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Perform a taylor expansion in 3 dimensions in time on the time compontent of of $T^{\alpha \beta}(t - r + n^{i} y_{i})$ given that $r$ is a contstant and $n^{i} y_{i}$ is the scalar product of a normal vector $n$ with a position vector $y$.

I believe the formula is $\phi (\vec{r} + \vec{a}) = \sum^{\infty}_{n=0} \frac{1}{n!} (\vec{a}\cdot \nabla)^{n} \phi(\vec{r})$ in general and that the first term is $T^{\alpha \beta}(t -r)$ but would appreciate some other opinion about what the proceeding terms should be up to order $n=3$.

Cheers

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1 Answer 1

Actually, your tensor depends in a very simple way on the coordinates $y_i$. You can see your tensor as a combination of two functions:

$$f:\mathbb{R}^3 \to \mathbb{R}: y_i \mapsto t-r+n^iy_i$$

and

$$T^{\alpha\beta}:\mathbb{R} \to \mathbb{R}^{3\times 3}: t \mapsto T^{\alpha\beta}(t)$$

All you really need is the basic Taylor expansion formula for the last function combined with the knowledge of the partial derivate of the first function. This will give

$$T^{\alpha\beta}(t-r+n^iy_i) \approx T^{\alpha\beta}(t-r)+(n^iy_i)\dot{T}^{\alpha\beta}(t-r)+\frac{1}{2}(n^iy_i)^2{\ddot{T}}^{\alpha\beta}(t-r)+\frac{1}{6}(n^iy_i)^3{\dddot{T}}^{\alpha\beta}(t-r)$$

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Stunning, thanks. I have the answer but your apprach in seperating the functions out was much more insightful. –  FuriousDee Jan 9 '11 at 20:57

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