7

I think your confusion comes from the fact that there are several different ways of looking at covariance and contravariance. The old-school way to treat this issue is to say that given a basis $\mathbf e_i$ for a vector space $V$, we can define a dual basis for $V$ which we write as $\mathbf e^i = g^{ij}\mathbf e_j$. In this framework, both $\mathbf e_i$ ...


6

The natural approach consists of starting by assuming that the physical space is an affine $3$-dimensional space $(\mathbb{A}^3,-, V^3)$ ($- : \mathbb A^3 \times \mathbb A^3 \to V^3$ is the map associating couples of points $P,Q$ with vectors $P-Q =: \vec{QP}$) whose vector space of translations $V^3$ is further equipped with a positive symmetric inner ...


4

As noted by @mmeent in the comments, the metric must be regular at the origin. This means, in particular, that $$ \mathcal{Q}(r) \approx r $$ in the limit $r \to 0$. Equivalently, we must have $\mathcal{Q}(0) = 0$ and $\mathcal{Q}'(0) = 1$. In addition, the metric components and their first derivatives must be continuous at the boundary $r = R$. We thus ...


4

First, to me it seems that the ansatz $(1)$ for this metric of thick cosmic string is wrong, since for a general $\mathcal{P}(r)$ the metric lacks the $SO(1,1)$ invariance under boosts along the string direction, i.e. Lorentz transformations in $(t,z)$ plane. My suggestion: $$ \tag{1*} ds^2 = \mathcal{P}^2(r) \,( dt^2 - dz^2) - dr^2 - \mathcal{Q}^2(r)\,d \...


3

I only managed to sort these things in my head after Halmos and Greub Lets talk about abstract vectors first. Lets define the contra-variant vectors as the 'usual' vectors: $$ \mathbf{V}=V^i\,\mathbf{e}_i\in \mathcal{V} $$ Where $\mathbf{V}$ is a vector, $V^i$ are the components, and $\{\mathbf{e}_i\}_{i=1\dots N}$ is the basis for this N-dimensional vector ...


3

A geodesic is a curve $\gamma$ which extremizes the path length functional $$S[\gamma]= \int d\lambda \ L(\gamma,\dot \gamma) =\int d\lambda \sqrt{g_{\mu\nu}(\gamma)\dot\gamma^\mu\dot\gamma^\nu}$$ The Euler-Lagrange equations are $$\frac{\partial L}{\partial \gamma^\alpha} = \frac{d}{d\lambda}\frac{\partial L}{\partial \dot\gamma^\alpha}$$ $$\implies (\...


3

Dot product of $u$ with itself is constant: $$u\cdot u=-c^2$$ Taking derivative with respect to proper time on both sides gives $$u\cdot \frac{du}{d\tau}=u \cdot a=0$$


2

$$\partial_m [e^{2 \alpha \phi} (H^{mnp}+H^{pmn}+H^{npm})] = 0$$ Shouldn't be? $$\nabla_m [e^{2 \alpha \phi} (H^{mnp}+H^{pmn}+H^{npm})] = 0$$


2

Spacetime doesn't come equipped with coordinates, so there is no reason to think that we automatically have anything called $\phi$ that works like an azimuthal angle, nor is there any guarantee that we can construct such a coordinate. This makes the problem different from the problem of defining something like the axial symmetry of a mass distribution in ...


2

The sign is the paper is correct. I do not know where is the problem so i'm gonna give you some hints. Using the fact that: $$δΓ^{α}_{νμ} = \cfrac{1}{2}g^{ασ}\big(\nabla_{ν}δg_{σμ} + \nabla_{μ}δg_{νσ} - \nabla_{σ}δg_{νμ}\big)$$ $$δΓ^{α}_{αμ} = \cfrac{1}{2}g^{ασ}\big(\nabla_{α}δg_{σμ} + \nabla_{μ}δg_{ασ} - \nabla_{σ}δg_{αμ}\big) =\cfrac{1}{2}g^{ασ}\nabla_{μ}...


2

You are mistaken. There is no fixed background geometry in GR. Calculations of curvature are made using only internal measurements (which is obvious, since we cannot go out of the universe and measure things from there). Gauss and Riemann showed that it is possible to measure curvature and other geometric features of a manifold only by doing internal ...


2

Spacetime curvature is not calculated with respect to a reference spacetime (Minkowski or other). Curvature only depends on the properties of your given spacetime (metric and connection). I don't know what you mean by "underlying spacetime". Do you mean that we calculate the extrinsic curvature of a spacetime embedded in some higher-dimensional ...


2

The "canonical" meaning of a vector in general, in a geometric sense, is a displacement: if you have a point $P$ in a suitably homogeneous space, such as Euclidean three-dimensional space, then $$P + \mathbf{v}$$ is a point displaced in the direction and through the distance encoded within $\mathbf{v}$. Likewise, if $Q$ and $P$ are two points, then ...


2

I do not have the book, but probably the global comoving frame simply demands, that the spatial grid is comoving at each point, while local comoving frame is probably frame where only the origin is comoving. For local comoving frames it makes sense to demand other properties, like locally reducing the metric to its standard Minkowski form, i.e. that locally ...


2

It is inherent in the postulate that spacetime is a pseudo-Riemannian manifold with signature (-+++). That postulate brings in the whole machinery of pseudo-Riemannian geometry, including the metric and the invariance of any scalar under any arbitrary coordinate transform.


2

$A^{\mu}$ should be a conformal Killing vector field. Various equivalent conditions are listed on Wikipedia.


2

I'm not sure what you mean by "Taylor's trick", but this is just linear algebra: if $A$ is a matrix with rank $1$, then $$\det({\rm Id}+A)=1+{\rm tr}(A).$$Write $\widetilde{G} = (\widetilde{g}_{\mu\nu})$ and $G=(g_{\mu\nu})$. Then $$\det(G)=\det(\widetilde{G}+\kappa\phi\ell\ell^\top) = \det(\widetilde{G})\det({\rm Id}+\kappa\phi \widetilde{G}^{-1}\...


2

I will offer an alternative way of looking at this. You care about transformations, right? More often than not these transformations form groups. For example rotation group is $\text{SO}(3)$, parity and time-reversal group is $Z_2$. Groups have representations over vector spaces, so different vectors simply are from different representation spaces for these ...


2

Well, one possibility is that this is just a matter of where attention is focused. Since, excluding the weak force, all physical laws are invariant under parity, it's interesting to note that pseudovectors transform differently than vectors under that operation. However, real physics isn't, as far as we know, invariant under spatial dilations, and so for ...


2

One way to see that the curvature tensor is zero is to start with the fact that there is no dependence of metric components on null coordinate $v$. So performing Kaluza–Klein reduction along the Killing vector $\partial_v$ we would obtain a Newton–Cartan spacetime with two spatial coordinates ($x$ and $y$), while $u$ now becomes Galilean time coordinate. ...


2

After some tips from @Andrew, I could solve the exercise: We start by writing Einstein's equation in vaccum ($T_{\mu \nu} = 0$): $$G_{\mu \nu} + \Lambda g_{\mu\nu} = R_{\mu \nu} - \frac{R}{2}g_{\mu \nu} + \Lambda g_{\mu\nu} = 0 \\ \Leftrightarrow R_{\mu \nu} - \frac{g^{\alpha \beta}R_{\alpha \beta}}{2}g_{\mu \nu} + \Lambda g_{\mu\nu} = 0 $$ Then, we ...


2

If I'm reading things correctly, it looks to me as though Carroll provides the components of the Ricci tensor in the $dx^\mu$ coordinate basis, while your instructor is providing the components of the Ricci tensor in the orthonormal $\omega^\mu$ basis: $$\mathbf R = R_{(x) \mu\nu} \big(dx^\mu\otimes dx^\nu\big) = R_{(\omega)\mu\nu} \big(\omega^\mu \otimes \...


2

This is not a completely well-posed question because there are some ambiguities in how you generalize a flat-space equation to curved space. The so-called "minimal coupling" prescription is to replace partial derivatives with covariant derivatives, and instances of the Minkowski metric $\eta_{\mu\nu}$ with a general spacetime metric $g_{\mu\nu}$. ...


1

Note that $\Lambda$ is not itself a tensor. It is a matrix of transformation coefficients. A tensor is well-defined in the absence of any specified frame, and its components can be defined in one frame or another. The Lorentz transformation, by contrast, is used to calculate how tensor components change from one frame to another. Contracting one up- and one ...


1

First a brief introduction to tensors. An $(r,s)$ tensor $t$ on a $K$-vector space $V$ is just a multilinear map from $s$ copies of $V$ and $r$ copies of the dual vector space $V^*$ to the underlying field $K$: $$t:\underbrace{V^*\times...\times V^*}_{r-times}\times\underbrace{V\times...\times V}_{s-times}\to K$$ The dual vector space is just the set of all ...


1

This took me quite a while to get at the beginning of relativism class but this may help: a general matrix multiplication like we all know it looks like this: $$[A][B] = \sum a_{ri}b_{ir}$$ Now the relativity version with einstein summation convenction would make this matrix multiplication look like this: $$[A][B] = a_{\nu\sigma}b^{\sigma\nu}$$ As you see ...


1

That a vector $V$ is timelike, lightlike/null, or spacelike indicates solely that $V^{\mu}V_{\mu}=V_{0}^{2}-V_{1}^{2}-V_{2}^{2}-V_{3}^{2}=V_{0}^{2}-{\bf V}^{2}$ is positive, zero, or negative. What that means in different specific case varies somewhat. For the four-momentum $p^{\mu}$: Since the momentum components are the canonical conjugates to the ...


1

$$ \mathbf{v} = q_i \mathbf{e^i} = q^i \mathbf{e_i} $$ Ouch! The basis vectors $\mathbf{e_i}$ and $\mathbf{e^i}$ belong to different vector spaces, namely a vector space and its dual. The use of an equal sign in this context is simply wrong! A vector space and its dual are mathematically different objects, but equals means they are mathematically the same ...


1

The article Axially symmetric spacetimes: numerical and analytical perspectives is a very nice reference and treats axially symmetric spacetimes in generality, showing how to construct the "cylindrical" slices (see also this). If your spacetime is not only "axially symmetric" but also "circular" (i.e. there are no meridional ...


1

"Invariance" can be an ambiguous term. If we speak of coordinate invariance, then today this invariance is understood to be somewhat trivial, and not a physical requirement at all. The first relativists, including Einstein, thought that imposing coordinate invariance was a physical requirement. With time some doubts appeared about this; I think one ...


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