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Del operator

  1. Jun 28, 2010 #1
    hi,

    I'm trying to follow a derivation in a paper and this equation is confusing me:

    (u'.[tex]\nabla[/tex])U = ([tex]\nabla[/tex]U).u'

    Where U and u' are velocities.

    The operation of del on the vector U without a dot or cross product is giving me some grief. Can someone explain how this works to me.

    Thanks

    Sean
     
  2. jcsd
  3. Jun 28, 2010 #2

    K^2

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    Is the quantity that's equated a vector or a scalar? I would read it like this.

    [tex](\nabla u)_{ij} \equiv \frac{\partial u_i}{\partial x_j}[/tex]

    In which case, the equivalence you show up there with indices in place would look like this.

    [tex]\left(u'_j \frac{\partial}{\partial x_j} \right) u_i \equiv \left(\frac{\partial}{\partial x_k} u_i \right) u'_k[/tex]
     
  4. Jun 28, 2010 #3
    both sides are scalars if that's what you're asking.

    My experience with index notation is kind of limited so i'm not sure exactly what that first equation you've written represents, could you expand it out in terms of Ux, Uy, Uz and dx, dy and dz so i'm certain i'm interpreting that correctly
     
  5. Jun 28, 2010 #4
    Del on it's own like that is the gradient operator (grad).
    It simply means the slope or rate of change with distance in the directions x,y,z. so d/dx, d/dy d/dz (I don't know how to show partials in this editor - sorry)
    Usually it operates on a scalar, (eg electric potential) to give a vector result (in that case, Electric field strength)
     
  6. Jun 28, 2010 #5
    yeah but in this case its acting on a vector not a scalar. In the text book i was looking at the only options for applying the grad operator to a vector field were a dot product or a cross product. The grad U term on the RHS of the equality i posted doesn't appear to be either and when i tried to verify the equality my results suggested that that grad U term should result in the vector (dUx/dx, dUy/dy, dUz/dz) if dotting it with u' is to make it equate to the LHS. That vector isn't the result of a cross product operation and, since its a vector, it obviously isn't the result of a dot product either. so i'm a bit confused as to what operation is being carried out there or if its just a convenient notation that i'm not used to.
     
  7. Jun 28, 2010 #6
    Well, if U is a vector, the result of the grad operation is a a tensor - which seems unlikely in the context.

    OTOH If U is a scalar it makes perfect sense.
    The dot product of u' with del is a vector operator u'x d/dx : u'y d/dy : u'z d/dz,
    acting on U gives (just for x) u'x dU/dx, which is the same as you would get for the second case - the dot product of vector u' with the grad of scalar U (also a vector).
     
  8. Jun 28, 2010 #7
    Assuming your equation is correct, then both sides of the equation are vectors. The LHS is actually much easier to interpret if you are not used to tensor analysis (in fact, a similar expression occurs frequently in fluid dynamics).

    The term in the parentheses on the LHS will give you u'_x d/dx + u'_y d/dy + u'_z d/dz (forgive the lack of Latex). This is simply a scalar operator that acts on each component U; remember, the x-component of U will be differentiated with respect to all three coordinates and so forth for the remaining components of U.

    On the RHS the gradient of a vector is indeed a tensor (rank 2). It's contraction with a vector (i.e. rank 1 tensor) results in another vector.
     
  9. Jun 28, 2010 #8

    K^2

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    That's what I was assuming, but OP says that both sides of the equation are scalars, which doesn't make sense with a tensor.
     
  10. Jun 29, 2010 #9
    ok thanks for all the help i managed to verify it. Turns out as well as not understanding exactly what gradU was i made a mistake in another part of my working so all good now!
     
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