Index notation matrix calculation (Intro to Relativity)

Question

Consider the next matrix:

$$M_{ab} = \left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array}\right)$$ and $$N_{ab} = \left(\begin{array}{cccc} 0 & 0 & \frac{-1}{2} & 0 \\ 0 & 1 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array}\right)$$

Calculate: $ I_M = M_{ab}\Delta x^a\Delta x^b $ and $ I_N = N_{ab}\Delta x^a\Delta x^b $ for the next subtractions:

• $a) \Delta x^a = (1,0,1,0) $ and

• $ b) \Delta x^a = (1,0,0,0) $

The problem is I don't really understand the notation nor, given $M_{ab}$ and $N_{ab}$ (as well as $\Delta x^a$), how to operate the matrix.

Any help on this particular example or a general procedure to operate this will be really appreciated.

Answer 1

Let's look at $M$ and $N$ first. Here, $a$ refers to the row, and $b$ to the column. So, $M_{11}$ corresponds to the upper left corner, and has a value of $-1$. $N_{13}$ corresponds to row $1$, column $3$, with a value of $-\frac{1}{2}$.

Now, lets look at $\Delta x$. This is very similar, but there's only one dimension to work with. So, for b), $\Delta x^{1} = 1$, while for $a = 2, 3, 4$ the value is $0$.

Finally, let's put everything together. The basic concept is that, when an index occurs in both the top and bottom of an equation, there is an implied summation over all values of that index. See below:

$ X_{\gamma} = \begin{bmatrix}1 \\ 2 \\3 \end{bmatrix}$, $Y^{\gamma} = \begin{bmatrix}4 & 5 & 6 \end{bmatrix}$

Then,

$$X_{\gamma}Y^{\gamma} = \sum_{\gamma = 1}^{3} X_{\gamma}Y^{\gamma} = X_{1}Y^{1} + X_{2}Y^{2} + X_{3}Y^{3} = (1*4) + (2*5) + (3*6)$$

Your problem is a bit more complex, as it includes a double summation.

$$M_{ab}x^{a}x^{b} = \sum_{a = 1}^{4} \sum_{b = 1}^{4} M_{ab}x^{a}x^{b}$$

So, now you're out of Einstein notation and in the somewhat more familiar realm of double summations. In case you need a refresher, first let $a=1$, and let $b$ cycle through $1 \to 4$. Now let $a = 2$, and repeat. Add up all these terms, and you have your answer. Alternatively, there are plenty examples of double summations online, the "Einstein notation" part is just understanding how to write them as such.

Answer 2

In the matrix $M_{ab}$, $a$ labels the rows and $b$ the columns. Now, Einstein's summation convention states that when an index appears twice, once as a subscript and once as a superscript, it is summed over.

So, $$I_M \ =\ M_{ab}\Delta x^a \Delta x^b\ =\ \sum_{a=1}^4 \sum_{b=1}^{4} M_{ab}\Delta x^a \Delta x^b$$

Now, consider the object $\Delta x_a\ =\ (1, 0, 1, 0)$. The numbers in the parenthesis are components of $\Delta x_a$ in the basis $(e_1,e_2,e_3,e_4)$.

$$\Delta x_a\ =\ 1.e_1\ +\ 0.e_2\ +\ 1.e_3\ +\ 0.e_4 $$

It is customary to write the bases as column vectors.

$$ e_1= \left( {\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array} } \right),\quad e_2= \left( {\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array} } \right), \quad e_3= \left( {\begin{array}{c} 0\\ 0\\ 1\\ 0 \end{array} } \right), \quad e_4= \left( {\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array} } \right)$$

Thus, $\Delta x_a$ can be written as a column vector $$ \Delta x_a\ = \left( {\begin{array}{c} 1\\ 0\\ 1\\ 0 \end{array} } \right)$$

Since your title says "Intro to relativity", let me just mention that the matrix $M$ is called the metric tensor, with which you can raise or lower the indices. So, $\Delta x^a$ and $\Delta x_a$ are related in the following way

$$ \Delta x^a\ = M_{ab} \Delta x_b,\ (c=1,..4) $$ Now, you can easily do the matrix multiplication to obtain $I_M$.