1)Metric tensor is used for lowering and raising indices. Does that means that it will always be the same matrix as long as it is in the same space(e.g. flat space)? $$g_{\mu\nu}=g^{\mu\nu}=g_{\alpha\beta}=g_{\alpha\gamma}=g_{\beta\delta}=...= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0& 0& -1 \end{pmatrix} $$

E.g. $A_{\alpha\beta}=g_{\alpha\gamma}g_{\beta\delta}A^{\gamma\delta}$<--------Does this means that $g_{\alpha\gamma}$ and $g_{\beta\gamma}$ are the same matrix(metric tensor)? But when I matrix product $g_{\alpha\gamma}$ and $g_{\beta\gamma}$, I get an identity matrix instead which doesn't seem to help in raising/lowering indices.

My confusion here is because there are lack of examples involving them in matrix.

2) To raise/lower indices, is metric tensor the only way to do it?

  • $\begingroup$ My advice is to play around with the metric for a 2-sphere. This might help you gain some intuition for the concept of the metric. $\endgroup$ – Crimson Feb 8 '17 at 15:01

In matrix notation you must write $$ (A_{\alpha\beta})= (g_{\alpha\mu})(A^{\mu\nu})( g_{\nu\beta})$$ \begin{eqnarray} =\begin{bmatrix} 1 &0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{bmatrix} \begin{bmatrix} A_{00} &A_{01}&A_{02}&A_{03}\\ A_{10}&A_{11}&A_{12}&A_{13}\\ A_{20}&A_{21}&A_{22}&A_{23}\\ A_{30}&A_{31}&A_{32}&A_{33} \end{bmatrix} \begin{bmatrix} 1 &0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{bmatrix} \end{eqnarray}

  • $\begingroup$ Tho positions of the indices tell us the structure of RHS the first index indicates row number and the second for the column number. $\endgroup$ – Saksith Jaksri Feb 8 '17 at 16:10

When the indices are at the same position the tensors are the same, thus:

$$g_{\mu\nu}=g_{\alpha\beta}=g_{\alpha\gamma}=g_{\beta\delta}\neq g^{\mu\nu}$$

One can get from one to the other by realizing that consecutive lowering and raising of indices should give the original result:

$$g_{\rho\nu} \cdot g^{\mu\nu} \cdot V_{\mu} = V_{\rho} = V_{\mu} $$


$$g_{\rho\nu} \cdot g^{\mu\nu} = \delta^{\mu}_{\rho} =\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

This allows one to get $g^{\mu\nu}$ when $g_{\mu\nu}$ is known.

For Minkowski spacetime we find that $g^{\mu\nu} = g_{\mu\nu}$, but this is not generally true.

Note that your strange result arose because the summation convention does not work when multiplying $g_{\alpha\gamma}$ by $g_{\beta\gamma}$ since the $\gamma$ indices are both low. Multiplying two second order tensors with lowered indices will give a fourth order tensor with lowered indices.

  • $\begingroup$ "Multiplying two second order tensors with lowered indices will give a fourth order tensor with lowered indices" How does it work? Isn't multiplying 2 second order tensors the same as matrix multiplication between the two matrices which will give me a matrix? $\endgroup$ – newbie125 Feb 8 '17 at 15:32
  • $\begingroup$ @newbie125 Multiplying 2 second order tensors can only be seen as matrix multiplication when the two tensors share one index that can be contracted. When all indices differ, the multiplication can be seen as a tensor product. see wiki $\endgroup$ – Crimson Feb 9 '17 at 13:11

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