# Tag Info

20

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 ...

20

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

15

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

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 ...

12

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 ...

10

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 ...

8

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 ...

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 ...

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

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. ...

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: $$u^j{}_{;i} = (\nabla_{\vec{e}_i} ... 5 In the Einstein convention, pairs of equal indices to be summed over may appear at the same tensor. For example, the formula {A_k}^k=tr~A is perfectly legitimate. But your formula looks strange, as one usually sums over a lower index and an upper index, whereas you sum over lower indices only, which doesn't make sense in differential geometry unless your ... 5 A covariant derivative on some manifold is a map which takes each differentiable tensor field of type (n,m) to a tensor field of type (n,m+1) with the following properties: Linearity Leibnitz rule Commutativity with contraction Consistency with notion of tangent vectors Additionally, one may wish to impose the condition of vanishing torsion. Using ... 5 No. Definitions: A rank-n tensor is a linear map from n vectors to a scalar. A symmetric tensor is one which in which the order of the arguments doesn't matter. An antisymmetric tensor in which transposing two arguments multiplies the result by -1. Suppose we have some rank-3 tensor T with symmetric part S and anti-symmetric part A so$$T(a,b,c) ...

5

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 ...

4

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 ...

4

Suppose $(M,g_{\mu\nu})$ is some spacetime and $p$ is a point in $M$. We can always choose coordinates so that at p, $g_{\mu\nu} = \eta_{\mu\nu}$ and $\partial_\lambda g_{\mu\nu} = 0$, where $\eta_{\mu\nu}$ is the ordinary flat metric. (Note that this only holds at $p$, not necessarily in a neighborhood). These are called exponential coordinates. In this ...

4

You are looking for the formalism described in the references listed here. The original article that got this line of research started is Samson Abramsky and Bob Coecke, A categorical semantics of quantum protocols , Proceedings of the 19th IEEE conference on Logic in Computer Science (LiCS’04). IEEE Computer Science Press (2004) (arXiv:quant-ph/0402130) ...

4

Use the anti-symmetry of $\epsilon$ to switch the indices $\epsilon$ and $\zeta$. Then relabel your dummy indices. EDIT: Allow me to expand on this answer on Bebop's behalf. You have two steps. In the first step, you use the fact that $\epsilon$ is anti-symmetric. That means you'll get a minus-sign if you exchange $\zeta$ and $\epsilon$. In the second ...

4

First of all, by interchanging $\Lambda_{\epsilon}^{\beta}$ with $\Lambda_{\zeta}^{\alpha}$ nothing changes: $$\Lambda_{\epsilon}^{\beta}\Lambda_{\zeta}^{\alpha}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa ... 4 First off, please don't use units with c\ne 1 in GR. It makes everything horribly messy. What we normally think of as a ruler or clock measurement is represented in GR by an upper index quantity like \Delta x^\mu. Therefore in a Cartesian coordinate system in the fluid's rest frame, we are guaranteed that u^\mu=(1,0,0,0), not (-1,0,0,0). This is ... 4 A variation of a tensor is always a tensor and the formula for the value above doesn't show otherwise. What you probably find surprising is that \delta g_{\mu\nu} and \delta g^{\rho\sigma} are not related to each other by simply raising the indices \mu,\nu or lowering the indices \rho,\sigma. Indeed, they're not related in this way. In this case, ... 4 Written explicitly, (assuming summation over indices from 0 to 3)$$a^{ij}b_{ij} = \sum_{i=0}^3 \sum_{j=0}^3 a^{ij}b_{ij}$$You can expand this to$$a^{ij}b_{ij} = \sum_{i=0}^3 \left( a^{i0}b_{i0} + a^{i1}b_{i1} + a^{i2}b_{i2} + a^{i3}b_{i3} \right) \implies a^{ij}b_{ij} = a^{00}b_{00} + a^{01}b_{01} + a^{02}b_{02} + a^{03}b_{03} + a^{10}b_{10} + ...

4

Start with the following form of the Bianchi Identities $$\nabla^\mu R_{\mu\nu} = \frac{1}{2} \nabla_\nu R$$ Contract both sides with $K^\nu$. We find $$\frac{1}{2} K^\nu \nabla_\nu R = K^\nu \nabla^\mu R_{\mu\nu} = \nabla^\mu \left( K^\nu R_{\mu\nu} \right) - R_{\mu\nu} \nabla^\mu K^\nu$$ The second term vanishes due to symmetry of $R_{\mu\nu}$. Now, ...

4

The books are correct. The statement is a definite relation that is being imposed between the 'old' metric structure and the transformed one, for the transformation to be conformal. The equation you're unhappy about, $$g_{\mu \nu}'(x') = \Omega(x) g_{\mu \nu}(x)$$ states that the transformed metric $g_{\mu \nu}'$ at the transformed point $x'$ can be ...

3

The gauge invariance problem of the Maxwell field in the presence of torsion has been known for many years. There is no known perfect solution for this problem. The importance of having the gauge invariance of the Maxwell action is that it implies charge conservation which is very well established experimentally. One type of suggestions (please see ...

3

In relativity there is an invariant called the proper time, $\tau$. It's an invariant in the sense that all observers will agree on it's value. In special relativity the proper time is defined as: $$d\tau^2 = ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2$$ or $$d\tau^2 = -ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$ You see both sign conventions and I've never been sure ...

Only top voted, non community-wiki answers of a minimum length are eligible