How do I find the metric ($d: X \times X \rightarrow \mathbb{R}$ that is positive definite, symmetric, and obeys triangle inequality) induced by a metric tensor? Or does a metric tensor not always induce a metric?

why is the field topological matter called topological

If it's Riemannian it does induce a metric

The metric is that of the shortest geodesic's proper length

@SillyGoose Given a positive definite metric $(v,w)\mapsto g(v,w)$, it induces a norm $||v||:=\sqrt{g(v,v)}$. You can now define a metric (distance) $d(v,w):=||v-w||$.

so i am trying to understand the relationship between metric tensor, inner product, and line element

in say a normal linear algebra class when dealing with euclidean space, the metric tensor is the identity matrix, the "line element" is the dot product, and the metric is |v - w| or whatever

is it analogous in relativity? idk if that makes sense

The metric tensor can be any symmetric bilinear form, linear algebra classes don't use only euclidean metric tipically

so the metric tensor is like a equivalent to the inner product?

@SillyGoose The line element appear when you want to measure distances along a curve

to my understanding you have some manifold. you look at a point and the associated tangent space (which is a vector space?). To this tangent space is associated a metric tensor. This metric tensor seems to be a slightly more general object than the inner product it induces

The metric tensor is a tensor field that assigns to each tangent space a scalar product (in a smooth way)

is a tensor field a field whose elements are tensors?

What I told you above for a vector space holds for each tangent space

@SillyGoose no, a tensor field is a function that maps each point to the manifold to a tensor in the corresponding tangent space

Oh

Vector fields, which you are familiar with, are an example of tensor fields $(1,0)$

Anyways, regarding the line element, it appears in the definition of lenght of a curve

so the metric tensor is defined over the manifold and given a point it outputs some tensor which tells us how to compute inner products in that tangent space?

@SillyGoose exactly and that "some tensor" is the kind of inner product you know from linear algebra

what is a curve? like a set of "continuous" points on the surface of the manifold?

A curve is a map from a real interval to the manifold

Take the definition of curve in $\mathbb{R}^n$ and put $M$ instead of it :P

errr I guess I don't know why we have a manifold and a tangent space associated with each point on the manifold--what physically does this tangent space correspond to

I don't know what the manifold is physically either. I naively think the manifold is spacetime and each point of spacetime is a minature spacetime

Well, tangent vectors at different points are completely different and unrelated entities, that's what having a tangent space for each point means. In the case of $\mathbb{R}^n$ the tangent space is the same of $\mathbb{R}^n$ itself and you don't care

@SillyGoose You should learn some differential geometry, because that's what it is about. A manifold is something that locally is like $\mathbb{R}^n$ for some $n$

That is a sloppy "definition" :P

More properly, a topological space which is locally homeomorphic to a $\mathbb{R}^n$. Or in the case of differentiable manifolds, locally diffeomorphic

one day i will learn diffe g :P

well so is the tangent space in its entirety useful or just say a basis for the tangent space

@SillyGoose When you embed a manifold into an ambient space (like thinking about a sphere as existing inside of 3d Euclidean space), then the tangent space at a point is literally just a set of vectors/arrows tangent in the ordinary geometric sense to the sphere at that point

i.e. it's the vectors in the plane that touches but does not intersect the sphere at that point

in the more abstract sense, you can think of tangent vector as "infinitesimal translations" on a manifold. On manifolds that are not vector spaces, there is no global notion of what a translation is, but in a neighbourhood around a point, there is a well-defined notion of what it means to step some distance $\epsilon$ into the direction specified by a tangent vector

flashbacks to SU(2) XD

in the formulation where you think about tangent vectors as coordinate derivatives, this is as simple as saying that $\partial_x$ corresponds to $x\mapsto x+\epsilon$

@SillyGoose This is not an accident!

it's all the same huh :)

Lie groups are manifolds and the Lie algebra of a Lie group is the tangent space at the identity

wait so tangent spaces are not vector spaces they are algebras in general?

They do have a natural algebra defined on them yes

defined by the Lie bracket of two vectors

@SillyGoose indeed there is a natural Lie bracket of vector fields

oh but you could work with just the vector space structure and think not of the additional algebra strcture

Yes it is pretty common to do that

this Lie bracket is also important in differential geometry, you will run into it sooner or later

well now it really sounds like i should take diffe g

well now how much is the theory of quantum angular momentum differential geometry and how much of it is representation theory

the beautiful thing about Lie theory is that it is the intersection of algebra and differential geometry!

You say that like they are independent theories

i see.... so Lie theory

Hm... this is the closest course there seems to be to Lie theory: Description: Selected topics in Riemannian geometry, low dimensional manifold theory, elementary Lie groups and Lie algebra, and contemporary applications in mathematics and physics.

sounds nice

Currently listening to a podcast of the entire history of western philosophy

Going through the medieval era is a bit of a slog

things are picking up a bit in the 14th century at least

@SillyGoose there are a bunch of different things in cond.mat that get thrown under the umbrella "topological"

one way is this: if you want to classify gapped phases of matter then you want to find equivalence classes of systems that can't be deformed into each other by "nice" local perturbations. To show this, you might characterise the system by a topological invariant (e.g. something that mathematically can only be an integer)

One big hurdle of mathematization of physics early on was apparently Aristotle

Assigning a quantity to a quality was considered ABSURD

Entirely different categories of being

@Slereah is it the one recently linked on Existential Comics? I know you read that one :P

It is historyofphilosophy.net

Why is the equation of state always a relation between scalars? I've never seen a vector equation of state?

Shower thoughts^

There are states related to magnetization

Wiki?

I mean why is it a dot product of vectors

Why can't the equation of state be a vector equation?

Isn't the equation of state a contact form

I forget

whats a contact form? I know of one form, etc from diff geo

@Mr.Feynman you are alive :D

@Slereah So what is considered absurd today that will appear medieval in a thousand years? :)

History of science isn't mathematical enough, we need the equivalence class of absurdities, one absurdity at a time won't do!! :D

@JohnRennie I have messaged you on fb, I know a lot of people dont look at that.

@user36093 I can't see the message. Did you just send it?

@ACuriousMind Oops :-)

I sent it yesterday

@user36093 I have not received it then. Can you resend it?

Ill try now

@JohnRennie no worries, it's a very hidden feature and I can't remember where I actually learnt about this

@JohnRennie can you see my message now?

Can someone help me with a question about the four vectors. If we consider the four momentum, the scalar product is : $p^\mu p_\mu=\frac{E^2}{c^2}+p^2$. But it is also said that $p^\mu p_\mu=(mc)^2$. How do we get the 2nd equation?

You are missing a minus sign in the first one

That may also clarify the second one :)

yes

We use the minkowski matrix to change an four vector from a contravariant to a covariant

and?

And that's where the minus sign in $E^2/c^2$ comes from (you have it as a +)

Ok I understood that part

It was a typo the +

but the 2nd equation

how do I get that?

Ah, that's just the definition of the proper mass then, which gives the second expression

Proper mass?

yes

that's what we get

when we equalize both eq. that I wrote above

But the first one has a wrong sign

You don't get that from the first one, because of the wrong sign...

I mean

with the correct sign

With the correct sign the second expression is immediate

the above two eq. once you equalize them, you get the relativistic expression

I simply wanted to know

$p^\mu p_\mu=(mc)^2$

Since I have no idea

I'm really not sure where is your difficulty at this point

$p^/mu=(\frac E c,\vec p)$. By knowing this, than for $p^\mu p_\mu$ we get $\frac{E^2}{c^2}-p^2$, which is understandable and expected. But this $p^\mu p_\mu=(mc)^2$, where does it come from?

@imbAF First, your notation is confusing. If you use $p^\mu$ to denote the 4-vector, then in ordinary notation everyone would understand $p^2$ to refer to $p^\mu p_\mu$, but you mean $\vec p^2$. Don't omit the arrow, it matters.

Ohh I see... well, I think it comes from first considering a mass that's at rest, so the four vector only has the energy component which is sometimes denoted $m_0c^2$.. then when you use a Lorentz transformation to a frame where it has momentum, we know that the magnitude of the 4-vector must be invariant

Second, at rest, $\vec p=0$ and $E=mc^2$, so $p^\mu p_\mu = (mc)^2$ in the rest frame and therefore in all frames.

But hold on

If the system has momentum, than it's total energy is that of the rest mass and whatever energy it has because of momentum

@imbAF The transformation doesn't really change momentum nor energy. It is just a Lorentz transformation. You need to imagine that the same length of the vector distributes differently among its components.

uhhhh

I don't think that's the right way to look at it

One second

a passive transformation doesn't change the 4-vector, but it does change energy and momentum when we view them as separate entities

right I'm sorry, it doesn't change the magnitude of the 4-momentum vector (which is a distribution on both, kind of) is what I should have said

$p^\mu p_\mu=\frac{E^2}{c^2}-\vec p^2$. Now you claim that $\vec p=0$ and by also knowing that the energy of the system at rest is $E=mc^2$ than, the above expression is ultimately $p^\mu p_\mu=m^2c^2$. And than, for this expression $p^\mu p_\mu=\frac{E^2}{c^2}-\vec p^2$ we don't make the same claim that $\vec p=0$. Where is the consistency and the logic behind equalized two expression where in one we assume no momentum and in the other we do?

@imbAF It doesn't seem consistent to you because you haven't really applied a lorentz transformation. You're just "out of the blue" claiming $\vec{p}\neq 0$ but a real transformation will also affect the energy component

@imbAF $E$ and $\vec p$ depend on the particular coordinate system you choose. The formula $p^2 = E^2 - \vec p^2$ (I'm omitting the annoying $c$) holds in all coordinate systems. I'm just saying that there is one coordinate system where $\vec p = 0$. In that coordinate system we arrive at $p^2 = m^2$ easily. Since both $p^2$ and $m^2$ are invariant under coordinate changes, this equation holds in all coordinate systems, even if we derived it only in a particular one

In particular in a Lorentz transformation $mc^2 \rightarrow \gamma mc^2$

I think a meta reason Gravity is fundamentalluly different from other forces wrt quantisation is that the physical time is determined by the field that we're quantising, rather than the minkowski metric that we perturb over

@ACuriousMind even if in an another reference frame $\vec p \neq 0$, you can still use that $p^2 = m^2$?

That is what I just said, yes.

Ok

I mean this about the physical time of the classical theory. This way, gravity is fundamentally different

So we shouldnt expect the usual tricks to give the correct quantum theory . This is just an opinion from a meta-perspective @ACuriousMind

@RyderRude As so often, I don't know what that means. The procedure of quantization - associating operators on a Hilbert space to classical observables - doesn't even need any notion of "time"

Yeah, but think about the classical theories on which that procedure is known to work. They all have one thing that gravity doesnt

we know in full generality that generally covariant theories like GR where there is no distinctive time coordinate generally become fully constrained when quantized

this does not, in general, have anything to do with the single technical issue we have in the quantization of GR, namely non-renormalizability

@ACuriousMind but we know of zero theories except GR where the corresponding classical theory has the field determining the physical time

I don't know what "the field determining the physical time" means

the whole point of relativity is that there is no universal time coordinate

In the classical GR, to determine the time measured along an oberver's worldline, u also need the dynamical component of the metric $h_{\mu \nu}$ @ACuriousMind

sure but what does that have to do with quantization?

This is y we shoudnt expect the usual quantisation tricks to work for GR. This is one way in which GR is unique

@ACuriousMind i'm just giving this as a meta-reason

"this is a meta-reason" is not an excuse for not actually giving any explanation

Like, think about all the theories on which that quantisation trick is known to work

what step of the usual quantization process cares about the proper time of observers?

@ACuriousMind lol i agree there :)

But isn't QFT basically a fully spacetime "embracing" theory? Including metric tensor?

and, again, the technical issue isn't that the quantization process somehow "fails" for GR, it just produces a non-renormalizable theory

@ACuriousMind u r right. The path integral quantisation doesnt care. Im just considering the nature of theories on which that trick has worked so far

producing a non-renormalizable theory is not "not working"

it's just "not producing a fundamental theory of nature"

Im saying that, even if it gave finite predictions, we shouldnt expect it to be physically correct

Becuz the underlying philosophy is missing. We are using that trick on a completely different beast

But i do have a concrete argument too other than this meta @ACuriousMind

I agree with literally nothing you just said

:)

I will try to give a concrete reason too

starting with the fact that you don't seem to have understood what non-renormalizability even means

it's not about "finiteness" of predictions

I thought it meant the infinities cant be gotten rid of by making the coupling constant dependent on cut off

What does it really mean

But i will try to put my concrete reasons into words

I need a few minutes tho

I'm not sure I can wait

I will respond in 30 mins :)

omg, ok... I am pending a juicy argument

@RyderRude in the Wilsonian view of renormalization (all theories are effective with an energy cut-off and renormalization is changing the cut-off) it means we can't lift the cut-off to infinity because the theory has operators/couplings that grow large at high energies and become small at low-energies, so extrapolating the high-energy behaviour from the low-energy behaviour is very difficult.

As you try to lift the cutoff to infinity, you find you'd need infinite amounts of information to actually compute that limit

in the "old" view of renormalization as getting rid of perturbative infinities, all theories have renormalization parameters, like mass or the fine structure constant, that you need as experimental input to the renormalization process. For non-renormalizable theories, the number of such parameters grows without bound as you increase the order of your perturbative computation, meaning you'd need infinite parameters for infinite accuracy

but that doesn't mean you can't use the theory in regimes where your cutoff isn't so high or where you only use the first few orders of perturbation theory

plenty of non-renormalizable theories can be used in practice to gain useful results, the textbook example is the Fermi theory of weak interaction

@ACuriousMind is all this related somehow to the popular'ish statement that at least it's partially because all of the integrals / limits treat space(time) as infinitely divisible?

I agree. In Gravity's case, by "u cant get rid of infinities", I meant that the quantum theory is completely useless beyond planck scale

:63278064 yes, but that is something very different from the quantization process itself somehow being defective, which is what you said

the quantization process is fine

it just doesn't produce a renormalizable theory

I got a concrete reason for why we shudnt expect the quantisation process to be fine either

that's why quantum gravity approaches aren't "let's try a different quantization method" but instead are genuinely theories beyond ordinary quantization

Yeah

So the approaches r already following my philosophy :)

Good thing you agree otherwise we would have to throw out all of quantum gravity

@Amit in some sense yes but probably not at all in the way you think when you hear that statement

Oh ok, devil is in the details as always

at its very root renormalization is related to idea that our field Lagrangian contains terms like $\phi(x)^2$, where we multiply two fields at the same spacetime point

Ok here's a concrete reason : First, we gotta agree that path integral quantisation isn't the deepest theory becuz it doesn't have finite time evolution built in. So we're actually looking at something like algebraic qft as our fundamental theory. Now, algebraic qft would do well with the quantisation of non-gravity theories. But it would face the "problem of time" in case of general relativity, becuz algebraic qft uses a spacetime, unlike path integral quantisation

The spacetime of algebraic qft has finite time built into it

I don't agree with any of this

@ACuriousMind We must always consider all quantities that are at the same point right? I mean, fields define quantities at a point so... it sounds reasonable but I don't know. Or are you saying that this only creates an issue when certain quantities "meet" that way?

I don't agree with your continued insistence that some formulations of QFT are "deeper" than others, it's very obvious you don't actually know what you're talking about with AQFT (AQFT on general spacetimes is a small but active research direction) and of course path integral quantization requires a spacetime

But path integral quantisation just integrates over all of it

the path integral does not integrate over spacetime at all

the problems of quantizing gravity are of a technical nature and I really don't think it is useful to try to talk about it with such disregard for accuracy

@Amit I'm saying this is a technical issue - there is no problem here when $\phi$ is a function

but for technical reasons the fields in QFT must be distributions, and trying to multiply distributions is problematic

one formal approach to renormalization traces the problems back to this root cause and phrases the choice renormalization parameters in terms of choosing how to actually define such products

ah, interesting. Distributions are almost certainly idealizations to physical quantities we want to "smear" over spacetime in a certain way to get consistent results... it does look suspiciously over idealized

and if we didn't have the infinitely divisible space, the canonical quantization wouldn't have a $\delta$ distribution but a discrete $\delta$ which is more like a function

is that the dirac delta?

that's why I'm not willing to say that the idea that the problems are related to the continuity of space is wrong, but I don't really want to agree either because without the technical background I think it produces a wrong picture

on a more general level, all the problems with QFT are just because the fields are "too many degrees of freedom"

QM with finitely many $x$ and $p$ is fine, but it all goes wrong once we want to have uncountably many of them

Haag's theorem, renormalization, it's all just punishment for not being content with finitely many d.o.f. :P

QM is certainly much tidier in its discrete form

It's the discrete automata giving us hints, again

@ACuriousMind The problem I have with path integral quantisation is that it merely allows mathematical derivations of non rel QM and classical field theory, but mathematical manipulations are not justified logically. e.g. when u derive rel. QM by dotting with the $\langle x |$, what is it about original path integral theory that allows this function of $x$ to be interpreted as probabilities, under some approximate physical conditions?

Becuz the original path integral theory only defines scattering probabilities as observables

All these problems go away if u derive rel. QM using finite time QFT. Especially, the derivation of Classical field theory which is logically justified becuz we're just taking expectation values of the probabilities

But the problem with path integral qft is that it defines scattering probabilities for "large time" experiments as ur observables. It doesnt even define how large the time should be. The math of it only talks about infinite time

@RyderRude I already pointed out at least once that the Schwinger-Keldysh formalism of path integrals allows investigation of time evolution of non-equilibrium states

and "non-equilibrium" really means just "states that change with time", you should not recoil from this just because it sounds like statistical mechanics

how do you quote an earlier chat message?

@Amit click on the little arrow to the left of the message when you hover with the mouse over it, then copy the link under "permalink" and just post a message that consists of only that link; chat automatically replaces it by a quote of the linked message

Ah, okay, thanks :) I thought it may work

Just checking it out lol

@ACuriousMind Is this formalism like a generalisation of the usual path integral formalism which only defines scattering probbailities as observables? I'm looking for a unified umbrella as the deepest known theory

Or is it like a separate formalism to deal with different physical situations

I'm looking for one formalism to derive all physical phenomena minus gravity

I'm not sure you can derive what a cat's meow sounds like using QFT

If it was possible the standard model wouldn't be so "ugly" as they say it is right?

@Slereah lol

@Slereah I mean all minus gravity and qualia :)

I mean more the practical side :p

you could theoretically do it

It's well known that cats mainly meow about gravity preventing them from reaching the cheese that's in the cupboard

But people don't use QFT to build skyscrapers

Qualia is a very good name for a cat

@ACuriousMind ok but if we take this formalism as the deepest theory to derive both rel. QM and scattering, then my technical problem with GR does pop up in this formalism. Becuz this formalism is not integrating over all of time. It is speaking of finite time. And in GR, the observed time would depend on the dynamical field

That, is unless this formalism is derivable from the original path integral formalism which just integrates over all of time and only defines "large time" scattering probabilities as observable

Isn't this one of the problems dirac solved to get the Nobel?

The first part is just a taylor expansion isn't it?

I'm speaking about the middle part

Relativistic QM... yes I think so

@nickbros123 yeah bur i think this particular question expects u to derive somethin like the klein gordon eqn, not dirac eqn

Ooh its also talking about the probability conservation issue

Ye its about the KG eqn

I don't think I'm ready for this lol

It does say you should look at another problem, this 1.36 thing, in tackling the second part. So it may subdivide the problem to simpler parts

It would b pretty crazy if a textbook exercise expected u to come up with dirac eqn lol

Why? Doesn't it just amount to applying a different H in the schrodinger eqn?

No, u gotta make up steps on ur own. Like u want a first order square root of the of the $\sqrt {p^2+m^2}$

The usual step wud b to Taylor expand it. Which gives crazy order derivatives

Yes, but that's already spelled out in the exercise

But Dirac took a non-trivial step here

That's how you get a nobel prize

you don't win a nobel prize for trivial steps

@Amit it's not. Lemme see

No it doesnt @Amit

It just asks u to make operator replacements to the einstein relation

This is so straightforward Schrodinger thought about this before the actual schrodinger eqn

Ah, so the Dirac equation is not just applying this replacement?

Yeah

It's non-trivial way to take a square root

That's how you get the spinors?

Yeah

That's how Dirac got spinors. There r better ways

Ok so he should get a Nobel prize

Like representations

lol I just heard a story about Dirac, when someone told him about parity violation... he goes "ohh.. parity conservation, that's not in my book, is it?"

I think that gossip is non renormalizable