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Given that $\vec{a}$ is a constant vector and $\vec{r}$ is a position vector.

We are asked to prove the following: $$\nabla\times(\vec{a}\times\vec{r})=2\vec{a}$$ I tried two ways. Could prove it from one method but from another I am getting another result.

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1st method(with correct result):

$$LHS=\nabla\times(\vec{a}\times\vec{r})$$ Let $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \,\,$ and $\,\,\vec{r}=x\hat{i}+y\hat{j}+z\hat{k},$ where $a_1,a_2,a_3$ are constants and $x,y,z$ is any arbitrary point.

$$Simplifying\,\,\,\, (\vec{a}\times\vec{r})$$

$$=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ x & y & z\end{vmatrix}$$ $$=\hat{i}(a_2z-a_3y)+\hat{j}(a_3x-a_1z)+\hat{k}(a_1y-a_2x)$$

Now performing the curl operation on the simplified result we obtained. $$=\nabla\times(simplified\,result)$$ $$=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} &\frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ a_2z-a_3y & a_3x-a_1z & a_1y-a_2x\end{vmatrix}$$

$$=2a_1x\hat{i}+2a_2y\hat{j}+2a_3z\hat{k}$$ $$=2\vec{a}$$ $$=RHS$$ $$\rule{20cm}{0.4pt}$$ 2nd method with incorrect result.

We know that $$\nabla\times(\vec{a}\times\vec{r})=2[(\nabla.\vec{r})\vec{a}-(\nabla.\vec{a})\vec{r}]$$

Solving RHS,

$$Also,\nabla.\vec{r}=3\,and\, \nabla.\vec{a}=0$$ $$=2[3\vec{a}-0\vec{r}]$$ $$=6\vec{a}$$ which is incorrect. Where did I make mistake in this method?

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The "vector calculus identity" you use in your second calculation is just false. You actually have \begin{align*} \nabla \times (\vec{a} \times \vec{r}) &= \vec{a}(\nabla \cdot \vec{r}) - \vec{r}(\nabla \cdot \vec{a}) + (\vec{r} \cdot \nabla)\vec{a} - (\vec{a} \cdot \nabla) \vec{r} \\ &= \vec{a}(\nabla \cdot \vec{r}) - (\vec{a} \cdot \nabla) \vec{r}. \end{align*} As you say, the first term on the RHS is then $3 \vec{a}$, but we need to calculate the other term. It is just the directional derivative of $\vec{r}$ in the direction of $\vec{a}$ multiplied by the magnitude of $\vec{a}$, so this is just $\vec{a}$. Thus $$ \nabla \times (\vec{a} \times \vec{r}) = 3 \vec{a} - \vec{a} = 2 \vec{a}, $$ as desired.

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  • $\begingroup$ photos.app.goo.gl/71CHGVa9Hkt2U7Hk8 is the proof for identity I have used. $\endgroup$ – Abbas Miya Mar 19 at 11:17
  • $\begingroup$ @AbbasMiya I can't see the image---you can just add it to your question. $\endgroup$ – Keeley Hoek Mar 19 at 11:30
  • $\begingroup$ But at any rate, the mistake must be there. $\endgroup$ – Keeley Hoek Mar 19 at 11:30
  • $\begingroup$ I am sorry for the link, it is fine now, you can see the picture. $\endgroup$ – Abbas Miya Mar 19 at 11:32
  • $\begingroup$ @AbbasMiya That "simplification" of the identity in red is not correct. The confusion seems to be that $\vec{a} \cdot \nabla$ does not equal $\nabla \cdot \vec{a}$, for example. Instead $\vec{a} \cdot \nabla$ is an operator which differentates in the $\vec{a}$ direction and then multiplies by the magnitude of $\vec{a}$. On the other hand, $\nabla \cdot \vec{a}$ is just a number (and in your case it is zero). $\endgroup$ – Keeley Hoek Mar 19 at 11:36

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