Albatross

That is an easily provable fact.

There is no B>0 such that A>B for all A>0.

@user1286486 I am not sure what you mean by "tricky."

@user1286486 I see the problem. Zero is neither positive nor negative, so when we say positive we mean strictly greater than zero, and when we say non-negative we mean either positive or zero. It is correct to say that there is no positive $B$ such that $A>B$ for every positive real number. But that is not what I, or the proof given above, said. When dealing with limits and their rigorous proofs, there is a crucial difference between $>$ and $\geq$ for precisely this reason.

@user1286486 Yes, that is true. And that is why, if $B<A$ for all $A>0$ and $B\geq 0$, the only possibility is that $B=0$.

@user1286486 I don't understand why you think that is a contradiction. There is no positive real number which is less than all other positive real numbers, precisely for the reason you say. But zero (which is not positive) is less than all positive real numbers.

@user1286486 Those propositions do not conflict at all - the former follows directly from the latter! If $B$ is a non-negative real number, then either $B>0$ or $B=0$. $B$ cannot be $>0$ precisely because your second proposition is true, leaving only the possibility that $B=0$.

@user1286486 Which two propositions are you referring to?

@user1286486 In any case, I'm sorry to hear that you have not understood the proof. I thought perhaps based on your comments to my answer to your other question, you might have accepted the possibility that you have a series of significant misunderstandings about this topic, but it appears that you are convinced that you are right, and are not actually trying to learn about the subject. I only answer questions asked in good faith. Best of luck.

@user1286486 There is a proposition that “the only non-negative real number which is less than every positive real number is 0”. This proposition is tricky because there is a conflicting proposition that “every positive real number has a smaller positive real number”. That is precisely why this proposition is not tricky - if you assume that there exists a positive real number less than all other positive real numbers, then you immediately arrive at a contradiction.

In any case - it was far from clear that you were talking about a conformal projection of the Schwarzschild spacetime. There are many such projections you could perform in general, but at least that specification narrows down the list.

I think you are confused about what extrinsic curvature is. Extrinsic curvature arises when you embed a space inside of a larger ambient space; intrinsic curvature is a property of the space itself, with no reference to an embedding. The 2-sphere possesses intrinsic curvature in its own right, and extrinsic curvature when you embed it in 3D Euclidean space. Similarly, the Schwarzschild spacetime has intrinsic curvature, and would have extrinsic curvature if you embedded it in a higher-dimensional Euclidean space too.

Can you explain to me how you might do this "projection" on the 2-sphere, given polar coordinates $(\theta,\phi)$? How do I "project" them onto a cartesian coordinate grid?

I don't know why you think that, the 2-sphere with the standard metric $\mathrm ds^2 = \mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2$ has intrinsic curvature as a 2D surface.

By Cartesian coordinates, one usually refers to a system of coordinates such that the unit vectors $\partial/\partial x^i$ are orthonormal. Such a system cannot exist in curved spacetime. This is precisely analogous to the fact that you can't put Cartesian coordinates on the surface of a sphere. You can use coordinates called $x$ and $y$ if you want, but those are simply names; there is simply no way to give them the properties that familiar, flat-space Cartesian coordinates have.

For example, you can apply the same transformation which you'd use to translate spherical coordinates in flat space to cartesian coordinates, but the resulting $(x,y,z,t)$ are not going to be Cartesian - in particular, they will no longer by orthogonal, which is presumably one of the properties you are trying to preserve.

What do you mean by $(x,y,z,t)$-coordinates? Cartesian coordinates do not exist on a curved spacetime, so you'll need to explain precisely what $x,y,z,$ and $t$ are meant to be.

Can you give an example of what you're trying to say? For example, if your manifold is $\mathbb R$, what is the acceleration field?

and velocity cannot be determined from the phase space or configuration space - it constitutes a decision being made by the physicist writing down the Hamiltonian. It is an independent input to a model, so there is no procedure for "getting the correct Hamiltonian" that does not at some point involve choosing which physics you want to model and working out which Hamiltonian to use in that specific instance.

@naturallyInconsistent Indeed the relationship between velocity and momentum is obtained from the Hamiltonian, but it remains true that this is an input to the theory. If you use the Hamiltonian $H = p^2/2 +x^2/2$, your system evolves like a harmonic oscillator and $p= \dot q$. If you use the Hamiltonian $H = (p-A(x))^2/2 + x^2/2$, then that proportionality is lost and your system evolves like a harmonic oscillator in a magnetic field. The nature of the relationship between (canonical) momentum

Given a generic symplectic space with no a priori physical interpretation for the coordinates, it is indeed not obvious how to pick a Hamiltonian. My example with the 2-sphere was, in part, meant to illustrate that fact. But it's not a problem which has a fixed solution - you have the freedom to construct any model you like, and deciding on the most appropriate one for your situation is why physicists get the big bucks (/s)

Of course, a randomly chosen function won't give you the dynamics you're looking for, so then we need to be more precise. Experience with systems which have equivalent Newtonian or Lagrangian formulations tells us that often - but not always - we can write H = T + V where T is quadratic in the p's and V depends only on the q's. You can impose the requirements of various symmetries on the system, you could add perturbations or variations to previously understood models, and the list goes on.

The original question was, in effect, how to get the "right" canonical momenta given some position coordinates. The answer is essentially trivial - if you construct the cotangent bundle, a set of position coordinates naturally induces a coordinate system on the contangent bundle, and those are your q's and p's. Somehow the question of how to write down a Hamiltonian got mixed in, and the answer to that is similarly trivial - literally write down any smooth function of the q's and p's.

Sometimes you can get one from the other. But at some point, you yourself need to prescribe the dynamics of the system you're trying to model.

@naturallyInconsistent The replacement for guessing a Lagrangian and performing a Legendre transform to obtain a Hamiltonian is simply guessing the Hamiltonian from the start. Of course, to the extent that the Lagrangian isn't a blind guess (e.g. you could impose symmetry constraints) then you can reason in analogous ways about the Hamiltonian. But there's no magic here - you need to provide L in Lagrangian mechanics, or H in Hamiltonian mechanics.

@naturallyInconsistent [...] prescription to make this choice for you purely from the kinematical phase space (or configuration space, if that is your starting point)?

@naturallyInconsistent [...] there should be some unambiguous way to determine a Hamiltonian from this data, which is absurd. Let $Q=\mathbb R$ so $T^*Q \simeq \mathbb R^2$ with natural coordinates $(q,p)$ and symplectic form $\mathrm dp\wedge \mathrm dq$ as described in my answer. Let $H_1=p^2/2m$ and $H_2= p^2/2m + m\omega_0^2q^2/2$. If you choose $H_1$, then you get the free particle, whereas if you choose $H_2$ you get the harmonic oscillator. You are free to choose literally any Hamiltonian you wish (though they may be physically uninteresting), so how could you possibly expect some [...]

@naturallyInconsistent I used the example of the 2-sphere to demonstrate how a symplectic manifold gives rise to Hamiltonian dynamics. If you wish to label my example as uninteresting (despite its connection to QM in finite dimensions), then that is your prerogative. The main point, however, is that there are two independent inputs to Hamiltonian mechanics - the phase space on which the dynamics plays out, and the Hamiltonian function which generates those dynamics. Given a configuration space $Q$, we can construct the phase space as $T^*Q$, but you continue to insist that [...]

@naturallyInconsistent In your comment to Qmechanic's answer, you refer to figuring out the KE and PE, but one never simply "figures out" these things - they ultimately supply them. Conventionally we might make the Hamiltonian of the form $H(q,p) = p^2/2m + V(q)$, but that's not necessary (e.g. for electrodynamics). The "right Hamiltonian" is uncovered in the same way as the "right Lagrangian" - by figuring out which prescription reproduces the dynamics of the system you're trying to describe.

@naturallyInconsistent I don't understand your objection at all. $(\mathrm S^2,\omega)$ is a symplectic manifold, and so you can do Hamiltonian mechanics on it - and indeed it is precisely what we should do if we want to obtain spin-$j$ systems via geometrical quantization. The "problem you want solved" is precisely what comes above that section of my answer - start from $Q$, construct $T^*Q$, and define the canonical symplectic form. You obviously can't get the Hamiltonian from that procedure because you have to specify the Hamiltonian yourself.

Likewise, always good to straighten out my own thoughts by talking to other people. Cheers.

I'm not a high energy guy, so I'm probably not the best one to talk to for an intuitive understanding of the issues at play. Things make more sense in condensed matter - to me, at least. But I'm glad I could help a bit.

The recognition that the infinities which arise in QFT are, in a large part, due to these quantization ambiguities is the key insight which takes renormalization from a dirty trick to a brilliant procedure for overcoming our fundamental ignorance of high energies and short distances.

this also ties in with renormalization, adding counterterms to the Hamiltonian to cancel infinities, etc

so the fact that we perturbatively expand the interacting vacuum in terms of the free particle eigenstates is ... problematic

Right.

as well as constants $(a a a^\dagger a^\dagger = a^\dagger a^\dagger a a + a^\dagger a + 1)$

So at the end of the day we find that $\lambda \phi^4 = :\lambda \phi^4: + $ terms like $aa, a^\dagger a^\dagger,$ and $a^\dagger a$

However, $\lambda \phi^4$ (no normal order) also contains additional terms which arise from the commutation relations, i.e. $a^\dagger a a^\dagger a = a^\dagger a^\dagger a a + a^\dagger a$

Then you find that $:\lambda \phi^4: = aaaa + 4a^\dagger a a a + 6a^\dagger a^\dagger a a + 4 a^\dagger a^\dagger a^\dagger a + a^\dagger a^\dagger a^\dagger a^\dagger$

and then examine the difference between that and its normal-ordered counterpart $:\lambda \phi^4:$

What I mean is that if you take a term like $\lambda \phi^4 \sim \lambda (a+a^\dagger)(a+a^\dagger)(a+a^\dagger)(a+a^\dagger)$

Okay, let me rephrase, what I wrote doesn't really make sense and isn't what I meant

Yes, one second I need to compose my thoughts on that final point

It's a bit like checking to see if a function belongs to the Hilbert space by checking to see that its norm is finite. Objects which don't belong to the Hilbert space don't even have norms, but we still go through the motions.

With respect to your earlier reservation about computing inner products between vectors of different Hilbert spaces, perhaps you could think of it as assuming the vectors belong to the Hilbert space, computing their inner product under that assumption, and obtaining a nonsensical/contradictory result.

Yes, that looks right to me

The first term is the free Hamiltonian $H_0$, and the latter term is the interaction Hamiltonian

It's a big of a pain in the ass to carry the calculation out (though it is a useful exercise) but notice that if we have something of the form $a^\dagger a + \lambda (a+a^\dagger)(a+a^\dagger)(a+a^\dagger)(a+a^\dagger)/4!$

But what I mean with that last line on normal ordering is the following