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00:11

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Q: Meta-commentary in questions should sometimes be allowed

I've edited in a 'meta-comment' into this question, which was rolled back by a moderator: Note: answer accepted for sake of closing the question, though parts remain unanswered, and will be explored in followup(s): Q1 Comment's self-explanatory: the accepted answer is not fully satisfactory...

ZeroTheHero

02:11

@Student404Mus @DavidZ do you have a minute?

David Z

03:06

@ZeroTheHero Not exactly, at the moment, but maybe if it's something quick?

John Rennie

06:39

@CaptainBohemian oops, yes, sorry.

Captain Bohemian

07:02

@JohnRennie Ok, thank you for confirming my question.

@JohnRennie In addition, I still have a question about the KS diagram, hoping you can help me clarify it. Wikipedia - White hole says In this spacetime, it is possible to come up with coordinate systems such that if you pick a hypersurface of constant time and draw an "embedding diagram" depicting the curvature of space at that time, the embedding diagram will look like a tube connecting the two exterior regions,...

known as an "Einstein-Rosen bridge" or Schwarzschild wormhole. Depending on where the space-like hypersurface is chosen, the Einstein-Rosen bridge can either connect two black hole event horizons in each universe (with points in the interior of the bridge being part of the black hole region of the spacetime), or two white hole event horizons in each universe (with points in the interior of the bridge being part of the white hole region).

John Rennie

I've never studied this in any detail, but the KS diagram has two regions of spacetime I and III connected by a non-traversable wormhole. It's non-traversable because the only curves that pass through it are spacelike.

You can immediately see this on the KS diagram because all lines with a gradient less than 1 are spacelike.

Captain Bohemian

is an "Einstein-Rosen bridge" or Schwarzschild wormhole a tube along any of the constant r curve in the II (blue) or IV (green) region?

John Rennie

I'm not sure to be honest.

Captain Bohemian

07:18

@JohnRennie The Wikipedia - White hole gives a different reason why the wormhole is non-traversable from yours. It says It is impossible to use the bridge to cross from one universe to the other, however, because it is impossible to enter a white hole event horizon from the outside, and anyone entering a black hole horizon from either universe will inevitably hit the black hole singularity.

John Rennie

That's the same reason :-) It is impossible to enter a white hole event horizon from the outside following a timelike trajectory.

Captain Bohemian

@JohnRennie maybe a tube along any spacelike curve, not necessarily one with constant $r$, can be an "Einstein-Rosen bridge" or Schwarzschild wormhole.

John Rennie

Constant r? Constant t isn't it?

Captain Bohemian

07:33

@JohnRennie but constant $t$ are straight lines passing the origin which don't connect the two universes I and III.

John Rennie

Every straight line passing through the origin with a gradient between -1 and 1 connects regions I and III.

I'm afraid I don't remember how to do the embedding to get the wormhole.

John Rennie

07:52

@CaptainBohemian I found this explanation and I think this does it very nicely:

Einstein-Rosen bridge

►

Captain Bohemian

08:23

@JohnRennie are those lines lightlike? I think they are in the Krusal coordinates but not in the Schwarzschild coordinates.

John Rennie

ds^2 is an invariant so if a curve is lightlike/spacelike in one coordinate system it is lightlike/spacelike in all coordinate systems.

Captain Bohemian

@JohnRennie you are correct. So they are lightlike?

John Rennie

Spacelike. On a KS diagram spacelike lines have a gradient between -1 and 1, and timelike lines have a gradient < -1 or > +1.

And any line with a gradient equal to +/- 1 is lightlike.

Yuvraj

@JohnRennie hi

John Rennie

@Yuvraj hi :-)

Yuvraj

08:33

@JohnRennieactually i was reading about some very old book .

John Rennie

@Yuvraj Yes ... ?

Yuvraj

at last pages authors have confirmed that what we understand the meaning of velocity is less what it actually is

@JohnRennie

John Rennie

What is the book?

Captain Bohemian

@JohnRennie sorry, I didn't read clearly that you mean between -1 and 1. I thought you mean the lines with slope -1 and 1 because I think only the lines with slope -1 and 1 connect I and III. You said Every straight line passing through the origin with a gradient between -1 and 1 connects regions I and III. I don't see how.

Yuvraj

i do not know! because i have very few papers with no cover

@JohnRennie

@JohnRennie book has papers on special and general velocity

Captain Bohemian

08:39

@JohnRennie also, these lines are constant t so are timelike.

John Rennie

That red line is spacelike, i.e. mod(gradient) < 1, and connects regions I and III

@CaptainBohemian are we getting mixed up about the terminology?

A line of constant t is spacelike and a line of constant r is timelike.

Captain Bohemian

@JohnRennie but this red line isn't a wormhole, is it?

John Rennie

@Yuvraj in general relativity coordinates don't have the physical significance that they have in Newtonian physics. They are just labels that we use to identify points in spacetime.

Captain Bohemian

@JohnRennie oh, you are correct. I didn't think clearly.

John Rennie

And if the times t and coordinates x don't have a physical significance then neither does the velocity dx/dt.

We call dx/dt the coordinate velocity to indicate it is the velocity according to some choice of coordinates and not a fundamental quantity.

Yuvraj

08:44

@JohnRennie ok can i show what pages say?

John Rennie

In GR the fundamental quantities are all tensors. The velocity we use is the four velocity.

@Yuvraj yes

Yuvraj

v, the principle of relativity goes back to Galileo, and plenty of people didn't agree with Newton that space was absolute. (And for Newton himself, Galilean invariance of physics is actually a theorem in Principia, but he thought that space must be absolutely anyway for conceptual reasons.) Note that having space relative is enough for the answer to this question to be morally the same, even if time was thought to beabsolute

Speed doesn't even exist by itself. It's always relative to a frame. It's not a property of an object or (even worse) a property of "a point in space". It's a relation between an object (the moving thing) and another object (the frame). "Points in space" don't have properties themselves, space is featureless.

@JohnRennie

John Rennie

I think it just means that space isn't a thing so a particular point in space isn't anything fixed.

Yuvraj

so velocity has no meaning in space?

John Rennie

Well velocity is always relative and is different in different frames. There is no absolute velocity.

If you choose a frame and measure all the velocities in that frame then the numbers you get have a physical meaning.

It's just that other observers using different frames would get different velocities.

This is less of a big deal than it seems. I would guess most students know that there is no absolute velocity in physics, and this is all the book is saying.

This is true even in Newtonian physics.

Yuvraj

08:52

yes sir ,i know sir you have said me these so many times ,but it is really hard to understand what these papers says

by the way look at this pic

@JohnRennie

John Rennie

41°C!!

It was 26°C in the UK yesterday, and that's unusually hot for May.

45°C next week.

Yuvraj

yes but the actual temperature is feel like 47 because of these dusty stroms

John Rennie

Right now it's 15°C in Chester

But it will get a lot hotter later in the day.

I don't know how people manage with temperatures that high. In the UK when it gets above 30°C I find it too hot to do anything except slump into my chair and read.

Yuvraj

my ac temp is 21

John Rennie

Ah OK, if you have air conditioning I guess it's not too bad.

Yuvraj

08:57

but i feel like it is giving up

you won't believe today I made omelet on sand

John Rennie

:-)

Yuvraj

@JohnRennie i have some more questions can i?

John Rennie

Yes, of course.

Rudi_Birnbaum

Hi there, anyone ides on this math.stackexchange.com/questions/3683396/… ?

Yuvraj

I'm under the impression that black holes in string theory are constrained to be extremal Nordstrom holes?@JohnRennie

John Rennie

09:02

I'm not sure anyone knows how to construct black holes in string theory. There are some rather special constructions that were used to derive the equation for the entropy of a black hole horizon, and I believe the AdS/CFT construction can be used to model a black hole, but I'm not sure how much relation these have to macroscopic black holes.

But then I don't know very much about string theory so I'm not speaking froma position of any great authority.

You need to take care when reading articles about what string theory predicts, because any prediction is based upon all sorts of assumptions of uncertain validity.

Yuvraj

@JohnRennie do you remember about VC i told you?

John Rennie

VC?

Yuvraj

video chat of our club?

John Rennie

Oh yes. Is that still going on?

Yuvraj

yes I just finished the video chat , tomorrow is a big day for me

John Rennie

09:07

What is happening tomorrow?

Yuvraj

the discussion will be on dark matter and gr

and all are very excited it is 5 hur chat

John Rennie

Cool :-)

Yuvraj

@JohnRennie

John Rennie

@Yuvraj the site for downloading books you mean?

Yuvraj

yes

John Rennie

09:10

(not that I would ever pirate books of course! :-)

Why What Where When How

10:00

@JohnRennie Can you give me some motivational quotes. I am losing interest in this theoretical thing since everybody is doing the same. Feels like being in majority make me men with no value. I want to do something which very few people does.

😢

John Rennie

There is one good reason for studying physics, and only one good reason, and that reason is because you enjoy it. If you are worrying about what there people are doing then you are allowing bad reasons to influence you.

4

Why What Where When How

:54429925 Though it doesn't make sense to you. What you said in second statement is how I feel. There are billions of people living on earth so I feel like I am not important.

JingleBells

How do EM waves get generated (Radio waves...)? If we look at RF toys, and basically wireless devices, is there some type of an antenna that vibrates super fast? If that's so, does that mean that EM is just quicker sound?

Does that mean that if I alternate my vocal chords super fast, I can produce EM waves? xD

Why What Where When How

Try not to become a man of success, but rather try to become a man of value.-Einstien but I can't be both lol.

John Rennie

@JingleBells the electrons in the aerial vibrate super fast.

Why What Where When How

10:07

(removed)

John Rennie

Basically you apply an oscillating voltage to the aerial and that creates an oscillating current.

An oscillating current means the electrons are oscillating to and fro, and oscillating charges generate EM waves.

Why What Where When How

@JohnRennie Thanks for quotes :○>

John Rennie

:-)

JingleBells

@JohnRennie So EM in RF toys are generated when there's alternating current in an antenna and that alternating current causes electrons in the air around it to oscillate as well, which causes the next outer electrons to alternate, which cause... and so on until it reaches the receiver (where of course the alternations of the electrons get weaker as they travel)

John Rennie

No, there are no electrons oscillating in the air.

JingleBells

10:11

Does "aerial" mean "antenna"?

John Rennie

The electrons in the aerial generate an EM wave and that wave travels through the air, or vacuum or whatever, to the receiver without needing any electrons in betwen.

@JingleBells yes

JingleBells

Lol got it. So it again basically breaks down to super quick oscillation of something (in this case, the antenna). That's really cool. I thought the antenna was mechanically alternated somehow (lol) but I'm surprised it's actually electrons that cause it to alternate.

@JohnRennie So does that mean if I get my vocal chords to oscillate super fast, I can open my mouth and light will shine through xDD?

Why What Where When How

Anyway bye guys! I am still studying group theory which is very interesting and intutive with topology and metric space to prepare myself to learn quantum Mechanics soon....

John Rennie

There is a question about that on the site somewhere ...

user434058

@JingleBells just eat Happydent, it's way more easier :P

JingleBells

10:19

From now on I'll be screaming all the time attempting to get my vocal cords to 500 THz

PM 2Ring

12:54

@JingleBells No, because your vocal cords aren't charged. On a related note:

@FourierFlux The wavelengths of EM cover a huge range, and different scales require different physical oscillation processes. At visible & UV wavelengths the oscillators are atomic electron transitions, radio oscillations happen at a larger scale. Crudely speaking, you can't build a single atom with a 1 metre radius, and you can't build a radio antenna that's a few hundred nanometres long. — PM 2Ring 4 hours ago

47

Q: Can I use an antenna as a light source?

Can I use a normal metal antenna to emit visible light?

electromagnetic-radiation visible-light antennas

@BioPhysicist Such answers are not useful, so it's valid to downvote them. OTOH, I don't like to DV answers that are technically correct, in case that confuses future readers. But I guess a comment can take care of that. And if the answer is literally a copy & paste from an existing answer, without attribution, then flag it for plagiarism.

BioPhysicist

13:18

@PM2Ring Yeah, I agree

Yuvraj

14:14

@JohnRennie hi

sir is it possible to make an electron stable in higher energy orbits?

John Rennie

@Yuvraj in general no it isn't.

As a rule higher and lower electronic states mix together and will always evolve towards the lower energy state eventually, but higher states can be stable for a surprisingly long time.

user434058

@PM2Ring Aren't moving charges enough for producing an EM wave (even if the net charge is neutral, there are constituent charges)?

ACuriousMind

@JohnRennie No one knows how to construct anything "in string theory", it's all just supergravity :P But in 10d SUGRA, so-called black branes as higher-dimensional analogues of black holes are well-known

John Rennie

The 21cm line is emitted from an excited state of a hydrogen atom, and the excited state has a lifetime of 10 million years!

But that's a special case :-)

Yuvraj

ah!

Jake Rose

14:25

user434058

Hey mods, if you handle my flags and see something like an order ("Delete it"), I am really sorry. I forgot to write "please" in both the cases (lazy me). I'll be softer from now on, thanks :-)

Yuvraj

There is also 11d sugra

@ACuriousMind

Jake Rose

This says to lowest order in $\partial \alpha$ and $ \partial \beta$ but isn't that cross product second order?

ACuriousMind

I'm well aware.

@JakeRose they count the orders separately - it's lowest (=first) order in $\delta \alpha$ and first order in $\delta \beta$.

Yuvraj

@ACuriousMind since you are the only one like me who accept the string theory well.

Jake Rose

14:28

Mhmm

Whenever I've used two different infinitesimals we haven't counted separate orders

Yuvraj

i posted a question in the afternoon, unfortunately, john Rennie is less interested in it!

5 hours ago, by Yuvraj

I'm under the impression that black holes in string theory are constrained to be extremal Nordstrom holes?@JohnRennie

Jake Rose

Intuitively how can you say $\delta \alpha \delta \beta$ isn't of the same order as $\delta a ^2$?

Yuvraj

@PM2Ring you almost forgotten me sir?

Jake Rose

bollocks

Dont know why I did patials

If anything, it could be smaller!

And even more negligible

John Rennie

@Yuvraj it isn't that I'm not interested, just that I don't know the answer. Where did you see it written that in string theory are constrained to be extremal Nordstrom holes? Have you got a reference?

Yuvraj

14:32

@JohnRennie sorry for typing wrong

actually, I was listening to my senior fellow on black hole entropy!

PM 2Ring

@Yuvraj Of course not! A few days ago, when I was not online, you said your birthday is coming. But you didn't say which day is your birthday.

Yuvraj

and he said about the thermodynamics of black hole in

John Rennie

@Yuvraj ah, in the chat session?

Yuvraj

@JohnRennie yup , sorry sir if anyhow I miss interrupt you

@PM2Ring yes in last of this month

ACuriousMind

@Yuvraj You first have to define what you mean by "black hole in string theory". As I just said the black hole-like solutions in higher-dimensional SUGRA are black branes where the singularity is not pointlike. One might call the BPS condition for a black brane to have unbroken supersymmetry the analogue to the extremality condition on Reissner-Nordstrom holes, but there is no rule that only BPS branes are allowed - they are just preferred because they preserve supersymmetry.

Yuvraj

14:38

if i am not wrong black holes in string theory is a collection of many microscopic ingredients in some generic complicated state?

ACuriousMind

That's pretty vague, I'd also say life is a collection of many microscopic ingredients in some generic complicated state ;P

Yuvraj

?????

??????

????

PM 2Ring

9

Q: Why does an accelerating neutral object not emit Bremsstrahlung radiation?

It is a well established fact that accelerating charges, positive or negative, emit radiation. Why then does a neutral object made up of equal amounts of positive and negative charge not emit any radiation when accelerated?

electromagnetism electromagnetic-radiation acceleration

ACuriousMind

@Yuvraj I'm saying what you just said is not a definition because is too vague.

Nothing about that distinguishes a black hole from any other complicated state

Yuvraj

oh!

John Rennie

14:41

Was there not something about obtaining a 4D black hole from a BPS state, but it ends up charged. I think extremally charged.

I thought I remembered hearing (and probably misunderstanding) something about this from the days when the derivation of the black hole entropy calculation was still new ...

Mid 90s ish

ACuriousMind

Yes, as I said you might view the BPS condition as the generalization of the extremality condition. However the black branes can not only carry electric but also magnetic charges, so it's not quite that easy

Yuvraj

@ACuriousMind please elaborate ,what you mean by elaborating?

@JohnRennie please sir can you join us tommorow?

ACuriousMind

And if you reduce in a direction such that the 4d universe intersects with the singularity of the brane only in a line, I guess you get an ordinary extremal hole.

Yuvraj

please, please please

John Rennie

I wonder if that's what Yuvraj had heard about.

ACuriousMind

14:45

@Yuvraj ...you want me to elaborate what I mean by elaborating? I didn't say anything about 'elaborating'.

Yuvraj

2 mins ago, by ACuriousMind

Yes, as I said you might view the BPS condition as the generalization of the extremality condition. However the black branes can not only carry electric but also magnetic charges, so it's not quite that easy

John Rennie

Apparently there was a physics dept chat session a few days ago and one of the PhD students was talking about it.

Yuvraj

this one ]

I was saying about this

@ACuriousMind

@JohnRennie you have not replied yet?

ACuriousMind

@Yuvraj What exactly do you want me to explain in more detail?

John Rennie

I would feel uneasy joining your college physics dept chat session.

Yuvraj

14:51

bps condition

John Rennie

@Yuvraj the problem is that it's impossible to explain further without spending hours and hours on all the background.

Even just explaining what a BPS state is requires more maths than 99.9% of physics graduates would have.

ACuriousMind

@Yuvraj It's a technical condition for preserving supersymmetry that also means "mass = charge", i.e the same as the extremality condition for RN holes. Anything more would require you to know supersymmetry and string theory at a technical level of detail.

John Rennie

I imagine ACM is reluctant to get dragged into an explanation that would take hours!

@Yuvraj but it does sound the sort of thing the guy speaking in your chat session might have been referring to.

Yuvraj

@JohnRennie we have no problem if you are ok?secondly you do not teach topic unlike you do with me because they are very good at math and physics

some of them are doing research

.

ACuriousMind

You really don't need to ping people again with a dedicated message if your message before that already pinged them.

Jake Rose

15:17

@ACuriousMind Do you have any intuition on why you can ignore square terms but not cross terms? (cross meaning $ab$ not a cross product)

ACuriousMind

15:30

@JakeRose You essentially just expand the two $R(...)$ separately to first order here. That you get a "second-order term" from that is expected - you're doing this to see how two successive rotations act, right? If you threw away the cross term you'd lose exactly the information you're looking for.

Jake Rose

But isn't it of order as the squared term? And so We should take into account that term too?

user434058

You should have an option to revert your "Leave Open" vote for about a 5 minutes after you reviewed that item...

ACuriousMind

@JakeRose Where would you get a squared term from? Again, the idea is that you're expanding both the $R(...)$ to first order.

(the real underlying math is that this is doing stuff with Lie groups and algebras without saying so and disguising it as ordinary Taylor expansions, so do not expect to find a fully rigorous explanation of why this works :P )

user434058

16:03

Just discarded the draft to a Meta question, 'cause I was too lazy to finish writing the question :-|

ⵃⴰⵎⵎⴰⴷ ⵣⵉⴷⴰⵏⵉ

@ZeroTheHero Hi!

user434058

(BTW, it wasn't related to that revert "Leave Open" feature request)

JingleBells

@PM2Ring So Em waves aren't generated from the oscillating antenna, but the oscillating electrons in it? coolio

ⵃⴰⵎⵎⴰⴷ ⵣⵉⴷⴰⵏⵉ

sorry I didn't notice your message until the moment.

user434058

@ⵃⴰⵎⵎⴰⴷⵣⵉⴷⴰⵏⵉ It's quite hard to ping you in the comments below any post :)

ⵃⴰⵎⵎⴰⴷ ⵣⵉⴷⴰⵏⵉ

16:05

yeah, that's right! maybe i'll change the first part of my name.

ooops I cannot change it until June!

Anyway, you can reply to my chat messages if you want to pin my name.

user434058

@ⵃⴰⵎⵎⴰⴷⵣⵉⴷⴰⵏⵉ no worries :-)

user434058

Does this answer come under the domain of "link-only answers"?

Krishna

@FakeMod no

user434058

@Krishna why so?

Krishna

If you strip the link off it still answers it

user434058

16:15

@Krishna it just says that if you somehow magically integrate, you would find it to be true.

Krishna

That's what DavidZ (I think) used as a criterion to define them

I stumbled upon it while searching for flag guidelines

There is a meta post on that

user434058

@Krishna alright, let's see what the mods say...

Krishna

Besides that, I just told what I'd read :)

user434058

@Krishna yup, I know.

Krishna

So, it is not my opinion

user434058

16:18

@Krishna I think the main problem is that does the answer really answer the question if you remove the link? IMO, no.

Krishna

And there is a different meta post, specifically for this

It entirely does not, I agree, but anyone who knows calculus should be able to do that.

Well, we'll leave it to the mods

user434058

@Krishna yeah, that's better :-)

Krishna

And I lack expertise in this, so I should become quiet now

Bye

user434058

Bye

BioPhysicist

Where can I file a complaint to remove the downvote altogether, like Facebook did? — DoctorNuu 8 mins ago

Good luck with that

ACuriousMind

16:24

@FakeMod It is perhaps not a particularly useful answer, but even without the link it points out the distinction between the field of a single charge being position-dependent and the field of the entire plate being position-dependent.

user434058

@ACuriousMind alright, so as long as the answer contains some relevant and useful stuff, along with a link, it isn't really a link only answer. Point taken!

user434058

@ACuriousMind I suppose you meant to write "independent" as your last word.

ACuriousMind

I debated it internally for some time and decided the sentence made equal sense with every combination of dependent and independent in the two positions :P

user434058

@ACuriousMind lol

ZeroTheHero

@ⵃⴰⵎⵎⴰⴷⵣⵉⴷⴰⵏⵉ hello.

user434058

16:33

@DavideDalBosco Hi, seen you for the first time here! :-)

user434058

16:48

Hi @DavidZ , any particular reason why this question was manually removed from the HNQ? I suppose that it wasn't really a "physics" question, or at least it wasn't the kind of question we want our site to showcase, that's it, right? Or is there something else? (Just curious)

BioPhysicist

@FakeMod Maybe this physics.meta.stackexchange.com/a/12723/179151 It's not exactly the same, but there are some similarities

user434058

17:16

@BioPhysicist yeah, I had seen that before, but in the case which I linked, the question had a definite answer and there wasn't really any speculative scope if one answered it correctly. But they are quite similar and maybe the same, if not similar, reasons apply here as well. Thanks!

bolbteppa

17:31

@Yuvraj I think this summarizes what you were trying to say about black holes:

> Black-hole solutions to general relativity carry a thermodynamic entropy, discovered by Bekenstein and Hawking to be proportional to the area of the event horizon, at leading order in the semiclassical expansion. In a theory of quantum gravity, black holes must constitute ensembles of quantum microstates whose large number accounts for the entropy. https://journals.aps.org/prx/abstract/10.1103/PhysRevX.10.021037

David Z

18:38

@FakeMod TBH I don't quite remember. I'm sure it was something mundane along the lines of what you're suggesting.

Jake Rose

@ACuriousMind But what I'm saying is, we have ignored all terms that are second order, so then when we get a term that is second order and no longer ignore it, it feels like there should be another bunch of terms from the expansion which are now significant?

ACuriousMind

18:57

@JakeRose The term is not "second order", it's first order in each of the variables. We're not changing the rule suddenly to "we keep some second order terms" - the rule is that terms first order in $\delta \alpha$ and terms first order in $\delta \beta$ are kept.

From a different viewpoint of infinitesimals, you would simply say that they square to zero. But $\delta \alpha \delta \beta$ is not a square, so it's not zero.

Jake Rose

Mhmm

This is quite at odds with how I usually think about this type of stuff

Hence why I'm asking

For example when you write the taylor series for a multivariable function, you ignore the cross terms as they are 'second order'?

Even though they are first order wrt to each variable

ACuriousMind

In principle, a Taylor series has infinitely many terms and by default we ignore none of them :P

Jake Rose

wdym?

David Z

@Krishna that's right BTW

Jake Rose

I'm talking in the of small $\delta x$ etc

ACuriousMind

19:03

@JakeRose I mean that the Taylor series is just (schematically) $\sum_i f^{(i)}(x_0)(x-x_0)^i$. When you're ignoring terms from it you're doing something more than just writing down this series

When we "ignore" terms, it's because we have a particular goal in mind and we want to get the dominant term relevant for that goal

In your case here, the goal is to examine the interaction of the two successive rotations

The terms that only contain $\delta\alpha$ but not $\delta \beta$ or vice versa are useless for that - they cannot tell you anything about the interaction between the two rotations. So you take the lowest order term in their product $\delta\alpha\delta\beta$

Jake Rose

Yes, but how can we be sure that $\delta \alpha \delta \beta$ is not more significant than $\delta \alpha ^2$

ACuriousMind

See the last thing I just wrote

Jake Rose

True, but why ignore $\delta \alpha ^2 \delta \beta$?

ACuriousMind

Because that's "order 3/2" in the product? :P As I said, the rigorous justification for why these terms and only these terms are the right ones to keep should come from Lie theory, not from thinking about this as power series expansions and arguing about significant terms.

I cannot offer you more than the heuristics I've already mentioned to explain what the author of your text likely thought was going on here

Jake Rose

Mhm

Fair enough

B. Brekke

19:16

Given two points on a two dimensional grid. How many paths of length N connects these points? What I really want to know, is if this problem has a name and if it is NP hard

user434058

@DavidZ no worries! Thanks :)

PM 2Ring

19:53

@B.Brekke Do the paths segments have to be horizontal or vertical? Or are diagonals permitted? What about self-intersecting paths? I don't know the answer offhand, but I bet Catalan numbers are involved.

Catalan number

In combinatorial mathematics, the Catalan numbers form a sequence of natural numbers that occur in various counting problems, often involving recursively-defined objects. They are named after the Belgian mathematician Eugène Charles Catalan (1814–1894). The nth Catalan number is given directly in terms of binomial coefficients by C n = 1 n + 1 (...

B. Brekke

@PM2Ring Yes, path segments can go both horizontal and vertical, but not diagonal. It can also be self-intersecting. It is not restricted to any direction, like some other problems are with only downward and rightward motion.

So my paths are not monotonic, as described in the wikipedia article

PM 2Ring

20:09

@B.Brekke Ok. I didn't expect your paths to be monotonic, but that would make things simpler. ;) Permitting self-intersection allows a lot more paths, but it might actually make it easier to count the paths. I think you should ask a question about it on the Mathematics stack, or one of the maths chat rooms.

Wikipedia has some stuff about this: en.wikipedia.org/wiki/Random_walk#Lattice_random_walk

You reminded me of this lattice walk question from a few years ago.

271

Q: Help with a prime number spiral which turns 90 degrees at each prime

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B. Brekke

20:31

@PM2Ring Haha, wow, cool! I will try at in some maths chat room. Thanks!

Nike Dattani

22:21

Hi guys, there is a bounty ending in 6 hours for this question and it still doesn't have any answers, does anyone know who might happen to work in this field?

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