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I am reading "An Introduction to the Mathematical Theory of Vibrations of Elastic Plates" by R. D. Mindlin in which he uses "summation convention" which I understand is the same as Einstein notation.

I'm a bit confused by the notation used for the differentiation operator and whether a summation is presumed to be embedded in it.

He writes out the traction equation as:

$${t_j} = {v_i}{T_{ij}} \equiv {v_1}{T_{1j}} + {v_2}{T_{2j}} + {v_3}{T_{3j}}$$

From what I understand, since $i$ appears twice it is the dummy index we are summing over and, since it appears once, $j$ is the free index.

Differentiation with respect to space coordinates is then defined as:

$$,i \equiv \frac{\partial }{{\partial {x_i}}}$$

The stress equation of motion is then given as:

$${T_{ij,i}} = \rho \frac{{{\partial ^2}{u_j}}}{{\partial {t^2}}}$$

This is expanded for the values of the free index j as:

$$\frac{{\partial {T_{11}}}}{{\partial {x_1}}} + \frac{{\partial {T_{21}}}}{{\partial {x_2}}} + \frac{{\partial {T_{31}}}}{{\partial {x_3}}} = \rho \frac{{{\partial ^2}{u_1}}}{{\partial {t^2}}}$$

$$\frac{{\partial {T_{12}}}}{{\partial {x_1}}} + \frac{{\partial {T_{22}}}}{{\partial {x_2}}} + \frac{{\partial {T_{32}}}}{{\partial {x_3}}} = \rho \frac{{{\partial ^2}{u_2}}}{{\partial {t^2}}}$$

$$\frac{{\partial {T_{13}}}}{{\partial {x_1}}} + \frac{{\partial {T_{23}}}}{{\partial {x_2}}} + \frac{{\partial {T_{33}}}}{{\partial {x_3}}} = \rho \frac{{{\partial ^2}{u_3}}}{{\partial {t^2}}}$$

So my confusion is in which quantity/operation the summation is implicitly embedded in.

In the quantity ${T_{ij}}$ by itself there is no double dummy index variable so I assume that no embedded summation is implied and that it just stands for a single element of a matrix as would be normally the case when not using the summation convention/Einstein notation.

Likewise there is no double dummy index in the operation $,i$ by itself so I assume there is no embedded summation in it either.

So it seems to me the implicitly embedded summation only exists when the two are combined:

$${T_{ij,i}}$$

Is this correct?

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  • $\begingroup$ Yes this is correct. This is actually the so-called "Einstein sum rule". See en.wikipedia.org/wiki/Einstein_notation $\endgroup$
    – Navid
    Nov 3, 2023 at 2:35
  • $\begingroup$ Hi, welcome to Physics SE. Just as the two ${}_i$s in the term $v_iT_{ij}$ contract despite having different origins, the same things happens in $\partial_iT_{ij}$, which is an alternative notation for the gradient you've asked about. $\endgroup$
    – J.G.
    Nov 3, 2023 at 11:55

1 Answer 1

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This is the aforementioned Einstein summation convention, you are correct. If in doubt, you can always reintroduce the explicit sum. Here that would yield

$$ T_{ij,i} = \sum_i \frac{ \partial T_{ij}}{\partial x_i}.$$

The implicit summation only exists where indices are repeated, that is the definition of the convention. Therefore, no sum exists over $T_{ij}$ or $_{,i}$ individually. To illustrate this further, you could consider the case of $T_{ij,j}$, where summation occurs over the other index. This becomes very handy as a shorthand when many indices are involved (e.g. in GR or QFT).

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