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Evaluate $\mathbf{u}\cdot\nabla\mathbf{u}$ (the directional derivative of $\mathbf{u}$ in the direction of $\mathbf{u}$)in cylindrical coordinates $(r, \phi,z)$, where $\bf{u}=e_{\phi}$.

The textbook that I am reading uses a purely vectorial approach and $$\mathbf{u}\cdot\nabla\mathbf{u}=-\frac{\mathbf{e_{\phi}}}{r} \tag{1}$$


I tried the index notation method as an alternative: $$[\mathbf{u}\cdot\nabla\mathbf{u}]_k=u^{n}\nabla_{n}u_{k}=g^{nr}u_{r}\nabla_{n}u_k$$$$=g^{nr}u_r(\partial_{n}u_{k}-\Gamma^{p}_{kn}u_p)= g^{nr}u_r\partial_n u^k- u_r u_p g^{nr} g^{pq} \Gamma_{kn,q}\tag{2}$$ $n,r,k,p=1,2,3$ and $\nabla$ is the covariant derivative.

Since $\mathbf{e_{\phi}}=\mathbf(0,1,0)$ in the cylindrical basis, the first term in the last equality of $(2)$ vanishes (i.e. $\partial_nu^k=0$), then $$[\mathbf{u}\cdot\nabla\mathbf{u}]_{k}=-g^{nr}u_r(\Gamma^p_{kn}u_p)=-u_2u_2g^{22}g^{22}\Gamma_{k2,2}=-(1)^2\frac{1}{(g_{22})^2}\Gamma_{k2,2}\tag{3}$$

Since $$\Gamma_{km,n}=\frac{1}{2}\left(g_{kn,m}+g_{mn,k}-g_{km,n} \right) \tag{4}$$

$(g_{kn,m}=\partial_{m} g_{kn})$

and $$g_{11}=g_{33}=1,g_{22}=r^2 \tag{5}$$ it follows that $$\Gamma_{12,2}=\frac{1}{2}(g_{22,1})=r, \Gamma_{2,22}=\Gamma_{3,22}=0 \tag{6}$$ so only $[\mathbf{u}\cdot\nabla\mathbf{u}]_1$ (the $r$ component) is non-zero $(6)$$\to$ $(3)$ gives $$[\mathbf{u}\cdot\nabla\mathbf{u}]=-\frac{1}{r^3} \mathbf{e_{r}} \tag{7}$$ which is clearly wrong. Can someone please explain where my conceptual errors lie?

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    $\begingroup$ I think part of the problem is how you are interpreting the gradient in index form. The first answer here might help: link. $\endgroup$
    – secavara
    Jun 13, 2020 at 11:07
  • $\begingroup$ So what is your physics question? $\endgroup$ Jun 13, 2020 at 11:52
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    $\begingroup$ @BioPhysicist This is GR related $\endgroup$ Jun 13, 2020 at 12:41
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    $\begingroup$ Right, but your question isn't about any physics concepts. Solving the equation $5=4.9t^2$ for $t$ is related to projectile motion, but if I were to ask about solving the equation then I would be asking a math question. $\endgroup$ Jun 13, 2020 at 13:53
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    $\begingroup$ @BioPhysicist at least at an advanced level, the boundary between mathematics and physics is more blurred, than, say in Newtonian mechanics. $\endgroup$ Jun 13, 2020 at 14:57

1 Answer 1

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There is a mistake in the first line - for $\vec{u}=\mathbf{e}_\phi$ should be (see ex. Wikipedia) $$\vec{u}\cdot\nabla \vec{u} = -\frac{\mathbf{e}_r}{r}$$

The difference in the two approaches comes from the definitions of the basis vectors, that is, whether you are using a coordinate or a non-coordinate basis.

In a coordinate basis, the basis of your vector space (basis of the tangent space) are partial derivatives w.r.t the corresponding coordinates $$(\partial_r, \partial_\phi, \partial_z)$$ and this is the basis used in the "index calculus" employed in the post.

While a non-coordinate basis often used in eg. classical electrodynamics is given by unit orthonormal vectors. $$(\mathbf{e}_r, \mathbf{e}_\phi, \mathbf{e}_z)$$

The difference between the two bases is in the length of the basis vectors. The second basis, being orthonormal, has vectors of unit length over each point in your space, while this is not true for one of the basis vectors of the coordinate basis, namely, the $\partial_\phi$ basis vector of the coordinate basis.

To see the exact difference, examine the following. The Cartesian basis $$(\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z)$$ is the same as the coordinate basis $$(\partial_x, \partial_y, \partial_z)$$

And we can connect $\mathbf{e}_\phi$ to the Cartesian basis $$\mathbf{e}_\phi = -\sin \phi \mathbf{e}_x + \cos\phi \mathbf{e}_y = -\sin \phi \partial_x + \cos\phi \partial_ y$$

By the chain rule, used to change coordinates, we have $$\mathbf{e}_\phi = -\sin \phi \frac{\partial \phi}{\partial x}\partial_\phi + \cos\phi \frac{\partial \phi}{\partial y}\partial_\phi$$

Using the connection between $\phi$ and cartesian coordiantes (assuming the first quandrant to simplify calculations) $\phi = \arctan\left(\frac{y}{x}\right)$ We get $$\frac{\partial \phi}{\partial x} = -\frac{y}{r^2} = -\frac{\sin\phi}{r},\quad \frac{\partial \phi}{\partial x} = \frac{x}{r^2} = \frac{\cos\phi}{r}$$

Which gives $$\mathbf{e}_\phi = \frac{1}{r}\partial_\phi$$ while $\mathbf{e}_r = \partial_r, \mathbf{e}_z = \partial_z$

Back to the example above

$$\vec{u} = \mathbf{e}_\phi = \frac{1}{r}\partial_\phi$$ which means that in a coordinate basis, of the contravariant componets $u^k$ only $u^\phi = \frac{1}{r}$ is non-zero. Since the only two non-vanishing Christoffel symbols are $$\Gamma^r_{\phi\phi} = -r, \Gamma^\phi_{\phi r} = \frac{1}{r}$$ The directed derivative, as a full vector is then $$(u^n\nabla_n u^k)\partial_k = \frac{1}{r}(\partial_\phi u^k + \Gamma^k_{\phi\phi}u^\phi) = - \frac{1}{r}\partial_r = -\frac{\mathbf{e}_r}{r}$$

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