Recent content by Roy_1981

I am not putting a dtdr term by hand, the metric with the dt dr term still respects spatial isotropy and homogeneity. It is easy to check it, just set dt=0, and the spatial section you are left with will be homogeneous and isotropic. Not sure what you mean by "usual coordinates", what I meant...

Thats actually not true, you can put a dtdr term, it does not violate spatial isotropy or homogeneity. dtdφ of course does, which is why I did not put in the metric I mentioned in my original post.

Choosing these coordinates for spatial slices was not my issue - I wrote them down explicitly as an example/illustration. I am well aware I can use other coordinates, say cartesian type coordinates x,y,z to label the spatial sections instead. I was more concerned about the dtdx pieces. That is...

Thanks but not what I am looking for. I get what you are trying to get at, that perhaps I can choose such a gauge regardless of the symmetry situation. But my question has to specifically do with symmetries, in particular global foliations So what I claimed in the question is correct, first...

Thanks but not what I am looking for. I get what you are trying to get at, that perhaps I can choose such a gauge regardless of the symmetry situation. But my question has to specifically do with symmetries, in particular global foliations of the metric in the presence of a certain set of...

Yes, I am aware of this as well, but he does not really make it any clearer. Instead, Just says it is a "powerful theorem" that such a choice can always be made and that "Proving the theorem is a mess", and then in turn refers to Weinberg's Chapter 13 which is on Maximally symmetric spaces...

Thanks. However it seems problematic to me that such a gauge (frame) can be chosen in all situations, as it would imply all stationary metrics ought to be static. E.g. if it can be chosen for the Kerr metric, then the Kerr metric would contain no dt dφ terms which would make it a static metric...

Thanks, I have looked at Carroll's notes in the past but I am not sure he explains why spatial homogeneity and isotropy implies the metric can be foliated R × Σ (to quote Carroll "this translates into the statement that the universe can be foliated into spacelike slices such that each slice is...

To arrive at the Robertson-Walker metric for a spatially homogeneous and isotropic cosmology, one first writes down the the metric for spatial sections i.e. constant t surfaces, dσ2 = d2 +f2(r) (dθ2 + sin2θ dφ2), where f(r) can take only 3 special forms, and then one promptly writes the...

"in QED you can (in principle, let us forget quantum anomalies for the moment) have particles of different charge. Those particles are going to transform differently under U(1) transformations, meaning that they belong to vector spaces with different representations of the gauge group." I see...

I agree that it does not matter what generators you choose, but given a particular set of generators the structure constants are fixed and so must be the same for each representation. Each fixed generator must map to a particular element of the Lie algebra (up to unitary equivalence) in order to...

Since the Lie algebra of an Abelian theory is commutative (every Lie bracket is zero), you can multiply the representation of the generator with an arbitrary constant and it will still satisfy the same commutation relations and vector space structure. Actually this is a bit more subtle. Even...

Thanks, this is what I was guessing as well (that the gauge self-coupling fixes the coupling with everything else and in the abelian theory there is no self-coupling to constrain other couplings) but the following sentence confuses me: Technically, particles with different charge in an Abelian...

Hi all, I am sure I am missing something really elementary, but I would really appreciate someone pointing it out to me. So, if we consider the situation in abelian gauge symmetry, say for fermion matter ψ, of charge q. The transformation law for ψ is, ψ→ψ' = e[- i q θ(x)] ψ. We then have to...

Hi George Jones, Indeed you are in the right direction. In case of gravity you can integrate Einstein equation across the shell to obtain junction conditions, relating metric derivatives to matter on thin shell. But what I asked is a bit different - its about a choice of gauge...