Covariant Definition and 361 Threads

I'm reading "Problem Book In Relativity and Gravitation". In this book there is a problem 7.5 Show that metric tensor is covariant constant. To prove it, authors suggest to use formulae for covariant derivative: Aαβ;γ=Aαβ,γ−AσαΓβγσ−AσβΓαγσ after that they write this formulae for tensor g and...

In Dirac's GTR. Sec. 12 (p. 22), he wants to show the equivalence of: (a) Vanishing of the curvature tensor ##R^\beta_{\sigma\nu\rho}=0##; or equivalently, the equality of mixed second covariant derivatives ##A_{\nu:\sigma:\rho}=A_{\nu:\rho:\sigma}##. (b) Path independence of parallel transport...

I am a graduate physics student currently studying electrodynamics as a core paper. I want to know why exactly do we use only covariant formulation for writing Maxwell's equations? Or do we also use contravariant formulation (i.e., if something like that even exists)?

I am a physics enthusiast reading Covariant Physics by Moataz H. Emam. In his chapter about Point Particle mechanics there is a transformation equation for a displacement vector. I don't see how he arrived at the final equation 3.6. Is it a chain rule or product rule? Can't seem to figure it...

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

Under the entry "Quantum electrodynamics" in Wikipedia, the Dirac equation for an electron is given by $$ i\gamma^{\mu}\partial_{\mu}\psi - e\gamma^{\mu}\left( A_{\mu} + B_{\mu} \right) \psi - m\psi = 0 ,\tag 1 $$ or $$ i\gamma^{\mu}\partial_{\mu}\psi - m\psi = e\gamma^{\mu}\left( A_{\mu} +...

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...

I am doing a project where the final scope is to find an extra operator to include in the proca lagrangian. When finding the new version of this lagrangian i'll be able to use the Euler-Lagrange equation to find the laws of motion for a photon accounting for that particular extra operator. I...

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...

What is often said in Covariant LQG is that the triangulation is a truncation, and is not what is responsible for the discrete volumes one ends up with in the theory. Rather, what is responsible is the discrete spectra of the volume operator acting on the nodes of a spin network. My confusion...

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...

This week, I've been assigned a problem about a 3-sphere. I am confused how to approach this problem and any comments would be greatly appreciated. (a) - would I be correct to assume the metric G is simply the dot product of two vector fields with dx^2 dy^2 du^2 and dv^2 next to their...

Is a connection the same thing as a covariant derivative in differential geometry? What Is the difference between a covariant derivative and a regular derivative? If you wanted to explain these concepts to a layperson, what would you tell them?

I am solving some GR problems. In one problem, some relation between a second covariant derivative and the Riemann tensor is to be proven. In the solution, the relation is first proven in a local flat coordinate system, followed by a statement that, since this relation is covariant it is true...

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...

We have a basis {##\mathbf{e}_1##, ##\mathbf{e}_2##, ##\dots##} and the corresponding dual basis {##\mathbf{e}^1##, ##\mathbf{e}^2##, ##\dots##}. I learned that a vector ##\vec{V}## can be expressed in either basis, and the components in each basis are called the contravariant and covariant...

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 have read many GR books and many posts regarding the title of this post, but despite that, I still feel the need to clarify some things. Based on my understanding, the contravariant component of a vector transforms as, ##A'^\mu = [L]^\mu~ _\nu A^\nu## the covariant component of a vector...

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.

If the comma means ordinary derivative, then ##(A_\mu A_\nu^{,\nu} - A_\mu^{,\nu} A_\nu)^\mu = A_\mu^{,\mu}A_\nu^{,\nu} + A_\mu A_\nu^{,\nu,\mu} - A_\mu^{,\nu,\mu} A_\nu - A_\mu^{,\nu}A_\nu^{,\mu} = A_\mu^{,\mu}A_\nu^{,\nu} - A_\mu^{,\nu}A_\nu^{,\mu} ##, where ##A## is some vector field...

I'm working with modfied gravity models and I need to consider the perturbation of field equations. I have problems with the term were I have two covariant derivatives, I'm not sure if I'm doing it right. I have: $$\delta(\nabla_\rho \nabla_\nu \left[F'(G)R_{\mu}^{\hphantom{\mu} \rho}\right])$$...

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...

Is there a purpose of using covariant or contravariant tensors other than convenience or ease in a particular coordinate system? Is it possible to just use one and stick to one? Also what is the meaning of mixed components used in physics , is there a physical significance in choosing one over...

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...

Summary: Failed find information on the internet, really appreciate any help. Can someone tell me what is ∇ϒ∇δ𝒆β? It seems to be equal to 𝒆μΓμβδ,ϒ+(𝒆νΓνμϒ)Γμβδ. Is this some sort of chain rule or is it by any means called anything?

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...

The first part I'm fairly sure is just the regular gradient in polar coordinates typically encountered: $$\nabla u= \hat {\mathbf e_r} \frac {\partial u} {\partial r} + \hat {\mathbf e_\theta} \frac 1 r \frac {\partial u} {\partial \theta}$$ or in terms of scale factors: $$=\sum \hat...

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 asked to show that: ## \frac{d^2x_\mu}{d\tau^2}= \frac{1}{2} \frac{dx^\nu}{d\tau} \frac{dx^{\rho}}{d\tau} \frac{\partial g_{\rho \nu}}{\partial x^{\mu}} ## ( please ignore the image in this section i cannot remove it for some reason ) Homework Equations The...

Depending on the source, I'll often see EFE written as either covariantly: $$R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = 8 \pi GT_{\mu\nu}$$ or contravariantly $$R^{\alpha\beta} - \frac{1}{2}Rg^{\alpha\beta} = 8 \pi GT^{\alpha\beta}$$ Physically, historically, and/or pragmatically, is there a...