The h Bar

3:44 AM

@ACuriousMind So, here, he is talking about the group that the Lorentz transformations make up, and he says that, initially, when we have that O(3) is the group of matrices that satisfy $R^TR = 1$ where 1 is the identity. then, he says "sometimes we choose to exclude parity transformations by demanding that the matrices have unit determinant, and this leaves us with SO(3)." [...]

[...] so then he says when we go to lorentzian group, we make this requirement into $R^TT = \eta$ and that this group then becomes O(3,1). Then, he says the quote i mentioned above. [...]

[...] and i have read that the universe does not exhibit symmetry under time translation or parity translation, so i suppose this is why we need to eliminate them, but i find it strange that they are solutions to begin with. and they are physically realizable solutions, yet we eliminate them? and i still dont see why there is only a discrete time reversal [...]

Silly Goose

4:10 AM

So I was told that in Quantum, we have our nice and fancy Lie group. At the identity, we have our tangent algebra or our Lie algebra. This Lie algebra together with the all mighty exponential map (usually) allows us to generate every element of the Lie group.

In GR it seems like we want to access the tangent space at every point on a manifold whereas in Quantum, we do not care so much about the tangent algebra at every point on our manifold (Lie group). Why? Are manifolds in GR not so nice as the manifolds in Quantum?

[1] The usually parenthetical: When we cannot generate every element of the Lie group with elements of the Lie algebra, is that when we have to compose two exponentials of hermitian operators: one that lives in the Lie algebra, and another that lives in some other tangent algebra?

or I guess maybe the quantum blurb is only applicable to SU(2).

nickbros123

4:46 AM

In books like kleppner and kolenkow, when talking about gyroscopes, and spinning cigars, why do they imply that a change in direction of the angular momentum vector (or a component of it) results in a change in orientation of the axis of rotation.but isn't this non trivial? Shouldn't the angular velocity not necessarily depend on angular momentum?

Mr. Feynman

@Relativisticcucumber No, that's not why we eliminate them. Although it's true that not all phenomena are parity preserving (weak interaction), the reason why we do that is mathematical. Those group you mentioned are not connected. Only the part of the group that is connected to the identity element is related to the Lie algebra, more precisely the exponential map takes the algebra into this part of the group

So we construct all the representation theory of such groups considering the connected component of the identity element i.e. $\mathrm{SO}(3)$ for $\mathrm{O}(3)$ and $\mathrm{SO}^+(1,3)$ for $\mathrm{O}(1,3)$

Relativisticcucumber

@Mr.Feynman can you elaborate on this "Only the part of the group that is connected to the identity element is related to the Lie algebra"

Mr. Feynman

Have you heard of the exponential map?

That's also what you consider in QM in most cases, the connected component of the identity element is really made up of those you call "infinitesimal transformations" (or rather, by composition of those :P)

In simpler words, it means there is a continuous path connecting every transformation to the identity, think of a $\mathrm{SO}(2)$ matrix for example: $\begin{pmatrix}\cos\theta&-\sin\theta\\ \cos\theta & \sin\theta\end{pmatrix}$. You can smoothly change $\theta$ to $\theta_F=0$ to make it the identity matrix, that is the whole group is the connected component of the identity and thus is connected

On the other hand, if you think of $\mathrm{O}(3)$, a parity transformation is represented by $\mathrm{diag}(-1,-1,-1)$ and there is no way to construct a continuous curve connecting it to the identity matrix

Anyways, once you know how to represent the connected component of the group, representing also parity is not a big deal. It's basically about multiplying by the above $\mathrm{diag}(-1,-1,-1)$ matrix above

@Relativisticcucumber Now I'm late for the bus so I'm leaving and I'll read the chat later, please tag me in case you reply otherwise I won't see the message.

Relativisticcucumber

5:41 AM

@Mr.Feynman i am looking into it now, but i have not before

@Mr.Feynman okay i get what you are saying about what it means that these do not smoothly connect to the identity. however, i am not sure why we have that requirement? i guess this is what you mean by "Only the part of the group that is connected to the identity element is related to the Lie algebra", but i do not see the physical meaning of this in this case?

by the way thank you for your response

Mr. Feynman

6:09 AM

@Relativisticcucumber It's Math, it is not Physics at this point. The point is that the Lie algebra is the tangent space at the identity, so it "approximates" things close to the identity, that is locally

Different groups can have (up to isomorphisms) the same tangent space at the identity. That doesn't mean in general that the groups are the same globally, but only locally. The Lie algebra doesn't catch (through the exp map) the whole group in general

Slereah

7:46 AM

It's like integrals being the same up to a constant

user36093

7:56 AM

@John Rennie Following on from our discussion the other day how would you write an equation that relates change in acceleration for some small change in tau (if the change of position is very nearly zero) I tried to do it with g=(G m1 m2/r^2)-(G m1 m2'/r^2) =G m1 m2'/r^2 but for higher values of m2' g is smaller which is clearly wrong.

John Rennie

Typically we look at the situation where the falling object is initially at rest. In that case the only non-zero component of the four-velocity is uᵗ.

The expression for the acceleration then simplifies to:
$$\frac{d^2x^\alpha}{d\tau^2} = \Gamma^\alpha{}_{tt} u^t u^t $$
OK so far?

user36093

no sorry I dont understand your way of expresing it

what is the $$ signify

Oh just the ends of the expression?

John Rennie

Ah, I used MathJax to write the equation. t should appear like this:

You'll need to install an addon to make MathJax work in the chat. Let me see if I can find the link ...

39

A: Any chance of MathJax in chat?

As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An alternat...

Silly Goose

i will forevermore think of fibers when i see integrals up to a constant (that then differentiate to the same derivative)

user36093

should I be able to reply in that new window you created?

John Rennie

8:11 AM

Just reply here

user36093

ok yes pleas go ahead

Silly Goose

wow I've just installed chatjax and how nice it is. Here's a celebratory $\mathcal{H}$

John Rennie

To get any further we need to know what the Christoffel symbols are, and they are tedious to calculate so we usually use a tool like Mathematica to calculate them or just look them up.

I have done the calculation somewhere on the SE. If you give me a moment I'll look for it.

user36093

righto

John Rennie

Found it! It's this answer:

41

A: How does "curved space" explain gravitational attraction?

If you have a look at my answer to When objects fall along geodesic paths of curved space-time, why is there no force acting on them? this explains how on a curved surface two moving observers will appear to exprience a force pulling them together. However two stationary observers will feel no fo...

Have a read through that and see what you think.

user36093

8:19 AM

ok thanks ps my message on fb the received tick is grey , apparently indicates you are not signed in. :-)

Silly Goose

@JohnRennie I've found the issue with my mathcals showing up as boxes business :). It turns out to be caused by me updating my MacOS. I found out a fix courtesy of this post: meta.stackexchange.com/questions/384924/…

John Rennie

@SillyGoose Aha :-)

Relativisticcucumber

@Mr.Feynman okay so is it like if we apply these transformations many times it should amount to doing nothing and should smoothly connect to the identity and we only want to consider the transformations that achieve this? sorry if im still missing smth

@SillyGoose what does this mean

Silly Goose

wait what does what mean ! @Relativisticcucumber

Relativisticcucumber

@SillyGoose wow can u help me install it

@SillyGoose this!

"i will forevermore think of fibers when i see integrals up to a constant (that then differentiate to the same derivative)"

Silly Goose

8:26 AM

Let's say you have a map $\varphi: G \rightarrow H$. Then, let $h \in H$. The fiber of $h$ is the set $\{g \in G: \varphi(g) = h\}$.

To get ChatJax, go to this page: math.ucla.edu/~robjohn/math/mathjax.html, and then book mark the "start ChatJax" and "render Chatjax". (like literally put the links in your bookmarks bar). And then come back to hbar and click "start ChatJax" from your bookmark bar and then be ready to have your pantst blown off by all the rendered latex

so the fiber of an element that lives in the codomain of some mapping is the set of all elements in the domain that collapse to said element in the codomain when sent through the map

Relativisticcucumber

@SillyGoose gotcha

Silly Goose

so in the context of integration/differentiation, the fiber of the function $f(x) = x$ could be $\{x^2/2 + C : C \in \mathbb{R}\}$ (if we sent this function through a differentiation map, I guess if that exists)

The concept of fibers comes up in group theory because fibers of homomorphisms are one way of defining elements of quotient groups I think

Relativisticcucumber

@SillyGoose I GOT IT WOW

@SillyGoose yeah yeah i see where ur going

Silly Goose

the kernal of a homomorphism is a particular fiber as well :) all elements of the domain that collapse to the identity of the codomain :D

Relativisticcucumber

kernal

Silly Goose

8:33 AM

pop

did you get chatjax :D @Relativisticcucumber

Relativisticcucumber

wait do you follow the replies to their original message

Silly Goose

OH

I see

Relativisticcucumber

LOL

Silly Goose

these notes follow dummit and foote it looks like pretty much exactly and are just more concise for the definition of fiber, how they build quotient groups, and then why elements of a quotient group can also be thought of as cosets: people.math.harvard.edu/~bullery/math122/…

Relativisticcucumber

nobody:

absolutely nobody:

Silly Goose

8:38 AM

LOL

Relativisticcucumber

silly goose: here is an infinite amount of abstract alg info

Silly Goose

now is it countably or uncountably infinite?

now we add analysis to the mix ;)

Relativisticcucumber

unctbly

Silly Goose

can you tell me the bijection into the reals please and thank you

Relativisticcucumber

my supervisor is asking me to get to work sorry toodaloo

Silly Goose

8:41 AM

tood a loo! @Relativisticcucumber

nobody:

undergraduate ENM: how well do you know geometry?

ACuriousMind

@Relativisticcucumber Why do you find it strange? I mean, there is no physical reason every random transformation that preserves the metric needs to be a symmetry of our full theory in the end

Relativisticcucumber

@ACuriousMind i guess i thought this is the definition of a symmetry

ACuriousMind

Physically, I think about this like this: The transformations continuously connected to the identity are those that are "realizable" - all the physical processes I know of are continuous, so I can see the physical realizability of a boost to velocity $v$ as first accelerating to velocity $\epsilon$, then to velocity $2\epsilon$ etc. until I arrive at velocity $v$

similarly for rotations - when there is an infinitesimal version of a transformation, then I can see how to "implement" it by repeating small steps, ultimately having a continuous process

but the discrete transformations like time-reversal or parity aren't like that - you either do them or your don't, you can't do a time reversal "halfway"

and so they aren't as "physical" as the continuous transformations: you can use them in the math but you can't implement them via a process

@Relativisticcucumber A symmetry of a theory is something that leaves my action invariant, no?

The very general idea of a physical theory doesn't even know what a metric is

the reason we care about the metric is that it turns out that since this metric seems to be what is realized on our spacetime, it is much more convenient for us to construct theories that are covariant/invariant w.r.t the isometry transformations of that metric

in principle you can write down any theory that breaks Lorentz symmetry that you want :P

Silly Goose

9:01 AM

i thought kant was the popularizer of the idea that human knowledge is limited. and i always assumed that acknowledging human knowledge to be limited necessitates there being no objective truth or at least an objective truth that can be meaningfully understood. but it seems this is not the case? kant believed in an objective truth (noumena), right?

the starred message of nigel made me think of this earlier :P

Relativisticcucumber

so the thing i am confused about is i was kind of (i think) taught this idea by looking at a shape -- maybe a triangle or a sphere, and the group is the elements that are a symmetry, and by this i refer to things that, when done, you have the identical thing that you started with (circle, triangle, etc.) so i see that in this context we are looking for things that leave the physics the same, and we are looking for ways to transform the coordinates to achieve this [...]

[...] but im confused about how we are doing anything that considers the action? because this section of the book just says we want things that satisfy this relation that $R^TR= \eta$

so bleh i think im still missing something here

@ACuriousMind the above messages are in response to this

ACuriousMind

@SillyGoose well...Kant "believed" in noumena but he didn't think they were knowable

Silly Goose

what is the rationale for believing in the existence of something that can never be known (well I suppose this may be asking you to distill a sizable chunk of quite a text) :P

ACuriousMind

One aspect of the "I had to get rid of knowledge to make room for faith" I alluded to somewhere else in that conversation is that precisely because we cannot know things-in-themselves (noumena) we are free to believe in humanity as a thing-in-itself

Silly Goose

hmm

@Relativisticcucumber want to read the critique of pure reason ;)

some may say you make a critique of pure reason every day

Relativisticcucumber

9:07 AM

@SillyGoose i dont feel the need to read kant

ACuriousMind

@SillyGoose I don't think it matters very much to Kant whether or not you really believe that noumena "exist". What matters is that you understand the concept of things-in-themselves and can therefore understand the formulation of the categorical imperative saying we must not treat humanity as a means to an end, but as an end-in-itself

Silly Goose

so kant wants to say noumena exists emphatically for the sake of saying that it cannot be known?

ACuriousMind

yes

very Socratic: The concept of the noumena is necessary so that we can know that we don't - and cannot - know everything

Silly Goose

maybe i shall try another swipe at kant soon :P i feel some education in physics and math may have helped improve my ability to read philosophy. i remember not being able to read some of a wittgenstein or something because it had a brief blurb about "going beyond Euclidean geometry" which I did not know what Euclidean geometry was and how one could go beyond it XD but stuck no longer I'd be on such a passage

ACuriousMind

I'm not sure I'd really recommend Kant to get into philosophy

pretty notorious for being hard to read, although I don't know how the English translations are

Slereah

9:29 AM

More like Emmanuel Can't

My advice is to not read any historical philosophers

Being first at a particular idea does not mean being the most pedagogical to write about it

just read an intro to whatever topic you wish to know about

This isn't the 12th century, we don't have to read the same 3 books of Aristotle for the rest of our lives

Silly Goose

how well established are interpretations of these source materials? Say the critique of pure reason. though, it doesn't really seem like you can win: you read secondary material and get someone else's idea of what is said which is wrong, or you read the primary material and understand nothing at all!

Slereah

I just use my pure reason to understand it

Silly Goose

:)

you use your pure reason to make your own critique of pure reason

Slereah

Kant thought Euclidian geometry was Fundamental so I wouldn't trust him tbh

Silly Goose

is that what he means by space?

ACuriousMind

10:20 AM

@SillyGoose I think a mix - as you do in an actual philosophy class - is best: Read the original (not all at once, in chapters or whatever), then read selected commentary on what you just read.

Mr. Feynman

11:14 AM

@Relativisticcucumber why doing nothing?

Amit

11:49 AM

Is there any advantage of ChatJax over the scripts? I was so glad finding out there's a way to render MathJax without having to install anything...

Mr. Feynman

@Amit You can put it in the "favorite" bar and render MathJax with a single click. I don't know what the scripts do but chatjax is really just this

Amit

Yes, that's what I mean, and that's what I'm doing :)

Got it from here: math.ucla.edu/~robjohn/math/mathjax.html

Ah I see -- ChatJax is also script based. I must have confused it with some browser extension

ACuriousMind

...what do you mean by "ChatJax"

there are several different things that are called that :P

user2236

re: ⭐ link

ACuriousMind

12:07 PM

@Amit 3 of the 4 options in the meta answer linked in the room description are called "ChatJax"

they're all "scripts", but for some of them you need one of the scriptmonkey extensions to run them

Amit

Ah I see

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