All Questions
54 questions
47
votes
4
answers
16k
views
What is the physical meaning of the connection and the curvature tensor?
Regarding general relativity:
What is the physical meaning of the Christoffel symbol ($\Gamma^i_{\ jk}$)?
What are the (preferably physical) differences between the Riemann curvature tensor ($R^i_{\ ...
- 13.7k
39
votes
5
answers
47k
views
Why is the covariant derivative of the metric tensor zero?
I've consulted several books for the explanation of why
$$\nabla _{\mu}g_{\alpha \beta} = 0,$$
and hence derive the relation between metric tensor and affine connection $\Gamma ^{\sigma}_{\mu \beta}...
- 929
68
votes
6
answers
48k
views
Laplace operator's interpretation
What is your interpretation of Laplace operator? When evaluating Laplacian of some scalar field at a given point one can get a value. What does this value tell us about the field or it's behaviour in ...
- 3,217
13
votes
1
answer
2k
views
Geometric meaning of spin connection
A very short question: Does the spin connection that we encounter in General Relativity $$\omega_{\mu,ab}$$ have a geometric meaning? I am supposing it does because it comes from mathematical terms ...
- 1,539
24
votes
4
answers
14k
views
Why is the covariant derivative of the determinant of the metric zero?
This question, metric determinant and its partial and covariant derivative,
seems to indicate $$\nabla_a \sqrt{g}=0.$$ Why is this the case? I've always learned that $$\nabla_a f= \partial_a f,$$ ...
- 749
20
votes
3
answers
22k
views
D'Alembertian for a scalar field
I have read that the D'Alembertian for a scalar field is
$$
\Box = g^{\nu\mu}\nabla_\nu\nabla_\mu = \frac{1}{\sqrt{-g}}\partial_\mu (\sqrt{-g}\partial^\mu).
$$
Exactly when is this correct? Only for $...
- 15.3k
107
votes
4
answers
11k
views
Why does nature favour the Laplacian?
The three-dimensional Laplacian can be defined as $$\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}.$$ Expressed in spherical coordinates, it ...
- 1,357
24
votes
5
answers
16k
views
How do I calculate the perturbations to the metric determinant?
I am trying to calculate $\sqrt{-g}$ in terms of a background metric and metric perturbations, to second order in the perturbations. I know how to expand tensors that depend on the metric, but I don't ...
- 541
5
votes
1
answer
302
views
Why do we do partial and not covariant differentiation with $x^{\nu}$?
Why when taking the velocity vector we make
$$u^{\nu}=\frac{d}{d\tau}x^{\nu}$$
and not
$$u^{\nu}=\frac{\nabla}{d\tau}x^{\nu}$$
where in the last equation I meant the covariant derivative. Why?
- 6,137
1
vote
2
answers
3k
views
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
7
votes
2
answers
3k
views
Covariant derivative of a covariant derivative
I'm trying to find the covariant derivative of a covariant derivative, i.e. $\nabla_a (\nabla_b V_c)$.
This is something I've taken for granted a lot in calculations, namely I though that by the ...
- 623
5
votes
2
answers
675
views
Is there a way to see that $ \nabla_\mu g_{\nu \rho} = 0 $ without explicit computation, where $\nabla_\mu$ refers to the covariant derivative?
In books, it is usually said that this is a consequence of the fact that parallel transport preserves dot product. How ?
4
votes
2
answers
2k
views
Difference tensor between two connections
I am using these supergravity lecture notes by Gary W. Gibbons. On page 18, the author claims that geodesics and autoparallels coincide for a theory with totally antisymmetric torsion, and proves it ...
- 330
3
votes
1
answer
593
views
Why aren't Christoffel symbols tensors? - asked from a fibre bundle perspective
I've been reading about connections on fibre bundles recently and it's made me think about the exact nature of the Christoffel symbols in GR.
If we have a vector bundle $E$ over $M$ and put a ...
3
votes
3
answers
2k
views
What is the physical meaning of the Levi-Civita connection?
I'm taking a course in General Relativity and I have studied the fundamental theorem of Riemannian geometry:
Let $M$ be a manifold with metric $g$. Then exists an unique torsion-free connection $\...
- 1,573
2
votes
1
answer
262
views
Exterior Derivative on Curved Manifold (SpaceTime)
Considering the 2-Form Gauge Potential $B_{\mu \nu}$, we can write $dB=H$.
In a flat manifold we have that $H_{\mu \nu \rho}=\partial_{\mu}B_{\nu \rho}+\partial_{\rho}B_{\mu \nu}+\partial_{\nu}B_{\rho ...
- 782
2
votes
2
answers
320
views
Is the contracted Christoffel symbol a tensor?
The coordinate transformation law (from coordinates x to coordinates y) for the Christoffel symbol is:
$$\Gamma^i_{kl}(y)=\frac{\partial y^i}{\partial x^m} \frac{\partial x^n}{\partial y^k} \frac{\...
- 613
1
vote
1
answer
404
views
Difference between covariant derivatives in general relativity and electromagnetism
There are many similarities between the gauging of a $U(1)$ symmetry to obtain the physics of electromagnetism and the gauging of diffeomorphisms to obtain the physics of general relativity. In ...
65
votes
4
answers
14k
views
Lie derivative vs. covariant derivative in the context of Killing vectors
Let me start by saying that I understand the definitions of the Lie and covariant derivatives, and their fundamental differences (at least I think I do). However, when learning about Killing vectors I ...
- 28.6k
18
votes
1
answer
3k
views
Is there a "covariant derivative" for conformal transformation?
A primary field is defined by its behavior under a conformal transformation $x\rightarrow x'(x)$:
$$\phi(x)\rightarrow\phi'(x')=\left|\frac{\partial x'}{\partial x}\right|^{-h}\phi(x)$$
It's fairly ...
16
votes
4
answers
6k
views
Covariant derivative for spinor fields
scalars (spin-0) derivatives is expressed as:
$$\nabla_{i} \phi = \frac{\partial \phi}{ \partial x_{i}}.$$
vector (spin-1) derivatives are expressed as:
$$\nabla_{i} V^{k} = \frac{\partial V^{k}}{ \...
- 14.8k
8
votes
3
answers
3k
views
Are indices conventionally raised inside or outside of partial derivatives in general relativity?
If $A_\mu$ is a one-form, then is there a widely accepted convention among physicists about whether the notation $$\partial_\mu A^\mu \tag{1}$$ means "the partial-derivative four-divergence of the ...
- 49.4k
7
votes
5
answers
729
views
What does the metric condition $\nabla_\rho g_{\mu\nu}=0$ in General Relativity intuitively mean for an observer measuring distances?
In General Relativity, the following condition hold: $\nabla_\rho g_{\mu\nu}=0$, where $g_{\mu\nu}$ is the metric of spacetime which has to do with measuring distances and angles and $\nabla$ is the ...
- 7,860
7
votes
1
answer
3k
views
Generalized divergence of tensor in GR
Although I've forgotten the proof (and cannot find it in, say, Carroll's book), the following formula holds for the covariant divergence in general relativity:
$$\nabla_{\mu} A^{\mu} = \frac{1}{\sqrt{...
- 2,022
6
votes
4
answers
751
views
Do $\vec r$ and $d \vec r$ have the same direction?
One question is bugging me for a long time but I couldn't make out anything nor could my friends. Here it goes:
We know, $\vec r$ is regarded as the position vector. So we can say,
$$\vec r \cdot\vec ...
- 4,984
6
votes
2
answers
4k
views
Metric determinant and its partial and covariant derivative
question : $\nabla_a \nabla_b \sqrt{g} \phi =\partial_a \sqrt{g} \partial_b \phi$ is true?
because $\nabla_a \sqrt{g}=0$ so we can write $\sqrt{g} \nabla_a \nabla_b \phi$ , but because the determinant ...
- 79
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}
\...
- 1,001
6
votes
2
answers
2k
views
Relationship between Connection and Material Derivative
Suppose $D\subset \Bbb R^3$ contains a fluid and that $f : D\times \mathbb{R}\to \mathbb{R}$ is a time dependent function defined on the fluid region. In that case, the material derivative is defined ...
- 37.4k
6
votes
5
answers
8k
views
Covariant Derivative of Kronecker Delta
I am reading Carroll's book on GR right now, and I ran into a little trouble in his chapter 3 on curvature. He is establishing the properties of the covariant derivative, and claims that the fact that ...
- 147
5
votes
2
answers
950
views
Gauge covariant derivative on form
Let $e$ be a one-form gauge field that belongs to the adjoint representation of the gauge group, that is SO(1,2). It is defined as
\begin{equation}
e = e_{\alpha}^{A}T_Adx^{\alpha}.
\end{equation}
...
- 295
5
votes
1
answer
1k
views
Is the inverse of the deformation gradient simply the deformation gradient of the inverse transformation?
If we have a continuum where the initial positions are denoted $X$ and the positions after some deformation are denoted $x$, the deformation gradient is defined:
$$ F = \frac{\partial x}{\partial X} $...
- 103
4
votes
3
answers
1k
views
Lie derivative of a vector along itself
The Lie derivative for a covariant and contravariant vector is:
$$\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$$...
- 469
4
votes
0
answers
1k
views
How is Infinitesimal coordinate transformation related to Lie derivatives?
I am reading the book "Gravitaion and Cosmology" by S. Weinberg. In section 10.9, while discussing Lie derivatives of tensors of different ranks, he makes a general comment:
The effect of an ...
- 437
4
votes
1
answer
3k
views
Differentiation of a vector with respect to a vector
Does differentiation of a vector with respect to a vector make any sense? Even if it makes sense, how does it make any physical meaning? I mean what is the physical interpretation?
- 4,984
3
votes
2
answers
639
views
The strange character of operator $\nabla$
I was first introduced to the mathematical operation gradient, divergence and curl not in Mathematics but during my studies of Electromagnetism. As you all know learning Maths from a Physics teacher ...
user240696
3
votes
1
answer
340
views
Definition of exterior derivative
In Sean Carroll's GR book, pg 84, the exterior derivative $d$ is defined as
$$(dA)_{\mu_1\mu_2...\mu_{p+1}} = (p+1) \partial_{[\mu_1}A_{\mu_2...\mu_{p+1}]}\tag{2.76}$$
where $A$ is a $p$-form and the ...
- 4,276
3
votes
2
answers
1k
views
Why is 4-velocity not defined as the covariant derivative of position instead of the regular time derivative? [duplicate]
The geodesic equation is usually written as
\begin{equation}
D_\tau u^\mu = 0
\end{equation}
where $D_\tau= u^\mu \nabla_\mu$ is the covariant proper time derivative and $u^\mu=\frac{dx^\mu}{d\tau}$ ...
- 4,737
3
votes
1
answer
1k
views
Can Yang-Mills field strength be defined as covariant derivative squared?
In Yang-Mills theory the field strength tensor $F_{\mu \nu}$ can be calculated as
$$
\begin{equation}
F_{\mu\nu} \equiv \frac{i}{g} [D_\mu,D_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu -ig[A_\mu,A_\...
- 4,833
3
votes
1
answer
2k
views
The role of the affine connection the geodesic equation
I apologise in advance that my knowledge of differential geometry and GR is very limited. In general relativity the equation of motion for a particle moving only under the influence of gravity is ...
- 155
2
votes
2
answers
339
views
Different definitions of exterior derivative
In Sean Carroll's GR book, pg 84, the exterior derivative $d$ is defined as
$$(dA)_{\mu_1\mu_2...\mu_{p+1}} = (p+1) \partial_{[\mu_1}A_{\mu_2...\mu_{p+1}]}\tag{2.76}$$
where $A$ is a $p$-form and the ...
- 4,276
2
votes
1
answer
346
views
Covariant derivative in field theory
I'm reading Physics from Symmetry by Jakob Schwichtenberg and in Chapter 7 equation 7.18 he introduces the covariant derivative when deriving the interaction Lagrangian density for the spin-$\frac{1}{...
- 432
2
votes
1
answer
261
views
Infinitesimal coordinate transformation and Lie derivative
I need to prove that under an infinitesimal coordinate transformation $x^{'\mu}=x^\mu-\xi^\mu(x)$, the variation of a vector $U^\mu(x)$ is $$\delta U^\mu(x)=U^{'\mu}(x)-U^\mu(x)=\mathcal{L}_\xi U^\mu$$...
- 372
2
votes
0
answers
86
views
Doubt on $SU(2)_{L} \times U(1)_{Y}$ covariant derivative and its action on a fermion
I) Introduction
I.1)
The mathematical structure is quite clear: given a spacetime $M$, and a Lie group $G$ (the gauge group), we can construct the Principal bundle $P^{G}_{M}$. The connection $1$-form ...
- 3,159
2
votes
2
answers
1k
views
Action of Lie derivative on 1-forms
In Sean Carroll's spacetime and geometry appendix B he derives the action of the Lie derivative on 1-forms. He finds that $\mathcal{L}_X Y^\mu = [X, Y]^\mu$, which I believe is meant as $\mathcal{L}_X(...
- 2,068
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=(...
- 49
1
vote
1
answer
270
views
Metric independent affine connections
Normally, the affine connections are objects that define parallel transport. In general Relativity they are the Christoffel symbols of the second kind. Consequently, they depend on the metric tensor. ...
1
vote
1
answer
197
views
What is the Lie derivative of the field describing the change of mass?
I'm trying to understand Ch. 3.2 of the paper On Bubble Rings and Ink Chandeliers by Padilla et al.. I'm trying to understand the derivation of equation (15). Right now I'm stuck at the point where ...
1
vote
3
answers
437
views
Question about Wald's example of a "derivative operator"
I am trying to study Wald's book on General Relativity. I thought the first two chapters were ok, but I'm completely stuck in chapter 3, on page 32 where he gives an example of a "derivative ...
- 213
1
vote
1
answer
1k
views
Covariant derivative ordering
I was working on a problem involving Bianchi identities, in a particular case I have to take the covariant derivative of the following, which indeed is the Ricci tensor in linearised limit
$$r^{\mu}_{\...
- 278
0
votes
1
answer
385
views
Dirac's book *General Theory of Relativity*: Doesn't this show the partial derivative of the metric tensor is zero?
See the bold text for my question.
This question regards Dirac's General Theory of Relativity page 13. In his demonstration that the length of a vector is unchanged by parallel displacement Dirac ...
- 2,052