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1. ## A toy Feynmann diagram

Write down the next-order diagrams for the equation of motion $$\Box h - \lambda h^2 -J =0$$. Check the answer using Green's function method.
2. ## Symmetry breaking Lagrangian

Q: How many constants $$c$$ are there so that $$\phi(x)=c$$ is a solution to the equation of motion? A: Three. $$c=0$$ and two solutions for $$c^2= \frac {3!} {\lambda} m^2$$ Q: Which solution has the lowest energy (the ground state)? A: The potential energy from the Lagrangian is $\frac {\lambda} {4!} \phi^4 - \frac 1 2 m^2 \phi^2$It is the lowest for the non-zero $$c$$: $$- \frac{3!m^4} {4 \lambda}$$.
3. ## Symmetry breaking Lagrangian

This is a multi-step exercise. It would be very helpful if somebody could check my step(s) as I go. @joigus, I'm sure it is a child play for you. I'd like to make sure that I've derived correctly the equation of motion for this Lagrangian: $\mathcal L=- \frac 1 2 \phi \Box \phi + \frac 1 2 m^2 \phi^2 - \frac {\lambda} {4!} \phi^4$ The EL equation: $\frac {\partial \mathcal L} {\partial \phi} + \Box \frac {\partial \mathcal L} {\partial (\Box \phi)} = 0$ The equation of motion: $\Box \phi - \frac 1 2 m^2 + \frac {\lambda} {3!} \phi^3 = 0$ How is it? P.S. As edit LaTex does not work, I add a typo correction here. The equation of motion is rather $\Box \phi - m^2 \phi + \frac {\lambda} {3!} \phi^3 = 0$

5. ## Where was the symmetry?

I think, I got it. The symmetry validates the equation (3), because this equation makes the variation of Lagrangian a total derivative, and this makes the variation of action vanish: IOW, without the symmetry, we can't go from (4) to (5).
6. ## Where was the symmetry?

Rather than trying to fix the OP, I've prepared the text elsewhere and just post its image: A minor correction: the equation (4) above should rather be
7. ## Where was the symmetry?

Here are steps of derivation of energy-momentum conservation: Consider a shift of the field ϕ by a constant 4-vector ξ : (1) ϕ(x)→ϕ(x+ξ)=ϕ(x)+ξν∂νϕ(x)+... The infinitesimal transformation makes (2) δϕδξν=∂νϕ and (3) δLδξν=∂νL Using the E-L equations, the variation of Lagrangian is (4) δL[ϕ,∂μϕ]∂ξν=∂μ(∂L∂(∂μϕ)δϕδξν) Using (2) and (3), (5) ∂νL=∂μ(∂L∂(∂μϕ)∂νϕ) or equivalently (6) ∂μ(∂L∂(∂μϕ)∂νϕ−gμνL)=0 The conclusion is, "The four symmetries have produced four Noether currents, one for each ν : (7) Tμν=∂L∂(∂μϕ)∂νϕ−gμνL all of which are conserved: ∂μTμν=0 ." My question: where in this derivation the assumption was used that the transformation is a symmetry? P.S. I am sorry that LaTex is so buggy here. I don't have a willing power to do this again. Ignore. Bye.
8. ## test

$\phi(x) \rightarrow \phi(x+\xi)=\phi(x)+\xi^{\nu} \partial_{\nu} \phi(x) + ...$ $\frac {\delta \phi} {\delta \xi^{nu}} = \partial_{\nu} \phi$ $\frac {\delta \mathcal L} {\delta \xi^{nu}} = \partial_{\nu} \mathcal L$ $\frac {\delta \mathcal L[\phi, \partial_{mu} \phi]} {\partial \xi^{\nu}}=\partial_{mu} (\frac {\partial \mathcal L}{\partial (\partial_{mu} \phi)} \frac {\delta \phi} {\delta \xi^{nu}})$
9. ## Metric or Kronecker delta?

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.
10. ## Metric or Kronecker delta?

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?
11. ## Metric or Kronecker delta?

They are different: $\delta=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$ $g=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$
12. ## Metric or Kronecker delta?

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 $$\delta$$.
13. ## Metric or Kronecker delta?

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 )
14. ## What is this unit step function for?

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. 🙃
15. ## What is this unit step function for?

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 $$\frac 1 {2 \omega_k}$$.
16. ## Hawking radiation is produced at the black hole horizon, and other pop-science myths

A light that is produced by hot infalling matter between the photon sphere and the event horizon can still escape radially, right?
17. ## What is this unit step function for?

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.
18. ## What is this unit step function for?

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?
19. ## Hawking radiation is produced at the black hole horizon, and other pop-science myths

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".
20. ## Hawking radiation is produced at the black hole horizon, and other pop-science myths

Yes, for an audience that thinks that Event Horizon is an actual physical structure, there is no difference between an approximation and a myth.
21. ## Derive Lorentz transformations in perturbation theory

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?
22. ## Hawking radiation is produced at the black hole horizon, and other pop-science myths

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?
23. ## QFT and the SM by Schwartz, checking

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?
24. ## Do we share a little bit more DNA with cousins from our mother side?

We too, just touched it.
25. ## Do we share a little bit more DNA with cousins from our mother side?

OTOH, if you are male, you have X and Y chromosomes, while if you are female, you have only one X, since the other X gets inactivated (see Barr body).
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