Recent content by Haorong Wu

Very sorry for the confusion. In my manuscript last year, I typed ",i.e." in LaTeX. However, in the proofs, I notice that they are all replaced by "-i.e.," by the editor, as in the following figure. Therefore, in my recent manuscript, I typed "---i.e.," in LaTeX. However, in the proofs, they...

Last year, I submitted a manuscript to PRD. I noticed that all ",i.e.," are modified into "-i.e.," by the editor. Therefore, in my recent manuscript, I typed "---i.e.," in LaTeX. However, all the "-i.e.," are modified again into "-i.e.," by the editor. I could not distinguish the difference...

Hi, @Ibix, and @PeterDonis. I am recently interested in boosting a scalar field, so I found this paper, Relativistic Hall Effect. I am sorry it is not open-access, so I am not sure whether you can access it or not. The mode functions, Eq. (9), are a little complicated as the authors use the...

Suppose we can boost from a frame ##S## to another frame ##S'## by using a Lorentz transformation ##\Lambda##. Also, ##\phi(x^\mu;\omega,\mathbf k)## is a mode function of a scalar field in frame ##S##. Then, how do we express this mode function in frame ##S'##? Here is my attempt. First, the...

In Peskin's textbook, the coupling factor is given by a constant in the interacting field theory. The scattering matrix ##S## is given by the time-evolution operator, ##\exp(-iHt)##, in the limit of very large t, i.e., ##t\rightarrow \infty##, as expressed in Eq. (4.71). In my mind, the...

@PeterDonis , from the image in the book, it appears that ##\mathbf r## is not the relative position vector. I may solve the problem. Denote ##\mathbf x=\mathbf r-\mathbf R##. Then, ##\nabla_{\mathbf R}\psi_n(\mathbf r-\mathbf R)=(\frac \partial {\partial R^1}\psi_n(\mathbf r-\mathbf...

In studying the Aharonov-Bohm effect, a model of an electron confined in a box is used, for example, on page 353 of Modern Quantum Mechanics by Sakurai et al. The box makes one turn along a closed loop surrounding a magnetic flux line. In the derivation, there will be an integration involving...

Thanks, @renormalize. I mistakenly thought the integral value was somehow related to the integration in the equation.

In the book, quantum fields in curved space, when calculating the vacuum energy divergence for scalar fields, it reads: I could get the answer by letting ##k=m\tan t ## and using the properties of Beta functions and Gamma functions, but I still do not understand what it means by saying "with...

Thanks, @Demystifier, @PeterDonis, @Sagittarius A-Star. Below is my attempt to understand this question. In the exponential, ##e^{-i\omega x^0}## of a mode function, the factor ##\omega## before the time coordinate ##x^0## had better be (or at least be related to) the frequency (energy)...

I am reading a paper, A Pedagogical Review of Black Holes, Hawking Radiation and the Information Paradox. On page 17, it reads that and I am not convinced that the two sets of coordinates are associated with different observers. I think the coordinate systems are independent of observers...

Thanks, @Philip Koeck and @pasmith. I will try to demonstrate the first expression. Suppose ##F(\omega)## is the Fourier transform of ##f(Q)##, i.e., ##f(Q)=(2\pi)^{-1/2} \int d\omega F(\omega) e^{-i\omega Q}##. Then the integral \begin{align} &~~\lim_{r\rightarrow \infty} \int_0^\infty dQ f(Q)...

Hi, there. I am reading this thesis. On page 146, it reads that I do not know how to calculate the limits when they are viewed as distributions. I am trying to integrate a test function with the limits. So I try (##Q## is defined as ##Q>0##) $$\lim_ {r\rightarrow \infty} \int_{0}^\infty dQ...

@Ibix Sorry, I thought that is not important, so I did not mention it. Here are the parameters (all are in Cartesian coordinates): mass of BH is ##1.988\times10^{30}~\rm{kg}=1.47\times10^3~\rm{m}## ; angular momentum per unit mass is ##0.9## (along z-axis); position of BH is ##(0,~0,~0)##...

Hi. I use Matlab to simulate that two parallel light rays pass near a Kerr BH. The angular momentum of the BH points to the ##z## direction. The ##z## components of the start points of the two rays are ## 1\times 10^3 ~\rm{m}## and ##- 1\times 10^3 ~\rm{m}##, respectively. The result, as shown...