Mathematics - 2016-07-26 (page 2 of 4)

I don't get what Zee is trying to get at p.31 @0celo7 with a mass pulled by a spring, $F(x)$ does not make
sense, only $F(x(t))$ does. The force exerted by the spring does not pervade all of space, and
hence is defined only at the position of the particle $x(t)$, not at any old $x$.

0celo7

the solution says to note it must be proportional to $\delta^{ij}$

I don't understand that

Semiclassical

4:28 PM

well, suppose you pick $i\neq j$. then the RHS is declaring that the integral vanishes

0celo7

And I'd prefer a direct argument

@Semiclassical We need to prove the RHS is equal to the LHS.

The LHS is "given"

Obliv

couldn't a mass on a spring be a function of only position? Time needn't be involved imo

Semiclassical

on symmetry grounds, the point is just to consider what happens if you do a coordinate change which swaps $i$ and $j$

0celo7

what

@Obliv uh

Obliv

I have encountered plenty of students who confuse these two basic concepts:
spatial coordinates and the location of particles. I may sound awfully pedantic, but when we
get to curved spacetime, it is often important to be clear that certain quantities are defined
only on so-called geodesic curves, while others are defined everywhere in spacetime.

Semiclassical

4:31 PM

Trying to remember the details, tbh.

0celo7

oh, you're only defining the force along the path of the particle

on on all of space

the path is $x(t)$

Obliv

what the heck is the difference between the location of particles and spatial coordinates

deostroll

Hi. Just noticed something about divergent series...

Semiclassical

@0celo7 Here's a brute-force approach: Suppose you can prove that the LHS vanishes when $i=x$ and $j=y$

0celo7

$\Bbb R^2$ vs. the curve the particle travels on

Semiclassical

4:33 PM

That's a simple direct calculation.

0celo7

...how

deostroll

If you subtract the series by itself...you can actually have that sum equal any real number...

0celo7

$\int p^xp^y\,\mathrm d\Omega=?$

How does that even work

Semiclassical

well, I'm figuring $p^x$ here is just the $x$-coordinate of $\vec{p}$?

0celo7

Yeah, so?

Is that a number, a function?

Semiclassical

4:35 PM

in that case, you can write it in spherical coordinates

it's parametrized by spherical coordinates. how else would it show up in an integrand?

deostroll

or is it just something trivial... :(

Vrouvrou

Hello, please if $F(t)>0,\forall t$ can we say tha $\int_{\mathbb{R}^N}F(u(s)) ds>0$ ?

Semiclassical

and in spherical coordinates, it should just be $\cos\phi\sin\theta$ (if I'm not mixing up my conventions)

0celo7

@Vrouvrou Yes

Vrouvrou

thank you

Semiclassical

4:36 PM

similarly, $\vec{p}^y$ would just be $\sin\phi \sin\theta$

so you'd just plug those expressions into the integrand and show that the resulting integral vanishes.

0celo7

I see

What the heck is $\vec p$ again

I'm just not sure what "averaging over direction" means

Obliv

@deostroll If you subtract $a$ from $a$ you get nothing, by definition, no? How could subtracting a divergent series from itself give you a real number $\ne 0$

deostroll

@Obliv its not a point-to-point or term-to-term subtraction...

The general process is, you add until the partial sum is greater than target, if it over shoots, you take one minus term, if this falls below the target, continue adding the positive terms, and likewise alternate...

Obliv

@deostroll if you mean to say $[\sum_{i=0}^{\infty} i] - [\sum_{i+4}^{\infty}i] = 1 + 2 + 3 + 4$ I think that is trivial.

deostroll

oh wait...not that divergent series...

harmonic series...or ones similar to that...

Semiclassical

4:44 PM

@0celo7 the best example in physics I know off the top of my head would be if you were to give a microscopic description of a gas

deostroll

@Obliv that is not quite the process I previously mentioned...

Semiclassical

In that case, you could consider questions like: If I repeatedly select a particle and measure its momentum $\vec{p}$, what is the expected value of $p_x$?

deostroll

YOu've to keep alternating between the positive and negative terms so that the partial sum is near about the target value...

Semiclassical

In the case of an ideal gas, you get the so-called Boltzmann distribution which is, crucially, isotropic in space. That is: While certain speeds of particles will appear more frequently than others, there's no direction in space that the particles prefer to travel.

In that case, a question like 'what's the expected x-component of momentum' should simply be zero because the particle is as likely to be moving left as right

Or, more symbolically, $$\langle \vec{p}^x \rangle_\Omega =\frac{1}{4\pi}\int d\Omega\,\vec{p}^x = 0$$ where the average is taken over all possible directions of $\vec{p}$

robjohn

$$
\begin{align}
\sum_{k=1}^\infty\frac1k-2\sum_{k=1}^\infty\frac1{2k}
&=1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots\\
&\phantom{=1\,\,}-\frac22\phantom{+\frac13\,\,}-\frac24\phantom{+\frac15\,\,\,}-\frac26-\cdots\\
&=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots\\[12pt]
&=\log(2)
\end{align}
$$

The problem is in the rearrangement of the terms. In general, if the series is not absolutely convergent, we must be very careful about rearranging the terms of the series.

Semiclassical

4:53 PM

the 1/4\pi is to ensure $\langle 1 \rangle_\Omega = 1$.

But one might be interested in things like $\langle \vec{p}^x \vec{p}^y\rangle_\Omega$, and part of what you're trying to show is that this in general vanishes as well

and now I'm suddenly having flashbacks to scattering cross sections. ew

Mary Star

Hello!!

Balarka Sen

Hi @Danu

Obliv

@0celo7 hey.. uh how do you take this integral $\int_{-\infty}^{\infty} \delta(x)f(x)dx$? I know the delta function is $0$ for all $x \ne 0$ and it approaches $\infty$ at $x = 0$ but I don't really get how to take this integral. hint?

fixed

Huy

@Obliv: it's f(0)

Semiclassical

@obliv What you've cited is a description of the delta function, not the definition

Obliv

5:00 PM

yeah @huy I'm trying to show that.

Semiclassical

Do you know the definition?

Obliv

@semiC I don't know the definition. I was given a description after all

Semiclassical

Then I have no idea how you're supposed to solve the problem.

Huy

^

Semiclassical

Do you have a problem statement in mind?

Balarka Sen

5:01 PM

Again, this goes back to the earlier discussion of trying to prove something about an undefined object. Unless you don't know the definition, it's hopeless to do mathematics.

Obliv

Show that for some suitably smooth function $f(x)$, the integral (mentioned above) is $f(0)$

Huy

@BalarkaSen: but one can do physics, in that case. :P

Balarka Sen

(FWIW, I don't know delta function business)

Obliv

@balarka It's not a math book.

Huy

of course it's not :D

Semiclassical

5:02 PM

Part of the problem is that the description you cited is simply not enough to characterize the delta-function

For one, it equally well describes $\delta(x)$ and $2\delta(x)$.

Balarka Sen

@Obliv Makes sense, all in all.

Obliv

after ~3 months learning algebra in a math book I can safely distinguish a math book from a physics book

deostroll

@robjohn Yes, what I am saying that is by careful rearrangement you can get any rational number...

Mary Star

I am looking at an exercise about the vertex cover. We are given the graph $G=(V,E)$ with $V=[100]$ and $E=\{(i, i+1)\mid i=1, \dots , 99\}$.
Before I use the approximation algorithm, I have to give the minimal vertex cover $C^\star$ of $G$ and the length of $C^\star$.
How do we find the minimal vertex cover without the approximation algorithm? The vertex cover has to contain at least one vertex of each edge, right? At the given graph, each vertex is connected with the consecutive one, so do we maybe take for example the first one, then the third one, etc?

Semiclassical

For another, one can find a sequence of functions which converge pointwise to zero away from zero and to infinity at zero, but for which the corresponding integral would fail to converge.

Balarka Sen

5:04 PM

I thought an essentially part of $\delta$ was that it integrated to $1$ over the real line, whatever that means?

robjohn

@deostroll yes, I was just giving a concrete example. Actually, you can get any number, rational or irrational.

deostroll

$\sum\limits_{k=1}^\infty \frac{1}{k}- \sum\limits_{k=1}^\infty \frac{1}{k}= c$

Semiclassical

@BalarkaSen Yeah. That's not in the "zero if x!=0, infinity at zero" description, though

Huy

@BalarkaSen: that means that the area enclosed by the graph of the function and the $x$-axis equals 1 :P

Balarka Sen

@Huy That doesn't make sense in this particular context without more formalism.

Huy

5:06 PM

@BalarkaSen: I know, that's the joke.

Balarka Sen

Oh. Ok.

Semiclassical

@Obliv Our point in all of this is that if one doesn't have some kind've definition here, even if a purely operational one (i.e. a delta-function is an object with X properties) then the problem is simply not solvable.

And the description you gave doesn't rise to that.

0celo7

@Semiclassical sorry, was getting lunch

@Obliv easy

Semiclassical

So unless there's some kind of dfn in your book (or if it refers to another book for the definition) then that problem is pretty silly.

0celo7

the delta kills $f(x)$ at every value besides $0$

So under the integral, $f$ might as well be the constant $f(0)$

Obliv

5:08 PM

I know that but mathematically how do you take the integral

Semiclassical

...

Huy

dat physics proof

0celo7

pull that out, then $\int f(x)\delta(x)=\int f(0)\delta(x)=f(0)\int\delta(x)=f(0)$

Balarka Sen

@0celo7 And at $0$, you have $f(0) \cdot \infty$

Wonderful.

0celo7

@BalarkaSen Physics.

Obliv

5:09 PM

lol..

0celo7

I've read the book, I know how the author wants @Obliv to do it.

@Obliv The rigorous proof uses the Lebesgue majorant theorem, IIRC.

Not fun.

Mary Star

@DanielFischer are you familiar with the vertex cover?

Obliv

What about what balarka said, though @0celo7

at $x = 0$ not only is there a constant $f(0)$ but $\delta(0)$ which is the maximum of $\delta(x)$ function

Semiclassical

Eh. I'm fine with not having a perfectly rigorous proof. But you need to have some source for $\int \delta(x)\,dx=1$.

0celo7

Notice that $f(x)\delta(x)=f(0)\delta(x)$

@Semiclassical He constructed the delta as a growing spike with area 1

Semiclassical

5:12 PM

If that's what's in the book, that's fine. But that's not what was given earlier.

0celo7

what was given?

I have the book memorized, I know how it's defined

Semiclassical

14 mins ago, by Obliv

@0celo7 hey.. uh how do you take this integral $\int_{-\infty}^{\infty} \delta(x)f(x)dx$? I know the delta function is $0$ for all $x \ne 0$ and it approaches $\infty$ at $x = 0$ but I don't really get how to take this integral. hint?

Obliv

I don't recall the area being constructed as $1$. He said Because we
included a multiplicative constant $\omega$, we could always normalize $\delta(t)$ by $\int \delta(t)dt = 1$

for a mass on a spring problem

Balarka Sen

@0celo7 Tell me what they say in the second paragraph of page 34.

0celo7

@Obliv what do you think $\int\delta=1$ means if not "area is one"

@BalarkaSen what

Semiclassical

5:14 PM

I should also note that, for me, the above quote was the entire context. I don't know what book is being used.

Balarka Sen

You said you have the book memorized

0celo7

Oh, it should be something about forces

Obliv

the context was of a different problem with different constants. How am I supposed to know he means $\int \delta = 1$ just by that

Semiclassical

Hence my harping on 'how does the book define $\delta$'

0celo7

Not word for word

But I know how things are defined

Semiclassical

5:15 PM

Which book is this, to be clear?

0celo7

Zee, Einstein Gravity

Where is the integral of $f\delta$ needed

Semiclassical

@Obliv glancing at google books, the point is that he's defining $\delta(x)$ there for further use

Obliv

@0celo7 what

exercise 1

Semiclassical

note that in the paragraph below (3), he specifically mentions that you could construct it by using a piecewise linear function which vanishes except on $[-1/\tau,1/\tau]$ and whose max is $1/\tau$

0celo7

Yes, just use that $f(x)\delta(x)=f(0)\delta(x)$

Semiclassical

5:18 PM

which you can check has an area of 1.

0celo7

then integrate

Obliv

i'm on the 2nd part now though. If what you said is true, then I suppose pulling out $\delta(0) = 1$ and $f(0) = c$ of the integral makes sense

Semiclassical

$\delta(0)\neq 1$

Obliv

:(

Semiclassical

Maybe try focusing on the case of a function $f(x)$ which vanishes at zero first.

Obliv

5:20 PM

@semiC I know it isn't rigorous, but $\delta(x)$ is extremely steep and for the interval $\frac{-1}{\tau},\frac{1}{\tau}$ the area exists mostly at $x = 0$

Semiclassical

sure. but in that case, the peak value of $\delta_\tau(x)$ would be $\delta_\tau(0)=1/\tau$ which diverges as $\tau\to 0$

Try taking $f(x)=|x|$ and imagine what happens to $\delta_\tau(x)f(x)$ as $\tau\to 0$.

Obliv

oh the interval was $-\tau,\tau$ not the inverse, btw.

Semiclassical

ah, yeah.

Obliv

the maximum is given by $\frac{1}{\tau}$ still

Semiclassical

careless of me

Obliv

5:25 PM

$\lim_{\tau \to 0}\frac{1}{\tau}\cdot f(\tau)$

Semiclassical

@0celo7 No. We're doing a triangular peak, not a square one.

Balarka Sen

length of [-t, t] is 2t. 1/2*(2t)*1/t = 1.

Obliv

lol..

Semiclassical

moving right along.

Obliv

@0celo7 can do morse theory but struggles with finding the area of a triangle. :^)

0celo7

5:27 PM

I struggle with Morse theory too

Semiclassical

Anyways. If you plot $\delta_\tau(x)f(x)$ with $f(x)=|x|$, you'll find that by construction it vanishes at $x=\pm\tau$ but also at $x=0$

Obliv

because of the limit i described above right@semiC

Semiclassical

not sure what you were getting at with that limit

Obliv

but isn't it dependent on whether $f(\tau)$ approaches $0$ faster than $\frac{1}{\tau}$ approaches $\infty$?

Semiclassical

$f(\tau)$ is the value at $x=\tau$, $1/\tau$ is the value at 0.

0celo7

5:29 PM

What's wrong with $f(x)\delta(x)=f(0)\delta(x)$

Semiclassical

@0celo7 I'm trying to show how to derive that :P

0celo7

oh, carry on

Obliv

what do you mean @semiC

0celo7

Semiclassical

Since thats equivalent to $f(x)\delta(x) = (f(x)-f(0))\delta(x)+f(0)\delta(x)$ and showing that the first term always vanishes under integration

Obliv

5:31 PM

i was mixing variables on accident

why does that vanish?

Semiclassical

that's what I'm trying to show :)

Obliv

$\delta(x)$ exists in the interval $-\tau,\tau$ and $f(x) - f(0)$ exists for all $x$ that it is defined on

Semiclassical

Just to make my strategy clearer, if you define $g(x)=f(x)-f(0)$ you've got a function which explicity vanishes at zero

Obliv

oh okay

Semiclassical

and then the point is to show that $g(x)\delta(x)$ should be zero

Obliv

5:34 PM

so at $x = 0$ $f(x)\delta(x) = f(0)\delta(x)$ but isn't $\delta(x)$ and $f(x)$ also defined on $-\tau$ and $\tau$ (and any real numbers in between)

Semiclassical

Part of the thing here is that we're using $\delta(x)$ like it's a function.

And strictly speaking it's not

Obliv

yeah in the book it said it's a limit of a sequence of functions

Semiclassical

The point isn't so much whether $g(x)\delta(x)$ equals something but what happens when you integrate it from -\infty to \infty

Yeah

Consider it like this for the example I suggested. What's the max value of $g(x)\delta_\tau(x)$ in that case?

Obliv

as @0celo7 pointed out earlier, $f(x)\delta(x)$ only exists on the interval $\delta(x) \ne 0$

$g(0)\delta_{\tau}(0)$

Semiclassical

No, it is certainly not.

By assumption, $g(0)=0$.

Obliv

5:38 PM

oh we're talking about $g(x) = f(x) - f(0)$

Semiclassical

Right. And I'm wanting to consider the specific case of $g(x)=|x|$.

Obliv

I would say it depends on what kind of function $f(x)$ is

oh okay

then the maximum is at $\frac{\tau}{2}$?

and $-\frac{\tau}{2}$

0celo7

This is the least mathematical discussion we've had here in weeks

Semiclassical

Yeah. Which gives a max value of... $\tau/2$, I think?

@0celo7 Pretty sure your frrustrations with equipment were less mathematical than this

Obliv

@0celo7 want me to ask about lattices of subgroups of groups instead?

0celo7

5:40 PM

@Semiclassical Those were nonmathematical

This is antimathematical

Obliv

physics is antimath

Semiclassical

pfft

Obliv

but this isnt even physics

0celo7

It's a physicist """proof"""

Semiclassical

Anyways. The point is that $|x|\delta_\tau(x)$ is bounded above by $\tau/2$, bounded below by 0, and vanishes outside of $[-\tau,\tau]$.

So what happens as $\tau\to 0$?

0celo7

5:42 PM

pretty sure you need uniform convergence to take limits like that

Obliv

mhm

uh

evaluates to $0$

Semiclassical

it goes to zero.

it vanishes outside of $[-\tau,\tau]$, and inside its squeezed to being zero.

Obliv

for $g(x) = |x|$ though, no?

Semiclassical

right

you can show it works more generally for $g(x)$ with $g(0)=0$

(Probably one needs some conditions on $g(x)$, but being continuous and finite should do it)

If that works for any $g(x)$ vanishing at zero, though, then for any $f(x)$ one can write $g(x)=f(x)-f(0)$ and therefore $f(x)\delta(x)=f(0)\delta(x)$.

Samuel Yusim

@Balarka arxiv.org/pdf/0807.0336v2.pdf looks like $n$-simplex embedding in $\mathbb{R}^n$ isn't characterizable in higher dimensions

0celo7

5:48 PM

@Semiclassical Oh, that's a nice way of seeing it

Obliv

The gist of this is, $\int_{-\infty}^{\infty} \delta(x)f(x)dx = f(0)$ because as $x \to 0$ $\delta(x)$ approaches $\infty$ and $f(x)$ approaches $f(0)$. In the case of $f(x) = |x|$ $f(0)\delta(0)$ gets squeezed to $0$. Is this correct so far @semiC

Balarka Sen

$n$-dimensional simplicial complexes hardly ever embed in $\Bbb R^n$.

Did you mean to say $m$ instead of $n$?

Samuel Yusim

$\mathbb{R}^{n+1}$

Semiclassical

I don't like the first sentence, because nowhere in there have you specifically used $\int \delta(x)\,dx=1$.

Balarka Sen

Hmm, OK.

Semiclassical

5:49 PM

Moreover, if you're going to talk about squeezing then you should be referencing $\delta_\tau(x)$ not $\delta(x)$.

Obliv

well that's assumed I think

what is $\delta_{\tau}(x)$ isn't that the same function

Semiclassical

No. It behaves the same in the limit $\tau\to 0$.

Balarka Sen

@SamuelYusin That's weird.

Semiclassical

But the nascent delta function and the true delta function are not the same. For one, the second isn't even a function.

@0celo7 Glad you like it, heh

Obliv

oh okay.

Samuel Yusim

5:52 PM

it's really disappointing honestly

Obliv

the book defines $\delta_{\tau}(x)$ then, not $\delta(x)$?@semiC

Balarka Sen

Why so? It just means you should restrict to some special subclass of simplicial complexes and try that same problem.

Semiclassical

What's plotted is a particular member of a sequence of functions converging to delta(x)$, yes.

I say "a sequence" because that's not the only one that would work.

A sequence of narrower and narrower gaussians would also work.

Obliv

Isn't the one defined as $\tau \to 0 ~~\delta_{\tau}(x)$ simpler for this exercise?

0celo7

@Semiclassical My proof is that it's "obvious" :P

Semiclassical

5:54 PM

Probably. I'm just saying :P

@0celo7 Now who's being a physicist? :P

0celo7

I'm an engineer

Obliv

you're a student

Balarka Sen

I don't know what they mean by Novikov's theorem on the 5-sphere.

Googling.

Semiclassical

I don't know how to prove $g(x)\delta_\tau(x)\to 0$ as $\tau\to 0$ for generic $g(x)$, tbh

Obliv

I don't think there is a proof for that

Samuel Yusim

5:57 PM

@BalarkaSen the problem is that the subclass will inevitably be matroids

Semiclassical

Oh, I'm sure there is.

Otherwise people wouldn't use it in generality.

0celo7

@Semiclassical uh, what

do you mean for $x\ne 0$

Balarka Sen

@SamuelYusim I am unfamiliar with matroids. What are those?

Semiclassical

I meant under the assumption of $g(0)=0$, woops

Obliv

@semiC if $g(x) = \tau$ then $g(0)\delta_{\tau}(0) ~~\tau \to 0$ evaluates to $1$

oh ok

Semiclassical

5:59 PM

...sure, but then you're doing a sequence of $g(x)$ as well

which you wouldn't do.

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