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7

No, you also need an initial condition for the field, and a boundary condition for the field. A plane wave solution to the Maxwell equations has the same 4-current as vacuum, after all.


6

This axiom is a "smoothness" property of manifolds. You could define a manifold without this property simply as a space which is locally Euclidean. However, then something which looks smooth in one chart may be highly non-smooth in another chart. For example a smooth curve in one chart may be discontinuous in another chart (since the representation in the ...


3

An equivalent way of defining the 3-covariant derivative $D$ is to say it projects the 4-covariant derivative $\nabla$ onto the hypersurface. This is worked out explicitly in an appendix to Carroll's Spacetime and Geometry, where he writes $$ D_\mu V^\nu = P^\alpha{}_\mu P^\nu{}_\beta \nabla_\alpha V^\beta $$ (D.48), given the projection tensor $$ P_{\mu\nu} ...


3

First, the definition above is a right definition to include things like (the surface of) a smooth torus – and any similar ... manifold ... with an arbitrary topology, but exclude objects that wouldn't be locally equivalent to smooth space. If you changed something important in the definition, you would get a difference concept that doesn't agree with our ...


3

We interpret OP's question (v3) as essentially asking Can a Lorentzian manifold (with Minkowski signature) by coordinate transformations be redefined as a Riemannian manifold (with Euclidean signature)? The answer is No since the metric signature of a pseudo-Riemannian manifold is invariant under general coordinate transformations. This follows e.g. ...


3

I think that your description that the points of the configuration manifold are possible states of the system is as close to a precise definition as one will find. So for $n$ particles in three dimensions, the configuration manifold is just $(\mathbb{R}^3)^n$. As for how this relates to constraints, consider the simplest example: two particles attached with ...


3

I think Jerry's point is that if you have a metric that looks like: $$ ds^2 = a(dx^0)^2 + f(dx^1, dx^2, dx^3) $$ where $a$ is a constant and $f$ is any function, then an observer moving in the $x^0$ direction follows a geodesic of the type you describe. An obvious example is the FLRW metric where a static observer follows the geodesic $t = \tau$, $x = y = ...


3

Check out V.I. Arnold's Mathematical Methods of Classical Mechanics. This book is pretty terse and can have hard to follow notation. However, it is rigorous and contains mathematical explanations and proof of a wide array of topics in mechanics. It is also filled with very interesting examples. He introduces the concepts needed from differential geometry; ...


3

Matt Visser's How to Wick rotate generic curved spacetime is a great reference on this subject, which basically summarizes a lot of folklore on the subject. Addendum (Summary of Paper). This turns out to be an important problem in quantum gravity and QFT in curved spacetime for the obvious reason ("How do we know the usual tricks still work in curved ...


3

I understand force to be a 1-form, through the following reasoning. Given a time-independent, conservative lagrangian $L$, its differential (a 1-form in the purest sense) is $$ \mathrm{d}L = p_a ~\mathrm{d}\dot{x}^a + f_a~\mathrm{d} x^a $$ where $$ p_a = \frac{\partial L}{\partial \dot{x}^a},~f_a = \frac{\partial L}{\partial x^a}. $$ So the components of ...


2

For $1.$, you have, by applying the Leibnitz rule for covariant derivatives : $\nabla_{d}(\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}) \\= \nabla_{d}(\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l} \delta_{c'}^{c}) \\= \nabla_{d}(\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l} g^{ca}g_{ac'}) \\=(\nabla_{d}\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l}) ...


2

Just like $(r,0)$ are "multilinear functionals" assigning a group of $r$ co-vectors i.e. $r$ $(0,1)$ tensors a scalar, and you seem to accept this visualization, the $(r,s)$ tensors are "multilinear functionals" assigning a group of $r$ co-vectors i.e. $r$ $(0,1)$ tensors and $s$ vectors i.e. $(1,0)$ tensors a scalar. This is just one way among many how to ...


2

Ok, here's my own (pedantic) argument that the answer is "No." Three background steps: Take $(M,g_{ab})$ and $(\tilde{M},\tilde{g}_{ab})$ to be two pseudo-Riemannian manifolds related by an isometry $\varphi:M\rightarrow\tilde{M}$. I will consistently use a tilde ($\tilde{}$) to refer to objects on the second manifold and no tilde to refer to objects on ...


2

$$\nabla_{\mu}\left[U,V\right]_{\nu} = \nabla_{\mu}\left(g_{\nu\lambda}\left[U,V\right]^{\lambda}\right)$$ You have two terms inside the parentheses, and you have to apply the derivative to both of them. Myself, I'd just remember that: $$\left[U,V\right]^{a} = U^{b}\nabla_{b}V^{a} - V^{b}\nabla_{b}U^{a}$$, so, we have: $$\begin{align} ...


2

First of all, the first equation for $ds^2$ is only valid if $f$ is nothing else than the azimuthal angle $\phi$. Second, if you are evaluating $X_i X^i$, the squared distance from the origin without any infinitesimals, then it is exactly equal to $-t^2+r^2$ and nothing else. The polar coordinate $r$ is chosen as $\sqrt{x^2+y^2}$ so its square already ...


1

OP is essentially asking about terminology. As usual, be prepared that different authors call different notions differently. Well, here is a suggestion: Call the configuration space before (after) the constraints are implemented for the extended (physical) configuration space, respectively. More generally, if an author is talking about a configuration ...


1

Since an $(k,\ell)$ tensor can be viewed as linear combinations of pertinent tensor products of $(1,0)$ tensors (=vectors) and $(0,1)$ tensors (=covectors), and by linearity and Leibniz rule for the covariant derivative, it is enough to consider the latter. The condition (1) in manifestly covariant notation then reduces to $$ ...


1

A manifold should be thought of as a topological space that locally looks like a $\mathbb R^n$. This is made precise by saying that every point has a neighborhood that is homeomorphic to an open subset of $\mathbb R^n$. If you don't require anything else (except maybe Hausdorffness and the existence of a dense countable subset to make everything easier ...


1

Consider any given spacetime $M$ and a given timelike geodesic $\gamma$ therein, exiting from an event $p\in M$, and parametrized by means of its proper time $\tau$ with origin fixed at $p$ itself. Next consider the coordinate system constructed as follows (it is possible to prove that it is well defined, see e.g., O'Neill's textbook). Fix a pseudo ...


1

In these particular cases, the authors are interested in the conformal structure, i.e. lightcone structure, of the manifold. A conformal structure can be defined by an equivalence class of metrics, all of which are related to each other by a conformal transformation, $$g_{ab}\sim e^{\omega(x)}\bar{g}_{ab}$$ A nice way to characterize a conformal structure ...


1

Let me first refer you to three references pedagogically treating Instantons in quantum mechanics: 1)Riccardo Rattazzi's lecture notes treating instantons in nonsupersymmetric quantum mechanics. In these notes the anharmonic oscillator model is elaborated with great detail 2) Philip Argyres lecture notes treating instantons in supersymmetric quantum ...


1

Consider an $n$-manifold $M$. A curve is simply a continuous map $\gamma : \mathbb{R} \to M$. For simplicity, suppose $M$ is covered by a single coordinate chart (diffeomorphism) $\varphi : M \to \mathbb{R}^n$. Putting these together, we have \begin{eqnarray} \mathbb{R} & \stackrel{\gamma}{\longrightarrow} & M & ...


1

There are two kinds of mathematical situations considering vectors and covectors. They are those: In the space without scalar product (dot product) they are very different entities and one should never confuse them. In the space with scalar product they are the same thing viewed from two perspectives. One can freely convert one entity to another with ...



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