I don't see that he's saying that. AdS/CFT never made any claim that our universe should be AdS. Their claim is that the duality defines a quantum gravity theory, which happens to live in AdS. Hopefully, gravity in our world works more-or-less similarly to how it works in AdS, at least as far as...
It is actually quite similar to the standard Transverse Traceless (TT) gauge. It might even be identical; I'm not sure about that. Kind of ironic that Weinberg calls his gauge "too ugly to deserve a name"! Was TT gauge in use for gravitational waves, back in 1965?
Anyhow, none of the classical...
Hey, good to see you're still around! I've been away from PF for a while, but when I come back I get my first response from an old friend!
DeWitt's work is certainly very central and powerful, but I'm specifically interested in the "Coulomb-gauge" approach developed by Weinberg in that paper. I...
In this 1965 paper by Weinberg, https://journals.aps.org/pr/abstract/10.1103/PhysRev.138.B988, he describes a quantum field theory of the graviton in a Coulomb-like fixed gauge, where the free graviton has only space-space components and is traceless. This of course makes the field dynamics...
In the article, you mention the hope that a "suitable summation scheme" will be found for Causal Perturbation Theory, thus proving the rigorous existence of ##S(g)##. To me this hope seems unsupported and wildly optimistic. Remember that this is a power series in ##g##, so we need is a summation...
I think my post does belong on this thread. The original post was a paper saying that we should consider "the real QFT" to mean QFT-with-cutoff, rather than a supposed continuum theory. That is the point I am trying to argue here: that it is likely that QFT does not have a rigorous UV limit at...
Because some comments have been moved, I will reproduce the reevant part of the conversation before responding:
Okay, I admit that this axiom-based approach is more successful than I gave it credit for. It's not just "here is a well-defined algorithm that that reproduces what physicists...
What? I am talking about processes like ##e \rightarrow e+ \gamma.## When ##g=1## this cannot happen because of conservation laws, but otherwise it should occur already at first order in perturbation theory.
But this raises another issue: Per the axioms, ##S(g)## should not take us out of single-particle subspaces. But without 4-momentum conservation, won't a single fermion have an amplitude to spontaneously emit photons?
##g## is the test function that switches on the interaction, correct? So anything first-order in ##g## is also first-order in ##e##.
Oh, I think I see. You are saying that since ##g(x)## is not translation-invariant, energy and momentum are not conserved by ##S(g)##, and so first-order...
This statement needs to be corrected. It indicates that the S-matrix for QED includes a term of first order in ##e##, when in fact the first term is of order ##e^2##. There are no one-vertex processes, because it isn't possible for all three particles (two fermions and a photon) to be on-shell...
What I would like from a theory of QFT-with-cutoff is a rigorous formulation and proof of the idea that the details of the cutoff do not matter. For that, I think we need a cutoff-independent formulation of the concept of a cutoff itself, and of the fields subject to it.
I don't know very much on these topics, so please enlighten me if I'm wrong. But if I understand correctly, Epstein-Glaser and the like merely describe in precise terms how renormalization is to be done, without making any attempt to justify the procedure "from first principles". They don't...
How long must a research program run without results, before we are allowed to say it has "failed"?
The basic problem is by now well-known: if the fields are operator-valued distributions, then defining interactions as products of fields at a point doesn't make sense, because distributions...
His basic point is that our working assumption about "reality" should be that there is some UV cutoff, with QFT as an effective theory per Wilson. This because the long effort to construct a mathematically sensible, UV-complete QFT (in 4D, with interactions) has failed, so perhaps such a theory...
You don't need an equivalence relation, you just need an equality relation. And the equality relation is a logical primitive - ##x\ne y## is treated as a well-defined proposition automatically. You do need axioms to tell you which things are or are not equal, but the equality relation itself is...
In "pure" set theory, the elements of sets are themselves sets. In that case the issue of distinguishability is settled by axiom: if every member of A is a member of B and vice versa, then A=B. This is the Axiom of Extensionality. So there cannot be "indistinguishable" sets: sets can be...
Yes, of course. That was a simple mis-type.
I am not trying to add anything new. I'm sorry if my notation made this unclear. I am just asking whether we can get a counterterm that depends on ##p^2## by simply multiplying the ##\partial^\mu \phi \partial_\mu \phi## term in the Lagrangian by an...
Well, I think I just provided a derivation where this is not necessary...
This is indeed an issue that also bothers me quite a bit. The ##Z## factors are defined in terms of a real inner product between normalized states, so ##0\le Z\le 1## is a physical restriction, but in fact ##(1-Z)## is...
If we have that much freedom, couldn't we just add an interaction term proportional to ##p^2+m^2##, without also multiplying the vertices by ##\sqrt Z## factors, and without the restriction ##0\le Z\le 1##? I think we cannot change the factor before the derivative term. In a theory with only one...
I'm afraid I didn't really understand any of your answers, so here is my own attempt. Please let me know whether I'm making sense.
It seems to me that we don't need to talk about rescaling the fields at all. We can produce the counterterms using only the canonically normalized fields...
I'm sorry, I still don't understand. Would you mind spelling this out in detail?
Yes, this agrees with my current understanding. But it brings us back to my original question: the free part of the Hamiltonian equals the sum of frequencies of particles, if and only if it is written using fields...
So let's just say that the standard convention ##\hbar=1## eliminates the freedom to change the unit of time without changing the unit of energy, and that if you change both, you don't get new factors in the Hamiltonian.
At any rate, in renormalization theory we certainly are keeping ##\hbar##...
I don't understand. You are saying that as long as we treat ##Z## as finite, the renormalized fields do satisfy the CCR without numerical factors? How does this make sense? We "rescale" the field without changing the commutator?