# Tag Info

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The covariant derivative of a field have values in the Lorentz algebra in the representation of the field, spinorial representation in this case, I.e., $$\mathcal{J}^{ab} = - \frac{\imath}{4}[\gamma^a,\gamma^b].$$ Thus the covariant derivative for spinors is $$\nabla_\mu \epsilon = \partial_\mu \epsilon + \frac{\imath}{2} \omega_\mu{}^{ab}\mathcal{J}_{ab},$$ ...

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TL;DR: It is the wedge product $\wedge$ and the exterior derivative/differential $d$ (which squares to zero $d^2=0$) that give rise to Grassmann-odd elements and supersymmetry. More concretely, Ref. 1 first writes down a (non-relativistic) SUSY algebra ${\cal A}$ $$\tag{10} \{Q_1,Q_1\}_+~=~2H~=~\{Q_2,Q_2\}_+, \qquad \{Q_1,Q_2\}_+~=~0,$$ spanned by two ...

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I think your limiting procedure should work. For the homogeneous Maxwell equations, when we restrict to a surface of constant $x$, the pull back of the field strength looks something like $$\iota^* F = E^\perp_i dt \wedge dx^i + B_x dy \wedge dz,$$ where $E^\perp$ is the component of $E$ in the $yz$-plane. Now your timelike pillbox integral will have ...

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The variational operator $\delta$ satisfies the product rule $$\delta (A \wedge B) = (\delta A)\wedge B + A \wedge (\delta B)$$ To find the variation of $G \wedge \star G$ where $\star G$ is the Hodge dual of $G$, note that $$G \wedge \star G = \frac{1}{p!}G_{\mu_1 \ldots \mu_p} G^{\mu_1 \ldots \mu_p} \sqrt{|g|}d^{D}x$$ From this, it is immediately ...

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Let $\mathcal{M}$ be our spacetime. Then, a gauge theory is given by a connection form $A$ on a principal bundle over it (that locally projects onto the spacetime in a way compatible with gauge transformations), which is the gauge potential. Maxwell's equations1 (in vacuum) are the equations of motion for the gauge field for the Yang-Mills action coupled to ...

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To my mind, Newton's equation makes the most sense as an equation of vector fields. Let $(M,g)$ be a (Pseudo-)Riemannian manifold with Riemannian connection $\nabla$. Then the equations of motion for a position-dependent conservative force $F$ are given by $$m {}^{\gamma}\nabla_{\frac{d}{dt}} \frac{d\gamma}{dt}=F\circ \gamma=-(\nabla V) \circ \gamma,$$ where ...

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You can write $$\text dV = - m v \text d v$$ where $v$ is the velocity.

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Yes, especially in research-level topics. There are several research groups that work with finding ways to apply differential geometry concepts to solid state systems (although condensed matter seems to be the preferred term nowadays). See for example the book by Altland and Simons, Condensed Matter Field Theory, Chapter 9 "Topology". This book is suitable ...

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Differential Geometry has been useful in Mechanics of Crystalline Solids with Dislocations. See for example: http://arxiv.org/abs/1212.5125

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Take your drawings, and imagine moving a vector around the loops you draw. Specifically, move the vector using parallel transport i.e. during each infinitesimal movement keep the vector parallel to its previous direction. If the space(time) is curved then when you have travelled around the loop you'll find the direction of the vector has changed. If we take ...

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Well, the very notion of curvature relies on the Riemann curvature tensor, in fact, the curvature tensor arises from the fact that in a general curved space, the second covariant derivatives may not commute: $A_{v;ps} - A_{v;sp} = A_{b}R^{b}_{vps}$. That being said, however, I think at minimum to determine whether something is curved or not, in a general ...

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I do not agree with the answer given by @ACuriousMind. @Scardenalli has asked for a compact Ricci-flat Riemannian manifold $M$ having as isometry group $U(1)\times SU(2)\times SU(3)$. This does not imply that $M$ must be a symmetric Riemannian manifold. However, the answer to @Scardenalli's question is still no, and it follows from a classical result in ...

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I believe you and the author are referring to different "gradients". You mean the gradient along the manifold defined by the particle trajectory; since this is a one dimensional manifold, the gradient would indeed have only one component. The authors are referring to the full three-space gradient. Using the formulae  r = \frac{1}{\kappa}+x \\ \phi = \kappa ...

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Here is a bare bones easy way to see that coordinate tuples are not 4-vectors. Start in an inertial coordinate system in flat spacetime. Change the coordinate system with a constant translation: $x' = x + A$ $y' = y$ $z' = z$ $t' = t$ Even in this idealistic case, 4-vectors and coordinate tuples transform differently. The components of the 4-vectors ...

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Great reasoning: as in Uldreth's fantastic answer but I would add one more thing that may help cement your good understanding in place. Co-ordinates are absolutely not vectors, they are labels on charts and are no more vectors than your street address is a vector. Almost certainly the reason people make the implication that you have correctly identified as ...

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You are correct. Position is a vector when you are working in a vector space, since, well, it is a vector space. Even then, if you use a nonlinear coordinate system, the coordinates of a point expressed in that coordinate system will not behave as a vector, since a nonlinear coordinate system is basically a nonlinear map from the vector space to ...

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