Hot answers tagged tensors
14
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 ...
7
The tensor equations you mention are not invariant, they are covariant. Big difference. Both are differential equations, which transform linearly under nonlinear transformations from one manifold to another because they are differential equations at a point. The nonlinear transformation from one manifold to another induces a linear transformation of the ...
7
The question seems to conflate many different things:
the invariance of a mathematical quantity (usually a scalar such as $ds^2$ for the separation of two events in special relativity)
covariance of tensors (the values of components of tensors may be calculated from those in another frame but they're not the same thing)
universality of equations in ...
7
Are those square brackets standard notation in Physics?
Yes. See, for example Sean Carroll notes. At least I can tell you from two other classic references using that notation, "General Relativity" by Wald (1984) and "A First Introducion to General Relativity" by Schutz (2009 for the most recent edition)
$ $
If I am in a non-curved $\mathbb M$ ...
7
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 ...
5
This depends on what you mean by "pass from one manifold to another". In General Relativity one generally considers a single manifold $\mathcal{M}$ and diffeomorphisms $\phi: \mathcal{M} \rightarrow \mathcal{M}$. I think the idea you are trying to get at is that if you consider a geometry on $\mathcal{M}$, that is a pair $(\mathcal{M} , g)$ where $g$ is a ...
5
The composition law for quantum systems is always a tensor product. Your problem arises from a confusion over what the tensor product is applied to: you are trying to tensor product the spatial coordinates together, when it is in fact the basis vectors of the Hilbert space you should be tensoring together.
More formally, take two quantum systems A and B, ...
5
Yes, it's just the second derivative of some function, it doesn't matter that this function is organized as a component of a tensor, $h_{\mu\nu}$. The identity above – assuming the function is differentiable and smooth etc. (add some "niceness" conditions on the function) – follows from the rules of calculus and is formally proven by the ...
4
Every term contains one $\lambda$ in the superscript and one in the subscript, so you sum over those. The only indices which don't appear in both superscript and subscript in the same term are $\mu$ and $\nu$.
Example:
$$\Gamma_{\lambda\sigma}^\lambda\Gamma_{\mu\nu}^\sigma = \Gamma_{00}^0\Gamma_{\mu\nu}^0 + \Gamma_{01}^0\Gamma_{\mu\nu}^1 + \cdots + ...
4
We start with the definition
$$\tag{1} S^{\alpha \beta}~:=~u^\alpha v^\beta-u^\beta v^\alpha.$$
Indices are raised and lowered with the metric.
Up to an overall factor, one has
$$\tag{2} \bar{S}_{\alpha \beta}~\propto~ \epsilon_{\alpha \beta \gamma \delta} S^{\gamma \delta},$$
so that the matrix trace
$$ \mathrm {Tr} (\mathbf{\bar{S}\cdot S })
~=~ ...
4
If p is a point of the manifold then F at p is equal to ϕ∗F at ϕ(p), since they are related by the tensor transformation law, and tensors are independent of coordinate choice.
This is roughly true. Initially, there is no meaning when one says that tensors at different tangent spaces are equal. However, the diffeomorphism induces an isomorphism between ...
4
The antisymmetric second-rank tensor being referenced is the electromagnetic field tensor. It is defined as follows. Let $\varphi$ be the electrostatic potential (a scalar field), and let $\underline{A}$ be the magnetic potential (a 3-vector) from classical E&M. Concatenate them into a 4-vector $\vec{A}$. Now define the tensor of interest as the exterior ...
4
Note; I'll use the summation convention throughout here.
In the context of differential geometry, the indices on tensorial objects are raised and lowered with the metric on the space (manifold) being studied. So for example
$$
T^i_{\phantom i j} = g^{ik}T_{kj}
$$
and
$$
T^{ij} = g^{ik}g^{jl}T_{kl}
$$
Notice that if the metric is simply that of ...
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, ...
3
Why do Maxwell's equations contain each of a scalar, vector, pseudovector and pseudoscalar equation?
In 3-space, one can interpret the 4 Maxwell equation as determining the relationship between the fields (the electric field vector and the magnetic field bivector) and all four types of possible sources.
But this is rather illusory. In relativity, the equations look quite different:
$$\begin{align*} \nabla \cdot F &= -\mu_0 J \\ \nabla \wedge F &= ...
3
From this transcript, the full quote is
And so, if you go to Berkley, where I got my PhD, you can buy a t-shirt which says, “In the beginning God said, the four-dimensional divergence of an antisymmetric, second rank tensor equals zero, and there was light, and it was good. And on the seventh day he rested.” Ladies and gentlemen, this is the equation ...
3
$\partial_t\equiv\frac\partial{\partial t}$ and $\partial^\mu\equiv g^{\mu\nu}\frac\partial{\partial x^\nu}=\left(\sum_{\nu=0}^3g^{\mu\nu}\frac\partial{\partial x^\nu}\right)_{\mu=0}^3$ are differential operators. $\partial^\mu$ is formally contravariant (upper index) and obeys the corresponding transformation laws. $\partial_t$ has a lower index and is (up ...
3
The doublets – I assume that you mean 2-dimensional representations of $SU(2)\equiv Spin(3)$ – are spin-1/2 representation. Tensor products mean the addition of the angular momentum. The tensor product is 4-dimensional and under $SU(2)$, it decomposes to a $j=0$ multiplet, a scalar or singlet, and a $j=1$ multiplet, a vector:
$${\mathbf 2}\otimes {\mathbf 2} ...
3
First, if you're going to keep proper track of covariant and contravariant components, you should lower the index on $B$ and make sure the dummy indices are always of opposite types: $B_k = \varepsilon_{kij} \phi^{ij}$. The reason we can be sloppy in Euclidean space is because of how trivial the metric can be. We can always consider our equations in the ...
3
First, some corrections.
The left side is a tensor, but each component of the right side is not by itself a tensor
You meant "each term on the right hand side", right? Components of tensors (or nearly tensors) of course never transform as tensors themselves. A component is one number that you may obtain by choosing a value of the index $\mu$ etc., for ...
3
Let's define $I=I^k_{~k}$ to make things look nicer. We have:
$$J_{ij}J^{ij}=(I_{ij}-\frac{1}{3}\delta_{ij}I)(I^{ij}-\frac{1}{3}\delta^{ij}I)$$
$$=I_{ij}I^{ij}-\frac{1}{3}I^{ij}\delta_{ij}I-\frac{1}{3}I_{ij}\delta^{ij}I+\frac{1}{9}\delta_{ij}\delta ^{ij}I^2$$
The second two terms are equal (they're just scalars), so:
...
3
Start with (iii) $ T^\mu{}_\mu = g_{\mu\nu}T^{\mu\nu}$ I don't think this can be correct because both indices appear twice.
What's wrong with $ g_{\mu\nu}T^{\mu\nu}$? Both indices are contracted. Explicitly it means
$$ \sum_{\mu=0}^3\sum_{\nu=0}^3 g_{\mu\nu}T^{\mu\nu}$$
which is a perfectly good scalar.
$g^\mu{}_\mu =2$ here I summed all the ...
3
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 ...
3
A perfect fluid is defined by the property that, in the local rest frame, it allows no energy fluxes and no anisotropic stresses. Thus, at a given space-time point, in the local rest frame [in which the components of the 4-velocity are $u^{\alpha} = (1, 0, 0, 0)^{\mathsf{T}}$], the energy momentum tensor components are $T^{\alpha\beta} = \mathrm{diag}(e, p, ...
3
The rules are simple but long, as stated in Wikipedia. I'll elaborate on them here.
We start with a rank $(r,s)$ tensor $T$ in $d$ dimensions.
We seek the $d^{r+s+1}$ components of $\nabla T$, a rank $(r,s+1)$ tensor.
We can think of these components as collected into $(r,s)$ sets of $d^{r+s}$, one for each of the $d$ values of the free index $\gamma$ in ...
2
A tensor is a generalization of the notion of scalars and vectors. A tensor of rank 0 is a scalar (it has $3^0$ compenent), while a tensor of rank 1 is a vector (which has $3^1$ components). In general, a tensor of rank $n$ has $3^n$ components.
See http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf for a nice introduction.
2
The method is to contract every index with the $\sigma$ four vector:
$$ \sigma^\mu_{\alpha\dot{\beta}}$$
Where $\sigma^0$ is the identity, and $\sigma^i$ for i=1,2,3 is the Pauli spin matrix.
If you have a symmetric tensor with all lower indices, you contract each $\mu$ index with a sigma, and you get the dotted-undotted form.
$$ M_{\mu\nu} ...
2
You should be contracting the following two objects
$$
F_{\mu \nu}=
\begin{pmatrix}
0 & E_x & E_y & E_z \\
-E_x & 0 & -B_z & B_y \\
-E_y & B_z & 0 & -B_x \\
-E_z & -B_y & B_x & 0
\end{pmatrix}
\quad \text{and}\quad
F^{\mu \nu} =
\begin{pmatrix}
0 & -E_x & -E_y & -E_z \\
E_x & 0 & -B_z ...
2
A two-body state $|\Psi_{12}\rangle$ can be written as a tensor product $|\Psi_{12}\rangle = |\Psi_1\rangle \otimes |\Psi_2\rangle$ only when there are not correlations between both bodies: e.g., two separated non-interacting bodies. The same happens in classical mechanics, where the two-body state factorizes as $\rho_{12} = \rho_1 · \rho_2$ only in absence ...
2
You have the right idea--by the time you're at your last step, everything is just numbers. At this point, just einstein sum your indices.
I will say that most working people, unless they are deriving general results, will start from the tensor in an explicit coordinate basis, and infer the mathematical object. This way is more formally "correct", but a ...
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