Understanding Del Operation on Vector U Without Dot or Cross Product

In summary, the equation u'\frac{\partial}{\partial x_j} = (\nabla u)\frac{\partial}{\partial x_k} is confusing Sean because it is unclear what the operation is. The equation can be simplified to u'\frac{\partial}{\partial x_j} = u'_j \frac{\partial}{\partial x_k} if U is a vector, or u'\frac{\partial}{\partial x_j} = u'_j \frac{\partial}{\partial x_x} if U is a scalar.
  • #1
seanl
4
0
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
 
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  • #2
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]
 
  • #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
 
  • #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)
 
  • #5
yeah but in this case its acting on a vector not a scalar. In the textbook 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.
 
  • #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).
 
  • #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.
 
  • #8
AJ Bentley said:
Well, if U is a vector, the result of the grad operation is a a tensor
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.
 
  • #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!
 

1. What is the meaning of del operator on vector U without dot or cross product?

The del operator, also known as the nabla symbol (∇), is a mathematical operator used in vector calculus to represent the gradient of a scalar field. When applied to a vector, it represents the divergence of that vector field.

2. How does the del operator affect vector U without using dot or cross product?

The del operator does not directly affect the vector U without the use of dot or cross product. However, when applied to a scalar function, it produces a vector field that is perpendicular to the isosurfaces of the function and points in the direction of increasing values of the function.

3. Can the del operator on vector U be simplified without the use of dot or cross product?

No, the del operator cannot be simplified without the use of dot or cross product. In vector calculus, the dot and cross product are essential tools for simplifying vector operations, and the del operator relies on these operations to produce meaningful results.

4. What is the significance of understanding the del operator on vector U without dot or cross product?

Understanding the del operator on vector U without dot or cross product is crucial in many areas of physics and engineering. It is used to solve problems involving vector fields, such as fluid flow, electromagnetism, and heat transfer. It also plays a role in understanding the behavior of functions and their derivatives in higher dimensions.

5. Are there any real-world applications of the del operator on vector U without dot or cross product?

Yes, there are many real-world applications of the del operator on vector U without dot or cross product. Some examples include predicting weather patterns, designing aerodynamic shapes, and optimizing heat transfer in industrial processes. The del operator is also used in computer graphics to model vector fields and simulate realistic fluid behavior.

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