The h Bar

1:24 AM

@Claudio writing that answer confused myowself quite a lot too. But it is correct. substitute n = 0 and 1 and 2 upwards and see that the results are correct. I might have miscomputed the last thing though.

naturallyInconsistent

2:01 AM

I have updated the answer. Was too late last night working on this

@Bml The result is true when you already know that it is quadratic and thus can be proved by expansion. However, the point of the whole argument is to establish the quadratic-ness and thus you cannot use the general $u$ right at the beginning.

Nairit Sahoo

2:29 AM

Can someone please help answering my question?

$[x,p]= i \hbar$. Can someone tell me how to find $p$?

naturallyInconsistent

2:40 AM

@NairitSahoo arent you already deep into QFT? This should be way beneath you

Nairit Sahoo

2:54 AM

@naturallyInconsistent I thought so. But the thing is my prof at QM class asked this. Idk how to find it

He didnt even mention whether to find it in position rep. or momentum rep. or the matrix form

He said explicitly that we shouldn't assume the form and prove it

We must derive it :(

naturallyInconsistent

@NairitSahoo That's a horrible prof. This is left unstated, but is actually something you have to pick.

@NairitSahoo I know what this is about. It is not as general as you think it is.

But you should be able to guess what is being meant.

PM 2Ring

4:23 AM

@Loong (I'm still reading the article). Yes, you must use ionized thorium because the isomer transition in neutral thorium is very rapid. en.wikipedia.org/wiki/Isotopes_of_thorium#Thorium-229m has some good background info. This clock experiment uses a thorium-doped calcium fluoride crystal. It's excited by 7th harmonic UV femtosecond pulses from a laser that's stabilised by a strontium-87 optical atomic clock.

John Rennie

Aha, I read about this yesterday in my weekly physics blogs catchup.

Though the interest appears to be in using the frequency to check for changes in fundamental constants rather than in using it as a clock.

Bml

5:25 AM

@naturallyInconsistent For "The result is true when you already know that it is quadratic and thus can be proved by expansion.", see this and let me know: math.stackexchange.com/questions/2937331/…

@naturallyInconsistent "However, the point of the whole argument is to establish the quadratic-ness and thus you cannot use the general $u$ right at the beginning.": I didn't understand this part. Why can't I use general $u$ at the beginning? Could you expand on this?

naturallyInconsistent

5:54 AM

@Bml $$T(v_0+v)+T(v_0-v)=2T(v)+Q$$ is correct but I think it is as yet still unjustified at the point it is introduced.

Bml

@naturallyInconsistent Why?

naturallyInconsistent

@Bml Because if you already had this, then the quadratic-ness is already trivial. (Also, there is a typo in there. Those are $v\pm v_0$ not $v_0\pm v$)

@Bml But they have assumptions 4 and 5, and together, that is sufficient.

Bml

@naturallyInconsistent Sorry, what does it change between $v\pm v_0$ and $v_0\pm v$? Also, why is the quadratic-ness already trivial?

@naturallyInconsistent Does that mean that the proof is consistent and leads to quadratic-ness, or are you trying to say me something other I don't grasp?

naturallyInconsistent

6:09 AM

@Bml in the CoM frame as treated in the question, the balls are coming in as $\pm v_0$ and the lab frame is moving with $-v$ relative to that, so that the balls are, in lab frame moving with $v\pm v_0$. That is, the question is being inconsistent with its own notation.

@Bml yes, it will work, given the stuff that is already assumed.

Bml

@naturallyInconsistent The CoM frame is moving with velocity $v$. In the CoM frame, a mass is moving with velocity $v_0$, the other with velocity $-v_0$. So, in the lab frame, a mass is moving with velocity $v+v_0$ and the other with velocity $v-v_0$. Yes?

naturallyInconsistent

@Bml correct

Ryder Rude

@NairitSahoo hi. the representation of P isn't determined even after u fix a representation of X. e.g. if $ [X,P]=ih$ then $[X, P+ f(X)]=ih$ for an arbitrary function $f(X)$

so this question is incorrect

Bml

@naturallyInconsistent I don't understand: why did you say we cannot use general $u$ right at the beginning to prove the quadratic-ness?

Ryder Rude

6:24 AM

@Bml hi. a crucial assumption in the proof is $E(0)=0$. u get to use this assumption when u set $u=v$

naturallyInconsistent

@RyderRude The question is correct. It is literally asking for the student to be aware that there is $f(X)$ in the general case.

Ryder Rude

so u r not using all the assumptions if u use a general 'u' @Bml

@naturallyInconsistent it says to find $P$, which is undetermined

Unless it is a trick question, it is incorrect

Bml

@RyderRude Why do I use the assumption $E(0) = 0$ when I set $u=v$? Why if I use $u \neq v$ I discard the assumption $E(0) = 0$?

Ryder Rude

@Bml when u use u=v, u get the term E(u-v)=E(0). now u can set this to 0

when u dont use u=v, u have to write E(0)=0 as an additional assumption, and then try to incorporate it

Bml

@RyderRude Yes... In what sense?

Ryder Rude

6:32 AM

what do u mean

Bml

@RyderRude If I set $u=v$, then $E(u-v) = E(0)$, and now I can use the assumption $E(0) = 0$ to set $E(u-v) = 0$.

Ryder Rude

yes. thats the proof in the answer

Bml

@RyderRude If I don't set $u = v$ (so $u \neq v$), what is the problem? Why shouldn't I use the assumption $E(0) = 0$?

Ryder Rude

u can use it. now u have two equations to work with

123

Hello Everyone...

Bml

6:36 AM

@RyderRude What are the two equations?

Ryder Rude

The first is the general equation u got involving both v and u

the second is E(0)=0

using both of these, u may derive qudraticness

nvm E(0)=0 is not an assumption..

PM 2Ring

@JohnRennie I think the plan is to eventually build a fully-fledged clock, but there's some way to go before that's possible. But in the meantime, it can still do some frequency-related tasks while it's a complicated benchtop system, not a portable timepiece.

Ryder Rude

E(0)=0 follows by taking u=v=0 in the first equation

so it is not independent

PM 2Ring

One of the benefits of a nuclear clock is that it's relatively immune to external electromagnetic fields. The thorium ions are still affected by the charge gradient inside the crystal, though.

OTOH, even if we had a thorium clock tomorrow, there wouldn't be much point in incorporating it in the network of atomic clocks that TAI is derived from. We don't even have any optical lattice clocks in the TAI network. It just uses caesium, rubidium, and hydrogen masers. We probably need to redefine the second before it makes sense to base TAI on the more modern clocks, like strontium.

naturallyInconsistent

7:08 AM

@PM2Ring It would sound so mundane to layfolk but these improvements are so exciting!

PM 2Ring

7:20 AM

@naturallyInconsistent Maybe. But I expect that plenty of layfolk have some vague awareness that precise timing is essential to the working of GPS. OTOH, nanosecond precision is adequate for GPS. Sub-attosecond precision is just mind-boggling. :)

PM 2Ring

7:37 AM

In today's episode of Stupid Arithmetic Tricks, $33\pi \approx 2\sqrt{2687}$ (103.6725576, 103.6725615)

Ryder Rude

@123 hello

123

@RyderRude Hi

Ryder Rude

@123 how r things going

123

@RyderRude Good. What about you?

Bml

7:55 AM

@naturallyInconsistent Given what I and @RyderRude said, is this argument still valid?

Jacob Miller

8:07 AM

@PM2Ring Hello, How're you? Did you see my earlier messages?

Ryder Rude

@123 i am good too :)

123

@RyderRude Nice :-)

Ryder Rude

@Bml u should link that other thread maybe. The induction one

the general proof works too. In the general proof, we only have to assume continuity

naturallyInconsistent

@PM2Ring That would not be a great assumption for layfolk, but yes, many nerds would know about this.

Ryder Rude

while in the other proof, we have to assume continuity of the second derivative

John Rennie

8:18 AM

Is there a way I can search a chat room (this room) for all messages by a specific user? The search page allows me to search for posts by a specific user in a specific room, but only if I enter some search text. If I leave the text field blank the search does nothing.

2

naturallyInconsistent

@Bml The argument is starting with kinetic energy being converted to heat energy; with kinetic energy, it is trivial that we kinda want $E(v=0)=0$ (the other natural option, in view of SR, is $m_0c^2$, but we just are working in NR right now). So yes, things will work that way. I was just trying to say that in the original argument, we don't know what $E(u\pm v)$ would do, only $E(0),\ E(v)\ \&\ E(2v)$, and we establish quadratic-ness first before showing that $E(u\pm v)$ does what it does

PM 2Ring

@JacobMiller Hi Jacob. I haven't seen any messages from you for a while. I think I saw a couple several weeks ago in "our" room.

naturallyInconsistent

@JohnRennie this thing also annoyed meow to no end. sometimes the exact wording is just forgotten and then it gets difficult to search

Jacob Miller

@PM2Ring Yes, It is about those messages. I was asking why the graph was not adding up correctly as expected. :)

PM 2Ring

@JohnRennie I don't think so. But maybe ask in The Tavern chat.meta.stackexchange.com/rooms/89/tavern-on-the-meta Some of the regulars there have vast knowledge on how stuff works. And balpha occasionally drops by. He knows how chat works, since he wrote it. :)

Bml

8:27 AM

@naturallyInconsistent Sorry, I got lost. What was the original argument in which we don't know what $E(u\pm v)$ would do, but only $E(0),\ E(v)\ \&\ E(2v)$? Why is $E(u\pm v)$ an unknown form, while the others are not? Sorry if I don't understand.

Jacob Miller

@JohnRennie Maybe this will help: meta.stackexchange.com/questions/274106/…

PM 2Ring

@JacobMiller Sorry. I don't know. I'm not quite sure what those graphs are doing, and I don't feel like trying to figure it out.

Jacob Miller

@PM2Ring But you did write the code.

PM 2Ring

@JacobMiller I don't think those graphs were produced using my unmodified code. Besides, I wrote that stuff months ago. I've written a lot of code since then, and I don't remember all the fine details of code I wrote at the start of the year.

Some of that sunrise code is fairly general, but it's not totally general. Eg, it tries to do something sensible in the arctic & antarctic circles, but I wouldn't trust it 100% whenever it has to take arccos of values with magnitude >1.

Obviously, there are no real angles with |cos(theta)|>1. So we have to work around that problem. ;)

Ryder Rude

@PM2Ring hi

PM 2Ring

8:41 AM

Anyway, I spent a lot of time on that stuff, and I'm kind of burnt-out on that topic. I might get back to it at some time in the future. But currently I want to do other stuff.

Jacob Miller

@PM2Ring You mean it is a result of that workaround in the code that inconsistency happens around 66º and up?

PM 2Ring

OTOH, I have been working on some related stuff, when I've been well enough. My new analemma code is looking pretty good. It does both 3D and 2D analemmas.

@JacobMiller Probably.

@JacobMiller When cos goes out of range, it means that the sun doesn't rise. Or doesn't set. My code detects that, but it doesn't bother with trying to be totally precise at the transitions, i.e., on the equinoxes.

Jacob Miller

@PM2Ring That's good to hear.

@PM2Ring I guess, then. I better use it only outside the Arctic Circle or Antarctic Circle for accurate results.:)

PM 2Ring

9:02 AM

@JacobMiller Yes. Here's another related thing I've been working on.

Aug 1 at 21:28, by PM 2Ring

My diagram uses a contour plot to show the altitude angles of the body, in 5° steps. It also plots contour bands for the Sun, showing the "twilight zones". Civil twilight begins/ends when the centre of the Sun is on the horizon (0° altitude). Nautical twilight begins/ends at -6°, astronomical twilight at -12°, and full night is when the Sun is below -18°.

It treats the Earth as a sphere, though, not an ellipsoid. So the contours aren't 100% accurate. And it totally ignores refraction.

@RyderRude Hi.

Ryder Rude

@PM2Ring would u like to friendly debate on consciousness philosophies

naturallyInconsistent

@Bml We are originally supposed to not know what $E(v)$ and $E(2v)$ are supposed to be, but we can impose $E(0)=0$. Our ignorance of $E(v)$ makes it difficult to argue what $E(u\pm v)$ should be; although as you have seen in the other form of the argument, you can supply this if you really want. The point is that if we can relate $E(2v)$ to $E(v)$ with the constraint that $E(0)=0$, then you have everything.

PM 2Ring

Unfortunately, I won't be able to publish that code. It pulls the map from Wikimedia. And Wikimedia doesn't like it when a bot fetches the same data over & over again.

@RyderRude No, thanks.

Jacob Miller

@PM2Ring Red dot the sun? Then what is the black dot?

PM 2Ring

@JacobMiller Yes, the red dot is the Sun. The black dot is the body of interest. In that plot, it's the near-Earth asteroid Apophis.

Bml

9:14 AM

@naturallyInconsistent Now the first part is clear to me, thanks. What I don't understand is: "if we can relate $E(2v)$ to $E(v)$ with the constraint that $E(0)=0$, then you have everything." What I know is that $E(2v) + E(0) = 4E(v)$, and with the assumption $E(0)=0$, we have $E(2v) = 4E(v)$. Is this what you are trying to say me?

PM 2Ring

That plot is for when Apophis reaches its maximum magnitude (brightness), which is ~1 hour before its closest approach. IIRC

Jacob Miller

@PM2Ring So the Apophis is only visible within the contours at that given time for an observer at sea level?

PM 2Ring

@JacobMiller Pretty much. But the observer's altitude (height above sea level) doesn't make much difference to the altitude angle (what Horizons calls elevation).

Apophis should be visible for several hours, though. If you have good eyes, and you're in a region without light pollution. Its ecliptic longitude changes a fair bit over that time span. I have some graphs here: astronomy.stackexchange.com/a/58153/16685

Jacob Miller

@PM2Ring What If I am 300 million kilometres above Earth? Will that make an observable difference?

PM 2Ring

@JacobMiller Of course! :D

At its closest approach, it's only ~38,000 km from the centre of the Earth.

Besides, 300 million kilometres is ~2 au. You aren't going to see an asteroid with the naked eye at that distance.

Ryder Rude

9:32 AM

@PM2Ring it's fine :)

Jacob Miller

@PM2Ring Then the orbit of Apophis must be very elongated considering that as of now, it is approximately 313 million kilometres (about 2.09 Astronomical Units) away from Earth.

@PM2Ring Wow, You drew all those figures using just a phone? Impressive!

naturallyInconsistent

@Bml correct. And again, this is an expression for kinetic energy; it is extremely logical and intuitive to define $E(0)=0$; this is WLoG.

Bml

9:49 AM

@naturallyInconsistent The oddity I find, however, is that: using $u \neq v$, that is, $E(u+v) + E(u-v) = 2E(u) + 2E(v)$, $E(0) = 0$ is not a constraint, but follows from $u=v=0$. Using $u=v$, and thus $E(2v) = 4E(v)$, $E(0) = 0$ is a constraint, because it does not follow from $v=0$. If we substitute $v=0$, we have $4E(0) = 4E(0)$, which is true for all values of $E(0)$. So, what is the real difference?

naturallyInconsistent

@Bml As mentioned above, the moment you write that $E(u+v)+E(u-v)=2E(u)+2E(v)$, you have already shown that it is quadratic. This is more than sufficient to obtain both $E(v)\propto v^2$ and $E(0)=0$. The point is that we have to argue why this is true, and it is not obvious that it should behave this way. If you put in the effort to argue it, then sure. If not, you can use the other argument avoiding it.

PM 2Ring

@JacobMiller No, its eccentricity is only 0.19, so its orbit looks fairly circular. There's an orbit diagram in the linked post.

Ryder Rude

@Bml E(0)=0 is not a constraint in Ron's derivation either. The equation we get there is : E(2v)+E(0)=4E(v), for all real v. Now set v=0 to get 2E(0)=4E(0) which gives E(0)=0

Bml

@RyderRude Yes, you are right. I was mistaken.

Ryder Rude

it's just that in Ron's proof, we have to assume continuity of the second derivative to get the quadratic

while in ur general eqn, we just use continuity

Bml

10:00 AM

@naturallyInconsistent "The point is that we have to argue why this is true, and it is not obvious that it should behave this way. If you put in the effort to argue it, then sure. If not, you can use the other argument avoiding it.": in what sense? I'm not sure I understand what you are saying here.

PM 2Ring

Here's a lively Latin song from one of my favourite Australian singers, Emma Pask.

Cosita Divina

►

Sorry, no video, just audio.

Ryder Rude

@PM2Ring fun song

Batzorig Vaanchig- Mongolian Throat Singing

►

PM 2Ring

@RyderRude It's interesting, but I can only take it in small doses.

Jacob Miller

10:16 AM

@PM2Ring Apophis orbit around what?

PM 2Ring

Dolphins do a similar trick with overtones to get their huge frequency range, going right up into the ultrasonics they use for echo-location, and down to the frequencies that humans can hear.

@JacobMiller It's an asteroid. It's orbiting the Sun. But its orbit is fairly close to ours.

So when it's close, it can get very close. But ~3.5 year later, it's about 2 au from us, on the other side of the Sun.

> Apophis makes ~79 orbits of the Sun in 70 Earth years. Its synodic period relative to Earth is almost exactly 2840 days, which is ~7.77536 years.

Jacob Miller

@PM2Ring Why is Earth's orbit depicted as an increasing spiral?

PM 2Ring

@JacobMiller No, that's a plot of Apophis relative to Earth. Earth is the immobile pale blue dot in the middle.

So Apophis spirals out, from orange to yellow, etc. Then it spirals back in, from green, to blue, purple, magenta.

Jacob Miller

10:33 AM

@PM2Ring You mean that it hit earth some day?

Or is that the 2029 near-earth pass?

@PM2Ring When it was first discovered in 2004, it was calculated that there would be a 2.7% chance of it hitting Earth in the near future. But later observations ruled that out. Will the future observations change that trajectory much again as more data comes in?

Bml

@naturallyInconsistent Also, I did not understand what is perhaps the crux of your argument. Several times you have said that if $E(u+v) + E(u-v) = 2E(u) + 2E(v)$, then you have already proved that $E$ is quadratic, as if it were an upstream assumption. If, on the other hand, $E(2v) = 4E(v)$, then you have not already proved that $E$ is quadratic, but you must prove it. I do not understand why this is so and where the difference lies. Why is it already proved in one and not in the other?

naturallyInconsistent

@Bml Either you proceed with 1) $E(m,v)=mE(v)$, 2) $E(0)=0$, 3) $E(2v)=4E(v)\qquad\implies\qquad E(v)\propto v^2$, or you have the same first step, but obtain instead b) $E(u+v)+E(u-v)=2E(u)+2E(v)$, and from there the same results

micsthepick

sciencedirect.com/science/article/pii/S0927650524001130/…

PM 2Ring

10:48 AM

@JacobMiller The 2 plots at the end of that answer are between 2 close approaches, covering the timespan between the last close approach and the next one, 2021-Mar-6 to 2029-Apr-13.

micsthepick

sorry if this was brought up already and I couldn't find it: thoughts?

it was shared to me by someone else who is not a physicist, and I'm not sure that it's a serious paper, or if it just needs a little peer review magic and work done to it

PM 2Ring

@micsthepick What is it? I'm not going to download some random anonymous PDF.

micsthepick

actually, let me try linking to the article page

sciencedirect.com/science/article/pii/…

"On the same origin of quantum physics and general relativity from Riemannian geometry and Planck scale formalism"

PM 2Ring

@JacobMiller It should miss Earth for the foreseeable future. But it's hard to predict its trajectory into the far future. Small uncertainties in its position & velocity vectors grow over time. Especially because it gets deflected slightly whenever it makes a close approach.

@micsthepick I'm sceptical. But my QM & GR skills aren't good enough to evaluate the proposal.

naturallyInconsistent

@micsthepick there have been quite many links to this article for a week now.

@PM2Ring lol, you jest.

micsthepick

10:59 AM

@naturallyInconsistent here, or perhaps on scattered main site posts?

PM 2Ring

@micsthepick Thanks. In general, always link to the article or abstract page, not the PDF, especially on the main Physics.SE site.

Bml

@naturallyInconsistent OK, so we have 1) E(m,v) = mE(v), 2) $E(u+v)+E(u-v)=2E(u)+2E(v) \qquad \implies E(v) \propto v^2$, not the contrary, right? Could you see also my last reply?

PM 2Ring

@micsthepick My first instinct is to reject it as sophisticated numerology

Sine of the Time

numerology sucks

naturallyInconsistent

@micsthepick both

PM 2Ring

11:02 AM

Buongiorno, Sine of the Time

naturallyInconsistent

@Bml yes

Sine of the Time

long time no see @PM2Ring

naturallyInconsistent

@PM2Ring it isn't even numerology? The stuff that is inside is so... lacking...

Sine of the Time

how is life going?

PM 2Ring

We get numerological stuff regularly on Physics.SE. But the really crazy stuff gets posted on Astronomy.SE. ;)

Sine of the Time

11:05 AM

also on Math.SE

a lot of users got suspended for questions related to specific prime numbers and similar nonsense

micsthepick

@PM2Ring what immediately sticks out to me is 8.1. Electron mass and quark masses: do they not just simply drop a c?

PM 2Ring

@SineoftheTime I often lurk in the Math chat, but I usually dont post unless I've got something to say.

micsthepick

actually, no it's just weird notation

Sine of the Time

maths chat is very busy lately

micsthepick

multiplying both sides by e=mc^2 or something

Sine of the Time

11:07 AM

it's nice to see a lot of active users

PM 2Ring

@SineoftheTime If you go looking for number patterns, you'll probably find number patterns. But they may not be very significant. ;)

Bml

@naturallyInconsistent And isn't it a contradiction with "the second to last" reply I sent?

naturallyInconsistent

@Bml The 2nd form, $E(2v)=4E(v)$, along with $E(0)=0$, is already sufficient to prove that it is quadratic. I'm saying that prior to invoking some physical argument, you haven't yet proved quadratic-ness in the steps before those.

Sine of the Time

if you go looking for number patterns, you see them everywhere. Like the "fanatics" of the golden ration

naturallyInconsistent

You have to be extremely clear what your premises are, and what you have actually argued for. If you are not careful about your ingredients, then you can easily lead to something wrong.

PM 2Ring

11:12 AM

Eg, $\pi \approx \sqrt{10}$ IIRC, some ancient geometers (before Archimedes), were convinced it was exact. They may have even tried to prove it. It's interesting, and kind of useful, but mostly just a coincidence of small numbers.

Bml

@naturallyInconsistent It is sufficient to prove that is quadratic if we assume continuity of the second derivative. It is not a trivial assumption.

PM 2Ring

OTOH, this is even closer, but I doubt there's much use for it: $\pi \approx (1067\sqrt{10} - 1263) / 672$

naturallyInconsistent

@Bml I don't know what you are talking about. The $v$ inside the function is arbitrary; that should be more than enough to prove that it is quadratic.

Bml

@naturallyInconsistent Also the 1st form, $E(u+v)+E(u-v)=2E(u)+2E(v)$, is already sufficient to prove that it is quadratic, if we assume continuity of $E$ and solve this functional equation. Yes?

Sine of the Time

@PM2Ring yeah, some people see patterns everywhere :D

PM 2Ring

11:16 AM

@SineoftheTime The Golden Ratio is pretty bad for that, since it does have a lot of fascinating properties & patterns. But because its continued fraction has the slowest convergence it has lots of ratios that approximate it. So fanatics can easily find ratios that are kinda close to the Golden Ratio.

Bml

@naturallyInconsistent I don't think it is so obvious. See for example this:

7

Q: Prove $f(x)$ is quadratic if $f(2x)=4f(x)$ and $f(x)$ is increasing over positive $x$

The problem arose in the context of kinetic energy, where it can be proven from symmetry principles that $E(2v)=4E(v)$ without assuming $E=mv^2/2$ (see e.g. physics stackexchange). While one may do further physics from this point to prove the desired result (that $E$ is quadratic in $v$) -- cons...

naturallyInconsistent

@Bml yes. You definitely can prove it is quadratic with this, but you do require a physical argument to get this form. That is not trivial either.

Bml

@naturallyInconsistent What physical argument do I require?

Sine of the Time

I agree that $\varphi$ have a lot of patterns also in nature, but it's really nonsense seeing it everywhere. I remember once reading about a castle in a city and some random "mathematicians" started to measure the lenght of the walls, etc and concluded that the proportion is the golden ratio

naturallyInconsistent

@Bml See velut luna's answer.

@Bml the very argument that supplies $E(u+v)+E(u-v)=2E(u)+2E(v)$ itself. This is highly non-trivial.

Bml

11:21 AM

@naturallyInconsistent I saw it. Why is it relevant?

Sine of the Time

the feature new feed items is nice btw

naturallyInconsistent

@Bml because that did not assume 2nd derivative continuity

Bml

@naturallyInconsistent Mhm. How can we search the very argument that supplies this equation?

PM 2Ring

$46\varphi + 173\pi \approx 160e + 183$

(617.925092553529, 617.925092553447)

Sine of the Time

-_-

PM 2Ring

11:34 AM

I've got a million of 'em. ;)

Slereah

isn't there a wiki page of those

naturallyInconsistent

@Bml the one that had it, supplied it

PM 2Ring

There's this:

Almost integer

In recreational mathematics, an almost integer (or near-integer) is any number that is not an integer but is very close to one. Almost integers may be considered interesting when they arise in some context in which they are unexpected. == Almost integers relating to the golden ratio and Fibonacci numbers == Some examples of almost integers are high powers of the golden ratio ϕ = 1 + 5 2...

Slereah

Mathematical coincidence

A mathematical coincidence is said to occur when two expressions with no direct relationship show a near-equality which has no apparent theoretical explanation. For example, there is a near-equality close to the round number 1000 between powers of 2 and powers of 10: 2 10 = 1024 ≈ 1000 = 10 3 . {\displaystyle 2^{10}=1024\approx 1000=10^{3}.} Some mathematical coincidences are used in engineering...

that's the one

Bml

@naturallyInconsistent What?

Ryder Rude

11:40 AM

@naturallyInconsistent this equation only follows by galiliean addition of velocities, which Ron's original argument is assuming anyway

PM 2Ring

These sorts of relationships can be useful. Eg, when evaluating tricky definite integrals numerically. If you can subtract out a big chunk of the area, then you only need to integrate the relatively tiny remaining slice.

naturallyInconsistent

@RyderRude just go away

@Bml the question statement in this link did supply an argument for why that works, except for the typo I already pointed out

Bml

@naturallyInconsistent Ah, are you saying "$T+Q$ remain the same before and after the collision" and "$Q$ is frame-independent"?

naturallyInconsistent

@Bml yes

Bml

@naturallyInconsistent But these are also assumptions we made for $E(2v) = 4E(v)$...

naturallyInconsistent

11:47 AM

@Bml the argument was not seeking generality at the point it was made. That is, it is slightly stronger an assumption to make that $u\pm v$ argument, whereas the $2v$ case is a weaker assumption. However, as you already know, the special case is sufficient to prove the general case, so in the end both work.

Ryder Rude

again, the original argument may only use a special galiliean transform, but it also has to pay for that by assuming stronger mathematical properties of E(v)

Bml

@naturallyInconsistent OK... And why does this question assume that $E(v)$ Is increasing over positive $v$? With that assumption, it can be guaranteed that $E(v) = a v^2 \geq 0$ because $a>0$. But why is that true?

Ryder Rude

The second argument uses a general Galiliean transform, but it only assumes continuity of E(v)

so it is a trade-off. Also, there is no reason to accept a special Galilean transform while not assuming the general Galilean transform

naturallyInconsistent

@Bml I'm not sure what you want to get more out of this. You already know that $E(0)=0$ and $E(-v)=E(v)$ and at least $\exists v_0\in\mathbb R\ |\quad E(v_0)\neq0$, and now you have $\forall v\in\mathbb R\quad E(2v)=4E(v)$; then it all depends upon $E(v_0)$; if this is positive, then the whole thing is positive, and if it is negative, the whole thing is negative. For convenience, we had already identified that the energy turns into heat energy, and we wanted positive heating.

Bml

12:09 PM

@naturallyInconsistent So is this the reason why the OP assumed $E(v)$ increasing over positive $v$?

naturallyInconsistent

@Bml more than good enough. We all know the correct final expression. Don't be so painfully pedantic

Bml

12:23 PM

@naturallyInconsistent Yes, you are right: I am often pedantic. But the problem is that $E(v) \geq 0$ is OK, but an increasing kinetic energy over positive $v$ sounds strange. Maybe it is because we want faster objects to have more energy because they raise the temperature higher upon collision (as @RyderRude suggested)?

Ryder Rude

12:50 PM

@Bml if assuming E>=0 is ok, then u do not need to separately assume increasing KE. In cmv^2, the only way to have positive energy is to have c>0, and then cmv^2 is increasing without further assumptions

so increasing KE is subsumed in the sum of all the assumptions we have made so far. Really fascinating how reasoning works

Sine of the Time

since you've already solved the fe and conclude that $E$ is quadratic, I don't understand why is it relevant that $E$ is increasing

you've found the solution and the function is increasing, so this discussion is pointless

Ryder Rude

@SineoftheTime their point is that we have found E=cmv2 which cud be a decreasing function if c<0

so some additional "physical arguments" related to temperature are made

but it is all circular imo. It all comes down to definitions

Sine of the Time

@RyderRude if we assume $E\ge 0$, then $c>0$ right?

Ryder Rude

@SineoftheTime yes. thats what i wrote. If we already assumed this, then no need to assume increasing

Sine of the Time

it's automatic that it's increasing

Bml

12:56 PM

@SineoftheTime I, too, was trying to understand this, since I cannot understand the argument that the OP makes in support of increasing $E$.

Sine of the Time

I don't understand what you're trying to understand

if you solved the functional equation and you found $E=cmv^2$, with $c>0$ for obvious reasons, the function $v\mapsto cmv^2$ is increasing

Bml

7

Q: Prove $f(x)$ is quadratic if $f(2x)=4f(x)$ and $f(x)$ is increasing over positive $x$

The problem arose in the context of kinetic energy, where it can be proven from symmetry principles that $E(2v)=4E(v)$ without assuming $E=mv^2/2$ (see e.g. physics stackexchange). While one may do further physics from this point to prove the desired result (that $E$ is quadratic in $v$) -- cons...

functions physics functional-equations

Sine of the Time

again?

you've sent it 10 times

Bml

The OP said: consider a system with other prime numbers of balls, then do algebra to prove the result for rational scaling in $v$, then use the fact that there are rational numbers between any two real numbers and assume the function is increasing to prove it for all real scaling -- it seems intuitively obvious from this point immediately, that if $E$ is increasing in $v$, $E=kmv^2$.

@SineoftheTime Am I bothering? If so, I apologize, it is not my intention.

Sine of the Time

you've solved the general FE from a mathematical point of view, didn't you?

and you got a "quadratic" function

now since $E(2v)=4E(v)$ is a particular case of what you've already solved, adding the assumption $E>0$ you have an increasing function

you don't need everything said in the post you linked, since you've taken a different route

@Bml you're not bothering me but this discusssion is no longer productive

Bml

1:04 PM

@SineoftheTime Yes. What I want to try to say is that from a physical point of view in my judgment we cannot assume the increasing function on positive $v$, and from there derive that the scaling factor is positive, so the kinetic energy is positive. Instead, the reasoning should be the reverse: assuming by convention that $E(v) \geq 0$, so $c>0$, then the first derivative of the function is positive, so $E$ is increasing. This is what I was trying to say.

Ryder Rude

either implication works. increasing is implied and implied by positive here

i feel "more speed = more temperature" is maybe natural to assume. but again, it is all sign conventions

Bml

@RyderRude Yes, it is all convention. What you said is interesting, but @SineoftheTime is right: this discussion is becoming no longer productive, it is all circular.

@RyderRude By the way, we have to prove that kinetic energy is proportional to temperature... We can't assume it a priori.

Ryder Rude

1:34 PM

@Bml yeah maybe

Bml

1:48 PM

@RyderRude So?

Ryder Rude

@Bml whatever it is, it's irrelevant. i dont wanna discuss now :)

Bml

@RyderRude Completely agree.

zeta space

6:26 PM

So fields interact (sometimes). Mathematically speaking are these fields like superimposed upon eachother or what?

I can't wrap my head around it

Are they disjoint like leaves in a foliation? Do they intersect in some way?

fields in terms of quantum field theory

nevermind I figured it out

Ryder Rude

7:29 PM

@zetaspace hi. Different fields in qft are sections of different bundles of spacetime

ACuriousMind

@RyderRude ...how on earth do you think this answers the question of how they interact?

Ryder Rude

sorry... i meant to answer the part "do they intersect in some way"

i mean that they kinda don't intersect as they live on different bundles, but they do intersect on the base manifold

ACuriousMind

@zetaspace When physicists talk about fields interacting, they mean that the action (usually in terms of a Lagrangian) describes interactions between them. This is not a quantum notion, this is also how one would describe interactions in classical field theory

that is, the notion of interaction is about the dynamics, i.e. the differential equations that govern the time evolution of the system, not about the nature of the fields themselves as mathematical objects

Ryder Rude

and via the intersection on the base manifold, they interact

ACuriousMind

@RyderRude what "intersection"? You're once again using terms that have a clear technical definition in math in completely wrong ways - what sets do you think "intersect" here?

Ryder Rude

7:40 PM

e.g. suppose we take two field configurations of different fields and take their "address" on the manifold, and then take the intersection of those addresses

ACuriousMind

as I just said, the interaction is via a term in the Lagrangian; the fields in an interacting Lagrangian are exactly the same mathematical objects as in a corresponding free Lagrangian, so their nature as mathematical objects has nothing to do with them interacting or not

Ryder Rude

if this is non empty, then they interact

Ryder is not Rude.

\o @Semiclassical

Ryder Rude

@ACuriousMind yes. but the Lagrangians are local physically

Semiclassical

o/

Ryder Rude

7:41 PM

so locality implies interaction via intersection

ACuriousMind

@RyderRude A free Lagrangian is as local as an interacting one for all ordinary field theories, you're talking nonsense. Also, what on earth is an 'address'? If you want to discuss the formal nature of fields - as the approach via bundles would suggest - you need to use proper terminology.

Ryder Rude

okay the precise idea i mean is : take two fields on the same space slice, and take the subsets of the space slice on which field 1 is non zero and field 2 is non zero. If these two subsets have zero intersection, then the fields evolve like free fields at the moment. @ACuriousMind

ACuriousMind

@RyderRude still wrong

Ryder Rude

this follows from Euler Lagrange equation, i think

ACuriousMind

Two classical scalar fields are functions $\phi_1 : M\to \mathbb{R}$, $\phi_2 : M\to \mathbb{R}$. A local free Lagrangian would be $L(x) = (\partial\phi_1(x))^2 + (\partial\phi_2(x))^2$. A local interacting Lagrangian would be $L = (\partial\phi_1(x))^2 + (\partial\phi_2(x))^2 + \phi_1(x)^2 \phi_2(x)^2$.

Even in the free case, you can have that both $\phi_i$ have as support all of $M$, and still they are not interacting.

Ryder Rude

7:47 PM

yeah...in the free Lagrangian case..there isn't even a possibility of interaction as there is no coupling

but in ur second case, take two configurations like i wrote.

ACuriousMind

no, you don't get to move the goalposts

2

Ryder Rude

if that intersection is empty, then they behave like free fields at the moment

ACuriousMind

that doesn't mean they are free fields

Ryder Rude

okay.... but tbf, i didn't say that if that intersection is non emtpy, they must not behave like free fields !!

i made a different claim, which is true for both lagrangians

ACuriousMind

you did initially not mention Lagrangians at all, and you answered this to someone asking about what physicists mean when we say that fields interact

Ryder Rude

7:49 PM

@ACuriousMind they behave free only at the moment. I think this follows from EL equations. Lemme check

ACuriousMind

do you not see how that was an incredibly misleading statement to make even if you might have had this technically correct interpretation that no physicists would come up with when asked to talk about interacting fields?

additionally the nature of the fields as sections of bundles - which was your very first statement - is completely irrelevant to this, you can talk about their support just looking at them as functions

Ryder Rude

@ACuriousMind i genuinely thought this was an intuitive way to think about what "interaction via intersection" could mean

and i realise the msg about bundles is mostly irrelevant.when i wrote that, i was contemplating writing the bit about local lagrangians

but i didn't for some reason

but i just mean the idea of local Lagrangians here

and local interactions

but i still need to check if it follows from EL equations. I think it's not obvious

$\frac{d\pi _j (x)}{dt}= f(\psi _i (x), \pi _i (x))$ where $i$ indexes all the fields

so the other field terms in $f$ should go to zero when the other fields are zero at $x$

i think it's not that obvious. For one, we don't just need the other fields to be zero, but also their momenta

Silly Goose

8:13 PM

are labels on Hilbert space redundant structure in quantum mechanics?

ACuriousMind

@SillyGoose what's a "label on Hilbert space" :P

Silly Goose

Suppose I have a finite-dimensional Hilbert space specified by $\dim \mathcal{H} \in \mathbb{N}$ and a Hamiltonian $H: \mathcal{H} \to \mathcal{H}$ specified by its spectrum.

Then, all computations in quantum mechanics are going to eventually take an inner product, which is defined on $\mathcal{H}$ and usable without any additional structure.

ACuriousMind

@SillyGoose A Hamiltonian is not fully specified by its spectrum.

Silly Goose

@ACuriousMind An isomorphism $U: \mathcal{H} \to \mathcal{H}'$. The more familiar of the like would be like $U: \mathcal{H} \to \otimes_i \mathcal{H}_i$

@ACuriousMind oh

what is a hamiltonian specified by

ACuriousMind

It's a linear operator on Hilbert space

so you can specify it by anything you can specify a linear operator by, e.g. its action on a basis

Silly Goose

8:16 PM

oh i see

ACuriousMind

or, in the finite-dimensional case, you could just write down a matrix :P

Silly Goose

okay hm well let's remove the Hamiltonian from the picture for now i guess

ACuriousMind

but the spectrum is not enough, most elementarily because it does not contain information about the multiplicity of the eigenvalues

Silly Goose

oh i see

ACuriousMind

@SillyGoose I don't really know what you mean by "redundant" here. Usually we write a space as a tensor product because those factors are physically meaningful subsystems, and then we want to talk e.g. about the resultant states of these subsystems via the partial trace. How could this be "redundant"?

Silly Goose

8:20 PM

hm well maybe the question becomes is there a way to talk about the partial trace in the "abstract" i.e. without any reference to a tensor product factorization of hilbert space

ACuriousMind

no

the construction is explicitly about the factorization

Silly Goose

i feel like that is strange :P

ACuriousMind

why?

Silly Goose

because the objects going in and the objects coming out can be talked about in the "abstract"

ACuriousMind

Again, the factorization is a physically meaningful thing: It represents how the subsystems embed into the total system

This information is not part of just writing down a Hilbert space

Silly Goose

8:22 PM

hm well maybe what I feel is strange is this notion of "factorization" is not natural it is quite artificial

ACuriousMind

I would contest the idea that anything in theoretical physics is not artificial :P

so what do you mean by "artificial"?

Silly Goose

it is something that needs to be done ad hoc every time

ACuriousMind

it's not ad hoc

Silly Goose

hm how so

ACuriousMind

"ad hoc" implies that somehow we have a choice here

but again, the factorization is physically meaningful: It represents our colloquial idea of subsystems

arguably, in almost all cases you don't really start with the total Hilbert space and then choose a factorization: Rather, what happens is that we argue once that the tensor product is the correct quantum realization of the notion of combining subsystems, and then you construct e.g. the Hilbert space of three particles, each of which has the one-particle Hilbert space $H_1$, as $H_3 = H_1\otimes H_1 \otimes H_1$

Silly Goose

8:26 PM

well consider a classical case. i can see a chair as my system, or i can see each leg as a subsystem of the chair system, or etc. all these map to "colloquial idea of subsystems", but there is no unique choice, so it is practicality or arbitrarity that drives the choosing of subsystems

ACuriousMind

I know not of a single example off the top of my head where you'd start with a Hilbert space for a composite system without at least implicitly having obtained it this way

Semiclassical

@ACuriousMind memories of annoying Clebsch-Gordan computations for three angular momenta are coming back to me

Ryder Rude

@ACuriousMind i had this question, that if we represent $su(2)$ via left invariant vector fields on $L^2(S^3)$, does the decomposition of the representation into irreps include spinorial reps

ACuriousMind

(I say "implicitly" because you might of course first classically combine $X$ and $Y$ into $X\times Y$ and then quantize, but this is just $L^2(X\times Y) = L^2(X)\otimes L^2(Y)$, i.e. the naive quantization functor turns the Cartesian product of classical configuration spaces into the tensor product of quantum state spaces)

Semiclassical

there's also a small subtlety here in that you can also do subsystems by demanding that the operators for each subsystem commute, but for finite-dimensional systems that doesn't give anything new

ACuriousMind

8:28 PM

@SillyGoose But there the choice is not in "choosing subsystems", it's in how granular you want your physical description to be

you might model the chair by its center-of-mass position, or you might model it by the four center-of-mass positions of its legs and the center-of-mass position of the rest

but that's not a choice of "factorization", that's choosing two entirely different state spaces (with different dimensions even classically!) because your model is supposed to do two different things (e.g. you might choose the latter if you're interested in processes that might destroy the structural integrity of the chair :P)

Semiclassical

(for infinite-dimensional systems you get Tsirelson's problem)

hmm, that's not a very useful link. let me find a better one

eh, i'll just go with this: arxiv.org/abs/0812.4305

ACuriousMind

oh no not another subtlety of infinite-dimensional QM :P

Silly Goose

well say my wooden chair is composed of dark planks and light planks. then i can partition my chair into the light plank parts and the dark plank parts. i wouldn't say this is getting more granular, it is a fundamentally different way of partitioning the chair

ACuriousMind

@SillyGoose But again, unless your initial description already contained e.g. the the positions of all planks, then you can't just partition the state space like that

Most people would start with $\mathbb{R}^3$ as the state space of the chair, i.e. its center-of-mass position

you can't partition that

instead, in order to do your partition, someone needs to say "well, I'm really interested in modelling the $N$ individual planks that make up the chair", and then your configuration space is $\mathbb{R}^{3N}$. Then saying "let's group by plank color" is just using associativity of the configuration space Cartesian product and writing this as $\mathbb{R}^{3m} \times \mathbb{R}^{3(N-m)}$

Silly Goose

@ACuriousMind well what if an experimentalist went into the lab blind, unknowing and started to probe some quantum system there

ACuriousMind

8:35 PM

but see how this was dependent on first making the choice to build the state space of the chair as the product of the plank spaces in the first place?

@SillyGoose generally I assume that's what all experimentalists do; I do not see the relevance to the discussion at hand :P

Silly Goose

well how would they know when they encounter the signature of a distinct "single particle" so that they can jot down Hilbert space one

Semiclassical

the more interesting practical point might be "how does an experimentalist notice that their system can be divided into subsystems" (or separate degrees of freedom, a la J=L+S for angular momenta)

Silly Goose

@ACuriousMind i think so. i suppose this point stands for systems that are physically visible at the least

Semiclassical

which...well, there's a reason why understanding quantum systems was hard :P

Silly Goose

@Semiclassical yeah i think this is along the lines of the question indeed

ACuriousMind

8:38 PM

@SillyGoose you don't find Hilbert spaces in the lab :P

If I just hand you "a quantum system" then you can't do anything to find "subsystems"

Silly Goose

then where would one pull out a notion of subsystems from :P (if not arbitrarily or practically)

ACuriousMind

The subsystems come either from it being obvious from the setup (e.g. Alice and Bob each have one particle) or from some theoretician having modeled something as a composite system and the predictions of this composite model being accurate (e.g. the proton as three quarks)

there's no magic trick you can perform to find out if something "has subsystems"

notoriously, we had to collect a lot of complicated evidence related to rather deep QFT lore to become pretty certain elementary particles like the electron really are probably not composite

Semiclassical

i think the case of angular momenta here is interesting, in that they're not really "subsystems" as such but yet the problem of disentangling different kinds of angular momentum was very important in the history of quantum physics

iirc the main way that was explored was via applying different magnetic field strengths and seeing the different splitting patterns you get

weak Zeeman vs strong Zeeman etc etc

ACuriousMind

yeah, this is more related to the notion of "commuting sets of observables" you mentioned, I think, but it also illustrates that "subsystem" is a bit of a vague notion

the angular momentum is special in that the quantum state space is a tensor product $L^2(\mathbb{R}^3)\otimes S$ where $S$ is some spin representation, but there is no corresponding classical Cartesian product of configuration spaces, aka "spin is really quantum"

Semiclassical

8:54 PM

yeah. what i typically associate subsystem is the connection with locality, and that doesn't map onto "angular momentum" terribly well

part of this is just the fact that two kinds of phenomena can have similar representations in QM despite being quite different in origin

one of these days i do want to get a better handle on stuff like C* algebras but i dunno if it's actually that valuable for my purposes

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