# Tag Info

3

$\Lambda_{\mu\nu} = {\Lambda_\mu}^\sigma\eta_{\sigma\nu}$. It doesn't "do" anything. $\delta_{\mu\nu}$ and $\delta^{\mu\nu}$ are not tensors, as I explain at length in this answer of mine. The matrix elements of the identity are $\delta_\mu^\nu$, which you could have determined by thinking about the fact that the identity must send vectors $v^\mu$ to other ...

1

To add to Ocelo7's answer, the transverse part is, as I have seen it used by e.g. Ellis, used to refer to components that are orthogonal to a future-pointing and geodesic null vector field spanning the past light cones of a particular world line -- the central observer (e.g. our world line). They are defined on parts of spacetime that is observeable by the ...

1

Take the covariant derivative of the equation $X^aX_a=-1$. The RHS becomes zero so we have $$2X_a\nabla_b X^a=0\implies X^aB_{ab}=0.$$ The other equation, $X^bB_{ab}=0$, is the geodesic equation, so it doesn't hold for just any $X^a$. Let's consider the situation at some point $p\in M$. Then $X^a$ is a prime candidate for the timelike basis vector of $T_pM$....

0

for getting Christofel symbols we should notice that if two vectors are parallel transported along any curve then the inner product between them remains constant under parallel transport. so you should write your sentence for 3 parameter a, lambda and nu cyclical,and then you obtain the correct Christofel.

2

In the 'strict' sense, you should only apply the summation convention to a pair of indices if one is raised and another is lowered. For example, consider a vector $v$ and a dual vector $f$ (i.e. a map from vectors to numbers). Then one can compute $f(v)$, the number that results from $f$ acting on $v$. In components, this would be written as $f_i v^i$, ...

1

There is nothing wrong by summing up indices when both indices are either up or down. It is just a matter of convention. However the meanings can be different if you are in a Relativistic theory. When you sum one up and one down indices in Relativity it means you have a Lorentz invariant quantity because you are combining covariant and contravariant ...

1

Here, a term spin should be understood as a quantum number you get after doing dimensional reduction of a higher dimensional theory to (1+3)-dimensions. If you follow this assumption, you find a spin of Kalb-Ramond field is 1. I think most people have this in mind, when they say a spin in higher dimensions. In general, you get more quantum numbers than you ...

3

I believe it should be: $$h^\prime_{\mu\nu} = h_{\mu\nu} + 2\epsilon\,\partial_{(\mu}\xi_{\nu)}$$ where the parentheses denote the symmetric part of the tensor in the $\mu$ and $\nu$ indices: $$h^\prime_{\mu\nu} = h_{\mu\nu} + \epsilon\,\left(\partial_{\mu}\xi_{\nu} + \partial_{\nu}\xi_{\mu} \right)$$ It is a generalization of the usual definition: ... 2 Notation-wise, you can use either of \begin{align} A\rho & = \sum_{ijk} p_k \Big(|i \rangle \langle j | \otimes |i \rangle \langle j |\Big) \Big(| k \rangle \langle k | \otimes |k\rangle \langle k |\Big) \\ & = \sum_{ijk} p_k |i \rangle \langle j |k \rangle \langle k | \otimes |i \rangle \langle j |k \rangle \langle k | . \end{align} As you well ... 2 It seems OP's main question is how to understand the representation of the matter fields of YM theory. The matter fields can in principle transform in any representation \rho:G\to {\rm End}(V) of the local gauge group G=SU(N), e.g. the fundamental, or adjoint representation. Here {\rm End}(V) denotes the algebra of endomorphisms on the vector space ... 1 Both of the forms you propose,\sigma^{(1)}_x \otimes \sigma^{(2)}_x \ldots \otimes \sigma^{(n)}_x \tag1$$and$$\left(\sigma_x \otimes \mathbb{I}_{n-1}\right)\cdot \left(\mathbb{I}\otimes \sigma_x\otimes \mathbb{I}_{n-2}\right) \cdot \left(\mathbb{I}_2\otimes \sigma_x\otimes \mathbb{I}_{n-3}\right)\cdot \ldots \cdot \left(\mathbb{I}_{n-1}\otimes \sigma_x\...

0

It's both - those two "long form" expressions you wrote are equivalent. Matrix multiplication distributes over tensor products, so when you multiply them together the $i$th factor in the tensor product just becomes $\sigma^{(i)}_x$ times a bunch of identity matrices.

0

They do decompose as symmetrized tensor products of Killing vectors. So nothing new here. I don't really remember the details but there are inequalities which tell how many "Killing objects" can a spacetime have, and this number is saturated for Minkowski and other maximally symmetric spaces. Class of spacetimes where you do expect to see nontrivial Killing ...

0

The usage of tensors in Physics, came about as a fortuitous "adaptation" to deal with the increased complexity of the Physics problems being investigated. The need to express more and more complex equations as succinctly as possible required a "shorthand" - and tensors came to the rescue.

0

I misunderstood what Menzel had intended by "covariant nature of the differential operator". He did not mean that the differential operation is synonymous with covariant differentiation. As Menzel is wont to do, he proceeded to expose a series of non-trivial equivalences, and then tacked on a final equivalence which does not obviously follow from the ...

1

Note that $\delta^i{}_j=\delta^i{}_k\delta^k{}_j$. Then $$\nabla_l\delta^i{}_j=\nabla_l(\delta^i{}_k\delta^k{}_j)=C(k,m)\nabla_l(\delta^i{}_k\delta^m{}_j)=C(k,m)[\delta^i{}_k\nabla_l\delta^m{}_j+\delta^m{}_j\nabla_l\delta^i{}_k]=2\nabla_l\delta^i{}_j$$ where $C(k,m)$ is the contraction of the $k$ and $m$ indices. Thus $\nabla_l\delta^i{}_j=0$.

2

The partial derivatives $\partial_\lambda \delta^\mu_\nu$ are clearly zero because the components of the Kronecker delta are constant functions of spacetime coordinates (one or zero). One may always go to a locally Minkowski frame where the Christoffel symbols vanish and there, the covariant derivative is equal to the partial one and vanishes, too. Because ...

1

You can write the Kronecker delta tensor as a product of the metric tensor $$\nabla_a(\delta^a_b) = \nabla_a (g_{bc} g^{ac}) = \nabla_a (g_{bc} g^{ac}) = g_{bc} \nabla_a g^{ac} + g^{ac}\nabla_a g_{bc}$$ As you may recall, the covariant derivative of the metric tensor is $0$ in general relativity. Version without using the metricity of the connection : ...

2

The trick is the following. Multiply both side of the equation by $\epsilon_{jkl}$. Then use the identity $\epsilon_{jkl}\epsilon_{ikl}=2\delta_{ji}$.That will give you the desired identity.

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