6:01 AM
@0xVikas it does not matter if it is thick or thin as long as it is an infinite conducting plane.
@0xVikas The plane is infinite. It is automatically grounded.

6:30 AM
Hello Everyone....

@naturallyInconsistent That makes sense but it's bothering me for some reason. Let's say the plane is not infinite but still large. How would the induced charges be distributed in this case? This depends on whether it has a thickness or not right?

@0xVikas yes, but it is so difficult that there is no known exact solutions in closed form. We have infinite expansions and numerical simulations and so forth, but nothing nice looking enough to present in a textbook.
As the textbooks themselves actually will tell you.

Found this paper on this same topic: https://math.mit.edu/~stevenj/papers/LevinJo11.pdf

Depending on the conductor's thickness the induced charges are distributed differently and that can affect the force between an uncharged conductor and a point charge (which is usually attractive, but the paper gives a counter-example using a conductor with no thickness).
@naturallyInconsistent True, I'm just looking for a qualitative description only

@0xVikas I am aware of that paper and its remarkable conclusions. Actually, it is not that surprising; a sphere can attract a point charge even if both are charged to the same sign, so the cutaway of a sphere, from the inside, should be investigated to see if it can repel a point charge, and that is what they have shown to be true.
@0xVikas Nothing much more than "if you are near the surface of the conductor, it will behave as if infinite plane, whereas if you are far enough away, then it looks like a localised metal sphere, say"
vague statements are all that can be offered

6:53 AM
@bolbteppa Is that like the Backlund transform
What did people use before the term "on shell" anyway
IIRC that's a relativistic term
for the mass hyperboloid

7:10 AM
@naturallyInconsistent I think I'm getting it now.. Can it be something like this? Is this a valid description?

I am outside the Earth looking the event someone drops the balls at some height, air to be neglected and consider perfectly elastic collision. From my point of view initial velocities of Earth and ball $U_E$ and $U_b = U_E + U$ . Where $U$ is the speed of the ball just before hitting the Earth.
From my frame of reference final velocities of Earth and ball $V_E = U_E$ and $V_b = 2U_E + U_b$
But if I converted my frame of reference to center-of-momentum frame then COM frame to Earth frame where Earth is stationary.
Then I found $V_E = U_E = 0$, and $V_b = U_E + U_b$ but in reality we don't see ball gains the speed of Earth.
Sorry not $U_b$ in final velocities of the ball, it is $U$

7:28 AM
@123 U is the speed of the ball from Earth's frame?

Yes it is the speed which earth observer see before striking the ground

@123 Can you show your calculations

Sure
Let me open my laptop
@0xVikas pls see calculation. In which i used different convention subscript.
$V_c$ is velocity of center-of-momentum
Initial velocities of the objects $U_1 = U_E$ and $\overline{U_2} = U_E + U = U_1 + U_2$ and final velocities $V_E = V_1$ and $V_b = V_2$. These are the velocities taken by me outside the Earth.
$U_{1c}$ , $U_{2c}$ , $V_{1c}$ , $V_{2c}$ are the velocities transformation in Center-of-Momentum frame.
$U_{1L}$ , $U_{2L}$ , $V_{1L}$ , $V_{2L}$ are the velocities transformation in Earth frame. Where L subscript i represent laboratory frame.

8:06 AM
@0xVikas the positives should be roughly constant. it is a shielding.
@0xVikas dont get roped in. Ive already presented the full calculations. There is no learning with this one.

@123 COM frame is the same as earth's frame. In the "perfectly elastic collision" equation, shouldn't it be V2 = 2U1 - U2?

@0xVikas There is no frame in which both initially and finally the Earth is at rest. The Earth will move infinitesimally, but it moves. His thing is just mathematically inconsistent.
again, I've already worked out the full CoLM frame equations and the Lab frame equations and showed that everything works out. He isnt accepting that. There is not much more we can do.

I think the equation he wrote for the elastic collision is wrong

@0xVikas it obviously is wrong somewhere...

@naturallyInconsistent as per my understanding shielding occurs inside the conductor, and in case of an infinite plane, everything on the other side can be considered as shielded (like in the case of a conducting shell with infinite radius, the inside space is shielded) but can that wont apply here right? Since the plane is clearly finite?

8:25 AM
@0xVikas Well, if the conductor completely surrounds the charge, it would shield this charge from being seen by the rest of the universe, except by having a constant potential difference from the rest of the universe, i.e. it will present the charge as a monopole (maybe higher multipoles, IDK) centred on the centre of the conductor, rather than where the actual location of the charge inside. Here, the shielding is incomplete, so there is some leakage, but it is still attempting to shield.

Agreed, so the positive charge density will be more at the edge of the plane and less (or 0) at the center. This can explain why there won't be any positive charge distribution in the case of an infinite plane, since the "edge" is at infinity.

@0xVikas no. It will look like a uniform distribution because only then can it produce the illusion of as little charge being there as possible.
"no charge is being hidden behind..."

@0xVikas yes but i took $U_2$ negative initially head-on collision opposite direction.

@123 either assume the sign (& direction) everywhere, or do not assume the sign anywhere

@0xVikas Also COM frame and Earth frame is same. I found it because due to heavier mass of Earth COM lies on Earth. May be i am wrong.

8:36 AM
@naturallyInconsistent uniform in case of a finite plane? or both?

@0xVikas both

@0xVikas Yes i assumed the sign (& direction) for initial velocities. But final velocities sign should comes up naturally. Earth initial velocity positive and ball initial velocity negative because of head-on collision.

@123 that is true. Just check again which frame of reference you're using when you're writing the collision equation. I think it should be V2 = 2U1 - $\overline{U_2} @0xVikas I am writing equations in a frame outside the Earth. @naturallyInconsistent Why is that the case in finite plane? If it has a significant thickness (like a cuboid) then it won't be uniform right? That should be the case even if we gradually shrink the thickness @123 Your initial velocity for the ball in a frame outside the earth was$\overline{U_2}

8:41 AM
@0xVikas yes, it will fail to be uniform, but remember, the infinite charged plane tends to be uniformly charged. The finite one will be dominated by this behaviour too, with slight variations modifying the distribution.

Where i can see Earth initial velocity $U_E$ and and ball initial velocity $U_b = U_E - U$
i took ball initial velocity in term of earth initial velocity and the velocity at which ball hit the earth surface. I think it will be $U_b = U_E - U$. Where $U$ velocity at which ball hit Earth surface by the Earth observer.

@naturallyInconsistent Got it. And in the infinite case, the charge will be uniform. I'm claiming this uniform charge density on the other side will be zero. If this fits in nicely and works out for all the conditions, should this be the solution based on the uniqueness theorem?

If it is infinitely large, then the uniform charge density on the other side will be zero. For the finite one, the dominant behaviour is a uniform charge density of very low magnitude, and then some corrections for finiteness.

@123 To avoid any confusion, stick to a single notation. Rewrite the equations with proper notation for the velocities in each frame and you'll get this.
@naturallyInconsistent Makes sense, thank you so much!

@0xVikas Do i use the notation which i shared on page?

8:49 AM
In the image you shared, Ub was U2 with a bar, and U = U2. And in some equations you are assuming the sign and in some places you're not.

Pls share your result of the final velocity of the which can be seen by the earth observer. So i try to match the result. And check for my mistakes.

If you can keep every algebraic thing positive, so that whatever signs that appear tell you precisely which way things go, it is easier to be correct.

The final velocity of the ball in earth's frame is obvious, it bounces back with the same speed but in opposite direction. So if initial velocity of the ball in earth's frame was U, final will be -U

Basically naturallyIncon shared his calculation and video with me. But this result in my calculation appear when observer outside the Earth. And this calculation does not match by Earth observer in my calculation. May be i am doing mistake somewhere.
@0xVikas Yes this is what wanted to calculate and comes in my calculation. But naturallyInc said ball will go more higher where it dropped. Let me share the calculation and equation in Latex. It will be easier to discuss.

How do you solve such equation?

8:57 AM
For perfectly elastic head-on collision (Earth +ve direction, ball -ve direction). Final velocity of the Earth $V_E = \Big(\frac{m_E - m_b}{m_E + m_b}\Big)U_E - 2\left(\frac{m_b}{m_E + m_b}\right)U_b$
$V_b = 2\left(\frac{m_E}{m_E + m_b}\right)U_E - \left(\frac{m_b - m_E}{m_E + m_b}\right)U_b$
$U_b$ taken negative due to negative direction that's why negative sign in between
$m_E >> m_b$
$V_E = U_E$ and $V_b = 2U_E + U_b$
@0xVikas If at this moment on the above equation i decide frame of reference outside the Earth. Then above equation holds as it is. But how can we justify $V_b$ has 2 times the Earth velocity plus the initial velocity of the ball. Why "2" here , how to explain it.

@123 How did you write these equations?

If the same above equations of final velocities i decided to use within Earth frame of reference, then Earth to be considered as stationary $U_E = V_E = 0$ then $V_b = U_b$. Here $V_b$ shows ball will bounce back at same height if gravity is in action. It is practically seen everyday life.

@123 It doesn't, Vb has Ue + Ub. I suggest drawing a little diagram of this scenario if you're going to assume the sign of velocities.
@123 Is Vb here the velocity from a frame outside the earth?

@0xVikas Let me share you picture of my calculation. and above equation are for perfectly elastic collision. Which is also available on internet as i have written
@0xVikas It depends i have notified in comments at which frame i used the same equations.

9:13 AM
Thats the problem, you're using same notation for different frames, don't do that.

@0xVikas I have problem with "2" in the above equation where i took frame of reference outside the Earth. $V_b = 2U_E + U_b$,
@0xVikas But i did this other way to go to the Earth frame. Because above original equations are valid in the frame outside the Earth. In first step i transform my outside Earth frame all velocities to center-of-momentum frame and calculate all velocities there.

Outside the earth, Vb = Ue + Ub, don't you think? For example, you're in a bus and you throw a stone towards the driver. From your frame of reference (bus) if the velocity of the stone is u, for someone standing on the road, the stone will have a velocity of Vbus + u.

Then i found center-of-momentum lies on the Earth.
@0xVikas Yes this is why i am having problem with "2" in final velocity of the ball $V_b = 2U_E + U_b$. Which was given by equations.

@123 This is the mistake. Everything works out now.
Since you used U2 bar here, you should take U2 bar in the collision equation too.

@0xVikas Yes i have noticed i made a mistake in sign of $\overline{U}$

9:23 AM
Now you'll get V2L = -U2, as expected

@0xVikas Okay thanks let me calculate everything again. But outside the Earth frame why "2" with the $U_E$

9:59 AM
@0xVikas Pls see the above calculation now it given me the correct calculation.

10:17 AM
Now frame transformation from outside_the_Earth-frame to Earth-frame. Also shows accurate results.

11:02 AM
how do i make this fit?

11:17 AM
@lucabtz Make one of the subscripts on the Qs a superscript and name the exponential with a single letter?

@ACuriousMind oh it used to be a superscript, but then its confusing because sometimes the superscript +/- is used in this context to mean something else
the exponential could be an idea though

You could do something like the notation used for Weinberg's angle. $s_J:=\sin(2\pi\lambda_J)$

@SirCrackpot i already have $\sigma_J = -2\cos(2\pi\lambda_J)$
i could also put the letters at superscript
but i have to change it in so many places

Are you using overleaf?
There is that search and replace function

@SirCrackpot i know but sometimes i mess up badly with it
also im not sure i like the letters at the superscript

12:23 PM
i went with superscripts in the end

2 hours later…
2:21 PM
@Mad The idea is that, when you have such a partial differential equation, where the derivative w.r.t. two independent variables are just a constant factor away from each other, then you attempt the ansatz that the two independent variables combine into one variable that essentially the function is dependent upon. Namely, let $Q=q-\frac pmt$ so that
$$\frac{\partial\rho}{\partial t}=\frac{\partial\rho}{\partial Q}\frac{\partial Q}{\partial t}\qquad\bigwedge\qquad\frac{\partial\rho}{\partial q}=\frac{\partial\rho}{\partial Q}\frac{\partial Q}{\partial q}$$ satisfies that equation exactly. That is, $\rho(q,p,t)=\rho(Q,p)$ is turned from a 3-independent variable function to a 2-independent variable function, with one of the independent variable itself a variable of 2 others.

3:07 PM
Dealing with categories of manifolds, fiber bundles and principal bundles the objects of such categories are sets (or, well, tuples including also set), so are all morphisms just functions?

@SirCrackpot i mean not all functions are morphisms, but the morphisms are functions

You have to remember that no categories are actually "made of" sets (except the category of sets), but in some sense yes, since the category of manifolds is a concrete category, you can map faithfully every object in it to a set
and the same is true for the morphisms, which map onto functions
And those functions are diffeomorphisms
Well, smooth maps, anyway

They don't have to be invertible

@Slereah well, the category of bundles is made of sets in a sense, just that those sets are with an additional structure

3:11 PM
Yes, that is true of a lot of the important categories
that's the meaning of a category being concrete
Being made of sets, in some sense
But you have to beware that it's not true of all of them

@lucabtz yup that's what I skimmed but I'm scared to get too deep inside cat theory
@Slereah that I know, that's why I asked for the case of bundles or manifolds

For bundles the maps are bundle morphisms

@SirCrackpot its not really to deep of a concept, it is just the categorical way of saying what you correctly understood yourself

Smooth maps that preserve the projection

concrete categories intuitively are categories made of sets with extra structure and the morphims are the functions between the underlying set to the objects of the category which are compatible with the extra structure

3:14 PM
Yes, I guess that is all that's useful in physics (?)

then the definition is done saying there is a faithful functor into Set, but i guess the intuition you got is right

One interesting thing you can look at btw is the "category of elements" of a concrete category
That's basically the formal version of what a category looks like as sets

@lucabtz is that the functor that makes e.g. a bundle $(B, M,\pi)$ correspond the sets $(B,M)$?

You can dissect a concrete category and turn it into its sets and elements and maps between those elements
Like this
That's the objects $X,Y$ with morphisms $f,g$

@SirCrackpot i think to only the total space for bundles but im not so sure how it works there

3:17 PM
@SirCrackpot Basically yeah
It's the "forgetful functor", as it is called, because it forgets the structure of the category
You remove all the informations about bundles, manifolds, topology, etc

@lucabtz Given that bundle morphisms are made up of two maps, one in the base and one in the total space, I'm not sure

Until you're left with just a bag of points

Lol
Category theory is both the most straightforwrard and the most twisted math I know

@SirCrackpot i know but the objects of set are sets, not pair of sets

It is nice to talk about very broad ideas but if you want more specific details it gets quite nasty

3:19 PM
im not sure really how this works with bundles

Like you could do 1+1=2 in category theory, but it would mean a lot of diagram chasing

I'm fine with a sound use of category theory to remember the definition of principal bundles with a nice diagram

@SirCrackpot what definition do you use?

Well, thinking about also most of the DG I know boils down to that with all the commutative diagrams
@lucabtz I think the standard one, just that it's easy to remember the relation between the group action and trivilizations and so on

@SirCrackpot hello. what books do u use for diff geo and stuff?

3:35 PM
The book I'll write using ACM'S answers
More seriously, there are many books I skim

Lee's book for DG is pretty good

thanks

Yes, Lee's book is good. In fact, they are all rather good.

I was reading this book "Geometry and Topology" by Glen Bredon and it's too many definitions with no examples
one definition was even wrong. it says a neeighborhood $N$ satisfies $N\supset S$ for an open set $S$

I don't have much background in algebra (besides a little bit of group/ring theory). Is it advised to learn representation theory directly from a book "for physicists", or to approach both fundamental algebra + representation theory from pure mathematics books?

3:44 PM
That is true

but it shud b $N\supset = S$
for some open set S

People are usually pretty loose with using $\subset$ v. $\subseteq$

oh. this really annoyed me
as in a later proof, he implicitly used the = property

4:27 PM
lee has four books
introductions to : smooth manifolds, topological manifolds, riemannian manifolds, curvature
@Slereah u meant the first one, right?

1. There is no such thing as a wrong definition

they're all good
Probably won't need the one on topological manifolds tho

2. For some people $\subset = \subseteq$, for others $\subset\subset \subseteq$

@SirCrackpot lol
i used loring tu's an introduction to manifolds
and also the other one he wrote

@Slereah in terms of relevance to physics, how are they ordered
@lucabtz thanks
@SirCrackpot yes, but he shud specify it and the former is a horrible convention becuz it doesnt leave room for proper subsets

4:44 PM
@RyderRude in maths it is common to use $\subset$ instead of $\subseteq$
to denote proper subset they put a small bar on the equal piece of $\subseteq$
$\subsetneq$
like that

the reason is that $\subseteq$ is way more common and its quicker to write $\subset$ i guess

makes sense

Hello, everyone. A few days ago one of my professors said that we could represent the whole universe with 800 qubits sampled as Planck length, and that this testifies how quantum computers are the future, give the great possibility that a new world will open up, and be able to develop unlimited functions. What are your thoughts on this? Does this seem like an overreaching and meaningless statement or do you think it is legitimate, and why? Thank you.

@Bml there are more than 800 electrons in the universe and the spin of each is basically a qubit
so idk how you would encode the data like that or what exactly that means

4:50 PM
Maybe he means in the sense that the dimension of the resulting Hilbert space would be $2^{800}$ or something

@lucabtz there's just one no cap

according to the holographic principle, the information reduces somewhat, I think
and then maybe u hav to divide the surface area by the Planck area
im not familiar with the details of the argument
but no way it's 800

@SirCrackpot sus
@RyderRude according to that its a gravitational theory that can be described by the fewer degrees of freedom of a boundary gauge theory

@Slereah Yes. $2^n = volume of the observable universe/(Planck length)^3$. I obtain 615. But it's not the entire universe, it is the observable universe. I think it is really different...

however the boundary theory still has an infinite number of dofs

5:00 PM
@lucabtz yes, so does this principle not hold for other forces?

@RyderRude i mean a good part of the point are black holes
so no

oh
theres also AdS CFT which is for string theory
so this one shud cover all forces, but it's speculative

from my thesis
but its arguments from a maldacena review on arxiv basically
@RyderRude i mean in the bulk you also have the other forces
but the point is that there has to be gravity
the piece i posted is still a bit poorly written but the idea is there

really cool
why is entropy so fundamental to these arguments

@RyderRude entropy basically counts the degrees of freedom

5:07 PM
yes, but entropy depends on ur definition of the macro variables
so it's just a measure of ignorance?

the argument here is of course handwaving, but gives the right conclusions

if u define micro variable to be the macro variable, then entropy is always 0

the macroscopic variables are well defined
it is stated that we work in the microcanonical ensemble

but still its handwaving

5:10 PM

but clearly entropy is the amount of information you need to specify one specific microstate, so it counts the dofs

Schreiber will be the death of me

@lucabtz yeah

@Slereah wait why negative dimensional spheres and quantum?

Is apparently a trick used in homotopy where you use "negative dimensional spheres" for the inverse operation of suspension
Suspension is an operation on topological spaces and in particular, the suspension of a sphere is a sphere of higher dimension
It has an inverse operation, which by analogy is said to produce spheres of negative dimensions

5:16 PM
why is this quantum

No clue

sounds really deep

Something homotopy related, but then again according to Schreiber everything is homotopy related

OK, could you explain me the point of this problem?

5:40 PM
I fail to see how the fact that one can simulate the observable universe by qubits is an attestation that quantum computers can perform unrestricted functions...

5:56 PM
What the hell is a negative dimension

6:22 PM
Locally true seems to be expressible as a modal operator apparently

6:56 PM
@Slereah Do you find this stuff easier than normal physics?

I do not

7:28 PM
Could any of you clarify my doubt?

@Bml Look, it is clear that your prof was just throwing out a back-of-the-envelope calculation that he would not be defending in a court of law. It is meaningless to seek deep meaning in it. Let it go.

7:53 PM
@naturallyInconsistent What you say is absolutely correct, but the problem is that he was the one who characterized the issue that way, not me. He literally asserted that the universe is a quantum computer by motivating it with this "back-of-the-envelope calculation," without giving any detailed explanation. So, is it true and certified that the universe is a quantum computer?

@Bml The universe is full of quantum theory-obeying objects. There is no classical computer fast enough to simulate all of these things. However, what does "universe is a quantum computer" even mean? If we are a program running on a computer, why cannot the computer take pauses here and there and update slowly? It makes no sense. Don't ask us to dig a stupid hole.

@naturallyInconsistent Sorry, I can't understand. There are several researchers who support this thesis, or am I wrong? For example MIT professor Seth Lloyd said that "everything in the universe is made of chunks of information called bits" and that universe is a giant quantum computer. If it is a stupid hypothesis, why is it within an academic paper by a renowned professor? What is the true degree of reliability? I'm just confused.

@Bml Again, none of these renowned professors would be defending these if their lives depended upon it. That is not to say that it is a stupid hypothesis---instead, it is a good hypothesis that is strongly suggested by evidence we see around us. What I am saying is stupid is you asking us to defend it to you. There is no known true degree of reliability. The professors are free to entertain an idea of this level of interestingness. But we cannot be chasing a rabbit hole for you. DIY.

8:26 PM
@naturallyInconsistent I have the impression that I expressed myself wrongly. I apologize for any misunderstanding. I do not want you to defend any hypothesis. I would just like to understand what you think about the question.
In your opinion what is the most convincing hypothesis? Is the fact that it is possible to estimate a number of qubits for which it is possible to represent the sampled universe at the Planck length indicative of the fact that quantum computers can perform unlimited and not restricted functions (as the professor suggests)? I would just like to understand how you feel about this; I don't want you to defend anything.

I think the hypothesis is neat and interesting but there is not much we can do about it. If you do not accept their explanations, then there is nothing much more to it than that and so it is not meaningful to have more conversations about what meaning can be ascribed to it. There is no convincing what is more than what is already stated.
> indicative of the fact that quantum computers can perform unlimited and not restricted functions (as the professor suggests)?

is false, whether the professor suggests that or not. A quantum computer cannot compute faster or more than what a quantum computer is limited to be able to do.

2 hours later…
10:25 PM
@Slereah How do you even transition from normal physics to this stuff

10:41 PM
@DIRAC1930 You start trying to figure out GR

Lol

And you end up cursing yourself everyday as soon as you wake up

From a quick skim it looks like a lot of algebra to me

Is that a good summary @Slereah ?

Being interested in real-world physics and theoretical physics seem mutually exclusive at the moment
Either I have a crisis that what I'm working on has nothing to do with the real world or I have a crisis that what I'm working on is not mathematically interesting