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
@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
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
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?
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
@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
in the formulation where you think about tangent vectors as coordinate derivatives, this is as simple as saying that $\partial_x$ corresponds to $x\mapsto x+\epsilon$
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.
@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)
I just submitted this question, and I don't like the way my .png image renders. I often make diagrams in questions and answers using Microsoft paint, which produces a .png. In Physics stack exchange it displays always as big as humanly possible. Is there a way to make the picture display smaller?
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?
$p^/mu=(\frac E c,\vec p)$. 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
@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.
$p^\mu p_\mu=\frac{E^2}{c^2}-\vec p^2$. 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
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
@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"
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
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
@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?
that's why quantum gravity approaches aren't "let's try a different quantization method" but instead are genuinely theories beyond ordinary quantization
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
@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
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
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
@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
@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
@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
@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
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?"
