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    I'm reading a paper where the symbols are throwing me off.

    For the 'dot' symbol, I see \nabla\cdot u which makes sense, but u\cdot\nabla also appears. I realize the dot product is commutative, but
    1) Why change the order?
    2) \nabla\cdot u = 0 is given as an initial condition, yet u\cdot\nabla shows up in the second initial condition. If the operation is in fact commutative, then why place u\cdot\nabla in the second condition when it would just zero out the term.

    Note - I doubt this is indicative of normal matrix multiplication since (\nabla\cdot u)u is seen.

    Does anyone know of another meaning for \cdot when talking vectors?

    For the \wedge symbol, I've seen A\wedge(A\wedge B) and the only thing I've stumbled across is the exterior product. Does anyone know of other possibilities or am I on the right track with the exterior product?
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    You know that \nabla is a differential operator, right? Which means that \nabla . u \neq u.\nabla, just as \frac{d}{dx}u \neq u \frac{d}{dx}.

    In this context, A \wedge B is almost certainly the same as A \times B; it's just the vector cross product. (In 3-space, this is basically also the same as the exterior wedge product, but you probably don't want to think of it like that).
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    (Original post by DFranklin)

    1) You know that \nabla is a differential operator, right? Which means that \nabla . u \neq u.\nabla, just as \frac{d}{dx}u \neq u \frac{d}{dx}.

    2) In this context, A \wedge B is almost certainly the same as A \times B; it's just the vector cross product. (In 3-space, this is basically also the same as the exterior wedge product, but you probably don't want to think of it like that).
    1) I know that the first is just the divergence, and I'm aware of the faulty notation since it 'views' as a dot product but no true product is taking place. Rather, the operators are acting on each component of the vector, respectively. I'm still lost on the meaning of u\cdot\nabla

    I'm working in 3D, so could it be that I multiply the matrices as normal but the dot is to let me know that I allow the operators to act on the entries rather than true multiplication. Meaning \nabla\cdot u would give me a 1\times1 matrix but u\cdot\nabla would give me a 3\times3 matrix? That's all I've been able to come up with so far.

    2) I googled 'cross product' with the suggestion, and I didn't realize the wedge is used in physics notation. I'm good on this one now. Thanks.
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    Be careful with A \wedge B \wedge C though - if it's the exterior product, then it means A \cdot B \times C. But in A \wedge (A \wedge B), it's probably just the normal cross product, because the exterior product would give zero.

    The operation of taking two vectors and forming a matrix is called the outer product (not to be confused with the exterior product), usually denoted \otimes; I really don't think it would be appropriate to denote it by the same symbol as the inner product. Now I'm just guessing, but I think (u \cdot \nabla) f is \displaystyle u_x \frac{\partial f}{\partial x} + u_y \frac{\partial f}{\partial y} + u_z \frac{\partial f}{\partial z}.
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    (Original post by monty1618)
    1) I know that the first is just the divergence, and I'm aware of the faulty notation since it 'views' as a dot product but no true product is taking place. Rather, the operators are acting on each component of the vector, respectively. I'm still lost on the meaning of u\cdot\nabla

    I'm working in 3D, so could it be that I multiply the matrices as normal but the dot is to let me know that I allow the operators to act on the entries rather than true multiplication. Meaning \nabla\cdot u would give me a 1\times1 matrix but u\cdot\nabla would give me a 3\times3 matrix? That's all I've been able to come up with so far.

    2) I googled 'cross product' with the suggestion, and I didn't realize the wedge is used in physics notation. I'm good on this one now. Thanks.
    Well, \nabla \cdot u = \frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3}, whereas u\cdot \nabla = u_1\frac{\partial }{\partial x_1} + u_2\frac{\partial }{\partial x_2} + u_3\frac{\partial }{\partial x_3}. The vital point is that \nabla does two things - it acts as a vector, and it acts as a differential operator. It differentiates u if u comes to the right of it, but not to the left - notice that \nabla \cdot u is a perfectly legitimate expression, but u \cdot \nabla is still an operator. You'd only ever see (u \cdot \nabla)a or similar, of which the meaning should be obvious.

    Also note that \nabla operates on a vector. but (u \cdot \nabla) operates on a scalar.
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    I've seen some good ideas so far, but I haven't given enough info to rule out certain things.

    \frac{\partial u}{\partial t} + (u\cdot\nabla)u is an expression from the paper.

    u\cdot\nabla must be able to act on a vector, and the result must be of the same dimensions as \frac{\partial u}{\partial t}, being 3\times 1. So u\cdot\nabla must be either a scalar or a 3\times3 matrix.
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    (Original post by Zhen Lin)
    The operation of taking two vectors and forming a matrix is called the outer product (not to be confused with the exterior product), usually denoted \otimes; I really don't think it would be appropriate to denote it by the same symbol as the inner product. Now I'm just guessing, but I think (u \cdot \nabla) f is \displaystyle u_x \frac{\partial f}{\partial x} + u_y \frac{\partial f}{\partial y} + u_z \frac{\partial f}{\partial z}.
    I haven't ran it yet, but I think that comes out the same way as treating u \cdot \nabla as a 3\times 3 matrix, even though your way is quicker by far.

    So basically, treat it as a dot product as usual, but it will just remain an operator until applied to another vector is what I'm getting from it.
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    (Original post by generalebriety)
    Also note that \nabla operates on a vector. but (u \cdot \nabla) operates on a scalar.
    I'm almost completely sold on treating it as the normal dot product if u \cdot \nabla can act on a vector. Wouldn't it be ok to just apply u \cdot \nabla to each component of a vector or would that present a problem?
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    Ok. I have it now, so disregard the last question(s). And Franklin, you had it pegged from the start, but for some reason the non-commutativity wasn't registering.

    Anyway, thanks to you all.
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    (Original post by monty1618)
    I'm almost completely sold on treating it as the normal dot product if u \cdot \nabla can act on a vector. Wouldn't it be ok to just apply u \cdot \nabla to each component of a vector or would that present a problem?
    Oh, yes, that'd be fine too. Good point. :p:
 
 
 
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