Einstein Summation convention with vectors

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Perhaps the epsilons go to some Kronecker delta? Though I don't see how that might be possible.

$\epsilon_{ijk} \epsilon_{kmn} = \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm}$

[edit: nvm, DFrank got there first]

...do I need to write out ALL these (non-zero) terms....

Couple of things:

I assume r is supposed to equal $(x_1, x_2, x_3)$ i.e. it is the spatial position.

In which case $r_m = x_m$ and then $\dfrac{d}{dx_i} x_m = \delta_{im}$.

The other thing I would do (*) is replace $\displaystyle \mathbf{c} \times (\mathbf{r} \times \mathbf{c})$ by $c^2 {\bf r} - ({\bf c . r}) {\bf c}$;

(*) if you want to do this using summation convention you will basically need to use the identity $\epsilon_{ijk}\epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}$ (which should have all been covered in lectures).

$\epsilon_{ijk} \epsilon_{kmn} = \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm}$

[edit: nvm, DFrank got there first]

Just for info: with practice you can often do these directly from the vector notation, or if you do need to go to summation convention, it's very rare to need to write out individual terms.

For instance, I can tell by looking that ${\bf \nabla . } (c^2 {\bf r}) = 3c^2, {\bf \nabla . (r.c)c} = c^2$, from which the result follows.

It is probably a good exercise to prove $\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = \mathbf{v}(\mathbf{u}\cdot \mathbf{w}) - \mathbf{w}(\mathbf{u}\cdot \mathbf{v})$, which is essentially what you do, in the special case u = c, v=r , w = u = c by using the identity in DFrank's post.

It is probably a good exercise to prove $\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = \mathbf{v}(\mathbf{u}\cdot \mathbf{w}) - \mathbf{w}(\mathbf{u}\cdot \mathbf{v})$, which is essentially what you do, in the special case u = c, v=r , w = u = c by using the identity in DFrank's post.

Although one feels something has gone very wrong with the planning of the course if this hasn't been explictly covered in lectures.

So I seem to have lost one of them along the way but I can't pinpoint where. I think it's the part at the end where I say (not explicitly) that $\frac{\partial }{\partial x_i} x_j x_j c_i=0$ but isn't this true?

No; implicitly i and j range from 1 to 3; and $\dfrac{\partial}{\partial x_i} x_j = 1$ when i = j.

More explictly, $\dfrac{\partial}{\partial x_i} x_j = \delta_{ij}$ (*) (as I posted upstream).

I know you're just learning, but this is an absolutely fundamental relationship - you'll be using (*) more than any other rule (other than how to use deltas at all) as you do these questions.

Edit: noticed another mistake. In the same line, it looks like you have thought that $\dfrac{\partial}{\partial x_i} x_i = \delta_{ij}$ (somehow!). Again if you use (*) you see that it actually equals $\delta_{ii}$, which ends up being a constant.

Ah, got it - I was slightly misunderstanding the delta thing there.

$... \displaystyle = \frac{\partial}{\partial x_i} (r_i r_j c_j-r_j r_j c_i)=\delta_{ii} r_j c_j - \delta_{ij}r_j c_i = 3\mathbf{c} \cdot \mathbf{r}- \mathbf{c} \cdot \mathbf{r}$

Thanks.

Actually, I kind of mislead you on this; what you've done still isn't right. (Note that I'm using x_i instead of r_i everywhere; to be honest I think using r_i is "essentially" wrong - that is, it's only right because r_i = x_i, and you're implicitly using this throughout, so you should just write x_i (or differentiate w.r.t. r_i, equivalently)).

It's true that $\dfrac{\partial}{\partial x_i} x_j = \delta_{ij}$, but it is not true that $\dfrac{\partial}{\partial x_i} x_j x_j = \delta_{ij} x_j$

What you need to do is use the product rule (for differentiation):

$\dfrac{\partial}{\partial x_i} x_j x_j = \dfrac{\partial x_j}{\partial x_i} x_j + x_j \dfrac {\partial x_j}{\partial x_i} = 2 x_j \delta_{ij}$

You need to do similarly with the $\dfrac{\partial}{\partial x_i} x_i x_j c_j$ expression - it's a product, and the partial derivative acts on both terms ($x_i$ and $x_j$).

@DFranklin


With $\phi(x_1, x_2, x_3)$ and $\mathbf{F}(x_1, x_2, x_3)$ being scalar and vector fields respectively, prove that $\displaystyle \nabla \cdot (\phi \mathbf{F}) = (\mathbf{F}\cdot \nabla) \phi + \phi (\nabla \cdot \mathbf{F})$ using E.C.

So I began by saying that:

$\displaystyle \nabla \cdot (\phi \mathbf{F}) = \frac{\partial}{\partial x_i} (\phi_i F_i)$

which by the product rule gives $\displaystyle \frac{\partial \phi_i}{\partial x_i} F_i + \frac{\partial F_i}{\partial x_i} \phi_i$

I can see how the second term comes to give $(\nabla \cdot \mathbf{F})\phi = \phi (\nabla \cdot \mathbf{F})$ but I can't quite see how the first term goes to $(\mathbf{F} \cdot \nabla) \phi$

I think the question you need to ask yourself is what does $(\mathbf{F}\cdot \nabla)$ actually mean? (At which point you'll realise you don't have much more to do...)