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  2. Nowhere. You made no mistake. None whatsoever. You are just realising that Schwartz made a mistake, not you. So it was a typo. He probably meant to write something like, \[ \partial^{\mu}\left(\sum_{n}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi_{n}\right)}\partial_{\nu}\phi_{n}-g_{\mu\nu}\mathcal{L}\right)=0 \] which is correct, and consistent with your derivation. That's the problem with books that don't follow the covariant/contravariant convention. Index gymnastics does that for you automatically. Sorry, I thought I'd told you:
  3. So, as I said, I think the piezo draws water in and forces it through the perforated piece as it oscillates
  4. 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?
  5. Today
  6. As matrices, they are. But as tensors, they aren't. They are one and the same basis-independent object, coding the same physical information. This is exactly the same as a vector in one basis looking, eg, like matrix (0 1 0 0) but looking like (0 -1 0 0) in another basis. You are confusing the tensor with its coordinates. IOW: All metrics, no matter the dimension and signature, look exactly like the identity matrix (the Kronecker delta) when the scalar product is expressed under the convention that the first factor is written in covariant components, and the second one in contravariant ones. A tensor is a physical object. A matrix is just a collection of numbers used to represent that object. Let me put it this way: \[ U_{\mu}V^{\mu}=U_{\mu}\left.\delta^{\mu}\right._{\nu}V^{\nu}=U^{\mu}\left.\delta_{\mu}\right.^{\nu}V_{\nu}=U^{\mu}g_{\mu\nu}V^{\nu}=U_{\mu}g^{\mu\nu}V_{\nu} \]
  7. They are different: \[\delta=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\] \[g=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}\]
  8. The PZT is of annular shape about 5mm id and 12mm od;, its center void has attached the stainless disc which found to have microperforations and emits/sprays the mist on its top; contacts the water at its bottom. The misting image shows the PZM on top of a wet sponge instead of floating on water. Overview of the little white buoy ---> https://www.youtube.com/watch?v=PlRgi6Mn6CQ
  9. All the time Hayley is in the news dilutes and delays Trump's desire to just focus on everything else and, of course, himself. Her best shot is to be a persistent arsefly until she really is out. She should look at this as training and building credibility for 2028.
  10. Oh. Got you. Yes, you're right. It should be what you say. Classic books in QFT tend to be rather fast-and-loose with the indices. \( \partial_{\nu}\mathcal{L}=\partial_{\mu}\left(g_{\mu\nu}\mathcal{L}\right) \) is not a tensor equation. \( \partial_{\nu}\mathcal{L}=\partial_{\mu}\left(\left.\delta^{\mu}\right._{\nu}\mathcal{L}\right) \) is. Although I should say there is no fundamental difference between \( g \) and \( \delta \) really. \( \delta \) is just \( g \) (viewed as just another garden-variety tensor) with an index raised (by using itself). Bogoliubov is similarly cavalier with the indices if I remember correctly. Is it from the 50's?
  11. Yesterday
  12. 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. No, no typo. It is actually a theorem (or lemma, etc) of tensor calculus that the gradient wrt contravariant coordinates is itself covariant. It is just a fortunate notational coincidence that the "sub" position in the derivative symbol seems to suggest that. Proof:
  14. 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 )
  15. Is the perforated material part of the PZT, or above it? I could envision water going into a gap and being forced through the holed by an oscillating piezo, producing a mist. I think stainless steel is a substrate or housing, and not a piezo itself.
  16. Hi all. Bought several tiny nebulizers to humidify a germinating room. Testing one for two weeks, ceased to produce the fog. Do not know how to revive. By cleaning sort of worked at half production for an hour. Read about and found : ---> https://www.electroschematics.com/mist-maker/ But the surprising specification says "5um" "740 holes" So I projected a laser to the stainless piezodisc and yes, shined trough the supposed-to-be holes in metal (5 micrometres per above) invisible to eyes or magnifying lens. Confirming it is a microperforated disc. Never saw a piezoelement perforated. How is the principle of operation of such nebulizer ? I have other piezo nebulizers with solid discs.
  17. I don't anymore read books that don't have letters..
  18. Facts don’t pierce the cults bubble. There is only one godhead, his skin is the color or Doritos, and his enemies are trying to hurt me and my family. All praise. Jonestown koolaid.
  19. That's pretty good advice for mortals like you, except for the Harry Potter series..
  20. This is fairly recent, Feb 4: (my bold to emphasize an example in a key areawhere Biden is not getting as much credit as he might deserve) "Despite a growing economy and little opposition for his party’s nomination, President Joe Biden confronts a dissatisfied electorate and a challenging political climate nine months before he faces re-election, according to a new national NBC News poll. Biden trails GOP presidential front-runner Donald Trump on major policy and personal comparisons, including by more than 20 points on which candidate would better handle the economy. And Biden’s deficit versus Trump on handling immigration and the border is greater than 30 points. The poll also shows Trump holding a 16-point advantage over Biden on being competent and effective, a reversal from 2020, when Biden was ahead of Trump on this quality by 9 points before defeating him in that election. And Biden’s approval rating has declined to the lowest level of his presidency in NBC News polling — to 37% — while fewer than 3 in 10 voters approve of his handling of the Israel-Hamas war. All together, these numbers explain why the poll shows Trump leading Biden by 5 points among registered voters in a hypothetical 2024 general-election matchup, 47% to 42%. While the result is within the poll’s margin of error, the last year of polling shows a clear shift." https://www.nbcnews.com/politics/2024-election/poll-biden-trump-economy-presidential-race-rcna136834
  21. It would. !!! The 1/2 factor doesn't change Lorentz invariance of the metric measure, but it's quite essential to the formalism that comes later. One would think we're done with negative energies/frequencies, and such. But no. They keep biting our buttocks later with the Fourier transform. That's where the Stueckelberg-Feynman prescription for antiparticles comes in.
  22. I watched the SC polls and really didn't see the bump I hoped to see, but then in the actual primary Haley seemed to go up by 5% and Trump down by about 2. Those were around the numbers I hoped to see going into the primary which I would have then hoped they'd go a little higher for Haley, and a little lower indicating trouble for Trump. Not sure what to make of it other than though disappointed I'm glad to see Haley seem upbeat and remaining in the race and continuing to campaign at least toward Super Tuesday.
  23. 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. 🙃
  24. I wonder, homework at your age and skills.. A new reincarnation has taken over your account..
  25. @Externet If you meant for the machine itself, there is a standpipe nearby that goes to the sewer and a separate(unused) pipe that runs through the wall to the backyard.
  26. But we’re not giving up covariance, are we? We’re simply considering how the EM field - a covariant object - decomposes in a particular frame. Is this not a form of non-locality? Basically you’re saying that whether or not a charge radiates in some local region depends on the existence of potentially distant sources (=external field). I need to think about this one first
  27. I guess what you mean is that the radiation field (and the EM field in general) will always be much larger than the local free-fall frame. There’s also the issue of the field “back-reacting” with the charge, which would make true free-fall impossible in the first place. These are good points, and I’m not sure how they influence the analysis of this situation. I should point out that I don't have a complete answer to the question. I'm just suggesting plausible possibilities as well as pointing out particular problems as I see them. That is, I'm not claiming that the radiation from a charge is a non-local phenomenon, but merely offer it as something that may nullify the equivalence principle. But it is also worth noting that equations in integral form may also have a differential form, a local form of otherwise non-local equations. For a charge in an external electromagnetic field, the acceleration of the charge must be accompanied by a back-reaction on the field. And this is both covariant and local. I’m struggling to understand this - why would the absence of a magnetic field contradict the charge not radiating? It would seem that I misspoke. What I meant was that an argument supporting the notion that a charged object sitting on a table doesn't radiate is the absence of a magnetic field. But this argument applies regardless of whether the observer is in the same accelerated frame of reference as the charge or is in an inertial frame of reference that is at instantaneous rest relative to the charge. I completely agree, and this insight should be all that’s needed to understand why some observers see radiation and others don’t. That's true. Give up covariance and the whole problem goes away. Seems like a Faustian deal to me. That’s fair enough - how would you yourself evaluate and understand this situation? As I said above, I don't have a complete answer. But I do see two possibilities that are not necessarily consistent: (1): Radiation only occurs if an accelerated charge is in an external electromagnetic field. That is, there is no radiation purely from an accelerated charge's own electromagnetic field. Such a hypothesis is not invalidated by cyclotron radiation, bremsstrahlung, nor Cherenkov radiation, but would explain in a covariant manner why some accelerated charges do not radiate. (2): The divergence of the quadratic expression in terms of the electromagnetic field tensor of the energy-momentum tensor gives not only the Lorentz force, but also some expression of the source in terms of an accelerated charge. This may actually be inconsistent with (1), but I think it is fair to assume that the Lorentz force law (along with its equivalent alternative expressions) is consistent with Maxwell's equations.
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