"As is well-known [17, p. 207] the existence of a continuous line element field on M is equivalent to the existence of a differentiable, covariant, symmetric, second order tensor field on M, with signature of one at each point, that is, a Lorentz "metric" on M."
@DanielSank I happen to have met Roger Penrose - he's anything but egotistical. He was remarkably modest and amiable, just as at ease discussing physics as at exchanging pleasantries and small talk.
Haven't tried Road to Reality, but really enjoyed his Emperor's New Mind.
In the book, he says that the reader doesn't need to know math to follow along.
He says that anyone who is reasonably intelligent and pays attention will be able to learn from scratch while reading the book.
Despite my entire life of education in math and physics I'm pretty sure I failed to complete the first "exercise" in the book.
This is either 1) Massive over-estimation of his pedagogical abilities. 2) Incredible misunderstanding of the abilities of other people. 3) An intentional lie meant to get people to try to read the book even if they are ill-equipped to do so. In this case, he's wasting my time.
Anyway, I gave up on the book and felt like I didn't understand why everyone made such a big deal about it :\
In a complex (non-hermitian) scalar QFT, is it correct that the creation/annihilation operators $a,a^\dagger$ (particle) and $b,b^\dagger$ (anti-particle) commutate, i.e. $[a,b] = [a,b^\dagger] = [a^\dagger,b] = [a^\dagger,b^\dagger] = 0$?
In a complex (non-hermitian) scalar QFT, is it correct that the creation/annihilation operators $a,a^\dagger$ (particle) and $b,b^\dagger$ (anti-particle) commutate, i.e. $[a,b] = [a,b^\dagger] = [a^\dagger,b] = [a^\dagger,b^\dagger] = 0$?
More generally asked, do different creation/annihilation...
Do I have to award the bounty even if the answer does not deserve it and is downvoted? This is in relation to this question of mine where there is only one and that too downvoted answer..
@Bass 1. It's commute, not commutate (although that way of strengthening a word has a long history) ;) 2. Any proper intro to QFT should just outright state they do :P
@DanielSank I'll try to find that book in a library. I'm surprised by that, as he was genuinely quite down-to-earth. That was what struck me the most about him. I expected him to be frighteningly smart and maybe a bit aloof.
Bounty is a popular coconut filled chocolate bar. He was the first eminent physicist I met. I expected an other-worldy mega-mind, but it wasn't the case at all.
@ACuriousMind 1. ok :) 2. he describes them as "two independent sets of c/a operators", I suppose that implies commutativity. I wanted to be sure, because of the equation $Q = \int d^Dk[a^\dagger(\vec{k})a(\vec{k}) - b^\dagger(\vec{k})b(\vec{k})]$ for $Q = \int d^DxJ_0(x)$ where $J$ is defined as $J_\mu=i(\varphi^\dagger\partial_\mu\varphi - \varphi\partial_\mu\varphi^\dagger)$.
I'm not able to deduce the result, I arrive at some $(a^2+b^2)+({a^\dagger}^2-{b^\dagger}^2)$ term. Do you see a quick solution? Never mind if not.
last term should have been $(a^2+b^2)-({a^\dagger}^2+{b^\dagger}^2)$
@Bass Where do the squares comes from? $\phi$ contains $a,b^\dagger$, $\phi^\dagger$ has $a^\dagger, b$. Plugging into $J$ and multiplying out doesn't give any squares.
@Bass I don't see it. that's (schematically) $(b+a^\dagger)(a+b^\dagger)$, the derivative does nothing but get us the $p^0$ to cancel the$\frac{1}{2\omega_p}$ in the measure of the mode expansion.
@ACuriousMind I get $(be^{-ikx}+a^\dagger e^{ikx})(-be^{-ikx}+a^\dagger e^{ikx})$, because the derivative produces one factor $i\omega_k$ and one $-i\omega_k$. So we get a term $b^2 e^{-2ikx}$.
I don't understand this stament by Geroch(1968), when discussing the definition of singularity:
We could not have required that all timelike world lines have infinite total length,
for this property does not obtain in any spacetime.
Can someone give an example of this?
I can only think th...
But it is quite common, at least I've seen it a couple of times now, so you should get used to asking yourself if someone didn't perhaps mean every instead of any when you see such a blatantly false statement.
(Following this answer on Chemistry.SE)
Calculated orbitals comprise a basis set, but they do not represent the actual "dwelling" of the electrons.
Are there actual "dwelling orbitals" which are the result of linear combination of these calculated orbitals? In other words, for a given set of ca...
Might be well-known, might be only mentioned in passing in an old Russian article that has not been translated
> Electrons don't "reside" in orbitals. Orbitals describe electrons. Saying that they reside in orbitals would be like saying that you reside in a box labeled "human". The reality is that you are a human.
@Slereah Do you realize that many mathematicians actually dislike using esoteric symbols? One should have good reasons for using new symbols. Using them without reason is seen a obfuscatory and an indication that you would not actually be able to eplain what goes on in the proof.
@0celo7 That reply was mainly a self-referential joke: "You're a contrarian" - "No, I'm not", but I generally have no other reason to disagree with you except that I think you're wrong :P
@0celo7: Rest easy in the knowledge that JohnRennie also was confused by the wording:
It beats me. If you consider Minkowski spacetime there are no worldlines that are everywhere timelike and have a finite proper length, and this seems to contradict Geroch. — John Rennie34 mins ago
@Slereah "Causality violation" is broader than "time travel", since e.g. PDEs can also violate causality, but unless they are equations for a movement, that's not time travel
@0celo7 Could neither give you a definition of what causality violating means for them, nor can I give an example, I merely know that notion exists. It would have been more untrue to say I know something.
@Fermiparadox My guess without doing a real simulation is that the ejecta is the most unrealistic. Rather than a few discrete blobs that just fly off, there should be more material spewed out at all velocities. At the km/s speeds we have, stuff will not behave as a viscous fluid with surface tension on such short timescales. The melting of the crust looks pretty accurate though.
@JokelaTurbine : Use the 'share' button under each post.
user54412
@Qmechanic I understand removing https, but what exactly is the difference between http://physics.stackexchange.com/questions/154198/can-gravitational-constant-be-changed and http://physics.stackexchange.com/q/154198/?
user54412
My understanding is the text after the numbers is ignored in redirect (so the first link won't rot if the title changes)
(A counterargument might be that unless/until the title changes, the longer link gives the user more confidence they are being redirected somewhere meaningful)
Do I have to award the bounty even if the answer does not deserve it and is downvoted? This is in relation to this question of mine where there is only one and that too downvoted answer..
I don't understand this stament by Geroch(1968), when discussing the definition of singularity:
We could not have required that all timelike world lines have infinite total length,
for this property does not obtain in any spacetime.
Can someone give an example of this?
I can only think th...
(Following this answer on Chemistry.SE)
Calculated orbitals comprise a basis set, but they do not represent the actual "dwelling" of the electrons.
Are there actual "dwelling orbitals" which are the result of linear combination of these calculated orbitals? In other words, for a given set of ca...
