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 - what's 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 - 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 $\mathbf{u}$, $\nabla \cdot \mathbf{u}$ is the divergence of u.

$\nabla$ is known as 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 http://i.imgur.com/VmbKS.jpg"in my hydrodynamics lecture, it is not what my professor claims. Hence the confusion.