Covariant derivative Definition and 174 Threads

I'm building a mental framework for the Levi-Civita connection that is intuitive to me. I start by imagining an arbitrary manifold with arbitrary coordinates embedded in a higher dimensional Euclidean space, then if I take the derivative of an arbitrary coordinate basis vector with respect to...

Hi, starting from a recent thread in this section, I decided to start a new thread about the following: Take a generic irrotational/zero vorticity timelike congruence. Do the 4-velocity and the direction of proper acceleration (i.e. the vector in that direction at each point with norm 1)...

I have read that we can define covariant derivative for spinors using the spin connection. But I can't see its proof in any textbook. Can anyone point to a reference where it is proved that such a definition indeed transforms covariantly ?

The perturbed line element: ##g = a(\tau)^2[-d\tau^2 + (\delta_{ij} + h_{ij})dx^i dx^j]## Expanding the covariant derivative with ##\nu = 0##, you get a few pieces. Here on keeping only terms linear in the perturbations, ##\partial_{\mu} T^{\mu 0} = a^{-2} \bar{\rho} \left[ \delta' -...

Hi, Penrose in his book "The Road to Reality" claims that Newton/Galilean spacetime has actually a structure of fiber bundle. The base is one-dimensional Euclidean space (time) and each fiber is a copy of ##\mathbb E^3##. The projection on the base space is the "universal time mapping" that...

Hi, a doubt related to the calculation done in this old thread. $$\left(L_{\mathbf{X}} \dfrac{\partial}{\partial x^i} \right)^j = -\dfrac{\partial X^j}{\partial x^i}$$ $$L_{\mathbf{X}} {T^a}_b = {(L_{\mathbf{X}} \mathbf{T})^a}_b + {T^{i}}_b \langle L_{\mathbf{X}} \mathbf{e}^a, \mathbf{e}_i...

What is the expression for the covariant derivative of a Weyl spinor?

In the 128 pages of 《A First Course in General Relativity - 2nd Edition》:"The covariant derivative differs from the partial derivative with respect to the coordinates only because the basis vectors change."Could someone give me some examples？I don't quite understand it.Tanks！

I need to evaluate ##\nabla_{\mu} A^{\mu}## at coordinate basis. Indeed, i should prove that ##\nabla_{\mu} A^{\mu} = \frac{1}{\sqrt(|g|)}\partial_{\mu}(|g|^{1/2} A^{\mu})##. So, $$\nabla_{\mu} A^{\mu} = \partial_{\mu} A^{\mu} + A^{\beta} \Gamma^{\mu}_{\beta \mu}$$ The first and third terms...

In a general coordinate system ##\{x^1,..., x^n\}##, the Covariant Gradient of a scalar field ##f:\mathbb{R}^n \rightarrow \mathbb{R}## is given by (using Einstein's notation) ## \nabla f=\frac{\partial f}{\partial x^{i}} g^{i j} \mathbf{e}_{j} ## I'm trying to prove that this covariant...

i) Let ##\pi : E \rightarrow M## be a vector bundle with a connection ##D## and let ##D'## be the gauge transform of ##D## given by ##D_v's = gD_v(g^{-1}s)##. Show that the exterior covariant derivative of ##E##-valued forms ##\eta## transforms like ##d_{D'} \eta = gd_D(g^{-1}\eta)##. ii) Show...

Hi all, I am currently trying to prove formula 21 from the attached paper. My work is as follows: If anyone can point out where I went wrong I would greatly appreciate it! Thanks.

Hello everyone, in equation 3.86 of this online version of Carroll´s lecture notes on general relativity (https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html) the covariant derviative of the Riemann tensor is simply given by the partial derivative, the terms carrying the...

First, we shall mention that it is known that the covariant derivative of the metric vanishes, i.e ##\nabla_i g_{mn} = 0##. Now I want tro prove the following: $$ \nabla_i A_k = g_{kn}\nabla_i A^n$$ The demonstration I encounter takes advantage of the Leibniz rule: $$ \nabla_i A_k = \nabla_i...

[Moderator's note: Thread spun off from previous thread due to topic change.] This thread brings a pet peeve I have with the notation for covariant derivatives. When people write ##\nabla_\mu V^\nu## what it looks like is the result of operating on the component ##V^\nu##. But the components...

Hey everyone, I was trying to learn in an unrigorous way a bit about making derivatives in the general manifold, but I'm getting confused by a few things. Take a vector field ##V \in \mathfrak{X}(M): M \rightarrow TM##, then in some arbitrary basis ##\{ e_{\mu} \}## of ##\mathfrak{X}(M)## we...

Apologies in advance if I mess up the LaTeX. If that happens I'll be editing it right away. By starting off with ##\nabla^{'}_{\mu} V^{'\nu}## and applying multiple transformation laws, I arrive at the following expression $$ \frac{\partial x^{\lambda}}{\partial x'^{\mu}} \frac{\partial...

My attempt: Realize we can work in whatever coordinate system we want, therefore we might as well work in the rest frame of the fluid. In this case ##u^a=(c,\vec{0})##. The conservation law reads ##\nabla^a T_{ab}=0##. Let us pick the Levi-Civita connection so that we don't have to worry about...

My try: $$ \begin{align*} \nabla^a T_{ab} &= \nabla^a \left(\nabla_{a} \phi \nabla_{b} \phi-\frac{C}{2} g_{a b} \nabla_{c} \phi \nabla^{c} \phi\right)\\ &\overset{(1)}{=} \underbrace{(\nabla^a\nabla_{a} \phi)}_{=0} \nabla_{b} \phi + \nabla_{a} \phi (\nabla^a\nabla_{b} \phi)-\frac{C}{2}...

[Throughout we're considering the intrinsic version of the covariant derivative. The extrinsic version isn't of any concern.] I'm having trouble reconciling different versions of the properties to be satisfied by the covariant derivative. Essentially ##\nabla## sends ##(p,q)##-tensors to...

Hi, I would like to ask for a clarification about the difference between parallel transport vs Lie dragging in the following scenario. Take a vector field ##V## defined on spacetime manifold and a curve ##C## on it. The manifold is endowed with the metric connection (I'm aware of it does exist...

This is from Griffiths particle physics, page 360. We have the full Dirac Lagrangian: $$\mathcal L = [i\hbar c \bar \psi \gamma^{\mu} \partial_{\mu} \psi - mc^2 \bar \psi \psi] - [\frac 1 {16\pi} F^{\mu \nu}F_{\mu \nu}] - (q\bar \psi \gamma^{\mu} \psi)A_{\mu}$$ This is invariant under the joint...

Hi, I've been watching lectures from XylyXylyX on YouTube. I believe they are really great ! One doubt about the introduction of Covariant Derivative. At minute 54:00 he explains why covariant derivative is a (1,1) tensor: basically he takes the limit of a fraction in which the numerator is a...

Can you take any non invariant quantity like components and take the covariant derivative of them and arrive at an invariant tensor quantity? Or are there limits on what you can make a tensor?

I read recently that Einstein initially tried the Ricci tensor alone as the left hand side his field equation but the covariant derivative wasn't zero as the energy tensor was. What is the covariant derivative of the Ricci tensor if not zero?

This is from QFT for Gifted Amateur, chapter 14. We have a Lagrangian density: $$\mathcal{L} = (D^{\mu}\psi)^*(D_{\mu}\psi)$$ Where $$D_{\mu} = \partial_{\mu} + iq A_{\mu}(x)$$ is the covariant derivative. And a global gauge transformation$$\psi(x) \rightarrow \psi(x)e^{i\alpha(x)}$$ We are...

I've stumbled over this article and while reading it I saw the following statement (##\xi## a vectorfield and ##d/d\tau## presumably a covariant derivative***): $$\begin{align*}\frac{d \xi}{d \tau}&=\frac{d}{d \tau}\left(\xi^{\alpha} \mathbf{e}_{\alpha}\right)=\frac{d \xi^{\alpha}}{d \tau}...

I am trying to derive the expression in components for the covariant derivative of a covector (a 1-form), i.e the Connection symbols for covectors. What people usually do is take the covariant derivative of the covector acting on a vector, the result being a scalar Invoke a product rule to...

My attempt so far: $$\begin{align*} (\nabla_X Y)^i &= (\nabla_{X^l \partial_l}(Y^k\partial_k))^i=(X^l \nabla_{\partial_l}(Y^k\partial_k))^i\\ &\overset{2)}{=} (X^l (Y^k\nabla_{\partial_l}(\partial_k) + (\partial_l Y^k)\partial_k))^i = (X^lY^k\Gamma^n_{lk}\partial_n + X^lY^k{}_{,l}\partial_k)^i\\...

Sean Carroll says that if we have metric compatibility then we may lower the index on a vector in a covariant derivative. As far as I know, metric compatibility means ##\nabla_\rho g_{\mu\nu}=\nabla_\rho g^{\mu\nu}=0##, so in that case ##\nabla_\lambda p^\mu=\nabla_\lambda p_\mu##. I can't see...

My question is shown in Summary section. Please see the attached file.

So I'm trying to get sort of an intuitive, geometrical grip on the covariant derivative, and am seeking any input that someone with more experience might have. When I see ##\frac {\partial v^{\alpha}}{\partial x^{\beta}} + v^{\gamma}\Gamma^{\alpha}{}_{\gamma \beta}##, I pretty easily see a...

I brought up this subject here about a decade ago so this time I'll try to be more specific to avoid redundancy. In chapter five of Bernard F. Schutz's A First Course In General Relativity, he arrives at the conclusion that in flat space the covariant derivative of the metric tensor is zero...

Can someone clarify the use of semicolon in I know that semicolon can mean covariant derivative, here is it being used in the same way (is expandable?) Or is a compact notation solely for the components of?

The energy-momentum tensor of a free particle with mass ##m## moving along its worldline ##x^\mu (\tau )## is \begin{equation} T^{\mu\nu}(y^\sigma)=m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}. \end{equation} Let contract...

I am reading I am reading Spacetime and Geometry : An Introduction to General Relativity -- by Sean M Carroll and have arrived at chapter 3 where he introduces the covariant derivative ##{\mathrm{\nabla }}_{\mu }##. He makes demands on this which are \begin{align} \mathrm{1.\...

In Carrol's gr notes the covariant derivative of a vector is given as ∇μAϑ=∂μAϑ+ΓϑμλAλ...(1) For a geodesic in 2-D cartesian coordinates the tangent vector is V=##a\hat x+b\hat y##(a and b are constt.)where the tangent vector direction along the curve is ##\hat n=\frac{a\hat x+b\hat...

For the simple case of a 2-D curve in polar coordinated (r,θ) parametrised by λ (length along the curve). At any λ the tangent vector components are V1=dr(λ)/dλ along ##\hat r## and V2=dθ(λ)/dλ along ##\hat θ##. The non-zero christoffel symbol are Γ122 and Γ212. From covariant derivative...

Homework Statement We are given a Lorentz four-vector in "isospin space" with three components ##\vec v^{\mu} = (v^{\mu}_1, v^{\mu}_2, v^{\mu}_3)## and want to express the covariant derivative $$D^{\mu} = {\partial}^{\mu} - ig\frac {\vec \tau} {2}\cdot \vec v^{\mu}$$ explicitly in ##2\times 2##...

Homework Statement Assume that you want to the derivative of a vector V with respect to a component Zk, the derivative is then ∂ViZi/∂Zk=Zi∂Vi/∂Zk+Vi∂Zi/∂Zk = Zi∂Vi/∂Zk+ViΓmikZm Now why is it that I can change m to i and i to j in ViΓmikZm?

If we are representing the basis vectors as partial derivatives, then ##\frac{\partial}{\partial x^\nu + \Delta x^\nu} = \frac{\partial}{\partial x^\nu} + \Gamma^\sigma{}_{\mu \nu} \Delta x^\mu \frac{\partial}{\partial x^\sigma}## to first order in ##\Delta x##, correct? But in the same manner...

##\int d^4 x \sqrt {g} ... ## if I am given an action like this , were the ##\sqrt{\pm g} ## , sign depending on the signature , is to keep the integral factor invariant, when finding an eom via variation of calculus, often one needs to integrate by parts. When you integrate by parts, with...

Homework Statement Hi I am looking at part a). Homework Equations below The Attempt at a Solution I can follow the solution once I agree that ## A^u U_u =0 ##. However I don't understand this. So in terms of the notation ( ) brackets denote the symmetrized summation and the [ ] the...

Hi, let ##\gamma (\lambda, s)## be a family of geodesics, where ##s## is the parameter and ##\lambda## distinguishes between geodesics. Let furthermore ##Z^\nu = \partial_\lambda \gamma^\nu ## be a vector field and ##\nabla_\alpha Z^\mu := \partial_\alpha Z^\mu + \Gamma^\mu_{\:\: \nu \gamma}...

What is the covariant derivative of the position vector $\vec R$ in a general coordinate system? In which cases it is the same as the partial derivative ?

On Riemannian manifolds ##\mathcal{M}## the covariant derivative can be used for parallel transport by using the Levi-Civita connection. That is Let ##\gamma(s)## be a smooth curve, and ##l_0 \in T_p\mathcal{M}## the tangent vector at ##\gamma(s_0)=p##. Then we can parallel transport ##l_0##...

Homework Statement Apologies if this is a stupid question but just thinking about the see-saw rule applied to something like: ## w_v \nabla_u V^v = w^v \nabla_u V_v ## It is not obvious that the two are equivalent to me since one comes with a minus sign for the connection and one with a plus...