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29

A (rank 2 contravariant) tensor is a vector of vectors. If you have a vector, it's 3 numbers which point in a certain direction. What that means is that they rotate into each other when you do a rotation of coordinates. So that the 3 vector components $V^i$ transform into $$V'^i = A^i_j V^j$$ under a linear transformation of coordinates. A tensor is a ...


28

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 $(x,y,z)$ ...


28

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 ...


25

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 ...


24

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 ...


22

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 ...


17

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 ...


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 ...


11

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 ...


10

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 ...


9

This is true - in fact you could define $\nabla^\sigma = g^{\sigma\rho} \nabla_\rho$. I assume this meant to say $$ g^{\sigma\rho} \nabla_\nu \nabla_\sigma = \nabla_\nu \nabla^\rho. $$ Again, this is true, but for a slightly less trivial reason than (1). To employ (1) to prove this, you need to be able to switch $g^{\sigma\rho}$ with $\nabla_\nu$, which you ...


9

A matrix is a special case of a second rank tensor with 1 index up and 1 index down. It takes vectors to vectors, (by contracting the upper index of the vector with the lower index of the tensor), covectors to covectors (by contracting the lower index of the covector with the upper index of the tensor), and in general, it can take an m upper/n-lower tensor ...


8

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. ...


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

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

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 ...


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

Let there be given a manifold $(M,\nabla)$ equipped with a (not necessarily torsionfree) tangent bundle connection $\nabla$. I got the (possibly faulty) impression from reading the first lines in OP's question formulation (v18) that OP is asking: Is it possible that the local coordinate expression for the covariant derivative of a co-vector/one-form ...


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

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

For doing matrix multiplication, yes, we usually put the dummy indicies together, but this is (assuming one is not working over $\mathbb{H}$) convention really. Don't forget that this is shorthand Einstein convention. When you write $$ \mathbf{A} \cdot \mathbf{B} = A_{ij} B_{jk} = \sum_{j=1}^n A_{ij} B_{jk} $$ you are writing a shorthand version in ...


5

There are two more points that can be made here. Sorry if I repeat someone. In a way you are right that if you have a vector space and its dual there is no intrinsic way to say which space is the original and which is the dual. This is because there is a canonical isomorphism between a vector space and the dual of its dual. In other words if $V$ is a vector ...


5

Gradient is covariant! Why? The components of a vector contravariant because they transform in the inverse (i.e. contra) way of the vector basis. It is customary to denote these components with an upper index. So, if your coordinates are called $q$'s, they are denoted $q^i$. Therefore, the gradient (or a derivative if you prefer) is $$\partial_i = ...


5

Equation (13) expresses the metric on an embedded hypersurface given by the relations $y^k = y^k(x^a)$. However, the equation for the inverse metric (4-th equation) is in general not correct. Take for example a hypersurface defined by: $y^1 = x^1$, $y^2 = x^2$, $ y^3 = x^2$. In our case, the partial derivative of $x^2$ with respect to $y^2$ or $y^3$ ...


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 $$ g^{\alpha\beta} \partial_\alpha f \partial_\beta f = \varepsilon\neq 0 $$ ...


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

Short answer. Example : The Palatini action , where the action is a functional of a tetrad/vierbien $e$ and a spin connection $\omega$. $S(e, \omega) = \int \epsilon_{abcd} e^a \wedge e^b \wedge \Omega^{cd}$ where $\Omega$ is the curvature associated to the connection form $\omega$ : $\Omega = D\omega = d\omega + \omega \wedge \omega$ Other examples ...


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

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 ...



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