All Questions
397 questions
3
votes
1
answer
118
views
Covariant derivative acting on Dirac delta function
Pardon my naive computational question. In my calculations, I encounter the following expression:
\begin{equation} \label{eq1}
\frac{\delta}{\delta g^{\gamma \epsilon}(z)} \left( g_{\mu \alpha}(x) \...
- 11.6k
3
votes
2
answers
340
views
Understanding the definition of the covariant derivative
I'm currently working my way through the book "Mathematical Methods for Physics - An Introduction to Group Theory, Topology and Geometry" and I think I have a very fundamental ...
- 3,358
0
votes
1
answer
80
views
The definition of the Lie Derivative
I am aware that an answer to an almost identical question already exist, however, I found the already existing answer not helpful (at least to my current question).
Carroll defines, in his book, the ...
- 541
-1
votes
0
answers
17
views
How to prove that a Lie algebra-valued differential form is exact for the covariant derivative [migrated]
Given a differential $p$-form $\omega^A$ over a smooth manifold with values on some Lie algebra, I wanted to know how could one prove that it can be written as an exact form for the exterior covariant ...
Buzz♦
- 17.1k
3
votes
1
answer
114
views
Relationship between covariant derivative and metric tensor
In general relativity, the covariant derivative of the coordinate vector is a tensor, equal to $$x^{\mu}_{:\rho} = x^{\mu}_{,\rho} + \Gamma^{\mu}_{\rho\nu}x^{\nu},$$ is it meaningful to equate this ...
- 3,358
1
vote
0
answers
62
views
A trick for derivatives of thermodynamic quantities [closed]
Starting from
$$dU=TdS-PdV$$
We can write, for instance $U(T,V)$ and $S(T,V)$ to obtain:
$$\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_T dV=T\left(\frac{\...
- 225
3
votes
1
answer
94
views
What happens to $g^{\alpha\beta}_{,\sigma}=-g^{\alpha\mu}g^{\beta\nu}g_{\mu\nu,\sigma}$ when $g_{\mu\nu}\rightarrow \eta_{\mu\nu}$ (weak field limit)?
The equation
$$g^{\alpha\mu}_{\,\,\,\, ,\sigma}\,g_{\mu\nu} + g^{\alpha\mu}\,g_{\mu\nu,\sigma} = (g^{\alpha\mu}g_{\mu\nu})_{,\sigma} = \delta^\alpha_{\nu,\sigma} = 0 $$
gives the useful relation
$$g^{\...
- 79
0
votes
0
answers
38
views
Four-divergence of a vector [duplicate]
The covariant derivatives of a four-vector is
$$
\nabla_{\nu}U_{\mu} = \partial_{\nu}U_{\mu} - \Gamma^{\lambda}_{\mu\nu}U_{\lambda}
$$
It has the following identity:
$$
\nabla_{\mu}U^{\mu} = \frac{\...
- 213k
0
votes
1
answer
69
views
Is 4-velocity a Vector in the Sense of Covariant Derivative along Worldline
The definition of 4-velocity $U^{\mu} \equiv dx^{\mu}(\tau)/d\tau$, however, we've learnt that the covariant derivative for a vector along a curve parametrized by proper time is,
$$\frac{DA^{\mu}}{D\...
- 16.7k
6
votes
1
answer
514
views
Covariant derivative of the vielbein determinant
The vielbein postulate says that
$$\nabla_\mu e_v^{\,a}=\partial_{\mu}e_\nu^{\,a}+\omega_{\mu\,\, b}^{\,\,a}\,e^b_\nu-\Gamma^\sigma_{\mu\nu}\,e^{\,a}_\sigma=0.$$
$\nabla$ is the coordinate covariant ...
CommunityBot
- 1
1
vote
1
answer
98
views
Proving a Superfunction Identity
I am trying to figure out the proof of the identity given between equations (1.11.7) and (1.11.8) in ref. [1], i.e.
\begin{align}
\Phi'(e^{-K}\,z\,e^K)=e^{-K}\Phi'(z) \tag{1}
\end{align}
where $z=(...
- 213k
5
votes
1
answer
330
views
Divergence of vector field term-wise
In a spacetime $(M, g)$ the following identity for the divergence of a vector field $X$ holds
$$ \nabla_{\mu} X^{\mu} = \frac{1}{\sqrt{-\det g}} \,
\partial_{\mu} \big( \sqrt{- \det g} \ X^{\mu} \big)...
- 213k
1
vote
1
answer
286
views
Covariant derivative with an upper index in terms of Christoffel symbols
I have encountered expression
$$\frac{1}{2}\left(2 \dot{g}_{\mu}{}^{\lambda ; \mu}-\dot{g}_{\mu}{}^{\mu ; \lambda}\right)$$
in a GR paper.
Here we assume to be working with the de Sitter metric $g$ ...
CommunityBot
- 1
1
vote
0
answers
62
views
Adjoint of the covariant derivative of a field?
Let's call $D$ the covariant derivative, $T$ the transposition of a field and $*$ its complex conjugate, so $T*$ is the "adjoint".
Is: $$(D_{\mu}\Phi)^{T*} (D_{\mu}\Phi)=D^{\mu}\Phi^*D_{\mu}\...
- 1,337
3
votes
3
answers
4k
views
Why is the covariant derivative of a one-form $\nabla_{i}v_j=\frac{\partial v_j}{\partial u^{i}}-\Gamma^k_{~ij}v_k$?
I understood that the covariant derivative of a vector field is
$$
\nabla_{i}v^j=\frac{\partial v^j}{\partial u^{i}}+\Gamma^j_{~ik}v^k
$$
Then why is the covariant derivative of a covector field
$$
\...
0
votes
0
answers
59
views
Covariant derivative with torsion
The covariant derivative is defined (on contravariant vectors) as:
$$\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu \rho} V^\rho \tag{1}$$
The purpose of the covariant derivative is to ...
- 3,672
2
votes
0
answers
46
views
Is a spin connection with torsion possible whereas the affine connection is only Levi-Civita (torsion-free) in Supergravity?
In the paper "Simple Supergravity" from G. Dall'Agata & M. Zagermann (arXiv:2212.10044v2 15 Feb. 2023) on page 8 when it comes to the antisymmetric part of the covariant derivative of ...
- 10.1k
2
votes
1
answer
190
views
Covariant derivative to the metric determinant?
I am reading the paper Alternatives to dark matter and dark energy, but cannot obtain one specific equation no matter how I tried. So I wrote an email to the author, the following is what he replies ...
- 635
0
votes
0
answers
159
views
What's the difference? $\nabla_\mu e_\nu=\Gamma_{\mu \nu}^\rho e_\rho~\text{ and }~\partial_\mu e_\nu=\Gamma_{\mu \nu}^\rho e_\rho~?$
What's the difference? $$\nabla_\mu e_\nu=\Gamma_{\mu \nu}^\rho e_\rho~\text{ and }~\partial_\mu e_\nu=\Gamma_{\mu \nu}^\rho e_\rho~?$$
In John Dirk Walecka's book 'Introduction to General Relativity',...
- 12.9k
1
vote
3
answers
98
views
Does (covariant) divergence-freeness of the stress-energy tensor ${T^{\mu\nu}}_{;\nu}=0$ follow from the Bianchi identity?
I'm working through Chap. $30$ of Dirac's "GTR" where he develops the "comprehensive action principle". He makes a very slick and mathematically elegant argument to show that the ...
- 918
3
votes
1
answer
133
views
A theorem on page 72 in The Large Scale Structure of Space-Time [closed]
In chapter 3 of the book, page 72, a static observer is defined as $V^{a}\equiv f^{-1}K^{a}$, where $K^{a}$ is a timelike Killing vector field and $f^{2}=-K^{a}K_{a}$. Then, Hawking & Ellis claim ...
- 7,425
0
votes
1
answer
81
views
Covariant Directional Derivative
How is the covariant directional derivative $\frac{D}{d\lambda}=\frac{dx^{\mu}}{d\lambda}\nabla_{\mu}$ in GR related to acceleration? I am motivated to ask this question because I’ve seen it stated ...
- 213k
3
votes
1
answer
163
views
Laplace-Beltrami operator for a vector field
For a scalar field $\varphi$, the "wave" operator is defined as follows:
$$\Box \varphi \equiv g^{ab}\nabla_a\nabla_b~\varphi = \frac{1}{\sqrt{|g|}}\partial_a\left\{\sqrt{|g|}~g^{ab}~\...
- 8,900
0
votes
2
answers
3k
views
Covariant Derivative of Metric Tensor
I'm an amateur studying General Relativity. I'm reading some notes of lectures by Susskind. In them, it is written that
"we know that [the covariant derivative of the metric tensor] is zero. ...
- 12.9k
0
votes
0
answers
58
views
Partial derivatives of Christoffel symbols to Covariant derivatives
I wanted to express this thing: $g^{ab}\partial_c\Gamma^c_{ab} - g^{ab}\partial_a\Gamma^c_{cb}$, in terms of a covariant derivative. I figured out that if you swap $a$ and $c$ in the $\partial \Gamma$ ...
- 213k
2
votes
1
answer
109
views
Spin connection for a vector field
Using Birrel & Davies convention, the covariant derivative for a field of arbitrary spin in curved spacetime is given by
$$\nabla_\mu=\partial_\mu+\Omega_\mu,\tag{1}$$
with
$$\Omega_\mu=\frac{1}{2}...
- 2,349
1
vote
0
answers
50
views
A covariant derivative computation in General Relativity [duplicate]
I am trying to compute $\nabla^\mu\nabla^\nu R_{\mu\nu}$.
I proceed as follows:
\begin{align}
\nabla^\mu\nabla^\nu R_{\mu\nu}&=g^{\mu\rho}g^{\nu\lambda}\nabla_\rho\nabla_\lambda R_{\mu\nu} \\
&...
- 372
1
vote
1
answer
72
views
Covariant derivative for spin-2 field
I have mostly seen the concept of covariant derivative with regard to spin-1 fields. Is it possible to define the covariant derivative for spin-2 fields as well?
- 213k
1
vote
1
answer
114
views
How is this deduced? (Differentiation of tensors)
In Schutz's An Introduction to General Relativity, he talked about how to differentiate tensors. Here is a step that I cannot understand.
$$\frac{d\mathbf{T}}{d\tau} = \left( T^{\alpha}_{\beta, \gamma}...
- 12.9k
1
vote
1
answer
77
views
Does the divergence theorem require the covariant derivative to be metric compatible?
I know this is more of a mathematical question, but it arises in the context of general relativity and uses its language so I thought it would be best to ask it here. I understand that the divergence ...
- 11.6k
0
votes
1
answer
83
views
What is the relation between gauge field and Levi-Civita connection?
In field theory, covariant derivative is something like
$$D_{\mu}\phi=(\partial_{\mu}-igA_{\mu})\phi$$
while in differential geometry, covariant derivative is something like
$$D_{\mu}V^{\nu}=\partial_{...
- 11.6k
0
votes
1
answer
75
views
Lie derivative: moving boat on a flowing river
Lie derivatives signifies how much a vector (Tensor) changes if flown in the direction of some other vector. I am thinking the typical moving boat on a flowing river problem where the river is flowing ...
- 66.3k
4
votes
1
answer
305
views
How to derive $\nabla_{\mu} T^{\mu\nu}=\frac{1}{\sqrt {-g}}\partial_{\mu}(\sqrt{-g}\,T^{\mu\nu})$?
This $$\nabla_{\mu} T^{\mu\nu}=\frac{1}{\sqrt {-g}}\partial_{\mu}(\sqrt{-g}\,T^{\mu\nu})\tag{3.39}$$ is from a textbook on general relativity on black hole Vaidya metric, where only non-zero term of ...
- 1,350
0
votes
1
answer
347
views
Isomorphism of the tangent space and the space of directional derivatives [closed]
I have already constructed the tangent space to a manifold, denoted $T_pM$, and I have a good basis for it $\{\hat e_{(\mu)}\}$. (I followed the method of equivalence classes of curves tangent at $p$....
- 2,029
1
vote
4
answers
445
views
How to find the double covariant derivative of a general vector?
I have been reading through Carroll's GR textbook and there is a line in the derivation of the Riemann tensor that I do not understand.
$$\nabla_\mu \nabla_\nu V^\rho=\partial_\mu(\nabla_\nu V^\rho) - ...
- 1,350
0
votes
0
answers
46
views
Application of Fermi-Walker derivative to specific problem
I am now reading about the tetrad formalism in GR and I am starting (how not) with the Wikipedia Article:
Frame fields in general relativity.
In this article, as an example, they show how tetrads can ...
- 213k
0
votes
3
answers
240
views
What is the correct term for $\nabla\phi$? Co-vector or 1-form or both?
In the olden days, $\nabla\phi$ was used to be called a covariant vector (Weinberg used this language in his book Gravitation & Cosmology). But this terminology is considered bad for several ...
- 13.4k
2
votes
1
answer
74
views
Closed interval in variation of a field
Let's suppose that $\psi$ is a field so that $$\psi\in\Gamma^\infty(\pi):=\{\psi\in C^\infty(M,E)\ |\ \pi\circ\phi=I_M\}$$ where $E$ is a spacetime bundle over spacetime $M$ and $\pi : E\rightarrow M$ ...
CommunityBot
- 1
0
votes
1
answer
115
views
Double covariant derivative of a mixed tensor
Let's say, we have a mixed tensor of type (2,1) denoted by $T^{mn}{}_p$ and the goal is to find the expression of $[\nabla_a, \nabla_b] T^{mn}{}_p$ in terms of fundamental tensors.
Firstly, I am ...
- 1
2
votes
2
answers
1k
views
Is curvature the exterior covariant derivative of the connection?
Let $P\to M$ be a $G$-principal bundle, $G$ a topological group, $\omega$ the connection and $V$ a vector space.
We define $d_\omega: \Omega^k_G(P, V)\to\Omega^{k+1}_G(P, V)$ the exterior covariant ...
- 130
1
vote
0
answers
113
views
Derivation of covariant derivative by means of parallel transport
I've studied covariant derivative in many courses so far, but I got stuck on the definition given by the teacher notes of the exam of Topological QFT.
I think that he improperly used the name "...
- 177
4
votes
3
answers
1k
views
Lie derivative of a vector along itself
The Lie derivative for a covariant and contravariant vector is:
$$\mathcal{L}_U V^\mu=U^\nu\nabla_\nu V^\mu- V^\nu\nabla_\nu U^\mu$$
$$\mathcal{L}_U V_\mu=U^\nu\nabla_\nu V_\mu+ V_\nu\nabla_\mu U^\nu$$...
- 5,244
6
votes
2
answers
4k
views
How to calculate the covariant derivative $\nabla_{\bf e_\beta}{\bf e}_\alpha$ of a basis vector along another basis vector?
So in my relativity course, we recently learned about the covariant derivative. it is defined as:
\begin{equation}
\nabla_{\mu}V^{\nu} = \partial_{\mu}V^{\nu}+\Gamma^{\nu}_{\mu,\lambda}V^{\lambda}
\...
1
vote
0
answers
94
views
Del operator confusion [closed]
The very first thing my textbook says is that the Del operator is defined as:
$$\vec{\nabla}=\vec{a}^i\nabla_i$$
Where $\nabla_i$ is the covariant derivative and " $\vec{a}^i$ is the curvilinear ...
- 16.6k
4
votes
2
answers
231
views
Looking for the geometric meaning of the curl of Killing vector fields
From Killing equation
$$\nabla_\nu \xi_\mu + \nabla_\mu \xi_\nu = 0$$
it can be shown that $\nabla_\nu \xi_\mu$ is antisymmetric.
From it we can construct an antisymmetric tensor $\mathcal{A}_{\mu\nu}$...
- 11.6k
1
vote
1
answer
102
views
Variation of Torsion-Free Spin Connection
In the book 'Supergravity' by Freedman and van Proeyen, in exercise (7.27) it is written
To calculate [the variation $\delta\omega_{\mu ab}$ of the torsion-free spin connection], consider the ...
- 98
0
votes
1
answer
113
views
What is the intuition or the derivation of covariant derivative?
I asked this question in mathematics but the answer I got was a bit too abstract for me so I hope that my fellow physicists can give me more of an intuition or an easier explaination of my question. ...
- 213k
1
vote
0
answers
118
views
Lie derivative of a one-form
I am going through Nakahara's textbook on geometry and topology in physics. Intuitively, I understand the definition of a lie derivative of a vector field
$$\mathcal{L}_xY = \lim_{\epsilon_\to 0}\frac{...
- 213k
2
votes
2
answers
152
views
How to calculate the rotation at a singularity?
An electrodynamics lecture asks me to prove that
$$
\nabla \times \left( \frac{\vec{M} \times \vec{x}}{ |\vec{x}|^3} \right) = \frac{8 \pi}{3} \vec{M} \delta^3(\vec{x})- \frac{\vec{M}}{|\vec{x}|^3}+ \...
- 1
8
votes
2
answers
915
views
How does the covariant derivative satisfy the Leibniz rule?
In Carroll's "Spacetime and Geometry", he states on page 95 (section 3.2) that the covariant derivative, $\nabla$, is a map from $\left(k, l\right)$ tensor fields to $\left(k, l+1\right)$ ...
- 213k