PROBLEM 43

The Matrix L is defined by:

L = ( \begin{matrix}l_{0}^{0} & l_{1}^{0} & l_{2}^{0} & l_{3}^{0} \\ l_{0}^{1} & l_{1}^{1} & l_{2}^{1} & l_{3}^{1} \\ l_{0}^{2} & l_{1}^{2} & l_{2}^{2} & l_{3}^{2} \\ l_{0}^{3} & l_{1}^{3} & l_{2}^{3} & l_{3}^{3} \end{matrix} ) .

(a)–Calculate −gL. (b)–Write down the transpose matrix L̃. (c)–CalculateL̃g. (d)–Compare (a) and (c) to find the general form of L (i.e. use L̃g = −gL).

(a)–The matrix g is defined by:

g = ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & −1 & 0 & 0 \\ 0 & 0 & −1 & 0 \\ 0 & 0 & 0 & −1 \end{matrix} ) .

–By matrix multiplication:

−gL = ( \begin{matrix}−l_{0}^{0} & −l_{1}^{0} & −l_{2}^{0} & −l_{3}^{0} \\ l_{0}^{1} & l_{1}^{1} & l_{2}^{1} & l_{3}^{1} \\ l_{0}^{2} & l_{1}^{2} & l_{2}^{2} & l_{3}^{2} \\ l_{0}^{3} & l_{1}^{3} & l_{2}^{3} & l_{3}^{3} \end{matrix} ) .

(b)–The transpose of L is:

L̃ = ( \begin{matrix}l_{0}^{0} & l_{0}^{1} & l_{0}^{2} & l_{0}^{3} \\ l_{1}^{0} & l_{1}^{1} & l_{1}^{2} & l_{1}^{3} \\ l_{2}^{0} & l_{2}^{1} & l_{2}^{2} & l_{2}^{3} \\ l_{3}^{0} & l_{3}^{1} & l_{3}^{2} & l_{3}^{3} \end{matrix} ) .

(c)–By matrix multiplication:

L̃g = ( \begin{matrix}l_{0}^{0} & −l_{0}^{1} & −l_{0}^{2} & −l_{0}^{3} \\ l_{1}^{0} & −l_{1}^{1} & −l_{1}^{2} & −l_{1}^{3} \\ l_{2}^{0} & −l_{2}^{1} & −l_{2}^{2} & −l_{2}^{3} \\ l_{3}^{0} & −l_{3}^{1} & −l_{3}^{2} & −l_{3}^{3} \end{matrix} ) .

(d)–Set −gL = L̃g and compare the components:

( \begin{matrix}−l_{0}^{0} = l_{0}^{0} & −l_{1}^{0} = −l_{0}^{1} & −l_{2}^{0} = −l_{0}^{2} & −l_{3}^{0} = −l_{0}^{3} \\ l_{0}^{1} = l_{1}^{0} & l_{1}^{1} = −l_{1}^{1} & l_{2}^{1} = −l_{1}^{2} & l_{3}^{1} = −l_{1}^{3} \\ l_{0}^{2} = l_{2}^{0} & l_{1}^{2} = −l_{2}^{1} & l_{2}^{2} = −l_{2}^{2} & l_{3}^{2} = −l_{2}^{3} \\ l_{0}^{3} = l_{3}^{0} & l_{1}^{3} = −l_{3}^{1} & l_{2}^{3} = −l_{3}^{2} & l_{3}^{3} = −l_{3}^{3} \end{matrix} ) .  (1)

–The resulting matrix is:

L = ( \begin{matrix}0 & l_{1}^{0} & l_{2}^{0} & l_{3}^{0} \\ l_{1}^{0} & 0 & l_{2}^{1} & l_{3}^{1} \\ l_{2}^{0} & −l_{2}^{1} & 0 & l_{3}^{2} \\ l_{3}^{0} & −l_{3}^{1} & −l_{3}^{2} & 0 \end{matrix} ) .  (2)

(e)–Obtain the same result by discussing the elements of g^{αβ} l̃_{β}^{γ} g_{γδ} = −l_{δ}^{α}. Performing the contractions we get

l̃_{δ}^{α} = −l_{δ}^{α} .

Using the definition

l̃_{δ}^{α} = l_{δ}^{α}

we have the equations

l_{δ}^{α} = −l_{δ}^{α} ,

which are identical with the equations (1), namely (no summations, a = 0, 1, 2, 3, i = 1, 2, 3 and j = 1, 2, 3):

l_{a}^{a} = +l_{a}^{a} ⇒ l_{a}^{a} = −l_{a}^{a} = 0 ,
l_{0}^{i} = −l_{i}^{0} ⇒ l_{i}^{0} = +l_{0}^{i} ,
l_{j}^{i} = +l_{i}^{j} ⇒ l_{i}^{j} = −l_{j}^{i}

and the resulting matrix is again (2).