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3rd Year Honours Mathematical Physics
Special Relativity
Summation Convention

Brian Dolan

1.
If an index appears twice in the same term, then it is summed over.
2.
If an index appears only once in a term, then it must also appear only once in every other term in the same equation. The equation then represents three (or four in four dimensions) separate equations, one for each possible value of the solitary index.
3.
If an index appears more than twice in the same term, then you have made a mistake.
For example:

\begin{displaymath}{ \underline {\hbox{a}} }=\sum_{i=1}^3 a^i { \underline {\hbo...
...2 { \underline {\hbox{e}} }_2 + a^3 { \underline {\hbox{e}} }_3\end{displaymath}


\begin{displaymath}\oalign{A\crcr\hidewidth
\vbox to.2ex{\hbox{\char'176}\vss}\...
...cr\hidewidth
\vbox to.2ex{\hbox{\char'176}\vss}\hidewidth}_0
\end{displaymath}


\begin{displaymath}{ \underline {\hbox{a}} }.{ \underline {\hbox{b}} }=\sum_{i=1...
...elta_{ij}
=a^i b^j \delta_{ij}
=a^1 b^1 + a^2 b^2 + a^3 b^3\end{displaymath}


\begin{displaymath}\oalign{A\crcr\hidewidth
\vbox to.2ex{\hbox{\char'176}\vss}\...
...
=A^a B^b \eta_{ab}
=A^1 B^1 + A^2 B^2 + A^3 B^3 - A^0 B^0.
\end{displaymath}

If we have two different sets of basis vectors ${\{{ \underline {\hbox{e}} }_1,{ \underline {\hbox{e}} }_2,{ \underline {\hbox{e}} }_3\}}$ and ${\{{ \underline {\hbox{e}} }_{1^\prime},{ \underline {\hbox{e}} }_{2^\prime},{ \underline {\hbox{e}} }_{3^\prime}\}}$ then we know that ${{ \underline {\hbox{e}} }_{1^\prime}}$ can be written as a linear combination of ${{ \underline {\hbox{e}} }_1,{ \underline {\hbox{e}} }_2}$ and ${ \underline {\hbox{e}} }_3$. Thus

\begin{displaymath}{ \underline {\hbox{e}} }_{1^\prime}={ \underline {\hbox{e}} ...
...2}_{1^\prime}
+ { \underline {\hbox{e}} }_3 {S^3}_{1^\prime}.\end{displaymath}

Similarly,

\begin{displaymath}{ \underline {\hbox{e}} }_{2^\prime}={ \underline {\hbox{e}} ...
...^2}_{2^\prime}
+ { \underline {\hbox{e}} }_3 {S^3}_{2^\prime}\end{displaymath}

and

\begin{displaymath}{ \underline {\hbox{e}} }_{3^\prime}={ \underline {\hbox{e}} ...
...2}_{3^\prime}
+ { \underline {\hbox{e}} }_3 {S^3}_{3^\prime}.\end{displaymath}

These equations can be summarised as

\begin{displaymath}{ \underline {\hbox{e}} }_{i^\prime}={ \underline {\hbox{e}} ...
...^j}_{i^\prime}
={ \underline {\hbox{e}} }_j {S^j}_{i^\prime}.\end{displaymath}

Under a change of basis, the components ai of the vector ${{ \underline {\hbox{a}} }=a^i { \underline {\hbox{e}} }_i}$ also change. We define ${R^{i^\prime}}_j$ by

\begin{displaymath}a^{1^\prime}={R^{1^\prime}}_1 a^1 + {R^{1^\prime}}_2 a^2
+...
...3
=\sum_{j=1}^3 {R^{1^\prime}}_j a^j
={R^{1^\prime}}_j a^j,\end{displaymath}

plus the other two equations for $a^{2^\prime}$ and $a^{3^\prime}$, i.e.

\begin{displaymath}a^{i^\prime}=\sum_{j=1}^3 {R^{i^\prime}}_j a^j = {R^{i^\prime}}_j a^j.\end{displaymath}

It is shown on the handout Summary of 3-vectors (equation (11)) that

\begin{displaymath}{\bf RS}={\bf 1}\qquad\hbox{or}\qquad
\sum_{k=1}^3 {R^{i^\prime}}_k {S^k}_{j^\prime}
={\delta^{i^\prime} }_{j^\prime}.\end{displaymath}

i.e.

\begin{displaymath}{R^{i^\prime}}_k {S^k}_{j^\prime}={\delta^{i^\prime} }_{j^\prime}.\end{displaymath}

It was also shown in Summary of 3-vectors (equation (13)) that, if ${\{{ \underline {\hbox{e}} }_1,{ \underline {\hbox{e}} }_2,{ \underline {\hbox{e}} }_3\}}$ and ${\{{ \underline {\hbox{e}} }_{1^\prime},{ \underline {\hbox{e}} }_{2^\prime},{ \underline {\hbox{e}} }_{3^\prime}\}}$ are both orthonormal (i.e. ${{ \underline {\hbox{e}} }_i . { \underline {\hbox{e}} }_j = \delta_{ij}}$ and ${{ \underline {\hbox{e}} }_{i^\prime} . { \underline {\hbox{e}} }_{j^\prime} = \delta_{{i^\prime}{j^\prime}}}$), then

\begin{displaymath}{\bf S}^{-1}={\bf S}^T.\end{displaymath}

In the summation convention, the transpose of a matrix is obtained by swapping the order of the indices, thus

\begin{displaymath}{(S^T)^i}_{j^\prime} = {S_{j\prime}}^i.\end{displaymath}

So for the orthonormal matrix, ${\bf S}$,

\begin{displaymath}{(S^{-1})^i}_{j^\prime} = {S_{j\prime}}^i.\end{displaymath}



 
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Brian Dolan
1998-12-03