Einstein Notation

The page that brought you here should have sent you to the page about Index Notation).

This page will be used as a work zone.

$x = e_i x^i$ implies $\displaystyle x = \sum_{i=1}^n e_i x^i$

We ran into a problem where the pages explaining Einstein Notation didn’t drive home the idea that one has to be “up” and one has to be “down”. We had examples using the idea in a scenario where what was being used was Matrix Notation, not Index Notation. We felt we had some some clarity when one author mentioned that Abstract Index Notation evolved out of the Einstein Summation Convention. It makes sense that something as useful as Index Notation evolved over time.

OK, enough of that. Below will be placed some examples:

Let there be F(a,b,c) with a rather dark secret–those seemingly independent variables are functions of three other variables, and of course, if you are a function, you are dependent on one or more variables…

a = a(d,e,f)
b = b(d,e,f)
c = c(d,e,f)

(season finale spoiler: d=d(a,b,c) and e=e(a,b,c) and f=f(a,b,c)…)

Below, we characterize the function in two ways, and since the two characterizations refer to the same function, they are equal:

f(d,e,f) = F(a(d,e,f), b(d,e,f), c(d,e,f))

$\dfrac {\partial f} {\partial d} = \dfrac {\partial F} {\partial a} \dfrac {\partial a} {\partial d} + \dfrac {\partial F} {\partial b} \dfrac {\partial b} {\partial d} + \dfrac {\partial F} {\partial c} \dfrac {\partial c} {\partial d}$

$\dfrac {\partial f} {\partial e} = \dfrac {\partial F} {\partial a} \dfrac {\partial a} {\partial e} + \dfrac {\partial F} {\partial b} \dfrac {\partial b} {\partial e} + \dfrac {\partial F} {\partial c} \dfrac {\partial c} {\partial e}$

$\dfrac {\partial f} {\partial d} = \dfrac {\partial F} {\partial a} \dfrac {\partial a} {\partial f} + \dfrac {\partial F} {\partial b} \dfrac {\partial b} {\partial f} + \dfrac {\partial F} {\partial c} \dfrac {\partial c} {\partial f}$

The work done above makes use of the Chain Rule.

Those symbols we chose, were chose for arbitrary (non-binding) reasons. If we can choose anything, we should consider choices that ultimately will provide an advantage:

{a,b,c} –> ${a^1, a^2, a^3}$
{d,e,f,} –> ${b^1, b^2, b^3}$

We will put these changes into the first equation above. Initially we will just replace {a,b,c} so you can see what that does:

$\dfrac {\partial f} {\partial d} = \dfrac {\partial F} {\partial a^1} \dfrac {\partial a^1} {\partial d} + \dfrac {\partial F} {\partial a^2} \dfrac {\partial a^2} {\partial d} + \dfrac {\partial F} {\partial a^3} \dfrac {\partial a^3} {\partial d}$

$\displaystyle \dfrac {\partial f} {\partial d} = \sum_{i=1}^3 \dfrac {\partial F} {\partial a^i} \dfrac {\partial a^i} {\partial d}$

If we use Index Notation we can eliminate the summation symbol:

$\dfrac {\partial f} {\partial d} = \dfrac {\partial F} {\partial a^i} \dfrac {\partial a^i} {\partial d}$ and all three equations are as follows:

$\dfrac {\partial f} {\partial d} = \dfrac {\partial F} {\partial a^i} \dfrac {\partial a^i} {\partial d}$

$\dfrac {\partial f} {\partial e} = \dfrac {\partial F} {\partial a^i} \dfrac {\partial a^i} {\partial e}$

$\dfrac {\partial f} {\partial f} = \dfrac {\partial F} {\partial a^i} \dfrac {\partial a^i} {\partial f}$

When we make the change from d,e,f to b with an index we can write the following:

$\dfrac {\partial f} {\partial b^i} = \dfrac {\partial F} {\partial a^j} \dfrac {\partial a^j} {\partial b^i}$

Another Example:

$A^i_k B^k_j = C^i_j$

In the above, something like $A^i_k$ is a component of the tensor rather than the tensor, so we can switch the order without changing the answer:

$B^k_j A^i_k = C^i_j$

Regarding Abstract Index Notation, an article by Roger Penrose on the topic makes for very interesting reading.

Share this: