What does it mean that the gradient is perpendicular/paralell to a vector?

In summary, for a solenoidal velocity field, the divergence of the velocity is zero, meaning that the del operator is perpendicular to the velocity vector. Similarly, for an irrotational velocity field, the curl of the velocity is zero, indicating that the del operator is parallel to the velocity vector. This means that the del operator can be thought of as a vector of differential operators. However, in some cases, it may be mistakenly referred to as a gradient, which can cause confusion.
  • #1
crocomut
17
0
For a solenoidal velocity field [ tex ] \nabla \cdot \mathbf{u} [ /tex ] which means that [ tex ] \nabla [/tex ] is perpendicular to [ tex ] \mathbf{u} [ /tex ].

Similarly, for an irrotational velocity field [ tex ] \nabla \times \mathbf{u} [ /tex ] which means that [ tex ] \nabla [/tex ] is parallel to [ tex ] \mathbf{u} [ /tex ].

So what exactly does it mean physically to have a gradient (of nothing) parallel/perpendicular to a vector?

PS - what's up with latex not working?
 
Last edited:
Physics news on Phys.org
  • #2
crocomut said:
For a solenoidal velocity field [tex] \nabla \cdot \mathbf{u} [/tex] which means that [tex] \nabla [/tex] is perpendicular to [tex] \mathbf{u} [/tex].

Similarly, for an irrotational velocity field [tex] \nabla \times \mathbf{u} [/tex] which means that [tex] \nabla [/tex] is parallel to [tex] \mathbf{u} [/tex].

So what exactly does it mean physically to have a gradient (of nothing) parallel/perpendicular to a vector?

PS - what's up with latex not working?

Don't put spaces in square brackets for tags. I fixed this issue in the quote above.

What you've posted doesn't make much sense. There seems to be many problems with it. For a solenoidal velocity field [itex] \mathbf{u} [/itex], [itex] \nabla \cdot \mathbf{u} [/itex] is the divergence of u.

[itex] \nabla [/itex] is known as http://mathworld.wolfram.com/Del.html" , and not "a gradient". It may be thought of like a vector of differential operators.
 
Last edited by a moderator:
  • #3


Sorry, let me correct and ask again:For a solenoidal velocity field [tex] \nabla \cdot \mathbf{u} = 0[/tex] which means that [tex] \nabla [/tex] is perpendicular to [tex] \mathbf{u} [/tex].

Similarly, for an irrotational velocity field [tex] \nabla \times \mathbf{u} = \mathbf{0}[/tex] which means that [tex] \nabla [/tex] is parallel to [tex] \mathbf{u} [/tex].

So what exactly does it mean physically to have [tex] \nabla[/tex] parallel/perpendicular to the velocity vector?
 
  • #4
Del may be thought of as a vector of operators. Claiming that del is perpendicular to a vector in ℝ3 makes no sense, like assigning a real number value to the plus sign.
 
  • #5
Your answer is exactly what I was thinking but, as you can see from http://i.imgur.com/VmbKS.jpg"in my hydrodynamics lecture, it is not what my professor claims. Hence the confusion.
 
Last edited by a moderator:

Related to What does it mean that the gradient is perpendicular/paralell to a vector?

1. What is a gradient?

A gradient is a mathematical concept that represents the rate of change of a function with respect to its variables. It is typically represented as a vector, with each component representing the partial derivative of the function with respect to a specific variable.

2. How is the gradient related to a vector?

The gradient is a vector itself, with each component representing the direction and magnitude of the function's rate of change with respect to a specific variable. This means that the gradient can be thought of as the direction in which the function is changing the fastest.

3. What does it mean for the gradient to be perpendicular to a vector?

If the gradient is perpendicular to a vector, it means that the two vectors are at a 90 degree angle to each other. This indicates that the rate of change of the function is changing in a direction that is orthogonal to the direction of the vector.

4. How is the gradient perpendicular to level curves?

Level curves are curves on a surface that represent points with the same value of the function. The gradient is perpendicular to these curves because it always points in the direction of greatest increase of the function. This means that it is always perpendicular to the level curves which represent points with equal values of the function.

5. Is it possible for the gradient to be parallel to a vector?

Yes, it is possible for the gradient to be parallel to a vector. This would mean that the two vectors are pointing in the same direction, indicating that the rate of change of the function is changing in the same direction as the vector. In this case, the gradient is not perpendicular to the vector, but they are still closely related in terms of direction and magnitude of change.

Similar threads

I Gradient With Respect to a Set of Coordinates
    • Calculus
Replies
3
Views
1K
A The ``kinematic equation'' of fluid flows
    • Calculus
Replies
13
Views
1K
I What's the physical meaning of Curl of Curl of a Vector Field?
    • Classical Physics
Replies
5
Views
487
I V * grad(V) = grad(V^2/2) - rotor(omega)
    • Differential Equations
Replies
6
Views
379
Integration by parts, changing vector to moment & divergence
    • Calculus
Replies
2
Views
2K
I Why does the curl of a vector field converge?
    • Calculus
Replies
3
Views
1K
I Vector field and Helmholtz Theorem
    • Calculus
Replies
20
Views
3K
Foucault Pendulum at the North Pole
    • Introductory Physics Homework Help
Replies
9
Views
772
A Gradient of higher rank tensor
    • Classical Physics
Replies
30
Views
2K
Showing that the gradient of a scalar field is a covariant vector
    • Advanced Physics Homework Help
Replies
5
Views
2K

Hot Threads

Recent Insights

Back
Top