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Vector cross product with curl

  • Thread starter hotvette
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  • #1
hotvette
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Homework Statement


Using index-comma notation only, show:
\begin{equation*}
\underline{\bf{v}} \times \text{curl } \underline{\bf{v}}= \frac{1}{2} \text{ grad}(\underline{\bf{v}} \cdot \underline{\bf{v}}) - (\text{grad } \underline{\bf{v}}) \underline{\bf{v}}
\end{equation*}

Homework Equations


\begin{align*}
\text{ curl } \underline{\bf{v}} &= \epsilon_{ijk} v_{j,i} \underline{\bf{e}}_k \\
\underline{\bf{v}} \times \underline{\bf{u}} &= \epsilon_{ijk} v_i u_j \underline{\bf{e}}_k
\end{align*}

The Attempt at a Solution


If I let [itex] \underline{\bf{u}} =\text{ curl } \underline{\bf{v}}[/itex] we get:
\begin{align*}
\underline{\bf{u}} &= \text{ curl } \underline{\bf{v}} =\epsilon_{prq} v_{r,p} \underline{\bf{e}}_q \\
u_j &= \epsilon_{prq} v_{r,p} \underline{\bf{e}}_q \cdot \underline{\bf{e}}_j \\
u_j &= \epsilon_{prq} v_{r,p} \delta_{qj} = \epsilon_{prj} v_{r,p} \\
\underline{\bf{v}} \times \underline{\bf{u}} &= \epsilon_{ijk} v_i u_j \underline{\bf{e}}_k \\
&= \epsilon_{ijk} \epsilon_{prj} v_i v_{r,p} \underline{\bf{e}}_k
\end{align*}
Is this correct so far? Trouble is, I'm not sure what to do next. I'm wondering if trying to simplify [itex]\epsilon_{ijk} \epsilon_{prj}[/itex] would be a fruitful approach. Appreciate hints/tips. Thanks!
 
Last edited:

Answers and Replies

  • #2
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Is this correct so far? Trouble is, I'm not sure what to do next. I'm wondering if trying to simplify [itex]\epsilon_{ijk} \epsilon_{prj}[/itex] would be a fruitful approach.
Looks correct to me. Yup, that approach is the way to go. This is a well-known identity,
[tex]\epsilon_{ijk} \epsilon_{\ell mk} = \delta_{i\ell} \delta_{jm} - \delta_{im} \delta_{j \ell}[/tex]
for which there are several different ways to prove.
 
  • #3
hotvette
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OK, I"m getting closer but (maybe) ran into another snag. Using the following:
\begin{equation*}
\epsilon_{ijk} = \epsilon_{jki} = \epsilon_{kij}
\end{equation*}
we get:
\begin{align*}
\epsilon_{ijk} \epsilon_{prj} &= \epsilon_{kij} \epsilon_{prj} = \delta_{ip} \delta_{jr} - \delta_{ir} \delta_{jp} \\
\underline{\bf{v}} \times \text{curl }\underline{\bf{v}} &=
(\delta_{ip} \delta_{jr} - \delta_{ir} \delta_{jp})(v_i v_{r,p}) \underline{\bf{e}}_k \\
&= v_i v_{r,p} \delta_{ip} \delta_{jr} \underline{\bf{e}}_k - v_i v_{r,p} \delta_{ir} \delta_{jp}\underline{\bf{e}}_k \\
\end{align*}
Looking at the LHS and comparing with the problem statement, we have:
\begin{align*}
v_i v_{r,p} \delta_{ip} \delta_{jr} \underline{\bf{e}}_k &= v_p v_{r,p} \delta_{jr} \underline{\bf{e}}_k \\
\frac{1}{2} \text{grad}(\underline{\bf{v}} \cdot \underline{\bf{v}}) &= \frac{1}{2} (v_i v_i)_{,j} \underline{\bf{e}}_j
= \frac{1}{2} ({v_i}^2)_{,j} \underline{\bf{e}}_j = v_j \underline{\bf{e}}_j
\end{align*}
Which means that [itex]v_p v_{r,p} \delta_{jr} \underline{\bf{e}}_k[/itex] needs to reduce to [itex]v_j \underline{\bf{e}}_j[/itex].
Providing the Kronecker delta can alter partial derivative indices(which I'm not sure), we get:
\begin{align*}
v_p v_{r,p} \delta_{jr} \underline{\bf{e}}_k &= v_p v_{j,p} \underline{\bf{e}}_k \\
&= v_j \underline{\bf{e}}_j \text{ ??}
\end{align*}
It isn't clear to me why the last step is valid. It looks goofy to me because [itex]v_{j,p} = \frac{\partial v_j}{\partial x_p}[/itex] which I think is a matrix of partial derivatives. Appreciate clarification/comment on that one. Now, to tackle the RHS...
 
  • #4
vela
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Your claim that ##\frac 12(v_i^2)_{,j} = v_j## isn't correct. If you correct that, you'll see you're essentially done.
 
  • #5
Charles Link
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Maybe it's the same thing, but I think the last term should read ## v \cdot \nabla v ##. (This identity can be found on the cover of J.D. Jackson's Classical Electrodynamics in the form ## \nabla (a \cdot b) =a \cdot \nabla b +b \cdot \nabla a + a \times \nabla \times b + b\times \nabla \times a ##.)
 
  • #6
hotvette
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Hmmm, is:
\begin{equation*}
\tfrac{1}{2} ({v_i}^2)_{,j} = v_i v_{i,j} \text{ ??}
\end{equation*}
If so, then I get:
\begin{align*}
v_p v_{j,p} \underline{\bf{e}}_k &= v_i v_{i,j} \underline{\bf{e}}_j && \text(a) \\
&= v_p v_{p,j} \underline{\bf{e}}_j &&\text{(b) for free index, ok to change}\\
&= v_p v_{j,p} \underline{\bf{e}}_j &&\text{(c) ok to switch indices on partial?}
\end{align*}
But then the basis vectors don't match :(
 
  • #7
vela
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We get:
\begin{align*}
\epsilon_{ijk} \epsilon_{prj} &= \epsilon_{kij} \epsilon_{prj} = \delta_{ip} \delta_{jr} - \delta_{ir} \delta_{jp}
\end{align*}
The free indices don't match. You're summing over ##j##, so it shouldn't appear on the righthand side.
 
  • #8
hotvette
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Ah, thanks for pointing out the error! OK, I now get:
\begin{equation*}
\epsilon_{kij} \epsilon_{prj} = \delta_{kp} \delta_{ir} - \delta_{kr} \delta_{ip}
\end{equation*}
which gives:
\begin{align*}
\underline{\bf{v}} \times \text{curl }\underline{\bf{v}} &= (\delta_{kp} \delta_{ir} - \delta_{kr} \delta_{ip}) v_i v_{r,p} \underline{\bf{e}}_k \\
&= \delta_{kp} \delta_{ir} v_i v_{r,p} \underline{\bf{e}}_k - \delta_{kr} \delta_{ip} v_i v_{r,p} \underline{\bf{e}}_k \\
&= v_r v_{r,k} \, \underline{\bf{e}}_k - v_p v_{k,p} \, \underline{\bf{e}}_k
\end{align*}
For the LHS I now get a match but I still can't get the RHS to match:
\begin{align*}
&\underline{\bf{u}} = \text{grad } \underline{\bf{v}} = v_{i,j} \, \underline{\bf{e}}_i \otimes \underline{\bf{e}}_j && \text{(a)}\\
&\underline{\bf{u}} \underline{\bf{v}} = u_r v_p \, \underline{\bf{e}}_r \otimes \underline{\bf{e}}_p && \text{(b)}\\
&u_r = v_{i,r} \, \underline{\bf{e}}_i && \text{(c)}\\
& (\text{grad } \underline{\bf{v}}) \underline{\bf{v}} = (v_{i,r} \, \underline{\bf{e}}_i) v_p \, \underline{\bf{e}}_r \otimes \underline{\bf{e}}_p && \text{(d)} \\
&= v_p v_{i,r} \, (\underline{\bf{e}}_r \otimes \underline{\bf{e}}_p)\underline{\bf{e}}_i && \text{(e)} \\
&= v_p v_{p,r} \, \underline{\bf{e}}_r = v_p v_{p,k} \, \underline{\bf{e}}_k \ne v_p v_{k,p} \, \underline{\bf{e}}_k && \text{(f)}
\end{align*}
What am I doing wrong?
 
Last edited:
  • #9
vela
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Ah, thanks for pointing out the error! OK, I now get:
\begin{equation*}
\epsilon_{kij} \epsilon_{prj} = \delta_{kp} \delta_{ir} - \delta_{kr} \delta_{ip}
\end{equation*}
which gives:
\begin{align*}
\underline{\bf{v}} \times \text{curl }\underline{\bf{v}} &= (\delta_{kp} \delta_{ir} - \delta_{kr} \delta_{ip}) v_i v_{r,p} \underline{\bf{e}}_k \\
&= \delta_{kp} \delta_{ir} v_i v_{r,p} \underline{\bf{e}}_k - \delta_{kr} \delta_{ip} v_i v_{r,p} \underline{\bf{e}}_k \\
&= v_r v_{r,k} \, \underline{\bf{e}}_k - v_p v_{k,p} \, \underline{\bf{e}}_k
\end{align*}
Using what you said in post #6, the first term is ##\frac 12 (\vec{v}\cdot \vec{v})_{,k}\hat{e}_k=\frac 12 \nabla(\vec{v}\cdot \vec{v})##. The second term is ##(\nabla \cdot \vec{v})\vec{v}##. So you're essentially done. (As Charles pointed out, the last term in the original post is wrong.)
 
  • #10
Charles Link
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I don't know this new notation very well= I write out all the partial derivative terms when I do a proof like this, but one suggestion would be to consider my post #5. The term should be ## v \cdot \nabla v ##, and not ## (\nabla v)v ## (or (grad v)v as you wrote it in the OP.)
 
  • #11
hotvette
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Ah, I understand now (I didn't know how to interpret post #5). I can now see that it works out if the last term is [itex] \underline{\bf{v}} \cdot \text{ grad} \, \underline{\bf{v}}[/itex]. I'll check with my professor. Thanks!
 

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