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Hi, I am a string theorist and a publicist.


Nov
24
comment Is this a simple Lie algebra?
I agree. The first commutator verifies that $\sigma_a$ are Pauli matrices. $\eta_1$ isn't a Pauli matrix, it is ${\rm diag}(+1,-1)$, as implicitly seen from the last two equations. So one deals with $4\times 4$ block diagonal matrices with a combination of Pauli matrices in each $2\times 2$ block, and that's clearly $SU(2) \times SU(2)$, with your right generalization. Just to answer the OP's question, it is a product so it is not simple.
Nov
24
comment Non-minimal coupling of electromagnetic field
For example, the covariant derivative $\nabla_\mu V^\mu$ of a vector may be written as a partial derivative $1/\sqrt{g} \partial_\mu (V^\mu \sqrt{g})$ - there should be $-g$ everywhere in the square roots, right? - which explains the first identity for the box of the scalar. A few more are needed to deal with the curvature etc. You will also have to learn why the Riemann curvature tensor is encoded in the "commutator" of two covariant derivatives as acting on a vector etc. Many identities but it's just maths.
Nov
24
comment Non-minimal coupling of electromagnetic field
This is straightforward maths - mathematical identities - and maybe it doesn't really belong to Physics SE. You should first try to understand why your first equation of motion for the scalar should have been written intelligently as $\nabla^\mu\nabla_\mu \phi$ in terms of covariant derivatives. This will require you to learn how to raise and lower indices by the metric tensor (you avoided this notation, suggesting you don't know this basic point), the Christoffel symbol, covariant derivatives, and various identities how to translate between those things.
Nov
18
awarded  Peer Pressure
Nov
18
comment What is the rationale behind representing a state function by a complex valued function in QM?
The commutator $[x,p]=i\hbar$ contains an $i$ - and it must contain an $i$ because the commutator of two Hermitian operators is anti-Hermitian - so it can't be represented by real operators on a real vector space. Equivalently, $\cos(kx)$ instead of $\exp(ikx)$ wouldn't remember whether one is moving left or right, or the sign of energy etc. One couldn't write first-order oscillating equations without $i$. See also Why complex numbers are fundamental in physics, motls.blogspot.com/2010/08/…
Nov
17
answered Inertial Mass of a scalar field
Nov
17
revised Is the spring constant k changed when you divide a spring into parts?
added 2 characters in body
Nov
16
answered Is the spring constant k changed when you divide a spring into parts?
Nov
16
answered Are there irreducible tensors of half integral degree in quantum mechanics?
Nov
16
comment Distinguishing two quantum states practically
Dear @Pratik, thanks for your interest. Quantum measurements end up being mundane things. For example $|0\rangle$ and $|1\rangle$ above refer to the electron with spin up or down. You send the electron into a magnetic field that deflects it up or down depending on the spin (the electron is a small magnet), and then you detect it BEEP up or down (after a second, or any amount of time). That's the measurement - a sort of destructive one. Most quantum measurements are "destructive"; to say the least, every measurement in quantum mechanics changes something about the state of the measured object.
Nov
16
comment Distinguishing two quantum states practically
Dear @lurscher, "yes, but then you have the problem..." - I personally don't have any problem related to these words. If you can't produce the exact same experimental situation arbitrarily many times, then you can't accurately measure any properties of the wave function describing the situation. The wave function simply isn't an observable, in the technical - and also informal - sense. This proposition isn't a "problem"; instead it is one of the "most fundamental postulates and deepest insights of [quantum] physics". At most, it may be someone's psychological problems - see Psychology SE.
Nov
16
comment Distinguishing two quantum states practically
Dear @Pratik, there's obviously no "general" prescription for how long time a measurement takes. Different experiments are different. Many of them can be fast but it is obviously nonsense to invent a general duration, like 2 nanoseconds, for "every" experiment.
Nov
16
comment What is the physical meaning of the terms in the multipole expansion?
Just to be sure, the dipole moment is called the dipole because the simplest way to realize a charge distribution with a nonzero dipole moment - but vanishing all other moments - is a pair (therefore "di") of nearby charges. Similarly for quadrupoles, octupoles etc. (powers of two). However, that doesn't mean that the number of charges always has to be a power of two (and they have to be pointlike): almost any charge distribution carries almost all the multipole moments. Just trust the formulae instead of words, and don't misinterpret the words.
Nov
16
comment What is the physical meaning of the terms in the multipole expansion?
In particular, it's not true that everything that isn't of the form of "the exact naive children's dipole" carries a vanishing dipole moment - which seems to be the (incorrect) assumption in Revo's question. @Revo, try this analogy: this assumption is analogous to saying that one kilogram is the mass of the platinum prototype in France, or whatever it is. Now, your reasoning is that the mass of a person has to be zero because the person isn't even made of platinum: she can't be the platinum stick. This sounds like a joke but your logic applied to the dipole moment is exactly the same.
Nov
16
comment What is the physical meaning of the terms in the multipole expansion?
SteveB's explanation is of course perfectly valid but it doesn't directly "attack" the core of Revo's misconception. Revo correctly knows that the "canonical dipole" is a pair of oppositely charged particles. However, that doesn't mean that every object that doesn't have this form carries a zero dipole moment. Quite on the contrary, generic objects (charge distributions) carry a nonzero amount of each multipole moment. One needs some "special" or "very special" distributions for some of these moments to be zero.
Nov
16
comment What is the physical meaning of the terms in the multipole expansion?
No, Revo, @SteveB didn't change the charge distribution. He took your very distribution - a single charged particle away from the center - and wrote it as a combination of a charge in the center and the "true dipole" consisting of a pair of oppositely charged particles. The latter clear has a dipole moment: it's the canonical representation of a "pure dipole". Well, except that this dipole isn't center at zero, so it will also carry a quadrupole moment (and maybe higher ones, too), but the latter is subleading in this expansion.
Nov
16
answered Relativistic transformation of the wave packet length
Nov
16
answered Distinguishing two quantum states practically
Nov
16
comment Partition Functions in (A)dS/CFT
There are many users with flawed arguments but lots of self-confidence in suggesting that they have invalidated quantum mechanics, relativity, or anything else. You're exactly the person who should get a lot of self-confidence because you got very far: but you end up as a dove. Isn't it you who should say more clearly that this is a fundamental inconsistency of the whole dS/CFT paradigm? This inconsistency is copied from the same one that spoiled the "S-branes"; it was always an incoherent union of "S-branes are continuations of D-branes" (instantons) and "they are mathematically independent".
Nov
16
comment A possible absolute reference system
Yes, this is a frame prominently used in cosmology. Still, one may neglect the weak CMB (or shield a system from it) to see that all other frames in relative motion with respect to the CMB frame still follow the same laws of physics.