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May
30
comment Problem understanding sign of volume integral in Minkowski space
@Josh Yeah that seems to resolve this paradox. More generally, it appears that the integrals of Lorentz covariant real integrands can never be finite on Minkowski space {in particular, integrals of the form $\int d^4 p f(p^2)^2 p^\mu p^\nu$ can not be finite}, and we necessarily have to go to Euclidean space to make sense of these integrals by assuming p0 to be a complex variable and making use of Wick rotation. The reason that they are not finite is may be because any function of the form f(p^2) doesn't go to zero at infinty on Minkowski space.
May
30
comment Problem understanding sign of volume integral in Minkowski space
Your reasoning is perfectly fine but there is still a paradox here that I am not able to understand. From the OP's integral we can see that for $m_1=m_2$ the integral is positive for all $\mu=\nu$, simply because the integrand is positive and also the measure $d^4 k =dk^0 dk^1 dk^2 dk^3$ is positive; whereas if the value of integral were equal to $g^{\mu\nu} B_2$ then the sign of integral for $\mu=\nu=0$ will be opposite to that for $\mu=\nu=i,\: i=1,2,3$!
May
13
comment Astronomical Constant in Astronomical units?
You forgot a factor of $10^{-11}$ in expression for $G$.
May
12
comment Self-dual Maxwell equations, the second homology group, and topological invariants of a four manifold
Now, since on a 4-manifold **=1 on 2-forms, so the space of 2-forms (and hence in particular space of harmonic 2-forms) itself can be written as a direct sum of the space of self dual and anti self dual forms. So we can say that the second cohomology class of a 4-manifold is the space of solutions of self dual and anti self dual Maxwell equations.
May
12
comment Self-dual Maxwell equations, the second homology group, and topological invariants of a four manifold
Not sure about the "intersection form", but i think the relation with second cohomology (and hence also with homology by Poincare duality) follows from the Hodge theorem. From Hodge theory it follows that the second cohomology class (which is usually defined as the quotient of closed 2-forms by exact 2-forms) of a four manifold is same as the space of harmonic 2-forms i.e. 2-forms $X$ which satisfy $dX=0$ and $d^{*}X=0$, where $*$ is the Hodge dual operation wrt a fixed metric. These are precisely the Maxwell's equations.
May
4
comment The properity of $\mathbb{R}^4$ that has infinitely many differential structures is related to Yang-Mills field?
Relevant : 1. Donaldson theory 2. Seiberg-Witten invariants
Apr
2
comment Why do we use functional integration in QFT?
@AlexNelson Its true that we choose a time coordinate but in all the expressions time and spatial coordinates appear on equal footing. There are no expressions such as Exp(itH) in which time 'appears' to be different from spatial coordinates.
Apr
2
comment Why do we use functional integration in QFT?
@AlexNelson You may be right; I am not sure. However its true that in path integral formalism all the spacetime coordinates t,x,y,z are treated on equal footing which is more in the spirit of relativity. In particular, path integral formalism doesn't require the spacetime to be of the form Space X Time (as is necessarily required in Hamiltonian formalism) and hence can be used to define theories on more general manifolds including compact ones.
Apr
2
comment Why do we use functional integration in QFT?
Some of the benefits of path integral formulation are - i) Perturbation series is easy to derive compared to Hilbert space formalism ii) All objects in path integral formulation are classical; one doesn't need to introduce quantum operatorts iii) In Hilbert space formalism one needs to choose a time slice which breaks the manifest Lorentz invariance; this is not a problem in path integral formalism iv) After Wick rotation to Euclidean space, the path integral formalism is completely equivalent to the partition function formalism of stat mech; hence use of RG methods etc is more intuitive.
Feb
17
awarded  Nice Question
Jan
21
comment How to prove $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$?
@joshphysics - There is a finite multiplicative group contained in the Clifford algebra generated by gamma matrices. E.g. in two dimensions and with metric of signature (1,1), this group consist of elements $1$, $-1$, $\gamma^0$, $-\gamma^0$, $\gamma^1$ ,$-\gamma^1$, $\gamma^0 \gamma^1$ and $-\gamma^0 \gamma^1=\gamma^1 \gamma^0$.
Dec
28
revised Irreducible Representations Of Lorentz Group
added 51 characters in body
Dec
12
comment A key relation in Di Francesco's book on Conformal Field Theory
You are right about the second term on r.h.s of equation 6.174. From the standard definition of a primary field this term should be $(n+1)h z^n$ instead of $(n+1)h$. However, I am not sure how the author replaced $L_n \phi$ with $[L_n,\phi]$ in the second step in equation 6.174 ? This can be done for n>0, but for n<0, $L_n$ will not annihilate the state $|h,\bar{h}>$.
Dec
12
answered Proof that higher genus surface admits a metric of negative Ricci scalar everywhere
Dec
10
comment Where does the divergence in the $g\phi^3$ $d=4$ 3 point one loop diagram (three external legs) come from?
@David Z (Comment 2, after 21 hours) I am still waiting for any response from you. Can this question be considered for reopening ?
Dec
10
comment Angular momentum conservation in pion decay?
Thanks for the clarification :)
Dec
10
comment Angular momentum conservation in pion decay?
I was imagining bound state of quark and antidown quark as forming a kind of atom (like the hydrogen atom) with various angular momentum levels of which pion is one state. But I am not sure if this picture make sense for strong interation
Dec
10
comment Angular momentum conservation in pion decay?
Hmm.. but does the state l=1, S=0 exist for a bound state of up and anti-downquark ?
Dec
10
comment Angular momentum conservation in pion decay?
Since pion is a bound state, so could it be that the pion first goes to a state of angular momentum 1 before going to the final state via W+ channel?
Dec
9
comment Where does the divergence in the $g\phi^3$ $d=4$ 3 point one loop diagram (three external legs) come from?
@David Z can it be considered for reopening? I mean its not so trivial homework problem, and the OP has also shown some effort