Jump to content

Genady

Senior Members
  • Posts

    4976
  • Joined

  • Days Won

    50

Genady last won the day on December 20 2023

Genady had the most liked content!

Recent Profile Visitors

8397 profile views

Genady's Achievements

Genius

Genius (11/13)

636

Reputation

  1. Thank you. All's well. Yes, I like the book otherwise, but it would be so much easier to follow if the indices were where they should be.
  2. Please, I really, really know this. I know this index gymnastics, lowering and raising indices, tensors vs. basis representations, etc. I appreciate your time, but there is no need to teach basics here. Let's focus. Back to my question. Let's take \(\nu=1\). If \(\partial_{\nu} \mathcal L = \partial_{\mu} (g_{\mu \nu} \mathcal L)\), then \(\partial_1 \mathcal L = \partial_{\mu} (g_{\mu 1} \mathcal L) = -\partial_1 \mathcal L \). Where is my mistake?
  3. They are different: \[\delta=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\] \[g=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}\]
  4. I understand this but I don't think it answers my question. This is what I mean: I can rewrite (3.34) so: \[ \partial_{\mu} (\sum_n \frac {\partial \mathcal L} {\partial (\partial_{\mu} \phi_n)} \partial_{\nu} \phi_n) = \partial_{\mu} (g_{\mu \nu} \mathcal L)\] Then, from this and (3.33), we get \[\partial_{\nu} \mathcal L = \partial_{\mu} (g_{\mu \nu} \mathcal L)\] I think, it is incorrect. It rather should be \[\partial_{\nu} \mathcal L = \partial_{\mu} (\delta^{\mu}_{\nu} \mathcal L)\] P.S. Ignore positions of indices; Schwartz does not follow upper/lower standard. The difference is between \(g\) and \(\delta\).
  5. My question is about the following step in a derivation of energy-momentum tensor: When the ∂νL in (3.33) moves under the ∂μ in (3.34) and gets contracted, I'd expect it to become \(\delta^{\mu}_{\nu} \mathcal L\). Why is it rather gμνL ? Typo? (In this text, gμν=ημν )
  6. It is not technically homework, but it could've been if I were technically student. Just a new textbook to work on. I don't anymore read books that don't have equations. 🙃
  7. Just to answer the OP question, It would not. Without the step function it would be \[\int dk^0 \delta (k^2-m^2) =\frac 1 {\omega_k} \] rather than \(\frac 1 {2 \omega_k}\).
  8. A light that is produced by hot infalling matter between the photon sphere and the event horizon can still escape radially, right?
  9. Thank you. I got it. My mistake was that when I replaced \(k^0\) with \(\omega_k\) I've missed that it can be + or - \(\omega_k\). The step function is needed to kill one of them.
  10. The question: Show that \[\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) \theta (k^0)=\frac 1 {2 \omega_k}\] where \(\theta(x)\) is the unit step function and \(\omega_k \equiv \sqrt {\vec k^2 +m^2}\). My solution: \(k^2={k^0}^2 - \vec k ^2\) \(\omega _k ^2 = \vec k^2 +m^2\) \(k^2 - m^2 = {k^0}^2 - \omega_k^2\) \(dk^0= \frac {d{k^0}^2} {2k^0}\) \(\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) \theta (k^0) = \int_{-\infty}^{\infty} \frac {d{k^0}^2} {2k^0} \delta ({k^0}^2 - \omega_k^2) \theta (k^0) = \frac 1 {2 \omega_k} \theta (\omega_k) = \frac 1 {2 \omega_k}\) However, the point of the unit step function there is unclear to me. Wouldn't the result be the same without it?
  11. I agree, the vagueness of such statements is an issue. Next time somebody says, "Hawking radiation is generated just outside the event horizon", I will ask first, how far is "just".
  12. Yes, for an audience that thinks that Event Horizon is an actual physical structure, there is no difference between an approximation and a myth.
  13. I've arrived to an expected answer, but I am not sure at all that the process was what the problem statement wants. First, I considered \(0=(t+\delta t)^2-(x+vt)^2-(t^2-x^2) \approx 2t \delta t - 2xvt - v^2t^2\). Ignoring \(O(v^2)\) gives \(\delta t=vx\), i.e., \(t \rightarrow t+vx\). Keeping \(O(v^2)\) gives \(t \rightarrow t+vx+\frac 1 2 v^2t\), which is the correct expansion of the full transformation to the second order. Now, taking \(x \rightarrow x+ \delta x, t \rightarrow t+vx\) gives by the similar calculation \(x \rightarrow x+vt+\frac 1 2 v^2x\). Is it what the exercise means?
  14. No, I don't call approximations, myths. They are different things. Is it an approximation to say that all life forms were created at once from scratch?
  15. Checking with the physicists here: On the p.17 it says, Shouldn't it say force rather than potential? Isn't any potential rather quadratic close to equilibrium?
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.