@JohnDuffield ::big sigh:: That really has nothing to do with my question. And your answer to my question was so poor it got a whole lot of downvotes and was deleted. Ouch.
@ACuriousMind Are we supposed to have a serif font now?
I mean like eddington says that newton just postulates that a certain force is acting whenever he sees that something is not maintaining constant velocity
No I've read feynman lectures he says that we should have a concept of force different than ma , if you define force to be ma then u have discovered nothing
Nope u r going much faar he says that u have a intuition of force as push or pull then you establish mass and then quantitatively get the number ma as measure of force
Arnold's definition is: A frame in which the laws of nature are the same as in any other inertial frame and any two inertial frames are related by rectlinear motion
this can be made precise by considering Galilean space
Definition I. ‘Inertial system’ is called any coordinate system of the kind that in relation to it three points P;P0;P00, projected from the same space point and then left to themselves— which, however, may not lie in one straight line—move on three arbitrary straight lines G;G0;G00 (e.g., on the coordinate axes) that meet at one point.
It goes over my head can anyone get it..
But still what so far I've got is that an isolated particle is the one which is sufficiently far away
@ACuriousMind In 2nd quantization, scalar/spinor/vector fields are replaced by "operator fields", i.e. functions that map each spacetime point onto an operator. Like, instead of mapping each spacetime point onto a number/spinor/vector, they are mapped onto operators that operate on the Fock space. Right so far?
@Bass Ehhh...no. It's only a Fock space for free theories. All your particle states and such live in the future/past asymptotic spaces, for which the theory is assumed to be free.
But the fields themselves are supposed to be operators on the space of states of the fully interacting theory
@ACuriousMind Oh okay. Haven't yet begun with interactions in 2nd quantization, but good to know.
@ACuriousMind So, my question is, what's the correct terminology or definition for such a field? A vector field is a section of a vector bundle, a scalar field is a section of a trivial line bundle etc., does that mean that a quantized field is a "section of an operator bundle" or something similar?
@Bass I suppose you could try to make that a bundle, but the space of operators will be infinite-dimensional, so that's not as straightforward a bundle as the others. Just say that the fields are operator-valued functions (actually distributions, but that's almost always glossed over)
@yuggib So $f$ is a classical solution of the KG/Dirac/Maxwell equation, right?
And in QFT, $f$ is interpreted like some measure that says "where do we need how much of $\phi$". If $f$ was "concentrated" at $x_0$ like $f(x)=\delta(x-x_0)$, we'd have $\phi(f)=\phi(x_0)$. Does that make sense?
@Bass the fact that it is a solution of the KG/Dirac etc is not strictly necessary, however yes, that's the idea; it is at least a function on a functional space where you can solve the suitable equation
and yes, if it would be allowed to take $f=\delta$, then you would have $\phi(x)$; sadly, it is usually not allowed, and you need to take $f$ that are much more regular
@yuggib Not strictly necessary meaning not a requirement on $f$? I just realized that would make sense, since $\phi$ satisfies the KG equation already, so we don't need $f$ to satisfy it, right?
@yuggib Regular you mean like smooth/continuous/not-a-distribution?
@Bass As far as I know, it is not a requirement to define the quantum fields; however there are people that take allowed $f$ to be only solutions of the equation
@Bass yes...for the usual scalar KG theory, the $f$ is usually taken to be in the Schwarz space of smooth functions
For a pseudo-Riemannian manifold, under the variation $g_{\mu\nu}\mapsto g_{\mu\nu}+\delta g_{\mu\nu}$, the determinant $g=\mathrm{det}g_{\mu\nu}$ varies as $$\delta g=gg^{\mu\nu}\delta g_{\mu\nu}$$
Nakahara proves this with the matrix identity $$\tag{1}\mathrm{ln}(\mathrm{det}g_{\mu\nu})=\mathr...
@ACuriousMind Suppose I have some submanifold $\Sigma\hookrightarrow\mathcal{M}$ and some vector field $l$ tangent to $\Sigma$. Let $u\in T_p\Sigma,p\in \Sigma$. How can I show that there is some vector field $\hat u$ that satisfies $\hat u_p=u$ and $\phi_{t*}\hat u=\hat u$ where $\phi$ is the flow of $l$?
$\hat u$ needs to be at least $C^1$ and defined in some open set around $p$.
@ACuriousMind Can I write the Lie transport equation $L_l\hat u=0$ as an ODE with initial condition $\hat u_p=u$ and somehow use the Star Trek: The Next Generation theorem to find a unique $\hat u$?
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@ACuriousMind Can I write the Lie transport equation $L_l\hat u=0$ as an ODE with initial condition $\hat u_p=u$ and somehow use the Star Trek: The Next Generation theorem to find a unique $\hat u$?
My "instinct" is by just defining $\hat{u}$ to be $u$ shifted along the flow, but that defined a field only along the integral curve through $p$, not on the whole $\Sigma$.
$\Sigma$ is a null hypersurface...but I'm not sure that's too important.
$\nabla$ is the Levi-Civita connection.
What I need to show is that $[\nabla_lv-\nabla_vl]_p=0$ for some $v$
So for the cov. derivatives to be defined I need some vector field extension of $u=v|_p$
@ACuriousMind The end result is that $\langle \nabla_ll,u\rangle=0$
@ACuriousMind But if I work with covariant derivatives along curves I don't need the vectors to exist in a neighborhood...just along a curve segment, right?
@yuggib What? My point would've been this isn't $C^1$ because it is non-zero along a curve and zero everywhere else, that doesn't look very differentiable or even continuous to me.
It's what being invariant means: It doesn't matter whether you take the vector field at one point and just push it along the flow to another point or if you look up its value at the target point.
Absolutely not.
In fact, \renewcommand really shouldn't be used at all.
Why? Because the title, question, answers, and comments on a page are not in separate MathJax environments. If you use \renewcommand (or even \newcommand, though that's not quite as bad) you risk clobbering others' contribu...
