@PM2Ring Absolutely by chance I’m re-reading the seminal paper by Fredkin and Toffolli…

Do yall have any recommendations for a linear algebra book that covers content beyond introductory linear algebra

@SillyGoose contents like what?

There’s a nice text on Matrix Analysis by Horne and Johnson that I like a lot.

cambridge.org/us/universitypress/subjects/mathematics/algebra/…

Maybe like a "definitive text" on linear algebra content applicable to quantum theory (not just textbook quantum mechanics) @ZeroTheHero

@SillyGoose But the linear algebra used in basic quantum theory is not a lot; if you go into density operators and so forth you might need to think of symmetries and all, but otherwise it is still not the kind of stuff that mathematicians would have to fret a lot about.

@Mad This is doomed. The H atom Coulomb potential is purely real valued, but the eigenfunctions are not real-valued in the Lz = m part.

Wait a minute, it's quite clear at this point to me that in the macroscopic situation, some self energy terms are lost. There may be a way to show that this is negligible for simple substances , perhaps non dielectrics?(clarification needed). (My reasons are simply that textbooks simply spam that formula everywhere) But I read in a paper arxiv.org/ftp/arxiv/papers/1312/1312.3383.pdf that one of the key macroscopic energy relations is actually postulated?

Basically the time derivative of energy density formula

@nickbros123 Link to the abstract page, not to the pdf itself. I told you before that we may not know in advance which of E D B H appears in Poynting theorem and thus have to postulate. It is much worse than this because potentials can also come in, and the very distribution of energy can change when you change the definitions.

I dont deserve this on a Sunday morning

@naturallyInconsistent my whole life has been a lie. Lorenz force law doesn't always work with that neat expression $ qE + qV\times B $ in case of macroscopic objects in ponderable media?

In link.springer.com/chapter/10.1007/978-0-387-34730-1_4 he derives in page no 178 the macroscopic Lorentz force relation , the averaged microscopic force density. It looks like a car accident with E B H P all in bits and pieces here and there

Sad thing is this is rather a neat looking formula in comparision to whatever's in the rest of the book

I think any author claiming to have solved this without first pointing out that there are a dozen incompatible schemes flying around, is just overconfident and not understanding the scope of the problem.

Oh fuck, I'm reading A&M on the topic and they said that the averaging procedure for covalent solids is different from ionic and molecular solids.

Lol

One should make a periodic table for averaging procedures

Hello guys

if i have for example a potential well

And find three solutions to three different regions

i want to apply the argument, that the soltuions are even or uneven due to symmetry

How can i apply this if i have three functions?

Say $\psi = \psi_1 $ if smaller than -a, $= \psi_2$ if between -a, a and $\psi_3$ if bigger than a

Does every function need to be even or uneven?

ok so i guess one can rerepesent $\psi$ as a sum of the three wave functions with step functions to get the zeros

Is the strike still ongoing? The close votes are 40+ by now, lol

@nickbros123 your newest question runs into the problem of defining the energy for the EM field. There is a tremendous amount of choice involved. Also, things might be easier in the quantum realm than in classical.

@naturallyInconsistent I just need an answer, or to know that an answer exists, or doesn't exist, what ever may be, so that I can put a closure for now, to this and continue my studies

in this question, qmechanic says that for any sufficiently nice function, we can use linearity to show that the covariant derivative of that function is zero, but $\sqrt{\vert g \vert}$ is not really "sufficiently nice" enough to apply linearity and leibniz? is there something im missing to this? physics.stackexchange.com/questions/249961/…

U can ask qmexhanic themselves. a nice person

@Qmechanic hehe

@Relativisticcucumber I think Ryan Unger and Petra Axolotl's answers are closer to what is actually being meant. When we say $\sqrt{|g|}$ in the basic scheme, it really should be $\sqrt{|g|}\mathrm dx^t\wedge\mathrm dx^x\wedge\mathrm dx^y\wedge\mathrm dx^z$

hm so how should i interpret $\partial_{x^{\sigma}}(dx^t \wedge dx^x \wedge dx^y \wedge dx^z)$

It is a little cube

Oh there is a derivative

It is the change in volume as you go along some direction, I guess?

yeah because ultimately i am trying to show (stupidly imo) that that covariant derivative of levi civita tensor is zero so i think i can use what naturally inconsistent said if i can show that the term i mentioned above is zero

even tho idk why one would formulate the levi civita symbol in this way in the first place bleh

@nickbros123 I don't think your question can be answered (at least in a short way) in the form it is being currently asked. Consider a system of charges similar to how a H atom is, with a positively charged finite ball of quasi uniform charge density, surrounded by a negatively charged finite ball of quasi uniform charge density, up to a certain bigger radius R, forming a neutral ball. Ignoring that the charge density actually would leak outside of R,

the $\frac12\varepsilon_0E^2$ integral puts the energy of interaction mostly in the space outside of the ball, whereas $\frac12\rho\phi$ integral says that you can ignore the integration outside of R because there is no $\rho$ there.

@Relativisticcucumber Actually, I would point out that the volume form has to come with $\sqrt{|g|}$ or else it is actually frame dependent, and that is a no-no. If you think about what the levi-civita symbol is really doing, it kinda makes more sense that it is coming from the volume form than not. Same thing with the Hodge dual and all.

@naturallyInconsistent they should give you the same answers right

@nickbros123 Only because we want it so by definition. You know what is the most convincing definition of what the energy should be, right?

Well in classical EnM the true energy is the summation of interaction energies of point particles, and the motivation to the $ \frac{1}{2} \rho \psi $ is that upon plugging the point charge rho and psi, we get the old summation (along with a divergent term that I ignore). From this pic one can go to the E^2 integral.

@naturallyInconsistent I don't know, to me apart from the summation of point charge interactions everything seems suspect now

@nickbros123 For the longest time I am one of those who prefer to talk about potentials. Until one day I saw that in our numerical simulations of the energy in quantum systems that the standard formalism is better. That made me realise that the correct definition of interaction energy is in the slowly bringing charges in from infinity, integrating the forces. You use that to define both the potential and the E^2 scheme. So, we cannot possibly know which is better or more correct, bcos they are same

Yes, bringing things from infinity and calculating work is the summation of interaction energy of the bunch of point charges. The $\frac{1}{2} \rho \psi $ integral gives us the old summation upon plugging the discrete variables (ignoring the blowing up self energy) so that's a decent motivation to write that down, and E^2 and this $\rho \psi $ thing are equivalent, right? Microscopically, it's perfect. I have no qualms

Extending the same integral to macro without proper explanation of the interpretation is what has led me here

Yeah, I think we all agree with what we should do microscopically.

I am rather prepared to just dump the concept of macroscopic fields.

That's what my string theory prof said when I askd him. But I didn't take him seriously

Not my prof, one of the prof in my place

The most important property that a scientific theory should be able to do is to stop arguments. This macroscopic shit just keeps the arguments going for centuries.

You did say the procedures are in good agreement with experiment

@naturallyInconsistent but if we differentiate a volume element dont we need to do product rule and thus we obtain some sort of a split? by split i mean a case where we differentiate solely the wedge product and then $\sqrt{\vert g \vert}$ separately

ah but i remember acm said this volume element is only defined when integrating i think so maybe that signifies what i am doing doesnt make sense

let me rethink my approach then blerp

@Relativisticcucumber Yes, we tend to think of it that way, but it is utterly unphysical. Just by a change of coördinates any n-form will have to give rise to multiplicative factors. It is actually kinda more natural to talk about things that are invariant under coördinate changes, and the volume form is one such thing.

@nickbros123 yes, they are, but that does not mean that they are able to dispel arguments.

@naturallyInconsistent what kind of arguments? Arguments regarding the procedures / some definitions / what r postulates, what are not / physical interpretation etc ?

@nickbros123 arguments like we have had: does the definition of macroscopic fields' energy correctly capture the energy of interaction as microscopic fields capture? The Abraham-Minkowski controversy is also another argument that refuses to die. Things like that

@naturallyInconsistent I see. I did read a bit about Avraham minkowski controversy, but I cannot speak on it till i fully finish introductory electrodynamics education. So how do I close this thing for now and move on with my studies? What do I tell myself

@nickbros123 I think you are more than ready to close this chapter for good until you are done with everything and actually want to slay this beast. Just tell yourself that we can get a good approximate value of D H P M if push comes to shove, but that we still do not know how the energy is going to be represented when we do averaging processes. You should be able to continue without such a thing, because we just need the energy to be captured correctly within the approximation scheme,

not that they should agree with reality.

@naturallyInconsistent thank you 👍

@naturallyInconsistent "just need the energy to be captured correctly within the approximation scheme" could u elaborate on this point I didn't quite get it

@nickbros123 You don't need the GPE in the "near Earth's surface" approximation to equal that to the GPE that is zero at infinitely far away that is what it should be in the exact scheme.

You just need each scheme to capture the energy of the parts you manually move (i.e. free charges) correctly.

The average Lorenz force law only allows me to move blobs right? I do know that the work it takes to move around these blobs of charge is exactly calculated in the scheme

Average force density I should say *

They always said that the Abraham-Lorentz force law is Lorentz force law plus radiation reaction. If we just move things quasistatically, then Lorentz force law should be exactly correct.

@naturallyInconsistent so do you think joseph f johnson's answer here is problematic? physics.stackexchange.com/questions/519396/…

@Relativisticcucumber I feel like Im not good enough to judge. Maybe it is time to summon Slereah

@Relativisticcucumber : For what it's worth I updated the answer.

Hello Everyone...

Currently looking into canoe construction in pacific islander cultures for the GR book

@Qmechanic absolutely no idea what you are writing, prior to update and after update.

@Qmechanic eek i am afraid i am still confused

This is how you do astronomy

@Relativisticcucumber @naturallyInconsistent : It would help if you could be more specific.

@Qmechanic ok maybe i see. i think i wasnt internalizing that metric compatibility says that the covariant derivative of every component of the metric is zero so now im thinking what you are saying is basically if we do anything to these components (combine them in anyway), what we get must be zero. is this correct?

and thank you for updating the answer !!!

Does anyoen know if there are any identities with determinant and partial tracing?

I am trying to compute $\text{det} (\text{Tr_2}(M))$ where $M$ is some $n \times n$ matrix and $\text{Tr}_2(M)$ is a $2 \times 2$ matrix.