# Tag Info

26

The simplest way to explain the Christoffel symbol is to look at them in flat space. Normally, the laplacian of a scalar in three flat dimensions is: $$\nabla^{a}\nabla_{a}\phi = \frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}$$ But, that isn't the case if I switch from the $(x,y,z)$ ...

24

This is not really an answer to your question, essentially because there isn't (currently) a question in your post, but it is too long for a comment. Your statement that A co-ordinate transformation is linear map from a vector to itself with a change of basis. is muddled and ultimately incorrect. Take some vector space $V$ and two bases $\beta$ and ...

23

You can decompose a rank two tensor $X_{ab}$ into three parts: $$X_{ab} = X_{[ab]} + (1/n)\delta_{ab}\delta^{cd}X_{cd} + (X_{(ab)}-1/n \delta_{ab}\delta^{cd}X_{cd})$$ The first term is the antisymmetric part (the square brackets denote antisymmetrization). The second term is the trace, and the last term is the trace free symmetric part (the round brackets ...

19

A second-order tensor can be represented by a matrix, just as a first-order tensor can be represented by an array. But there is more to the tensor than just its arrangement of components; we also need to include how the array transforms upon a change of basis. So tensor is an n-dimensional array satisfying a particular transformation law. So, yes, a ...

19

The connection is chosen so that the covariant derivative of the metric is zero. The vanishing covariant metric derivative is not a consequence of using "any" connection, it's a condition that allows us to choose a specific connection $\Gamma^{\sigma}_{\mu \beta}$. You could in principle have connections for which $\nabla_{\mu}g_{\alpha \beta}$ did not ...

14

The dual of a tensor you refer to is the Hodge dual, and has nothing to do with the dual of a vector. The word "dual" is used in too many different contexts, and in this case it is even used the same $*$ symbol. One usually specifies "Hodge dual", or "Hodge star operator", to avoid confusion. Both these "duals" are isomorphisms between vector spaces endowed ...

13

Matrices are often first introduced to students to represent linear transformations taking vectors from $\mathbb{R}^n$ and mapping them to vectors in $\mathbb{R}^m$. A given linear transformation may be represented by infinitely many different matrices depending on the basis vectors chosen for $\mathbb{R}^n$ and $\mathbb{R}^m$, and a well-defined ...

9

A (higher) $n$-rank tensor $T^{\mu_1\ldots \mu_n}$ with $n\geq 3$ cannot always be decomposed into just a totally symmetric and a totally antisymmetric piece. In general, there will also be components of mixed symmetry. The symmetric group $S_n$ acts on the indices $$(\mu_1,\ldots ,\mu_n)\quad \longrightarrow\quad (\mu_{\pi(1)},\ldots ,\mu_{\pi(n)})$$ via ...

9

In a class I'm lecturing, I mention to my students (in a very, very elementary way) that vectors and covectors do not live in the same space. It's a typical school phrase... "Do not add apples and pears", and it's true! If you keep in mind the custom column and row representation of a vector, you can prove that both of them (by themselves) satisfy the ...

9

I) Let us for simplicity discuss tensors in the context of (finite-dimensional) vector spaces and multilinear algebra. [There is a straightforward generalization to manifolds and differential geometry.] II) Abstractly in coordinate-free notation, the Kronecker delta tensor, or tensor contraction, is the natural pairing $$\tag{1} V \otimes ... 8 The covariant derivative is metric compatible, so \nabla_{\alpha} g_{\beta \gamma} = 0. This is the condition that the inner product is preserved under parallel transport. 8 Each of the indices in a tensor have a particular left-right ordering. This ordering cannot be changed unless the tensor has some particular symmetry that permits it (or rather, that equates different components on interchange). The up-down positions of indices tells us about whether the index is associated with using a basis vector (up) or a basis ... 8 A physicist would write your first equation x^a = x^\mu e_\mu^a. The notation x^a is invariant in your terminology. The a is an abstract index. It is ostensibly not supposed to be thought of as ranging over a set of numerical values, but is just a marker that indicates that x is a vector (i.e., rank 1,0 tensor.) Similarly for each \mu, e^a_\mu is ... 7 The notion of covariant derivative is equivalent to the notion of connection. More precisely, for every connection \nabla and vector field X, the operation \nabla_X is a covariant derivative. Connections \nabla on the tangent bundle TM of a manifold are usually induced by a metric, this is the so called Levi-Citiva connection. It is essentially ... 6 The earliest instance I have found is Minkowski's "Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern" in "Nachrichten von der Georg-Augusts-Universität und der Königl. Gesellschaft der Wissenschaften zu Göttingen" from 1908. A digitized version is found at ... 6 Physicists are always interested in what properties of a physical system are invariant under symmetries. If it's tricky to see the symmetry then they'll rearrange the system to make the symmetry more obvious. For example, consider a covariant rank two tensor like T^{ab}. In general the components of this tensor will change if the tensor is rotated in 3D. ... 6 I'll write this as an answer so that the math is more clear. So given an (p,q)-tensor T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}, this one transforms as:$$T'^{\mu'_1\cdots\mu'_p}{}_{\nu'_1\cdots\nu'_q}=\frac{\partial x^{\mu'_1}}{\partial x^{\mu_1}}\cdots \frac{\partial x^{\mu'_p}}{\partial x^{\mu_p}}\frac{\partial x^{\nu_1}}{\partial ...

6

See Edit below, the original answer is not completely correct. There is no gauge freedom in $F$. $F$ is gauge invariant. In fact, $F$ is completely measurable. It's components are the Electric and Magnetic fields, so you just go out with a set of test charges and measure $E$ and $B$ and you've got $F$. One hint that $T$ and $F$ do not contain the same ...

5

In the Einstein convention, pairs of equal indices to be summed over may appear at the same tensor. For example, the formula ${A_k}^k=tr~A$ is perfectly legitimate. But your formula looks strange, as one usually sums over a lower index and an upper index, whereas you sum over lower indices only, which doesn't make sense in differential geometry unless your ...

5

1) The confusion comes from an omission of parentheses in these notations. In the first case we do indeed have $$\nabla_{\vec{e}_i}(u^j) = \frac{\partial u^j}{\partial x^i},$$ since $u^j$ is just one non-specific component of $\vec{u}$. In the second case, they mean to take the component after differentiating the tensor: $$u^j{}_{;i} = (\nabla_{\vec{e}_i} ... 5 A covariant derivative on some manifold is a map which takes each differentiable tensor field of type (n,m) to a tensor field of type (n,m+1) with the following properties: Linearity Leibnitz rule Commutativity with contraction Consistency with notion of tangent vectors Additionally, one may wish to impose the condition of vanishing torsion. Using ... 5 Note that there is no physical meaning of Christoffel symbols as they are not tensors. It's always possible to choose local coordinates such that all of \Gamma vanish. But their mathematical meaning is that they form a pseudotensor. Technically, if we have two covariant derivatives \nabla_1 and \nabla_2 then their difference \Gamma := \nabla_1 - ... 5 An easy way to see that they are distinct is to consider what happens upon raising (or lowering) all indices. For example, upon lowering,$$ T_{ab}{}^{cde} $$becomes T_{abcde}, whereas$$ T_{a}{}^{cd}{}_{b}{}^{e} $$becomes T_{acdbe}, and similarly$$ T_{a}{}^{cde}{}_{b} $$becomes$$ T_{acdeb}. $$You need to "slant" the indices so as to keep track ... 5 No. Definitions: A rank-n tensor is a linear map from n vectors to a scalar. A symmetric tensor is one which in which the order of the arguments doesn't matter. An antisymmetric tensor in which transposing two arguments multiplies the result by -1. Suppose we have some rank-3 tensor T with symmetric part S and anti-symmetric part A so$$T(a,b,c) ...

5

First off, please don't use units with $c\ne 1$ in GR. It makes everything horribly messy. What we normally think of as a ruler or clock measurement is represented in GR by an upper index quantity like $\Delta x^\mu$. Therefore in a Cartesian coordinate system in the fluid's rest frame, we are guaranteed that $u^\mu=(1,0,0,0)$, not $(-1,0,0,0)$. This is ...

5

The (anti)symmetrization simply acts on all the enclosed indices (at the same "height" which are really enclosed between the brackets), regardless of their belonging to the same tensor or different tensors. For example, $$\delta^{[\alpha}{}_{[\gamma} R^{\beta]}{}_{\delta]} = \frac 12 \left(\delta^{[\alpha}{}_{\gamma} R^{\beta]}{}_{\delta} - ... 5 A variation of a tensor is always a tensor and the formula for the value above doesn't show otherwise. What you probably find surprising is that \delta g_{\mu\nu} and \delta g^{\rho\sigma} are not related to each other by simply raising the indices \mu,\nu or lowering the indices \rho,\sigma. Indeed, they're not related in this way. In this case, ... 5 What you said is only true if the hypersurface is space-like or time-like. If a non-null hypersurface is defined by f(x) =  constant, then the normal to the hypersurface is given by$$ n_\alpha \propto \partial_\alpha f $$The fact that the hypersurface is non-null implies$$ g^{\alpha\beta} \partial_\alpha f \partial_\beta f = \varepsilon\neq 0  ...

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