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    Just trying to become fluent with using these in questions, with the first one that I'm stuck on being:

    Using Einstein Summation convention, prove that \displaystyle \nabla \cdot ( \mathbf{c} \times (\mathbf{r} \times \mathbf{c})) = 2c^2

    I began with simplifying a bit and saying that \mathbf{r} \times \mathbf{c} = \mathbf{s} so that we then have:

    \displaystyle \nabla \cdot ( \mathbf{c} \times \mathbf{s}) = \frac{\partial}{\partial x_i}\epsilon_{ijk} c_j s_k and now I THINK (because I'm not certain) I can express this as \displaystyle \frac{\partial}{\partial x_i}(\epsilon_{ijk} c_j \epsilon_{kmn} r_m c_n) = \frac{\partial}{\partial x_i}(\epsilon_{ijk} \epsilon_{kmn} c_j r_m c_n) and now I'm not sure if there is anything here to be simplified or not, or where to move to - do I need to write out ALL these (non-zero) terms? Perhaps the epsilons go to some Kronecker delta? Though I don't see how that might be possible.

    ALSO: I believe c= | \mathbf{c} | here with \mathbf{c}=(c_1,c_2,c_3)
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    (Original post by RDKGames)
    ..
    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).
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    (Original post by RDKGames)
    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]
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    (Original post by Zacken)
    \epsilon_{ijk} \epsilon_{kmn} = \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm}

    [edit: nvm, DFrank got there first]
    Thanks for providing a version of the identity that matches RDK's indices...
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    (Original post by RDKGames)
    ...do I need to write out ALL these (non-zero) terms....
    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.
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    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.
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    (Original post by DFranklin)
    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}.
    Ah yes, that is mentioned at the top of the worksheet that I missed out. So \mathbf{r}=x_i \mathbf{e}_i

    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};
    Cheers, I'll have a go using this.

    (*) 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).
    (Original post by Zacken)
    \epsilon_{ijk} \epsilon_{kmn} = \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm}

    [edit: nvm, DFrank got there first]
    Thanks. Yes this formula was derived in the lecture notes, I just got stuck on applying it correctly as the first index did not start with i on both, then I quickly realised that \epsilon_{ijk}=\epsilon_{kij} which made the application to Kronecker delta simple.


    (Original post by DFranklin)
    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.
    I see. Currently I'm not a fan of this convention but I need to get used to it, and it took me a short while to see how you got those but they are relatively simple so I hope soon enough they'll just come out naturally for me.
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    (Original post by Zacken)
    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.
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    (Original post by Zacken)
    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.
    (Original post by DFranklin)
    Although one feels something has gone very wrong with the planning of the course if this hasn't been explictly covered in lectures.
    We've covered this and proved it, just completely forgot about it while doing this question.
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    What about this one? DFranklin


    Prove that \displaystyle \nabla \cdot ( \mathbf{r} \times (\mathbf{r} \times \mathbf{c})) = 2 \mathbf{c} \cdot \mathbf{r}

    \displaystyle \nabla \cdot ( \mathbf{r} \times (\mathbf{r} \times \mathbf{c})) =  \frac{\partial}{\partial x_i}(\epsilon_{ijk}r_j \epsilon_{kmn}r_m c_n)=...

    ...\displaystyle=\frac{\partial}  {\partial x_i} [(\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm})r_j r_m c_n]=...

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

    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} r_j r_j c_i=0 but isn't this true?
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    (Original post by RDKGames)
    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.

    Spoiler:
    Show

    \delta_{ii} = 3
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    (Original post by DFranklin)
    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.

    Spoiler:
    Show



    \delta_{ii} = 3


    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.
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    (Original post by RDKGames)
    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).
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    (Original post by DFranklin)
    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).
    Oh okay, thanks for clarifying. It makes sense what you've said, and I've adjusted my answer. The \delta_{ii}=3 caught me out as it's hard for me to keep track of the fact that these are sums, so I was stuck here confused as I kept getting =0 from me writing \frac{\partial}{\partial x_i} x_i = 1

    \displaystyle ... = \frac{\partial}{\partial x_i} (x_i x_j c_j-x_j x_j c_i)= c_j(3x_j+\delta_{ij}x_i)-c_i(2x_j \delta_{ij})

    \displaystyle  = 3c_j x_j+\delta_{ij}x_i c_j-2\delta_{ij}x_jc_i

    \displaystyle = 3 \mathbf{c} \cdot \mathbf{r} + \mathbf{c}\cdot \mathbf{r} - 2 \mathbf{c}\cdot \mathbf{r}
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    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
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    (Original post by RDKGames)
    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...)
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    (Original post by DFranklin)
    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...)
    Oh right, so it's just as simple as \displaystyle \frac{\partial \phi_i}{\partial x_i} F_i = F_i \frac{\partial \phi_i}{\partial x_i} = (F_i \frac{\partial}{\partial x_i}) \phi_i=(\mathbf{F} \cdot \nabla) \phi ?
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    Yes, basically.
 
 
 
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