Tag Info

New answers tagged

0

I guess you have some typos in your question (or your lecture notes), since the two lines you give just differ by the index names (as already commented) . Starting from $$ R^n_{ikl;m} +R^n_{imk;l} +R^n_{ilm;k} =0 $$ you can rewrite that with the symmetry relations $R^n_{ikl}=- R^n_{ilk}=-R^i_{nkl}$ to $$ R^n_{ikl;m} - R^n_{ikm;l} +R^n_{ilm;k}=0. $$ Now let ...


2

To contract a tensor is to set two of the indices equal and sum over them, so given a tensor $A^i_j$ the contraction is $A=A^i_i=A^1_1+A^2_2+A^3_3+A^4_4$ The Bianchi identities you list have five indices. To contract them, you would set some pair equal and sum over them. Your second version is the same as the first, it just has the indices renamed. ...


6

At the most basic level, you can just use the definition of the Christoffel symbols in terms of the metric: $\Gamma^i_{jk} = g^{is} (\partial_j g_{sk} + \partial_k g_{sj} - \partial_s g_{jk})$. Plugging this into the right-hand side of your expression will yield the left-hand side. However, one can obtain your expression directly from one of the ...


0

If you are working with the Newman Penrose null tetrad, i.e. with respect to which the metric looks like $g=\left(\begin{array}{cccc}0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 1 & 0\end{array}\right)$ then the standard notation is $\lbrace l,n,m,\overline{m}\rbrace$, where $l,n$ distinguish the ...


1

What you have done is correct, but it isn't yet in the smallest possible form. In fact it will eventually become $a_i\mu_{ij}b_jx_k\nu_{kl}y_l$. Do you see how this works? At least try expanding what I wrote to see that it agrees with what you have. I'll show you what some things become in index notation: Suppose $M$ and $N$ are matrices and ...



Top 50 recent answers are included