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

[crocomut]

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 - whats up with latex not working?

[MisterX]

For a solenoidal velocity field \nabla \cdot \mathbf{u} which means that \nabla is perpendicular to \mathbf{u} .

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

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

PS - whats 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 \mathbf{u} , \nabla \cdot \mathbf{u} is the divergence of u.

\nabla is known as del (http://mathworld.wolfram.com/Del.html), and not "a gradient". It may be thought of like a vector of differential operators.

[crocomut]

Sorry, let me correct and ask again:


For a solenoidal velocity field \nabla \cdot \mathbf{u} = 0 which means that \nabla is perpendicular to \mathbf{u} .

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

So what exactly does it mean physically to have \nabla parallel/perpendicular to the velocity vector?

[MisterX]

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.

[crocomut]

Your answer is exactly what I was thinking but, as you can see from this slide (http://i.imgur.com/VmbKS.jpg)in my hydrodynamics lecture, it is not what my professor claims. Hence the confusion.