Nabla operations, vector calculus problem

In summary, Nabla operations are mathematical operators used in vector calculus to perform calculations on vector fields. They include the gradient, divergence, and curl operators, which are used to find the direction and rate of change of a scalar field, the flow of a vector field, and the rotation of a vector field, respectively. These operations are essential for solving problems in areas such as electromagnetism, fluid mechanics, and differential geometry. Mastering the use of Nabla operations is crucial for understanding and solving complex vector calculus problems.
  • #1
Karl Karlsson
104
12
Homework Statement
A liquid flows in a cylindrical tube with the z-axis as the axis of symmetry and radius R. The flow rate of the liquid is given by $$ \vec v = \frac {p_0} {4lμ} (R^2-ρ^2)$$
where ##\frac {p_0} {l}## is the pressure drop in the pipe per unit length, ##μ## is the viscosity of the liquid and ##ρ## is the radial cylinder coordinate. Calculate ##∇\cdot\vec v## and ##∇^2\cdot\vec v##
Relevant Equations
$$ \vec v = \frac {p_0} {4lμ} (R^2-ρ^2)$$
Here is how my teacher solved this:
Skärmavbild 2020-08-31 kl. 00.10.27.png

I understand what the nabla operator does, ##∇\cdot\vec v## means that I am supposed to calculate ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## where ##x_n## are cylindrical coordinates and ##\vec e_3 = \vec e_z##. I understand why ##∇\cdot\vec v = 0##, I would get the same answer as my teacher if I used ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## to calculate it. However when I calculate ##∇^2\cdot\vec v## I don't get the same answer. I would get ##∇^2\cdot\vec v = (∇\cdot (∇v_1),∇\cdot (∇v_2),∇\cdot (∇v_3)) = \frac {p_0} {4lμ}\cdot (0,0, ∇\cdot (∇(R^2-ρ^2))) =\frac {p_0} {4lμ}\cdot (0,0, \frac {\partial^2 (R^2-ρ^2)} {\partial ρ^2})##. Which is not equal to what my teacher got. Did I miss something? Also could someone explain the steps he took to arrive at his answer?

Thanks in advance!
 
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  • #2
Karl Karlsson said:
Homework Statement:: A liquid flows in a cylindrical tube with the z-axis as the axis of symmetry and radius R. The flow rate of the liquid is given by $$ \vec v = \frac {p_0} {4lμ} (R^2-ρ^2)$$
where ##\frac {p_0} {l}## is the pressure drop in the pipe per unit length, ##μ## is the viscosity of the liquid and ##ρ## is the radial cylinder coordinate. Calculate ##∇\cdot\vec v## and ##∇^2\cdot\vec v##
Relevant Equations:: $$ \vec v = \frac {p_0} {4lμ} (R^2-ρ^2)$$

Here is how my teacher solved this:
View attachment 268604
I understand what the nabla operator does, ##∇\cdot\vec v## means that I am supposed to calculate ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## where ##x_n## are cylindrical coordinates and ##\vec e_3 = \vec e_z##. I understand why ##∇\cdot\vec v = 0##, I would get the same answer as my teacher if I used ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## to calculate it. However when I calculate ##∇^2\cdot\vec v## I don't get the same answer. I would get ##∇^2\cdot\vec v = (∇\cdot (∇v_1),∇\cdot (∇v_2),∇\cdot (∇v_3))
The last expression above doesn't seem right to me, but I am somewhat rusty with this stuff. The part that bothers me is where you're taking the gradient of the individual components ##v_1, v_2, v_3##.
Karl Karlsson said:
= \frac {p_0} {4lμ}\cdot (0,0, ∇\cdot (∇(R^2-ρ^2))) =\frac {p_0} {4lμ}\cdot (0,0, \frac {\partial^2 (R^2-ρ^2)} {\partial ρ^2})##. Which is not equal to what my teacher got. Did I miss something? Also could someone explain the steps he took to arrive at his answer?

Thanks in advance!
 
  • #3
Karl Karlsson said:
I understand what the nabla operator does, ##∇\cdot\vec v## means that I am supposed to calculate ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## where ##x_n## are cylindrical coordinates and ##\vec e_3 = \vec e_z##.
Not exactly. If you're using cartesian coordinates, then
$$\nabla \cdot \vec v = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} +\frac{\partial v_z}{\partial z}.$$ If you're using cylindrical coordinates, however, the divergence takes on a more complicated form. See, for example, https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates.

I understand why ##∇\cdot\vec v = 0##, I would get the same answer as my teacher if I used ##\sum_{n=1}^3\frac {d\vec v} {dx_n}## to calculate it. However when I calculate ##∇^2\cdot\vec v## I don't get the same answer. I would get ##∇^2\cdot\vec v = (∇\cdot (∇v_1),∇\cdot (∇v_2),∇\cdot (∇v_3)) = \frac {p_0} {4lμ}\cdot (0,0, ∇\cdot (∇(R^2-ρ^2))) =\frac {p_0} {4lμ}\cdot (0,0, \frac {\partial^2 (R^2-ρ^2)} {\partial ρ^2})##. Which is not equal to what my teacher got. Did I miss something? Also could someone explain the steps he took to arrive at his answer?
I think if you visit that Wikipedia page and see what the Laplacian is in cylindrical coordinates, it'll answer your question. Your notation is a bit off too. Note that your professor wrote ##\nabla^2 \vec v##, not ##\nabla^2 \cdot \vec v##.
 
  • #4
This is another consquence of [itex]\mathbf{e}_\rho[/itex] and [itex]\mathbf{e}_\phi[/itex] not being constant.

In this case it's not too difficult to calculate [itex]\nabla \cdot \mathbf{v}[/itex] and [itex]\nabla^2 \mathbf{v}[/itex] in cartesian coordinates using [tex]
\nabla (\rho^2) = 2x\mathbf{e}_x + 2y\mathbf{e}_y[/tex] and [tex]
\nabla \cdot (f\mathbf{g}) = f\nabla \cdot \mathbf{g} + \mathbf{g} \cdot \nabla f.
[/tex]
 

Related to Nabla operations, vector calculus problem

1. What are Nabla operations?

Nabla operations, also known as vector calculus operations, are mathematical operations used to manipulate vector fields. They involve the use of the symbol ∇ (pronounced "del") and include operations such as gradient, divergence, and curl.

2. What is the purpose of using Nabla operations?

Nabla operations are used to describe and analyze vector fields, which are quantities that have both magnitude and direction. These operations help us understand the behavior and properties of vector fields in various physical and mathematical contexts.

3. How do you perform a gradient operation using Nabla?

The gradient operation using Nabla is represented by ∇f, where f is a scalar function. To perform the operation, you take the partial derivatives of the function with respect to each variable and combine them using the ∇ symbol.

4. What does the divergence operation tell us about a vector field?

The divergence operation using Nabla is represented by ∇⋅F, where F is a vector field. It tells us the rate at which the vector field is spreading out or converging at a given point. A positive divergence indicates a spreading out, while a negative divergence indicates a convergence.

5. How is the curl operation related to rotation?

The curl operation using Nabla is represented by ∇×F, where F is a vector field. It tells us the amount of rotation or circulation of the vector field at a given point. A non-zero curl indicates the presence of rotation in the vector field.

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