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0celo7
0celo7
12:42 AM
@yuggib Should I do the exercises in Munkres on nets? I'm not sure we'll cover them in class.
 
 
2 hours later…
dmckee
dmckee
3:02 AM
@SirCumference Oy. Just looked at the times displayed for questions on astronomy.stackexchange.com/questions. I fear that site is having the same trouble as the last Astronomy.
 
0celo7
0celo7
@dmckee If $f$ is a vector valued function, would you believe $\int |f|\ge |\int f|$?
The norms are the vector norms, of course.
 
dmckee
dmckee
Looks reasonable. I mean, if you integrate first there is a chance of bits canceling out, but the integral of the norms must be non-decreasing. Right?
 
0celo7
0celo7
Right, that's how the proof goes for $f:\Bbb R\to\Bbb R$, basically.
but a rigorous argument in multiple dimensions is eluding me
trying to prove straight lines are actually the shortest curves is turning out to be hard.
 
dmckee
dmckee
@0celo7 Ah. For that I'd use the calculus of variations. Don't know if it can be done with simpler tools.
 
0celo7
0celo7
@dmckee Problem with that is you get an extremum, not necessarily a minimum.
You need some sophisticated tools to show it's a true minimum.
(either second variational derivatives or a topological argument via Riem. geo.)
 
dmckee
dmckee
3:13 AM
@0celo7 Then you find one other path for which you can compute the length exactly and show that it is larger then the found extreme. From that you conclude that the extreme is at least not a maximum.
 
0celo7
0celo7
@dmckee Saddle point.
 
dmckee
dmckee
Or is that inelegant?
 
0celo7
0celo7
No, it just doesn't show what you want.
 
dmckee
dmckee
@0celo7 Well, yeah. I haven't figured out how to exclude that possibility by contradiction.
Something to do in my copious spare time. When I'm done writing my book.
 
0celo7
0celo7
@dmckee I'll send you my proof when I'm done.
I already have a proof...but it's a mosquito nuke.
but I think I need the p<1 Minkowski inequality for this....not exactly trivial either.
 
Sir Cumference
Sir Cumference
3:25 AM
@dmckee What do you mean, "times"?
 
dmckee
dmckee
@SirCumference The posting times. As a way to judge how much activity the site is getting. And it's not much. I don't think the team is going to shut them down, but graduation is a long way off.
 
Sir Cumference
Sir Cumference
@dmckee Well we seem to be doing pretty well, relative to other beta sites
1079
Astronomy – astronomy.stackexchange.com

Beta Q&A site for astronomers and astrophysicists.

Currently in public beta.

Another 3 years and we might be graduating
@dmckee Er, what do you consider "good"?
We on the brink of shutting down?
 
dmckee
dmckee
The previous astronomy stack starting pretty strong and then faded. So, I'm kind of keyed up to worry about that. Do you have access to the analytics of the astro?
 
Sir Cumference
Sir Cumference
Not yet, need another 1k
 
dmckee
dmckee
As long as the activity is growing there is no danger of being shit down, but if the site stagnates (or worse, contracts by any significant degree) there is a danger.
 
Sir Cumference
Sir Cumference
3:39 AM
But we've been going for almost four years. That say anything?
 
dmckee
dmckee
That's what happened to the previous astronomy and to theroreticalphysics.
 
Sir Cumference
Sir Cumference
Previous astronomy lasted a year
@dmckee Oh, here ya go
Our stats
 
0celo7
0celo7
@dmckee Aha! Got that integral identity I needed.
Follows from the $p<1$ Minkowski inequality, as I thought.
I have no idea how to prove that, of course.
 
Sir Cumference
Sir Cumference
@dmckee Here's the more important parts
 
dmckee
dmckee
@SirCumference You really need to see the trends over the course of a few months, but you've got less than 4000 questions and only get about 6 per day. That's not a disaster, but it's not strong either.
 
Sir Cumference
Sir Cumference
3:43 AM
 
dmckee
dmckee
Those are at least trending up. Especialy the visits. But you haven't found critical mass yet.
 
Sir Cumference
Sir Cumference
@dmckee Hmm...how would you say space exploration is doing?
 
dmckee
dmckee
More visits but fewer questions? Not sure what to make of that. I don't think either site is in danger from the team, but growth seems to be slow.
 
Sir Cumference
Sir Cumference
Yep :/
Guess not that many people are into astronomy
Not to mention, the Astronomy chat is essentially dead, though at least the space ex chat is pretty active
 
0celo7
0celo7
@dmckee Ok: I have shown that if a curve goes from $x$ to $y$, then its length is $\ge ||x-y||$.
Now I have to show there is precisely one curve going from $x$ to $y$ with the minimal length.
@dmckee Ok, I have the proof w/o calculus of variations.
 
dmckee
dmckee
4:00 AM
@0celo7 Pathological case: antipodes on a spherical manifold. //because apparently nerds just can't help that kind of snark
 
0celo7
0celo7
@dmckee What?
I'm talking about curves on $\Bbb R^n$.
 
dmckee
dmckee
@0celo7 OK, that's cool.
@0celo7 I figured you were, but I couldn't help myself.
 
0celo7
0celo7
The only problem is that I use the reverse Minkowski inequality, which is nontrivial. I'm considering a more strightforward proof. Perhaps using polygonal approximations.
AH
Cauchy-Schwarz, come save me
 
John Rennie
John Rennie
Gosh a little spilled coffee goes a long way. What a mess!
 
DanielSank
DanielSank
4:18 AM
I don't suppose you guys know how to clean glazed ceramic tea cups?
0
Q: How to clean red tea stains from cracked glaze Japanese tea cups

DanielSankI have a set of green, cracked glaze Japane tea cups. A totally unused one is pictured in Figure 1. Figure 1: A clean, unused tea cup. As I have used one of these cups several times with African red tea, the cup has begun to take on a stain, as shown in Figure 2. Figure 2: A used and reddish...

cleaning tea japanese-cuisine
 
John Rennie
John Rennie
@BernardMeurer: if you're around I need the phone number for the concierge at your apartment. I've just tried to arrange delivery and they want a contact number and e-mail.
 
Bernard Meurer
Bernard Meurer
4:55 AM
@JohnRennie sent on Facebook
 
John Rennie
John Rennie
@BernardMeurer Ah, thank goodness you saw this! I was worried I'd have to delay another day. Right, I'll get the delivery off. It's a 3-5 day tracked delivery. Sorry it's slow, but the air deliveries won't accept anything with lithium batteries in.
@BernardMeurer Just sent a text. Have you got it?
Aha, we're in business :-)
 
Bernard Meurer
Bernard Meurer
5:13 AM
We got lucky! I can't sleep :)
That's alright, lithium is evil
 
Physics Meta
Physics Meta
5:48 AM
0
Q: Improper use of quote marks and straw man logical fallacies

CountTo10This is a request for quote marks to be used solely for their original purpose, that is to quote verbatim a statement made in a post. They should not to be used as part of a logical fallacy /straw man argument. This has happened to me twice in two days by users with undoubtedly the best of motiv...

discussion
 
yuggib
yuggib
6:40 AM
@0celo7 maybe; but you probably won't need nets anytime soon
 
John Duffield
John Duffield
7:25 AM
@2physics : no. I'm an IT guy who's always been interested in physics. I got even more interested about ten years ago when our then teenage children gave up all their science subjects. I found out that UK physics A-level numbers had fallen by 56% in 25 years, and I started taking an interest in where physics was going. To hell in a handcart, it would seem.
 
John Rennie
John Rennie
@2physics gravity is described by general relativity i.e. the phenomenon we call gravity is actually the spacetime curvature described by GR. So how can there be any incompatibility between gravity and GR?
Gravity does cause problems with special relativity, and it was because Einstein realised that SR and gravity had a problem that he went on to formulate general relativity.
 
John Duffield
John Duffield
@JohnRennie : I know why gravity is so weak.
 
John Rennie
John Rennie
@knzhou This happens to any high profile user i.e. anyone who answers lots of questions. It's happened to me in the past - I'm afraid there are some idiots out there. Don't let it get you down! Every upvote is someone you've helped understand physics better!
 
Secret
Secret
[Worldbuilding math idea]: Scale dependent derivative. Consider the position of a classical particle $r(t)=r(0)+@(t)$. Now let the time derivative of $@(t)$ be $\dot{@}(t)(h)=\frac{@(t+h)-@(t)}{h}$. This derivative has an usual property in that if $h < k$ where $k > 0$, then $\dot{@}(t)(h)=0$. Therefore the equation of motion of this particle become dependent on how accurately you measure its velocity.
One can then further generalised this by having $h=f(h)$ for the "scale derivative" to vanish, and f(h) can be a solution to some equations
To be checked: see if the usual laws of physics still holds in such framework
Typo: unusual instead of usual
Suppose all works out, then we will have an interesting world where there is a continuous map that relates a continuum of trajectories of that particle at given space and time, where the parameter of such map will be some scale such as time or length window
 
Bass
Bass
7:41 AM
@ACuriousMind I see what you mean. Tell me if I'm nitpicking, but in my opinion, you can be wondering about a scientific question even if you do have a plausible answer. After all, that answer might eventually turn out to be incomplete, just an approximation, or wrong (not that I think the answer at hand is wrong). I have read OP's sentence as "it's a good question why the dinosaurs...", and by linking the biology.SE question, I wanted to show that this is indeed a nontrivial question.
 
Secret
Secret
One can then formally replace "particle" with "state vectors", adding corresponding noncommutative relations to generalise it to the quantum case
 
Secret
Secret
8:09 AM
NB there is a mistake in my above equations, I will fix those later
 
John Rennie
John Rennie
8:25 AM
@KaumudiHarikumar: hi :-)
 
user228700
@JohnRennie: Hello :) Firstly, sir, I should tell you that I'm not familiar with the math involved in the probability distribution curve.
 
John Rennie
John Rennie
Everyone finds probability density functions hard to understand at first.
But they are used in lots of areas of physics so you'll get used to them eventually :-)
 
user228700
Okay, that's comforting :P My major doubt comes from this whole area being unity thing...
 
John Rennie
John Rennie
If we go back to the histogram I mentioned in my answer, are you happy with my explanation of what that means?
i.e. each column is the fraction of molecules with velocities in some range.
 
user228700
I'm so sorry but I still don't get that.
 
John Rennie
John Rennie
8:31 AM
Well suppose we have 1000 molecules, all with different velocities.
 
user228700
Okay, starting from the basics, can you tell what it is that we're plotting on the y-axis?
 
John Rennie
John Rennie
The $y$ axis is the number of molecules in each velocty range dividied by the total number of molecules.
Let's try and take a concrete example.
Suppose the velocities are all the range 0 to 100 m/s
 
user228700
So...the probability of finding a given molecule in that range?
 
John Rennie
John Rennie
Yes
 
user228700
Okay, go on sir...
 
John Rennie
John Rennie
8:33 AM
Suppose my first column is for velocities 0 to 10 m/sec
Suppose I count the number of molecules with velocities in the range, and get the number 15. So 15 out of the thousand molecules have velocities in the range 0 to 10 m/sec.
Then the height of the column, the $y$ value, is 15/1000 = 0.015.
So far so good?
 
user228700
Yes, OK...
 
John Rennie
John Rennie
The net column would be velocities in the range 10 to 20 m/sec, the next column would be 20 to 30 m/sec and so on giving me ten columns covering the while velocity range of 0 to 100 m/sec.
And all the 1000 molecules I started with are counted in one of these ten columns.
 
user228700
Yup, OK...
 
John Rennie
John Rennie
For each column the $y$ value is the number of molecules in that column divided by the total number of molecules, so if I add together all ten $y$ values I will get the number one.
Still OK?
 
user228700
Yeah, that's exactly where the confusion comes from!
 
user228700
8:40 AM
That's why I kept asking you about just adding $y$ values for all the points on the curve to get unity...
 
John Rennie
John Rennie
OK. The thing is that those $y$ values depend on the width of the column. In the histogram I've described I have ten columns so I add up ten $y$ values.
 
user228700
OH...
 
John Rennie
John Rennie
Suppose I narrow my velocity range to 0-5, 5-10, 10-15, and so one, so each column is now half as wide as they were.
In my example i said that 15 out of the 1000 molecules were in the 0-10 column, so if I split this into two I'm going to get for example 5 in the range 0-5 and 10 in the range 5-10 m/sec.
 
user228700
OK...
 
John Rennie
John Rennie
So the $y$ value for my new first column would be 5/1000 = 0.005 and the second column would be 10/1000 = 0.01. The $y$ values for my new columns are smaller than the original ones.
But if I add up the 20 $y$ values in my new histogram they still all add up to one.
OK so far?
 
user228700
8:46 AM
Yup :)
 
John Rennie
John Rennie
If I keep making the columns narrower and narrower then I'll have lots more $y$ values but because each $y$ value is smaller they still all add up to one.
 
user228700
OK...
 
John Rennie
John Rennie
To get to the continuous curves like the one you show in your question we, in principle, make the columns infinitely thin so we have an infinite number of them. And adding up an infinite number of infinitely thin columns is exactly what we mean by integration.
 
user228700
Yes, OK...
 
John Rennie
John Rennie
So when we integrate the curves we are in effect adding up lots of infinitely thin columns and we are going to get the number one.
That's why Resnick and Halliday say that the integral has the value one.
If you're OK with this so far I need to explain one more thing that I've kind of glossed over so far ...
 
user228700
8:52 AM
So, essentially, when we integrate it, we're not multiplying with the $dv$ or anything. We're just integrating the function, correct..?
 
John Rennie
John Rennie
This is the thing I glossed over ...
 
user228700
Okay..?
 
John Rennie
John Rennie
The way I've defined my $y$ values they are the number of molecules in each column. But if we reduce the column width to zero there aren't any molecules in each column because the columns are too thin. So all my $y$ values will be zero - and that isn't much use!
 
user228700
Yeah! Please go on...
 
user228700
Wait, you're doing this...why?
 
John Rennie
John Rennie
8:56 AM
No hang on, I got that wrong, let me try again.
 
user228700
OK, sure :)
 
John Rennie
John Rennie
OK, I'm going to define a new value P that is given by:
P = y/dv
That is, it is my y values divided by my column width dv.
 
user228700
Riight. Why?
 
John Rennie
John Rennie
I said above that as I make the columns thinner my values of $y$ are going to reduce. But as I make the columns thinner the column width dv also reduces.
 
user228700
So this value remains constant..?
 
John Rennie
John Rennie
8:59 AM
So in the fraction y/dv both the y and the dx reduce at the same rate and the ratio y/dx remains well defined.
@KaumudiHarikumar yes, the ratio y/dv remains constant.
 
user228700
Okay...
 
John Rennie
John Rennie
And that ratio, for any value of v, is how we define P(v)
 
user228700
OH! P(v) is THAT ratio?!
 
user228700
Oh God, I didn't know...
 
John Rennie
John Rennie
Yes. So now you should be able to see why we multiply P(v) by dv to get P(v)dv
Since I defined P as y/dv that means Pdv = y
And when we add up all the y's we get one, so that means when we add up all the P(v)dv's we also get one.
 
user228700
9:01 AM
Yes, I see that. Thanks so much! This was just less intuitive than say, integrating the velocity function to obtain the displacement or something like that...
 
John Rennie
John Rennie
And that's why when we integrate P(v)dv we are going to get the value one.
 
user228700
Less intuitive...and I didn't know how P(v) was defined.
 
user228700
Yes sir, I completely understand this now.
 
John Rennie
John Rennie
It is a bit unintuitive, and that's part of the reason it's so hard to grasp when you first encounter probability density functions.
Have another look at those curves you show in your question, and look closely at the units for the y axis.
 
user228700
Yeah, s/m. Makes total sense :D
 
John Rennie
John Rennie
9:05 AM
The units for the P(v) axis are shown as s/m, but really they are 1/(m/s).
i.e. they are a number (my $y$ number) divided by dv that has the units of speed.
That's why the units are 1/speed.
 
user228700
I see that now :) Thanks so so much sir. I've been trying to get my head around this for quite some time now and today, I think I finally did.
 
John Rennie
John Rennie
Like I say, this puzzles everyone when they first meet it :-)
It puzzled me as well about 40 years ago when I was a student :-)
 
user228700
OK, that's definitely comforting. Thanks, again! You just saved me a lot of hours.
 
John Rennie
John Rennie
When you learn quantum mechanics you'll find that the wavefunction squared is also a probability density function.
 
David Z
David Z
BTW now that you know, don't forget the difference between "probability" and "probability density"! That's a particularly confusing thing, especially because they're often denoted with the same letter.
 
user228700
9:09 AM
Oh God, yes, I was learning a bit of that and it didn't make sense but now it does!
 
John Rennie
John Rennie
Congratulations, you're now officially 40 years ahead of me when I was your age :-)
And now I'm going off to make a cup of coffee to celebrate.
 
user228700
All thanks to you! :D
 
user228700
Okay, enjoy your coffee sir! It's been great.
 
Secret
Secret
10:59 AM
0
Q: The Fate of the World Lines of Individuals According to String Theory

LijisheAccording to string theory, what happens to the finite world lines of individuals after they have ended and the Universe has entered (as it likely will) heat death? Is it correct to think of them as eternally existing at various 'chunks' of space-time in the past?

string-theory
$*$Calling for String theorists$*$
 
John Duffield
John Duffield
11:53 AM
@knzhou : can we have some examples please?
@Secret : world lines do not actually exist. They are abstract things.
 
John Duffield
John Duffield
12:16 PM
All: I've answered another question.
0
A: According to special relativity, has the future already happened?

John Duffield According to special relativity, a photon traveling at the speed of light, does not “experience” time. Nor would you if you were travelling at the speed of light. (Whilst you can't actually travel at the speed of light, you can in theory travel so close to the speed of light that we couldn'...

Anyway, here's why gravity is so weak.
The Coulomb force is really strong. Really, really strong:
 
bolbteppa
bolbteppa
@JohnDuffield can you calculate the ratio of the gravitational force to the Coulomb force for two electrons and tell exactly how much stronger one is to the other?
Hint, 10^39 appears somehow
 
John Duffield
John Duffield
@bolbteppa : no, I can't calculate it. But I can measure it.
 
bolbteppa
bolbteppa
Do you deny the equation F_gr = Gmm/r^2 (Newton's law of gravitation) or F_el = kee/r^2 (Coulomb's Law)?
 
John Duffield
John Duffield
The picture above is from Rod Nave's excellent hyperphysics website.
 
Secret
Secret
@JohnDuffield Reading the answers, from the block universe, to relativity, to uncertainty principle, to the nature of light, to relative motion. So cannot really test the existence of the future, only knew that it is somewhat random...
 
bolbteppa
bolbteppa
12:30 PM
The ratio F_e/F_g will give some number, say C, so that F_e/F_g = C implies that F_e = C F_g, that is, F_e is C multiples of F_g, going to be a huge number, so F_e is C times what we'd measure for F_g, of the order 10^39, what is that, 1,000 trillion trillion trillion or something, what is C Exactly?
 
John Duffield
John Duffield
Next, take a look at this article from 1929 where Einstein describes a field as a state of space: "The two types of field are causally linked in this theory, but still not fused to an identity. It can, however, scarcely be imagined that empty space has conditions or states of two essentially different kinds". He's talking about electromagnetism and gravity.
Now ask yourself this: How many states of space are there where an electron is?
Anybody?
One.
 
0celo7
0celo7
wtf are you talking about
No one has ever heard of this @JohnDuffield
guess why
Einstein was a crackpot
 
bolbteppa
bolbteppa
Rudin is literally the best analysis book ever, man he is so concise it's insane revising the things you skip over
Give me a set in $\mathbb{R}$ with exactly 3 limit points
 
John Duffield
John Duffield
OK, when you contrive your charged particles as the current-in-the-wire, the Coulomb forces nearly balance, but not quite. So there's a "residual" or trace force which is relatively weak. Again see Rod Nave's most excellent hyperphysics:
 
bolbteppa
bolbteppa
I can't tell if that's your answer to what $C$ is or not
 
Secret
Secret
12:38 PM
@bolbteppa (a,b) U (b,c) is this cheating?
where a,b,c are real numbers
 
bolbteppa
bolbteppa
but every element of $(a,b)$ is a limit point
 
Secret
Secret
oops
 
bolbteppa
bolbteppa
Yeah, it's cool, bit tricky to see
 
John Duffield
John Duffield
We call this residual or trace force magnetism, and we say a magnetic field is present. That's reasonable in that it fits with what Einstein was saying. The space around the current-in-the-wire has a particular state.
 
bolbteppa
bolbteppa
Try find a set with one limit point first, and remember that no finite set can contain any limit points, but what is the next best thing
Ah, the trace force, I'll just take the trace of a matrix and get magnetism so
 
John Duffield
John Duffield
12:41 PM
Remember, the magnetic force between the two wires is weak because all the very-strong Coulomb forces nearly cancel, but not quite.
 
Secret
Secret
I am thinking about the real number line with the origin dug out, but then any other points in the (0,infiniy) and (-infinity,0) are still limit points...
 
bolbteppa
bolbteppa
Yeah, so the problem with that example is that every element of the intervals is a limit point, so can you weaken things to stop that from happening
 
Secret
Secret
@bolbteppa How about the singleton {0}?
 
bolbteppa
bolbteppa
It's a finite set, no limit points
 
John Duffield
John Duffield
So, how do you make it to the next level? What do you do to make those inter-charged-particle Coulomb forces cancel each other out even more?
 
bolbteppa
bolbteppa
12:45 PM
Your examples are close
 
Secret
Secret
What happens if I use the set of integers ,which is infinite?
(intuitively every integer is like a singleton, which is finite, but the whole set of integer is infinite?)
 
John Duffield
John Duffield
You just switch off the current. That changes the state of space around the wires. The magnetic force is gone. But there is still a force pulling the two wire together. It's very very weak, because the Coulomb forces now almost cancel, but not quite. And the point to note is this: we don't call it magnetism any more. We call it gravity.
 
bolbteppa
bolbteppa
Well - you are using open sets on the real line, which are open intervals, and you want a subset which has only one limit point, yet every neighborhood of your limit point has to intersect this subset and still contain infinitely many points of that subset
What is an infinite set such that none of it's points are limit points, but it converges to a unique point
 
John Duffield
John Duffield
@SpaceOtter : see above, from here.
 
0celo7
0celo7
@bolbteppa $\{a\}\cup\{b\}\cup\{c\}$.
 
bolbteppa
bolbteppa
12:55 PM
That's a finite set, needs to be infinite
 
0celo7
0celo7
You didn't say that.
 
Secret
Secret
How about the set $\frac{1}{n^2}$ where integers $n>0$ Then each value for each n is a singleton, hence no limit points, but the set as a whole converge to 0?
 
bolbteppa
bolbteppa
I said it
 
0celo7
0celo7
I'd say union 3 Cauchy sequences with different limits.
 
bolbteppa
bolbteppa
Yeah that works
Or $1/n$
So you have an infinite set converging to the unique limit point $0$ where no other point is a limit point
 
0celo7
0celo7
12:56 PM
@bolbteppa What?
 
bolbteppa
bolbteppa
@JohnDuffield if you gave that as an answer in any college you would get 0 marks, you have to write down a numerical answer, just fill in the values for your diagram into F_c = kee/r^2 and F_g = Gmm/r^2 and then compare them by calculating F_c/F_g = C so that F_c = CF_g, all you need are the mass of an electron m, and some other stuff
 
John Duffield
John Duffield
@bolbteppa : see hyperphysics re the relative strength of the forces. I'm not denying $F=G{\frac {m_{1}m_{2}}{r^{2}}}\$ or $F=k_{e}{\frac {q_{1}q_{2}}{r^{2}}}$ at all.
 
bolbteppa
bolbteppa
$\{m \in \mathbb{R} | m = \frac{1}{n}, n \in \mathbb{N} \}$ is an infinite set with a single limit point of $0$.
 
0celo7
0celo7
@bolbteppa You're babbling, again.
 
Secret
Secret
ok that is easy to visualise
So for your original question, basically using 0celo7 answer, we just find 3 lots of such set (each with different limit points) and union them to get the required set of 3 limit points?
 
bolbteppa
bolbteppa
12:59 PM
@0celo7 I said the set needs to be infinite like 3 times, where is the babbling
Yeah, so write down an example of a set, in set-builder notation like I just did, of a set with exactly 3 limit points
 
Physics Guy
Physics Guy
@Ocelo7 How's your Ricci-Flow work going on ?
 
bolbteppa
bolbteppa
@JohnDuffield I don't see any calculations on that page, give it a shot trying to calculate it yourself, i.e. do real physics without quoting 15'th century papers as argument from authority
You should get an answer around $10^{39}$
 
Secret
Secret
@bolbteppa Let $S_i=\{m \in \mathbb{R}\lvert m=\frac{1}{(n)^2}+i,n\in {\mathbb{N}}\}$. Then a required set is e.g. $\{S_1\}\cup\{S_0\}\cup\{S_{-1}\}$
 
Jon McClung
Jon McClung
Would this question be on topic for physics.stackexchange? "Yesterday I tried on my friend's glasses and noticed that they made everything look blurrier and farther away. If they do this to my vision, how can they make things look clearer and closer for him?"
 
Secret
Secret
ok just fixed it, how's the new set?
 
John Duffield
John Duffield
1:05 PM
@bolbteppa : calculations don't tell you the relative strength of the forces. You have to measure the forces. That's real physics. Only then do you come up with a model and an equation that allows you to calculate the forces.
 
bolbteppa
bolbteppa
Yeah
That is right
Tricky but cool one haha
 
Secret
Secret
But it feels so cheating, are we supposed to have a nicely named set that has such properties?
 
bolbteppa
bolbteppa
@JohnDuffield Oh god, you can't do the basic algebra can you, just give it a shot, stop rationalizing it away, go for it, I gave you all you need to do it, measurements are a different question
3
 
John Duffield
John Duffield
@bolbteppa : LOL, of course I can do the basic algebra. But calculations don't explain why gravity is so weak, now do they?
 
ACuriousMind
ACuriousMind
@JonMcClung Yes, I'd say that is on-topic.
 
bolbteppa
bolbteppa
1:08 PM
I think the problem is just to make sure you understand what a limit point is, and he knows you'll get tripped up writing intervals straight away and stuff
 
Jon McClung
Jon McClung
@ACuriousMind, thanks
 
Secret
Secret
hmm, makes sense
 
bolbteppa
bolbteppa
I know Euclid has pons asinorum on page 5, but page 2 of Rudin is something else, understanding where his $q = p - \frac{p^2-2}{p+2}$ comes from when analyzing $\sqrt{2}$
 
John Duffield
John Duffield
@bolbteppa : are you a mathematician?
 
bolbteppa
bolbteppa
No comment
 
John Duffield
John Duffield
1:14 PM
@bolbteppa : only a mathematician would think you could explain the relative strength of the Coulomb force and the force of gravity by referring to a couple of equations.
 
bolbteppa
bolbteppa
:p
 
2physics
2physics
hello everybody
 
bolbteppa
bolbteppa
@JohnDuffield guess what, you will never measure any of those quantities accurately
But you know the answer from the algebra you don't want to do
 
John Duffield
John Duffield
@bolbteppa : Well I do. I'm always referring to hard scientific evidence. And people like you have scant regard for it.
@bolbteppa : a certain Charles-Augustin de Coulomb measured the force between charged objects using a torsion balance.
How old are you?
:32183727 : hi. We were talking about why gravity was so weak. See my comments starting from here. But I have to go I'm afraid. Bye.
 
bolbteppa
bolbteppa
Sorry order of magnitude of 42, ahh
Maybe this is why they say it's the meaning of life
In fact, since Einstein only believed in the gravitational and electromagnetic field, and as your quote from Einstein says there is only one true field in empty space, therefore 42 is the meaning of life - logic without math...
 
2physics
2physics
1:28 PM
@JohnDuffield how much is the amount of one sec's worth of charge from a 120 watt bulb?
 
yuggib
yuggib
@0celo7 did you already cover ordinal arithmetics in Jech?
@bolbteppa It's not the meaning of anything. It's just the answer.
 
0celo7
0celo7
@yuggib I returned the book to the library on the way to class this morning.
 
yuggib
yuggib
@0celo7 pavido
 
0celo7
0celo7
I don't speak French.
 
yuggib
yuggib
@0celo7 it's not french
 
0celo7
0celo7
1:36 PM
@yuggib For $f:\Bbb R\to\Bbb R^n$, do you believe $|\int f|\le \int|f|$?
 
yuggib
yuggib
and if you omit a letter, it's also english
 
0celo7
0celo7
@yuggib ...what
avido, pvido, paido, pavdo, pavio, pavid
none of those are real words.
 
yuggib
yuggib
yes, one is
 
bolbteppa
bolbteppa
@yuggib yeah calculating how many times stronger the coulomb force is than the gravitational force is such meaningless ivory tower empty vessel stuff, it's stronger because of flashing lightbulbs in a children's picture
 
2physics
2physics
@JohnDuffield oh I got it ..
 
Secret
Secret
1:38 PM
avido
Adjective: avido m ‎(feminine singular avida, masculine plural avidi, feminine plural avide)
  1. greedy
  2. eager, avid
  3. avidō
  4. dative masculine singular of avidus
  5. dative neuter singular of avidus
(2 more not shown…)
 
yuggib
yuggib
that is, one in english
more if you include other languages
 
0celo7
0celo7
Pavid is also a word, apparently.
 
yuggib
yuggib
@Secret that is not english...
 
Secret
Secret
ooops, should get back to write my research proposal, slacking off too much
 
yuggib
yuggib
@bolbteppa not exactly; 42 is just the answer to the ultimate question about the life, the universe, and everything
 
bolbteppa
bolbteppa
1:39 PM
Ah gotcha, I agree
 
2physics
2physics
With one mobile electron per atom, and with the electron charge of 1.6 x 10-19 Coulombs, this means there are about 13,600 Coulombs of potentially mobile charge in one cm3 of copper. :O
 
bolbteppa
bolbteppa
 
2physics
2physics
2:02 PM
@JohnRennie ok thanks
 
Secret
Secret
I found chat flooding is very effective whenever there are popcorn events and flame wars in the chat
For example, suppose there is a flame war. Choose a very bulky mathjax and paste it (which also double as a question). This split the chat so much that the flame comments get diluted too much and eventually the flame wll stop
(NB the flame war took place in anther chatroom, not here. As far I knew, there are no major flame wars in this chat)
 
John Rennie
John Rennie
2:26 PM
@2physics it's interesting to see how Einstein moved from a realisation that gravity and special relativity were incompatible to developing GR. This is covered in various histories of relativity. If you're interested I can look for a reference.
 
Secret
Secret
https://en.wikipedia.org/wiki/Uniqueness_theorem_for_Poisson%27s_equation
@johnrennie In the uniqueness theorem proof, how do we ensure the $\rho_f$ term for the $\psi_1$ and $\psi_2$ to be equal, is it because they are both solutions to the poisson equation for the same system thus there cannot be two different $\rho_f$ one for each $\psi$?
(sorry for the typos...)
 
John Rennie
John Rennie
@Secret It's been many years since I did this sort of thing and I forgot all the details long ago. If no-one else can help I might be able to skim enough of that article to remind myself how it all works, but I'm afraid right now I don't have the time available. Sorry :-(
 
Secret
Secret
ok
 
bolbteppa
bolbteppa
2:44 PM
@Secret the point is to solve $L(y) = f$ given a known $f$, you are given the density/distribution (of electric charge) $\rho_f$ and you want to find the potential $\psi$ generating such a density. It's like solving $y' + a(x)y = e^x = f(x)$ where the inhomogeneous term $f(x) = e^x$ is known.
 
Secret
Secret
ok that makes sense, thanks
 
John Duffield
John Duffield
3:30 PM
@bolbteppa : the point is that there's only one state of space where the electron is, so its electromagnetic field is not something distinct from its gravitational field.
 
John Duffield
John Duffield
3:55 PM
Ah, I see the serial downvoters are at work again.
 
0celo7
0celo7
@JohnDuffield I have not downvoted you today.
 
John Duffield
John Duffield
Good.
It's like what I was saying yesterday. Science is not a democracy. All the upvotes in the world won't make a wrong answer right, and all the downvotes in the world won't make a right answer wrong.
Sometimes the huge number of initial upvotes one sees on a wrong answer makes me wonder if some people have made a lot of friends on physics stack exchange.
If you catch my drift...
LOL!
@bolbteppa : noted:
 
David Z
David Z
4:12 PM
@JohnDuffield In the future, if you suspect serial voting and it doesn't get reversed by the automatic script within 24 hours, bring it to the attention of the mods privately. Or if you don't want to do that, you can try your luck notifying the SE team directly. No need to mention it in public.
And same goes for any time you're not satisfied with the voting on a post, whether it's suspected serial downvoting, or upvoting on an answer you believe is wrong, or anything like that. Your options are to comment on the answer pointing out why you think it's wrong, post your own answer, and/or link to it in chat along with an explanation of why you think it's wrong to ask people to go vote on it as they see fit.
But complaining about undesirable voting patterns in general is pointless.
 
John Rennie
John Rennie
I think I've discovered something interesting and unexpected about the Reissner Nordstrom metric ...
The RN metric is: $$ds^2=-\bigl(1-\frac{r_s}{r}+\frac{r_q^2}{r^2}\bigr)dt^2+\bigl(1-\frac{r_s}{r}+‌​\frac{r_q^2}{r^2}\bigr)^{-1}dr^2+r^2d\Omega^2$$
 
0celo7
0celo7
@JohnRennie Dammit
I was about to go to my QM lecture
now I have to sit and listen to you
 
John Rennie
John Rennie
So if: $$\frac{r_s}{r}=\frac{r_q^2}{r^2}$$ then the terms in $r$ and $r^2$ cancel and we just get a Minkowski metric.
 
0celo7
0celo7
What?
$r$ is a variable.
The other things are constant.
How are you gonna have that be equal?
 
John Rennie
John Rennie
4:27 PM
Yes, but at that particular value of r everything cancels.
 
0celo7
0celo7
Ok, but the Riemann tensor is probably nonzero.
 
John Rennie
John Rennie
i.e. at $r = \frac{r_Q^2}{r_s}$
 
0celo7
0celo7
You still have tidal forces.
Sigh, I'll just take my laptop with me. Brb gotta walk to the physics building.
 
John Rennie
John Rennie
Well, I calculated the point where the norm of the four-acceleration for a stationary observer is zero, and it isn't at $r = \frac{r_Q^2}{r_s}$.
Unless I've done the sums wrong, which is entirely possible, the point of zero acceleration is at $r = \frac{2r_Q^2}{r_s}$ i.e. twice the radius where the $r_s$ and $r_Q$ terms cancel.
This is probably a meaningless result, but I found it unexpected.
 
Secret
Secret
But what does this r physically mean, the ratio of the square of the characteristic length against the schwartzchild radius, so it is somewhere outside the event horizon?
Also if I have a point where the 4-acceleration is zero, is it the same as saying that there the riemann curvature tensor vanishes?
 
0celo7
0celo7
4:32 PM
@JohnRennie What is special about $2r^2_Q/r_s$?
 
John Rennie
John Rennie
@Secret I don't think so. having the four-acceleration vanish basically just means that the Christoffel symbol $\Gamma^r_{tt}$ is zero.
 
0celo7
0celo7
@Secret No.
 
John Rennie
John Rennie
@0celo7 I don't know. It's just the point where $\Gamma^r_{tt}=0$
 
Secret
Secret
ok then that is not enough to say spacetime is flat there
 
0celo7
0celo7
@JohnRennie So...what is your result?
@Secret Spacetime is flat in some open set iff the Riemann tensor vanishes in the open set.
 
John Rennie
John Rennie
4:34 PM
My result is that $\Gamma^r_{tt}=0$ at $r=\frac{2r_Q^2}{r_s}$
 
0celo7
0celo7
@JohnRennie ...ok?
What is $\Gamma^r_{tt}$
 
Secret
Secret
0celo7: No, I am talking about vanishing 4-acceleration does not imply riemann curvature tensor vanishes hence spacetime is flat there
 
John Rennie
John Rennie
@0celo7 if the particle is stationary then only $U^t$ is non-zero, so the only contribution to the four-acceleration is the term $\Gamma^r_{tt}U^tU^2$.
 
0celo7
0celo7
@JohnRennie No, what is the full expression you got for it?
 
John Rennie
John Rennie
$((-2 r_Q^2 + r_s r) (r_Q^2 + r (-r_s + r)))/(2 r^5)$ is the raw mathematica output.
Prettifying this gives: $$\frac{1}{2r}(\frac{r_s}{r}-\frac{2r_Q^2}{r^2})(1-\frac{r_s}{r}+\frac{r_Q^2}{r^‌​2})$$
 
0celo7
0celo7
4:39 PM
yeah
 
John Rennie
John Rennie
It's that factor of $2$ that makes the difference.
 
0celo7
0celo7
Gtg for real now
 
John Rennie
John Rennie
Maybe I should calculate the Christoffel symbol by hand and see where it comes from.
 
Secret
Secret
For me I am more interested on whether the riemann curvature tensor vanishes at that speical radius, as that might remind me of saddle points or something related. However since I have just finsihed my research roposal draft and sent to my supervisor, my brain is too crumbled to calculate 4th order tensors right now
 
0celo7
0celo7
@JohnRennie What is the Kretschmann for Reissner-Nordstrom?
 
John Rennie
John Rennie
4:48 PM
@0celo7 Don't know. I'm using Danu's notebook for calculating the Christoffel symbols, and it doesn't calculate the Kretschmann scalar. But I would lay odds that it isn't zero there.
I'm sure there is absolutely no significance to my result - it just surprised me.
 
Secret
Secret
well, unless the riemann curvature tensor does vanish at that r, then it is an interesting saddle point like structure
probablly it means the electrostatic (soemthing) counteracts the gravitational well or something
 
0celo7
0celo7
@JohnRennie Thats what I'm confused by.
Why?
@Secret what saddle point are you going on about
 
Secret
Secret
well, suppose the riemann curvature tensor does vanish there, but it does not for some r slightly away from that special radius, then isn't that analogous to the case of having a stationary point on a surface?
 
0celo7
0celo7
What do you mean by a stationary point
 
Secret
Secret
If the curvature tensor vanishes there, but not elsewhere, it means you have a spherical region where spacetime is flat there, but curves at other locations
so that kinda reminds of stationary points on saddles
 
0celo7
0celo7
4:55 PM
Uh, curvature of submanifolds doesn't work like that.
Are you saying that spherical region is flat?
 
Secret
Secret
yes, that spherical shell
unless the curvature tensor have a lot more things that are missed out by my low dimensional intuiton, as you might be poitning out now
 
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