dyn said:
.. if i am just given the element Λuv how do i find the corresponding element in the inverse ?
See bellow.
As regards the Faraday tensor equation F'uv = Λua Λvb Fab I realize these are just elements multiplied together so the order doesn't matter but if I want to do the matrix multiplication how do I decide the order of the matrix multiplication ?
Follow the rule of matrix multiplication: the second index of the first matrix is summed (or contracted) with the first index of the second matrix: \bar{F}^{ab} = \left( \Lambda^{a}{}_{c} \ F^{cd} \right) \ \Lambda^{b}{}_{d} = \Lambda^{b}{}_{d} \left( \Lambda \ F \right)^{ad}. Now, let \Lambda F = B, \bar{F}^{ab} = \Lambda^{b}{}_{d} \ B^{ad} = \Lambda^{b}{}_{d} \ ( B^{T})^{da} , or \bar{F}^{ab} = \left( \Lambda \ B^{T}\right)^{ba} = \left( B \ \Lambda^{T}\right)^{ab} . Therefore \bar{F} = B \ \Lambda^{T} = \Lambda \ F \ \Lambda^{T} . But, why do you want it in matrix form? The world isn’t made of only rank-2 tensors. Any way, here are some rules and conventions you need to follows when you treat rank-2 Lorentz tensors as matrices:
i) I have already given you the first rule which is the matrix multiplication rule above.
ii) \eta is a (0,2) tensor: \eta_{\mu\nu}. It is also the matrix element of the diagonal matrix \eta.
iii) \eta^{\mu\nu} is a (2,0) tensor and can be regarded as the matrix element of the inverse matrix \eta^{-1}.
iv) \Lambda is a Lorentz group element. Lorentz group is a matrix Lie group and \Lambda, therefore, has matrix representation. The convention for its matrix element is \Lambda^{\mu}{}_{\nu}, where \nu represents the rows (i.e. the first index on a matrix) and \nu numerates the columns (i.e. the second index on a matrix). This convention though makes it mandatory to represent \Lambda^{-1}, \Lambda^{T} and all other MATRIX OPERATIONS by the same index structure for their matrix element. So, like \Lambda^{\mu}{}_{\nu}, we must write (\Lambda^{-1})^{\mu}{}_{\nu}, (\Lambda^{T})^{\mu}{}_{\nu} and so on.
v) Even though \Lambda^{\mu}{}_{\nu} is NOT a tensor, we can raise and lower its indices by the metric tensor \eta. This becomes important when dealing with the infinitesimal part of \Lambda. Examples:
(1) The infinitesimal group parameters satisfy the following MATRIX equation, (\eta \ \omega)^{T} = - (\eta \ \omega) . \ \ \ \ (1) The \alpha \beta-matrix element is \left( (\eta \ \omega)^{T}\right)_{\alpha \beta} = - \left( \eta \ \omega \right)_{\alpha \beta} , or, by doing the transpose on the LHS, \left( \eta \ \omega \right)_{\beta \alpha} = - \left( \eta \ \omega \right)_{\alpha \beta} . Following the above-mentioned rule for matrix multiplication, we get \eta_{\beta \mu} \ \omega^{\mu}{}_{\alpha} = - \eta_{\alpha \rho} \ \omega^{\rho}{}_{\beta} . Thus \omega_{\beta \alpha} = - \omega_{\alpha \beta} . \ \ \ \ \ (2) You can also start from (2) and go backward to (1).
(2) The defining relation of Lorentz group is given by \eta_{\mu \nu} \ \Lambda^{\mu}{}_{\alpha} \ \Lambda^{\nu}{}_{\beta} = \eta_{\alpha \beta} . \ \ \ (3) Before we carry on with raising and lowering indices, I would like to make two important side notes on Eq(3): A) equations (1) or (2) are the infinitesimal version of Eq(3), and B) since the \Lambda’s form a group, Eq(3) is also satisfied by inverse element, \eta_{\mu \nu} \left( \Lambda^{-1}\right)^{\mu}{}_{\alpha} \left( \Lambda^{-1}\right)^{\nu}{}_{\beta} = \eta_{\alpha \beta} . \ \ (4)
Okay, lowering the index on the first \Lambda in Eq(3), we obtain \Lambda_{\nu \alpha} \ \Lambda^{\nu}{}_{\beta} = \eta_{\alpha \beta} . Now, raising the index \alpha on both sides (or, which is the same thing, contracting with \eta^{\alpha \tau}), we obtain \Lambda_{\nu}{}^{\tau} \ \Lambda^{\nu}{}_{\beta} = \delta^{\tau}{}_{\beta} . \ \ \ \ \ (5) Notice that Eq(5) does not follow the rule of matrix multiplication. This is because of the funny index structure of \Lambda_{\nu}{}^{\tau} which does not agree with our convention in (iv) above. However, we know the following matrix equation \left( \Lambda^{-1} \ \Lambda \right)^{\tau}{}_{\beta} = \delta^{\tau}{}_{\beta} . So, using the rule for matrix multiplication, we find \left( \Lambda^{-1}\right)^{\tau}{}_{\nu} \ \Lambda^{\nu}{}_{\beta} = \delta^{\tau}{}_{\beta} . \ \ \ \ \ (6) Comparing (5) with (6), we find \left( \Lambda^{-1}\right)^{\tau}{}_{\nu} = \Lambda_{\nu}{}^{\tau} . \ \ \ \ \ \ (7) We will come to the (matrix) meaning of this in a minute, let us first substitute (7) in (4) to obtain: \eta_{\mu \nu} \ \Lambda_{\alpha}{}^{\mu} \ \Lambda_{\beta}{}^{\nu} = \eta_{\alpha \beta} . This shows that we could have started with the convention \Lambda_{\mu}{}^{\nu} for the matrix element of \Lambda. The lesson is this, once you choose a convention you must stick with it.
v) Finally, Eq(7) means the following: giving the matrix
<br />
\Lambda = \begin{pmatrix}<br />
\Lambda^{0}{}_{0} & \Lambda^{0}{}_{1} & \Lambda^{0}{}_{2} & \Lambda^{0}{}_{3} \\<br />
\Lambda^{1}{}_{0} & \Lambda^{1}{}_{1} & \Lambda^{1}{}_{2} & \Lambda^{1}{}_{3} \\<br />
\Lambda^{2}{}_{0} & \Lambda^{2}{}_{1} & \Lambda^{2}{}_{2} & \Lambda^{2}{}_{3} \\<br />
\Lambda^{3}{}_{0} & \Lambda^{3}{}_{1} & \Lambda^{3}{}_{2} & \Lambda^{3}{}_{3}<br />
\end{pmatrix} ,<br />
the inverse is obtained by changing the sign of \Lambda^{0}{}_{i} and \Lambda^{i}{}_{0} components ONLY, and then transposing ALL indices:
<br />
\Lambda^{-1} = \begin{pmatrix}<br />
\Lambda^{0}{}_{0} & -\Lambda^{1}{}_{0} & -\Lambda^{2}{}_{0} & -\Lambda^{3}{}_{0} \\<br />
-\Lambda^{0}{}_{1} & \Lambda^{1}{}_{1} & \Lambda^{2}{}_{1} & \Lambda^{3}{}_{1} \\<br />
-\Lambda^{0}{}_{2} & \Lambda^{1}{}_{2} & \Lambda^{2}{}_{2} & \Lambda^{3}{}_{2} \\<br />
-\Lambda^{0}{}_{3} & \Lambda^{1}{}_{3} & \Lambda^{2}{}_{3} & \Lambda^{3}{}_{3}<br />
\end{pmatrix} .<br />