B A H ?

BAH

@naturallyInconsistent oh yesh

@ACuriousMind omg i think they are like an orange or brown? i need to check. wow they are at least a month old cant believe it. i guess maybe their mom would not let them wander around when they are too small

@SillyGoose it is gonna have to wait till next decade...

cripes

here, are we suppressing notation?

in particular, in Tu's notation, $X: U \to TU$ is a vector field so in local coordinates it should be written as $\sum_i a^i(p) \partial_i \lvert_p$ where $p \in U$?

but i guess we can get away with suppressing such notation?

@SillyGoose yes, extremely standard suppression. Note that the picture is based: It only makes sense to sum indices when one is covariant and the other contravariant (or vice versa), and definitely not two indices both up there or both down there.

hm so how to fix the index placement here? im a bit confused because there are a number of ways to write or rewrite the inner product at a point. why is it necessary to do it as it is done in physics with lower and upper indices contracted

@SillyGoose The result of a scalar product between two vector fields should actually be a scalar, i.e. invariant to all changes in, say, coördinate parameterisations. The usual physics covariant-contravariant contraction achieves this, but if you have two of the same type, then they don't cancel out their changes, and so you don't actually get a scalar

If you have two of the same type, then you use the metric to help you fix that and get a scalar.

Unrelated: These QFT notes are beautifully done: people.phys.ethz.ch/~nbeisert/lectures/QFT2-17FS-Notes.pdf

m i a o ~~

i am always open to new qft resources >:D

M I A O ~ ~

The QFT1 part is here: people.phys.ethz.ch/~nbeisert/lectures/QFT1-19HS-Notes.pdf

I trust yall can find the main page where I grabbed these pdf links from

for this theorem, is the partition of unity argument solely invoked to prove that the metric is smooth on $M$?

In particular, it seems like we can certainly construct a metric in the sense of a mapping $p \mapsto \langle -, - \rangle_p$ without anything fancy. But to turn this into a Riemannian metric, we need to prove smoothness.

er let me be more precise

let $(M, \tau, \{(U_\alpha, \phi_\alpha) \}$ be a smooth manifold. Let $(U, x^1, ..., x^n)$ be a chart in the atlas with coordinates. The Euclidean inner product on $\mathbb{R}^n$ induces a natural inner product over $U$ via the chart map $\phi$.

@SillyGoose You might be technically correct but I don't think this is the aim. The issue is that, on a generic manifold, it is not usual for there to just be one chart to map the entire manifold. You often have to have multiple charts, each taking care of a subset of the manifold, and then you would have overlapping regions, making sure that every point is at least covered by one chart. In the overlapping regions, there will be more than one chart, and the metric on each chart will differ.

this inner product is given at a point $p \in U$ by $\langle X_p, Y_p \rangle_p = \sum_i a^ib_i$ where $X_p = a^i \partial_i$ and $Y = b^i \partial_i$.

this natural construction is immediately an assignment of an inner product $\langle -, -\rangle_p$ to every $p \in M$

Now, each chart's metric gives the full contribution, say, if you want to compute a scalar product, at any point inside the chart. That means that at the overlapping regions, you have more than one metric giving full contribution. The partition of unity is needed to prove both smoothness and uniqueness as you transition from a region where there is only one chart, through the overlapping region, into another "only one chart", and guaranteeing that there is no overcounting nor undercounting.

@SillyGoose As long as you have $U$ stated without its labelling $\alpha$, then you arent going to be considering the partition of unity.

@naturallyInconsistent i mean $U$ to be a $U_\alpha$

Part of the issue is that you still have not yet have $b_i$ so that you can have $a^ib_i$ at all. Usually you have a hidden metric $\eta_{ij}$ doing that for you

right...

well at this point in the book we have established writing an inner product as a 2-form, which i could probably write it as without breaking any laws...

so is this like an accurate visual of the situation. consider for simplicity the intersection of two charts of a fixed maximal atlas of a smooth manifold $M$.

so we literally just use the partition of unity to write a new inner product that is just a particular sum of the intersecting inner products

@SillyGoose 2-forms are antisymmetric. Inner products are symmetric.

@SillyGoose yes, weighted sum that guarantees correct normalisation. That is how it is used in normal analysis too. This way, the newly defined metric will work for the entire manifold, especially the orientation preserving and positive-definiteness

@naturallyInconsistent thanks for sharing.

@user70432 M I A O ~ ~

:D

@naturallyInconsistent ah yes...i conflate the words for tensor products of covectors and forms...

S C H W E G ~

It's unfortunate terminology that we also call inner products bilinear forms :p

> Since it is not true in NG, and NG is a limit of GR, it cannot be true in GR.

@Qmechanic Is this true? For example frame dragging could be different for two otherwise similar bodies but since frame dragging does not exist in Newtonian gravity they would be the same in the Newtonian limit.

@JohnRennie where is this from?

@naturallyInconsistent This question

I see. Yall active after miao miao left that convo. I think I most concur with the commenter who said that the question is left too vague for us to judge. As you correctly noted, Birkhoff's theorem applies, so there is at least cases whereby what Qmechanic asserts so definitively would be wrong.

hi

en.m.wikipedia.org/wiki/Constructor_theory

What do u think about this theory

@JohnRennie and naturallyInconsistent: It should probably be stressed that OP did not explicitly assume spherical symmetry, so Newton's shell theorem and Birkhoff's theorem do not apply.

@Relativisticcucumber yeah if they don't have blue eyes these aren't babies anymore - all cats are born with blue eyes and somewhere between 1 and 2 months they get their "adult eye color"

(they're still cute though)

@ACuriousMind how did I survive without knowing this

Not a lot of kitten related deaths

what r some non manifold objects that spacetime cud emerge from

Kittens

Does the hairy ball theorem apply to a cat?

many worlders say that spacetime emerges from the hilbert space, but they forget that QFT Heisenberg picture is defined on spacetime

@Mr.Feynman it applies to male cats