"

I did a search but couldn't find any other uses of this term

I have only a basic understanding of atomic clocks but I have to write a summary of the paper this is from and want to understand as much of it as I can. Can anyone shed some light on this?

John Rennie

6:46 AM

@Simplex Suppose you measure some time T with your clock. T could be any time - a second, a year or whatever. The measured time T will have some error ∆T. The fractional timing precision is just ∆T/T. It's like a percentage error.

Simplex

Ahh ok, thanks. I was thinking ∆t referred to the time difference between two events

user400188

Are photons absorbed instantly? Or is the a timescale over which absorption takes place?

John Rennie

@user400188 no, the interaction with the photon takes a measurable time. The interaction timescale is calculated by an equation called Fermi's Golden Rule.

user400188

Thank you @JohnRennie.

PM 2Ring

7:10 AM

Wow. This question from 2015 got bumped yesterday. I don't think it's answerable (as I implied in a comment) but it has a -5 answer that's non-mainstream.

Akash. B

7:57 AM

@JohnRennie I have a question

John Rennie

@Akash.B hi :-)

Akash. B

@JohnRennie hi

You know that in sun there occurs nuclear fusion and fission

Hydrogen fuses to give helium and helium decomposes to hydogen

@JohnRennie do you feel something fishy here?

Well hydrogen has no neutrons and helium has neutrons

John Rennie

Helium doesn't decompose to give hydrogen ...

Akash. B

???

One min let me google it to find more information

@JohnRennie you there?

John Rennie

@Akash.B hi

Akash. B

8:09 AM

Hi again

@JohnRennie so hydrogen fuses to give helium in sun

John Rennie

@Akash.B yes

Akash. B

It's only the process happening there, right?

So let me ask

Hydrogen has no neurons right?

John Rennie

Correct.

Akash. B

So where did the hell it get neutron to form a helium atom with two neutrons?

Is energy being converted to mass?

John Rennie

A proton can form a neutron by electron capture.

Akash. B

8:13 AM

Wow a new term

John Rennie

In the Sun two protons form a very short lived diproton state. This normally immediately falls apart again, but occasionally one of the protons will capture an electron and form a deuterium nucleus.

Akash. B

Hmm @JohnRennie so how's helium formed?

PM 2Ring

Proton–proton chain reaction

The proton–proton chain reaction is one of two known sets of nuclear fusion reactions by which stars convert hydrogen to helium. It dominates in stars with masses less than or equal to that of the Sun, whereas the CNO cycle, the other known reaction, is suggested by theoretical models to dominate in stars with masses greater than about 1.3 times that of the Sun.In general, proton–proton fusion can occur only if the kinetic energy (i.e. temperature) of the protons is high enough to overcome their mutual electrostatic or Coulomb repulsion.In the Sun, deuterium-producing events are rare. Diprotons...

John Rennie

Snap :-)

PM 2Ring

Great minds think alike... or fools seldom differ. :D

In the p-p chain, diprotium converts to deuterium by beta-plus decay (emission of a positron). Electron capture is possible, but it generally doesn't have time to happen. And the beta-plus decay itself is very rare, normally diprotium just splits into a pair of protons.

There are questions relating to this on Physics.SE & Astronomy.SE. Even forming diprotium is hard, due to the electrostatic repulsion between the protons: they have to get really close for the nuclear force to be strong enough to overcome that repulsion.

But even when a diprotium manages to be formed it nearly always falls apart. The odds of it converting to a deuteron are in the order of 1 part in $10^{26}$. So in the Sun's core, a proton has a mean lifetime of around a billion years before successfully being incorporated into a helium nucleus.

Slereah

8:38 AM

@RyanUnger i used the form for outside people requestiing digitization of documentd

Akash. B

11:07 AM

@PM2Ring oh I see

But what about nuetrons

Oh one of them undergoes beta plus decay I see

bolbteppa

11:23 AM

6

A: Is there a useful way to visualize the symmetries of the relativistic Riemann curvature tensor?

Yes, the method is called Young tableaux. For a rank $n$ tensor, give its $n$ slots names. Then the possible symmetries of the tensors may be classified by Young taleaux, arrays of $n$ boxes filled with the names of the $n$ slots. The rule is that one symmetrizes indices in rows, then antisymmetr...

@knzhou how did you know the vertical diagram doesn't contribute because of $R_{[ab,cd]} = 0$ and how did you know the first one is linked to it

12:21 PM

12:31 PM

The biggest issue is knowing, just by looking at a diagram, that say $R_{[ab,cd]} = 0$

Ryan Unger

12:44 PM

@Slereah ok I don't see why it should be denied then

have you heard back from them?

Semiclassical

@PM2Ring the version my collaborator likes: “great minds stink alike”

Slereah

@RyanUnger not so far, but from what I've read it can take 2-3 weeks to process

yuvraj singh

1:07 PM

@JohnRennie old but still good

Novalium Company

So Amerigo found out that South and North America are not part of Asia. Did Columbus think his whole life that what he found was just Asia?

PM 2Ring

2:01 PM

@NovaliumCompany Probably. From en.wikipedia.org/wiki/Christopher_Columbus#Voyages

Columbus always insisted, in the face of mounting evidence to the contrary, that the lands that he visited during those voyages were part of the Asian continent, as previously described by Marco Polo and other European travelers. Columbus's refusal to accept that the lands he had visited and claimed for Spain were not part of Asia might explain, in part, why the American continent was named after the Florentine explorer Amerigo Vespucci and not after Columbus.

Novalium Company

@PM2Ring Thx

How can the geocentric theory be wrong? Everything is relative, so geocentric and heliocentric are equally correct depending on which point of reference you use?

Am I stupid?

Ahh I'm confused now that I think about it

PM 2Ring

@NovaliumCompany Yes, in GR you can assume that everything goes around the Earth, but it makes the calculations uglier.

Slereah

@PM2Ring I mean the same is true of all frames really

depending on what you're looking at

Heliocentric frame for the CMB is probably ugly

PM 2Ring

2:18 PM

@NovaliumCompany GR gives you a lot of freedom to choose weird & wonderful coordinate systems, but you normally choose coords that simplify the problem you're analyzing.

Novalium Company

Oh ok, thx

Slereah

bad news tho

for most problems, it will always be awful

PM 2Ring

@Slereah Agreed. But lets face it, doing GR on anything that doesn't have nice symmetries gets painful very quickly.

Slereah

alas

heather

3:08 PM

morning

Slereah

3:41 PM

hello

Sir Cumference

4:03 PM

howdy

Novalium Company

6:12 PM

sup

I think I'm 2 hours late

ACuriousMind

7:18 PM

2 hours late for nothing, it appears :P

Emilio Pisanty

7:53 PM

2

Q: Is there a magnetic attraction between two parallel electron beams?

I am refering to Ampere's force law, and to the beams accelerated after the cathode, so the deflection is not due to their respective cathode. In other words, do two electrons accelerating parallel to each other converge because of magnetic attraction? Does it apply to wires only or to a beam of ...