Material Derivative (Convective Derivative Operator)

[thehappypenguin]

Hi,

I've learned that material derivative is equal to local derivative + convective derivative, but can't seem to find out which way the convective derivative acts, like for example in velocity fields:

The equation my teacher gave us was (with a and v all/both vectors):
Acceleration = material derivative of velocity = local acceleration + convective acceleration
∴ a = Dv/Dt = dv/dt + v⋅∇v

My question is whether the convective acceleration term (v⋅∇v) works like:
1. (v⋅∇)v, which in my understanding is the (v⋅∇) operator working on the vector v
2. v⋅(∇v), which I take as the grad operator working on the vector v, dotted with the vector v outside the brackets
3. Or is it that Options 1 and 2 are the same thing anyway?

[Side note: Sorry, I'm new to PF and don't know how to use the equation symbols or LaTeX.]

Thank you in advance for your help! :)
- TheHappyPenguin

 

[member 428835]

Hi thehappypeguin!

1) and 2) are equivalent expressions, so technically 3) is correct. I'm sure someone else here can easily send you to a link where you can figure out latex, if that's what you're trying to figure out.

 

[dextercioby]

Even if you can't write in LaTex, you can still use a writing to distinguish vectors from scalars. So write v for the velocity vector and v for its modulus (or projection onto a coordinate axis). :)

 

[thehappypenguin]

Thank you! I'll look up latex when I have time :)

About the Material Derivative, I read some more about it online and found that in Pgs 1-3 of http://www.chem.mtu.edu/~fmorriso/cm4650/lecture_6.pdf, ∇v is a tensor which would make 2) different from 1) as 1) is a vector. Is this accurate?

Thank you again!
- TheHappyPenguin :)

 

[stevendaryl]

Thank you! I'll look up latex when I have time :)

About the Material Derivative, I read some more about it online and found that in Pgs 1-3 of http://www.chem.mtu.edu/~fmorriso/cm4650/lecture_6.pdf, ∇v is a tensor which would make 2) different from 1) as 1) is a vector. Is this accurate?

Thank you again!
- TheHappyPenguin :)

Yes, $\nabla \vec{v}$ is a tensor, but no, the two expressions are not different. The dot product of a vector with a tensor (of the right type) produces a vector. So both

$(\vec{v} \cdot \nabla) \vec{v}$ and $\vec{v} \cdot (\nabla \vec{v})$ produce vectors.