# Taking the derivative of a vector along a surface

Imagine that you have a surface x(p,q,t) where at any given point $$\mathbf e_p$$ and $$\mathbf e_q$$ are unit vectors along the surface and n is the unit vector normal to the surface, such that $$\mathbf n= \mathbf e_p \times \mathbf e_q$$. The unit vector, n, points in the direction that the surface is propagating. This surface is immersed in a velocity field, v.

I would like to take the tangential divergence of this vector field along the surface, and a (physics) book that I'm reading informs me that I should do this:

$$\nabla_t \cdot \mathbf v = \frac{\partial (\mathbf e_p \cdot \mathbf v)}{\partial p} + \frac{\partial (\mathbf e_q \cdot \mathbf v)}{\partial q} \\ = \left(\mathbf e_p \cdot \frac{\partial \mathbf v}{\partial p} + \mathbf e_q \cdot \frac{\partial \mathbf v}{\partial q}\right)+ \mathbf v \cdot \left(\frac{\partial \mathbf e_p}{\partial p} + \frac{\partial \mathbf e_q}{\partial q}\right)$$

So, there is one term that describes how the velocity changes along the surface, and another term that describes how the unit vectors change along the surface.

First, I'd like the argue that the dot product could be computed before the differentiation, so this could also be written

$$\nabla_t \cdot \mathbf v = \frac{\partial v_p}{\partial p} + \frac{\partial v_q}{\partial q}$$

Now, let's say that I want to find $$\nabla_t \cdot \mathbf v_t$$, where $$\mathbf v_t$$ is the velocity tangent to the surface. Then

$$\nabla_t \cdot \mathbf v_t = \frac{\partial (\mathbf e_p \cdot \mathbf v)}{\partial p} + \frac{\partial (\mathbf e_q \cdot \mathbf v)}{\partial q} \\ = \frac{\partial (\mathbf e_p \cdot (v_p \mathbf e_p +v_q \mathbf e_q + v_n \mathbf n))}{\partial p} + \frac{\partial (\mathbf e_q \cdot (v_p \mathbf e_p +v_q \mathbf e_q + v_n \mathbf n))}{\partial q} \\ = \frac{\partial v_p}{\partial p} + \frac{\partial v_q}{\partial q}$$

which implies that $$\nabla_t \cdot \mathbf v = \nabla_t \cdot \mathbf v_t$$.

But I know that this is not true because if $$\mathbf v = \mathbf v_n + \mathbf v_t$$ then

$$\nabla_t \cdot \mathbf v = \nabla_t \cdot \mathbf v_t + \nabla_t \cdot \mathbf v_n \\ = \nabla_t \cdot \mathbf v_t + \nabla_t \cdot \mathbf (\mathbf v \cdot \mathbf n)\mathbf n \\ = \nabla_t \cdot \mathbf v_t + (\mathbf v \cdot \mathbf n) (\nabla_t \cdot \mathbf n) + \mathbf n \cdot \nabla_t (\mathbf v \cdot \mathbf n)$$

where I argue that the last term is zero so

$$\nabla_t \cdot \mathbf v = \nabla_t \cdot \mathbf v_t + (\mathbf v \cdot \mathbf n) (\nabla_t \cdot \mathbf n)$$

How can I resolve this contradiction? I'm doing something wrong somewhere!

Edit: Fixed notation as suggested by answer.

• $\nabla$ is not a “real” vector and the operator denoted by $\nabla\cdot$ is not a “real” dot product. Be very careful when trying to apply the things you think you know about dot product to this.
– amd
Aug 29 '19 at 23:08
• @amd, yes this is something I have definitely been guilty of before! I don't see where I'm doing in here though, although maybe the use of the identity $\nabla \cdot (f \mathbf A) = f(\nabla \cdot \mathbf A) + \mathbf A \cdot (\nabla f)$ isn't appropriate b/c it applies to the full divergence, not just the tangential part? I feel like that must be the line you are talking about b/c it is the only place I do anything with the $\nabla_t$. Is that what you are refering to?
– EPH
Aug 30 '19 at 1:09
• If $\nabla_t \; \cdot$ means to first project the vector field onto the tangential components of the surface (that is, you create a new vector field with two components)...and then take derivatives of that new vector field with respect to the parameters of the parameterized surface...then $\nabla_t \cdot \vec{n} = 0$. First you make the new vector field, $(\vec{e}_p \cdot \vec{n})\vec{e}_p + (\vec{e}_q \cdot \vec{n})\vec{e}_q$....and then take derivatives. But this new field is 0. In my opinion, the two results should be the same value Aug 31 '19 at 17:31
• Given $\vec{v}_t = (\vec{e}_p \cdot \vec{v})\vec{e}_p + (\vec{e}_q \cdot \vec{v})\vec{e}_q$..first you project to create the new field: $(\vec{e}_p \cdot \vec{v}_t)\vec{e}_p + (\vec{e}_q \cdot \vec{v}_t)\vec{e}_q$, which when you do the parentheses, turns into the exact same function again $\vec{v}_t$ that we started with Aug 31 '19 at 17:35
• Hey @DWade64! Thank you for considering this question. I don't think that the $\nabla_t \cdot$ means just to project the vector field and then take the derivative, though. The $\nabla_t \cdot \mathbf n$ terms ends up being the curvature term which is important to the physics application that this math applies to (curved flames) so I don't think that it is always zero. There's some issue here though, and I still haven't figured it out!
– EPH
Sep 2 '19 at 21:42

First, be careful with your notation. When you did the dot product before the differentiation, you kept the components bold even though they shouldn't be because they are not vectors. As a sanity check divergence must give you a scalar, but those bold quantities are not scalars.

Second, your decomposition of $$\mathbf{v}$$ is not complete. It should read $$\mathbf{v} = v_p\mathbf{e}_p + v_q\mathbf{e}_q + v_n\mathbf{n}$$ When you compute $$\nabla_t \cdot \mathbf{v}_n$$, what are $$\mathbf{v}_n$$'s $$p$$ and $$q$$ components?

• Thanks, I fixed the notation! That was a copy-paste/sloppiness issue. I had assumed that v_n’s p and q components were zero since v_n is the velocity in the normal direction. Is that wrong?
– EPH
Aug 29 '19 at 22:36
• No, that is correct since $\mathbf{e}_p \times \mathbf{e}_q = \mathbf{n}$ the normal component is perpendicular to both $p$ and $q$. So what does that mean $\nabla_t \mathbf{v}_n$ is? Aug 29 '19 at 22:54
• I think it is zero, which is why I dropped that term. Is that wrong?
– EPH
Aug 30 '19 at 0:51
• @EPH Don't just think, prove! What happens when you plug it in with the hint I provided? Aug 30 '19 at 0:53
• I agree that the components are zero, but the normal vector could still change in the tangent direction, which is the term I kept. Are you suggesting that the curvature term is zero? I'm sorry that I'm not getting this faster, but I've got a major block here.
– EPH
Aug 30 '19 at 1:00