I think I ever read something about positive frequency in Sparling's paper and someone answering my question in physics forum mentioned negative frequency, but I didn't know what these mean.
I just thought: "frequency must always be positive, so where can one encounter negative frequency?"
@bolbteppa but positive energy condition is only for gravity field, isn't it?
Free particle solutions have terms of the form $e^{i p_{\mu} x^{\mu}}$, from $i p_{\mu} x^{\mu} = i (Et - \mathbf{p} \cdot \mathbf{r})$ you see energy $E$ acts as a frequency, $\omega t = E t$, and $E = \pm \sqrt{\mathbf{p}^2+m^2}$ produces both positive and negative values, however the minus sign can be seen as being an artifact of complex conjugation since we actually have $ i Et$ and then linked to a creation operator so that $E$ is ultimately always positive i.e. positive frequencies
@Slereah what are negative probabilities? I read in Wikipedia about probability theory that positive probability can be extended to negative but I don't know how to extend.
Alright, this is probably a weird question but sometimes I get two competing answers to a question both of which I think are very good. So, in case of a tie, how do I choose the answers?
we know that the wavefunction of a free particle is ψ(x)=Ae^(ikx)+Be^(i(-k)x) So can we let k have both positive and negative value where the positive represents traveling to the right and the other represents wave traveling to the left and then simply write ψ(x)=Ae^(ikx)
this is the definition of what canonical quantization does: You can't derive that it must work from some other principle, it just turns out - as an experimental fact about the world - that this indeed yields a good quantum description of many classical systems
but it doesn't always 'work', and not for all variables - there's both abstract reasons (Groenewold-van Howe theorem) and practical reasons (what do you do with classical expression like $xp + px$?) it's not as straightforward as "just replace all variables by operators"
A wave function takes the form $\psi \approx e^{iS/\hbar}$ as you approach the classical limit, it's time derivative is $\frac{\partial \psi}{\partial t} = \frac{i}{\hbar} \frac{\partial S}{\partial t} \psi = - \frac{i}{\hbar} H \psi$ i.e. the Schrodinger equation. The Lagrangian possesses Galilean symmetry thus the Hamiltonian is fixed to be $p^2/2m + V(x,y,z)$. You then define the Hamiltonian to be the quantum operator which reduces to this form on the classical limit.
The momentum operator is found the exact same way.
you're essentially defining the operator $\hat{T}$ there to be $\hat{p}^2$, and you've got $\hat{p}$ defined from e.g. the argument bolbteppa alludes to
if you think this can be "wrong", you need to exhibit an alternative definition of $\hat{T}$ first
This shows the basic issue with QM that freaked/s people out, it's existence is predicated on the existence of a theory it is more fundamental than yet must reduce to and without which would imply the theory is meaningless
If we're discussing contra/co-variance with regards to general coordinate transformations, are we asking whether an object transforms under the representation $d$ or $d^{-1}$ of the diffeomorphism group? So for example when we change from cartesian to polar coordinates, a covariant vector transforms as the group inverse of a covariant vector with respect to the diffeomorphism group?
if the Jacobian is an endomorphism, is it a representation of the diffeomorphism group?
That's basically what I was thinking, given a vector space and a representation of the diffeomorphism group, the Jacobians are a representation of this group
I'm basically just trying to reframe "coordinate transformation" properties of tensors in terms of how they transform under representations of groups
So when we change from cartesian to polar coordinate we use the Jacobian, and this Jacobian is a representation of some element of the diffeomorphism gropu
I was basically just wondering if, given what was discussed yesterday about contra/co-variant vectors transforming by elements of the Lorentz group that are group-inverses of eachother, whether when we talk about general coordinate transformations (ie. Jacobians) we would say that the endomorphism Jacobians (not sure what else to call them) can be a representation of the diffeomorphism group
@Charlie To expand on my "they're not": the Jacobians act on the (co)tangent spaces at every point (and hence by linear extension on all tensors) in the essentially unique $\mathrm{GL}(n)$ representation, so each diffeomorphism is mapped to infinitely many $\mathrm{GL}(n)$ values - one for each point. That's not a representation.
but even in a completely mathematical sense, the Jacobians that represent, say, cartesian to polar coordinate transformations, must correspond somehow to a group right
@Charlie yes - at each point. And you can certainly form the ("gauge") group of GL(n)-valued functions. But that doesn't mean that you get a representation of this (infinite-dimensional!) group on a single vector space from this
there is a "representation" of this large group in some colloquial sense here certainly, but not in the formal sense of having a pair of one vector space and one representation map.
the $\mathbb{R}^n$ examples are very poor to see this distinction because the manifold - $\mathbb{R}^n$ - is isomorphic to its tangent spaces $\mathbb{R}^n$ so you can get away with a lot of conflation
oh I think i see what was being said earlier then, that the jacobian from cartesian to polar coordinates has to act on every tangent space of the manifold
@ACuriousMind yeah this is almost certainly what's tripping me up
and it doesn't help that instead of teaching the math physicists tend to talk as long as possible about "free vectors" or "vectors attached to a point" or whatever rather than do this bit of differential geometry :P
I think there's a different between explaining the elementary notions of manifolds and tangent spaces and doing bundles. Of course the tangent bundle is there but you don't really need to bother anyone with this notion unless you're gonna do something with it :P
the bundle view is equivalent to just working with local functions in trivializations. For many applications it's entirely sufficient to just work locally and never mention to global bundle object once
@Slereah and on smooth manifolds they're all isomorphic, what's the problem? Just pick the one you like!
@ACuriousMind what you've said here I think i get now, if we want to talk about the jacobian of cartesian to polar coordinates, we need to map one element of the diffeomorphism group to a particular jacobian acting on each tangent space
I mean, you can certainly "freeze" the map at one point - the map that maps a diffeomorphism to its Jacobian at a particular point would form a representation on the tangent space at that point, but that's not a very natural viewpoint to take
at that point I'd really question whether we're looking for any insight or have become Representation Hunters (TM) that just reflexively want to find representations :P
quick question, the expansion of a qft operator as: $$\phi(x)=\int\frac{\text d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}[a_{\vec p}e^{ip\cdot x}+a^\dagger_pe^{-ip\cdot x}],$$ this integral is over the entire real number line right?
fwiw, it is a point of contention how exactly biological life started. it has to have started from somewhere, but exactly how this happened is afaik still not agreed upon
as in, exactly which chemical reaction created the first building block
maybe God spawned atoms with specific properties and bound-favorings so naturally they attract each other to form cells or entities that are under the forces of evolution.
Well, I imagine we know roughly which molecules have to have been present, but exactly what happened is difficult to answer, there are too many variables
well, "before evolution" would be the realm of geology, mostly :P
what's not known for certain is what actually happened between "there's no life" and "here's a bunch of cells with DNA/RNA in them" because unfortunately single cells don't leave many fossils
damn, the miller urey experiment worked, therefore by just having a bunch of elementary atoms and striking them with lightning we can get amino acids and therefore more complex structures from there? Could this be the answer to abiogenesis?
@sheltonBenjamin could you be more specific about what you mean by rocket equation? Assuming you are talking about the version of Newton's equation where instead of $ma$ you take $\\frac{d}{dt}(mv)$, yes you would need to take care of gravity
No actually i got little bit confused while reading the derivation of a rocket equation I didn't read the title properly that its side neglecting the effects of gravity
Haha. But still, you need to take care of gravity. What would differ is that one can simply use $mg$ for the fireworks rockeet since it stays sufficiently close to the surface.
but now I want to derive the equation in the presence of gravity so I was asking if I can apply conservation of momentum from the rocket frame of reference then use concept of relative velocity
@sheltonBenjamin as I recall the rocket equation calculates the delta v, which is the maximum change of velocity of the vehicle if gravity (or any other forces) are not acting.
Haha. But still, you need to take care of gravity. What would differ is that one can simply use $mg$ for the fireworks rockeet since it stays sufficiently close to the surface.
but now I want to derive the equation in the presence of gravity so I was asking if I can apply conservation of momentum from the rocket frame of reference then use concept of relative velocity
but now I want to derive the equation in the presence of gravity so I was asking if I can apply conservation of momentum from the rocket frame of reference then use concept of relative velocity
It all gets very complicated when you include gravity, because gravity changes with distance from the planet. To be honest I don't know a lot about doing these calculations.
"The space of degenerate states at a point k in the Brillouin zone provides a representation of the group of the wave vector k." So I don't really understand what this means, how does a set of states (vectors) provide a representation (matrix)?
@B.Brekke A representation is not a matrix, it's a tuple: A vector space $V$ and some map $\rho : G \to \mathrm{GL}(V)$ that maps abstract group elements to matrices. I don't know what "the group of the wave vector $k$" is but likely this statement is trying to say that for the space of degenerate states there is a way for this group to act on it and the space is closed under this action.
I use ublock origin and I haven't seen a youtube ad in ages. Sometimes YT changes something and you'll see them until the block lists are updated but it's pretty rare in my personal experience
Alright, this is probably a weird question but sometimes I get two competing answers to a question both of which I think are very good. So, in case of a tie, how do I choose the answers?
