Tag Info

Hot answers tagged

7

Those Greek letters are indices indexing the components of $g$. Generally if one expresses a rank-2 tensor like $g$ as a matrix, the first index indexes the rows, the second the columns. In your example, we have $g_{rr} \equiv g_{11} = 1$, $g_{\theta\theta} \equiv g_{22} = r^2$, $g_{r\theta} \equiv g_{12} = 0$, etc. As you can see, we sometimes use numbers ...


3

The double inner product expands to be (for second rank tensors that you encounter in hydrodynamics): $$ \mathbf{a}\mathbf{:}\mathbf{b} = a_{ij}b_{ij} = a_{11}b_{11} + a_{12}b_{12} + ... $$ So it behaves just like you would expect a vector dot product to behave. You add up the product of all of the values with the same indexing. You can do the same ...


2

$$g_{\mu\nu}=\begin{pmatrix}g_{00}&g_{01}&g_{02}\\g_{10}&g_{11}&g_{12}\\g_{20}&g_{21}&g_{22}\end{pmatrix}$$ $\mu,\nu=0,\ldots,N$ are the matrix indices of the metric (and of tensors in general) in $N+1$ dimensions.


1

It depends on the specifics what the notation means, but in general one might say: we pretend, mathematically, that we can label each particle $1, 2, \dots n$, and look at its individual distribution in space, which is (for pure states) some wavefunction $a_k(\vec r, t)$. The product state simply means "particle #1 is in state $a_1$, particle #2 is in state ...


1

I suppose the right function to consider for $R$ is $R=\Vert\mathbf x_i-\mathbf x_j\Vert$ and the usual differentiation operators for the $\nabla$s. So, if $\mathbf x_i = (x_i,y_i,z_i)$, then $$R = \sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2}$$ and $$\nabla_i = \left(\frac\partial{\partial x_i},\frac\partial{\partial y_i},\frac\partial{\partial z_i}\right)$$ ...



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