All Questions
397 questions
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Transporting Tangent Vectors when Taking Lie Derivatives
I am currently working on understanding the (intrinsic) differential geometry underpinning General Relativity, and I think I could benefit from a more intuitive grasp of the process of taking the Lie ...
- 55
3
votes
3
answers
291
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Different variations of covariant derivative product rule
This is a follow-up question to the accepted answer to this question: Leibniz Rule for Covariant derivatives
The standard Leibniz rule for covariant derivatives is $$\nabla(T\otimes S)=\nabla T\otimes ...
- 1,071
4
votes
3
answers
2k
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Laplace operator and tensor calculus:
I'm studying Tensor calculus and I found this interesting problem:
Show that:
$$ \Delta F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kF\right)$$
Here's some ...
user263457
8
votes
2
answers
4k
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Parallel transport: Lie derivative vs covariant derivative
Given a manifold, we can generalize the idea of derivatives in multiple ways: two of them being the Lie derivative and the covariant derivative. Whereas Lie derivatives do not require any additional ...
- 994
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4
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Deriving the Covariant Derivative of the Metric Tensor
First off, I did look through some other questions:
Covariant Derivative of Metric Tensor
Why is the covariant derivative of the metric tensor zero?
https://math.stackexchange.com/q/2174588/
But they ...
- 81
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votes
1
answer
150
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Is the Covariant Derivative a phenomenological attempt?
I am trying to self study QFT and i am very confused about the covariant derivative.
When we require our theory to be invariant under local gauge transformations we kind of "guess" that we ...
- 2,236
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votes
2
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3k
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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. ...
- 113
7
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2
answers
1k
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Covariant Derivative: What does changing direction mean in curved space?
I am on my way to general relativity, but I am struggling with the covariant derivative.
At this point I am trying to ignore the spacetime character of the world i.e. I am trying to understand what ...
- 570
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votes
1
answer
82
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Why $\Gamma^\mu_{\beta\delta;\gamma} =\Gamma^\mu_{\beta\delta,\gamma}$?
Gravitation Page 276 Exercise 11.3 solution indicated that
$$\nabla_\gamma \nabla _\delta e_\beta
=e_{\mu}\Gamma^\mu_{\beta\delta,\gamma} +(e_\nu\Gamma^\nu_{\mu\gamma}) \Gamma^\mu_{\beta\delta}$$
...
- 3,306
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1
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369
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Is Ricci's theorem can be simply deduced using covariant derivatives of fundamental tensors? [duplicate]
Well Ricci's theorem is given by:
$$\mathrm{D}g_{ij}=\mathrm{D}g^{ij}=0$$
I was wondering that if the theorem can be proved using covariant derivatives of $\delta_i^k$, $g_{ij}$ and $g^{ik}$.
I ...
user262095
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0
answers
216
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Lie derivative of the non-coordinate metric being 0
I'm trying to answer a question about a the Lie derivative of a metric in a non-coordinate basis.
Here, the $C^{a}_{bc}$ are from the Lie derivative of one basis vector with respect to another, or ...
- 420
0
votes
4
answers
95
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Replacing infinitesimals with full vectors in a differential relation. Is it legit?
I'm reading Leonard Susskind's "Special Relativity and Classical Field Theory". On pg. 138 he generalizes a differential relation by replacing infinitesimals with full vectors like so:
Is this ...
- 141
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votes
1
answer
57
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Basic index question on tensor calculus
after some calculations I obtain the next expression
$$\lambda^{c}\nabla_{a}g_{bc}=\lambda^{c}{}(\lambda_{c}\nabla_{a}\lambda_{b}-g_{be} \Gamma^{e}_{ad}\lambda^{d}\lambda_{c}) $$
So my question is ...
- 439
2
votes
1
answer
37
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Relation bewteen to different operators in different charts
Let $\phi$ be a coordinate system and $\partial/\partial^{\mu}$ and $dx^{\mu}$ be the associated coordiantes bases. Then in the region covered by these coordinates we can define a derivative ...
- 439
0
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1
answer
84
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Identity of covariant derivative
I was reading about Einstein-Hilbert action, and in some point in this page they use this identity
$$\sqrt{-g}A^{a}_{;a}=(\sqrt{-g}A^{a})_{,a}$$
I Know that $\nabla_{\sigma}g_{\mu\nu}=0$. And $g$ ...
- 439
-2
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1
answer
558
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What is the covariant derivative of the Ricci Tensor?
How do I take the covariant derivative of the Ricci Tensor? Could someone be so kind as to give me the process of how it is done?
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2
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173
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Basic question of General relativity about covariant derivative
I was reading the book of Wald on General relativity. And in the page number (33) he derives the equation for the action of $\nabla_{a}$ over a tensor of rank $(k,l)$.
This is the equation (3.1....
- 439
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votes
0
answers
252
views
Covariant derivative of a metric determinant
The covariant derivative of a metric is zero $g_{\alpha\beta;\sigma}=0$. Is the covariant derivative of a metric determinant zero following the assumption($g_{\alpha\beta;\sigma}=0$):
$$
g_{;\sigma}=...
- 638
2
votes
2
answers
540
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Inverse of the coordinate transform Jacobian
Assume a point's position is given by the coordinates $x_i$. Introducing a new set of coordinates $\Theta_i$, one can relate the differentials $d\mathbf{x}=(dx_1, dx_2, dx_3)$ and $d\mathbf{\Theta}=(d\...
- 1,026
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1
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2k
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Derivative of a metric tensor
I would like to ask you a question - maybe simple - but bothering me.
We have two four-position vectors product in curvilinear coordinates given by
$(1) \quad X^{\alpha}g_{\alpha \beta}X^{\beta} = \...
- 9
1
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1
answer
314
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What are Connection Forms in General Relativity?
I'm trying to follow an article by H. Ellis (1973), where he developed the first ever metric of a traversable Wormhole (more info here).
In pages 105-106 (the end of the 3rd page in the linked file ...
- 167
1
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2
answers
805
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A Question On Indices Notation In General Relativity
I am trying to make sense of this simple case in my book, but I am still baffled by the notation that is used in the indices with the commas and semicolons; I also do not understand how these are ...
- 55
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votes
0
answers
133
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Does the Lie derivative of a tensor satisfy the strong Einstein equivalence principle?
The Lie derivative of a tensor is a tensor of the same rank and type. But it is connection independent meaning it can be expressed in terms of covariant or partial derivatives. Since the Strong ...
- 1
6
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0
answers
242
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Defining the covariant derivative on bitensors
Bitensors (tensors defined on two different points) are an extension of tensors found in some applications of general relativity, where objects such as the world function, parallel transport operator, ...
- 16.7k
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1
answer
1k
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Proof that covariant derivative of contravariant components of metric vanish for metric compatabilit
In this excercise I want to show that $\nabla_\rho g_{\mu \nu}=0$ and $\nabla_\rho g^{\mu \nu}=0$
This should probably be very easy, but excuse me I'm completly new to GR.
So to do this I used that ...
- 153
-1
votes
2
answers
205
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Divergence of vector fields intuitive understanding
I am troubled by the idea of positive and negative divergence of a vector field. I understand that the idea of e.g. positive div is that a gas expands everywhere (for the velocity field of a gas ...
- 111
1
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1
answer
203
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Why does the covariant derivative of a $(p,q)$-tensor produce a $(p,q+1)$-tensor?
In the specific case of the covariant derivative acting on a scalar function:
$$\nabla_\nu f$$
it seems strange to me that this would return a covector. Am I wrong in thinking the covariant ...
- 7,008
3
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2
answers
639
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The strange character of operator $\nabla$
I was first introduced to the mathematical operation gradient, divergence and curl not in Mathematics but during my studies of Electromagnetism. As you all know learning Maths from a Physics teacher ...
user240696
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1
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90
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Two questions about these notations/operations in the covariant derivative?
For two vector fields V and T we can take the covariant derivative:
$$\nabla_V T=\nabla_{V^\mu \hat e_{\mu}}T$$
$$=V^\nu \nabla_{\hat e_\nu}T$$
What exactly are we doing when we take the vector ...
- 7,008
2
votes
2
answers
164
views
Is there a equivalent theorem for closed form/ exact form for derivative with respect to fields
For a vector (one-form) $A_\mu$, when
\begin{eqnarray}
\partial_{[\mu}A_{\nu]}=0
\end{eqnarray}
then, there exists a scalar $\phi$ such that
\begin{eqnarray}
A_\mu =\partial_\mu\phi
\end{eqnarray}
...
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1
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1
answer
198
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Derivative of the induced metric
2-metric $\gamma_{AB}$ induced on the world sheet by the spacetime metric $g_{\mu\nu}$ is $$\gamma_{AB}=g_{\mu\nu}X^{\mu},_A X^{\nu},_B$$
$$\gamma^{AB},_B=-\gamma^{AC}\gamma^{BD}\gamma_{CD},_B$$
How ...
- 638
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votes
2
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727
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Covariant derivative of a tangent vector
2-metric $\gamma_{AB}$ induced on the world sheet by the spacetime metric $g_{\mu\nu}$ is $$\gamma_{AB}=g_{\mu\nu}t^{\mu}_A t^{\nu}_B$$
where $t^{\mu}_A=\frac{\partial X^{\mu}}{\partial \xi^A}$.
...
- 638
1
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1
answer
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Path Coordinates: direction problem (doubt) in derivative of tangential vector
Why is the direction of derivative of tangential vector perpendicular to the direction of the tangential vector?
- 29
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0
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211
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What exactly is the Leibnitz rule in General Relativity?
In the Differential Geometry part of a course in General Relativity (for instance in David Tong's notes here in page 99, accessed 21 Nov, 2019), one frequently comes across the Leibnitz rule when ...
- 542
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1
answer
205
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Covariant derivative contracted with a metric
I would like to calculate $\nabla_\mu(g^{\mu\alpha}g^{\nu\beta}\nabla_\alpha \kappa_\beta)$. How would this expand?
Where $\nabla$ is the covariant derivative, g the metric and $\kappa_\beta$ a 1-...
- 2,083
2
votes
2
answers
257
views
Product rule of variations
I am deriving the Einstein equation using the Einstein-Hilbert action:
It is obvious that the variation in the Riemann Tensor is calculated from a variational product rule. What is not obvious to ...
- 431
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votes
2
answers
152
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Covector basis derivation
On page 65 of Schutz's A first course in General Relativity, he introduces the notation $\phi_{,\alpha}=\partial\phi/\partial x^\alpha$. He then says that $x^\alpha_{\ \ ,\beta}=\delta^\alpha _{\ \ \ \...
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2
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3k
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Leibniz Rule for Covariant derivatives
I recently came across a video by prof Fredrick Schuller on general relativity where he defines the leibniz rule to be,
$\nabla_X (T(\omega,Y))=\nabla_XT(\omega,Y)+T(\nabla_{X} \omega,Y)+T(\omega,\...
- 291
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0
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63
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Is my thinking correct for partial derivatives and tensors?
So I was transforming the affine connection and I ended up with a term like this:
$$
\frac{\partial^2 x'^a}{\partial x'^b \partial x^p}
$$
where $x$ and $x'$ are two different coordinate systems
...
- 1,953
2
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1
answer
807
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The equation of motion for a scalar field in curved spacetime in terms of the covariant derivative
The equation of motion for a scalar field in curved spacetime $$\frac{\partial\mathcal{L}}{\partial\phi}=\frac{1}{\sqrt{-g}}\partial_{\mu}\left[\sqrt{-g}\frac{\partial\mathcal{L}}{\partial\left(\...
- 12.3k
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1
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103
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Is there a generalization of ${{\partial }_{\alpha }}\left( \sqrt{-g}{{V}^{a}} \right)=\sqrt{-g}{{\nabla }_{a}}{{V}^{a}}$ for arbitrary connection?
I am studying general relativity and I am trying to understand how to perform variation of the Einstein–Hilbert action with respect to the metric ${{g}_{\mu \nu }}$ and an arbitrary connection ${{\...
- 13
3
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1
answer
225
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Derivative with respect to vector
How in Lagrangian and Hamiltonian mechanics we take derivatives with respect to velocity and momentum respectively if they are vectors? Can we take derivative with respect to a vector?
- 1,688
2
votes
3
answers
4k
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How to prove that the covariant derivative obeys the product rule [closed]
In General Relativity the covariant derivative of contravariant vectors $A^\mu$ is:
\begin{equation}
\nabla_\mu A^\nu=\partial_\mu A^\nu+\Gamma^\nu_{\mu\alpha}A^\alpha
\end{equation}
where $\Gamma^\...
1
vote
0
answers
796
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Covariant derivative with respect to commutator
I have some confusion with the notion of $\nabla_{[A, B]}\bf{v}$, that expression, with a commutator of vector fields as the subindex of the connection appears for instance in the definition of the ...
- 275
2
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0
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240
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Torsion form and exterior covariant derivative
The torsion form can be defined as the exterior covariant derivative of a solder form, $\Theta=d_\omega\theta$. This derivative is always in the fundamental representation of the algebra $\mathfrak g$ ...
- 258
2
votes
1
answer
413
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Is the Lie derivative along the normal well-defined?
This question is cross-posted at https://math.stackexchange.com/q/3274757/247251
Let $(\Sigma, q)$ be a non-degenerate submanifold of a Lorentzian manifold $(M,g)$. Let $N$ be the section of $T\Sigma ...
- 245
2
votes
1
answer
732
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Divergence of a tensor
On pg.70 of Dalarsson's "Tensors, Relativity and Cosmology"
For a mixed tensor of contravariant order 2 and covariant order 1 $(T^{mn}_{p,m})$, the divergence with respect to m is defined as:$$T^{...
- 1,047
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1
answer
77
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Covariant derivative in a basis
Reading through this paper, I saw that the energy momentum conservation:
$$\nabla_\mu T^{\mu\nu}=0$$
can be evaluated as:
$$\partial_t(\sqrt{-g}T^{t}_\nu)=-\partial_i(\sqrt{-g}T^{i}_\nu)+\sqrt{-g}T^...
- 43
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2
answers
1k
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Why is the partial derivative a contravariant 4-vector?
The contravariant partial derivative is defined as following:
$$\partial ^\mu = \frac{\partial}{\partial x_\mu}$$
where the index $\mu$ runs from 0 to 3. A contravariant vector under Lorentz ...
user63248
0
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0
answers
31
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General Relativity and cosmology [duplicate]
What is the physical meaning of Ricci scalar is a covariantly constant?
