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Sorry about the confusing title; it was hard to phrase this.

For some context, I have an exam very soon and these are mock questions, my teacher has not provided any solutions to theses. So I am working my way through the questions.

I think I have solved part a) although this is probably relatively easy, although I know I have the correct idea in solving this I am unsure whether my working out is completely accurate.

For b) and c) I am confused of how to apply the cross product to two vectors in the form of $\bf{m}$ and $\bf{r}$. Whilst I am comfortable showing that vector fields are conservative and calculating the curl and divergence of a single vector, i.e. only $\bf{r}$, I have not came across how to apply this for two vectors.

Any help would be most welcome. Please note I am not asking for a whole answer, rather how to better apply my working out for part a), and the format or a starting point for b) and c).

If $\bf{m}$ is a constant vector and $\mathbf{r}=x\hspace{0.5mm}\mathbf{i}+y\hspace{0.5mm}\mathbf{j}+z\hspace{0.5mm}\mathbf{k}$, show that

$\hspace{4mm}$ a) $\nabla(\mathbf{m}\cdot\mathbf{r})=\mathbf{m}$

$\hspace{4mm}$ b) $\nabla\cdot(\mathbf{m}\times\mathbf{r})=0$

$\hspace{4mm}$ c) $\nabla\times(\mathbf{m}\times\mathbf{r})=2\mathbf{m}$

My work:

$\hspace{8mm}\nabla\,\times\implies$ Matrix

$\hspace{8mm}\nabla\,\cdot\implies$ Diff

$\mathbf{r}=x\hspace{0.5mm}\mathbf{i}+y\hspace{0.5mm}\mathbf{j}+z\hspace{0.5mm}\mathbf{k}$

$\mathbf{m}=a\hspace{0.5mm}\mathbf{i}+b\hspace{0.5mm}\mathbf{j}+c\hspace{0.5mm}\mathbf{k}$

$\text{a)}\:\:\nabla(\mathbf{m}\cdot\mathbf{r})=\dfrac{\partial}{\partial x}(ax)+\dfrac{\partial}{\partial y}(by)+\dfrac{\partial}{\partial z}(cz)=a+b+c=\mathbf{m}$

$\text{b)}\:\:(\mathbf{m}\times\mathbf{r})=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\a&b&c\\x&y&z\end{vmatrix}$

$\hspace{2.3cm}=(bz-cy)\hspace{0.5mm}\mathbf{i}+(cx-az)\hspace{0.5mm}\mathbf{j}+(ay-bx)\hspace{0.5mm}\mathbf{k}$

$\hspace{7mm}\nabla\cdot(\mathbf{m}\times\mathbf{r})=\dfrac{\partial}{\partial x}(bz-cy)\hspace{0.5mm}\mathbf{i}+\dfrac{\partial}{\partial y}(cx-az)\hspace{0.5mm}\mathbf{j}+\dfrac{\partial}{\partial z}(ay-bx)\hspace{0.5mm}\mathbf{k}$

Thanks if you read all this.

-Alexis.

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    $\begingroup$ The answer is, just do it componentwise. You are nearly there. However in the last line there should not be any, i,j or k, since you already did the scalar product. $\endgroup$ – mlk Jul 3 '17 at 12:33
  • $\begingroup$ Omg ofc so since these are just some constants, i.e. 100, 3844 then by differentiating wrt to x the terms will all become 0? $\endgroup$ – Alexis Sanchez Jul 3 '17 at 12:51
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    $\begingroup$ Exactly, to be more precise, constants with respect to the variable you are differentiating in. $\endgroup$ – mlk Jul 3 '17 at 12:57
  • $\begingroup$ Excellent comment, I should be able to tackle c now. $\endgroup$ – Alexis Sanchez Jul 3 '17 at 12:59
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You're nearly there for part b). Note, however, that $\nabla \cdot$ denotes the divergence, not the gradient. So, $$ \nabla \cdot (\mathbf m \times \mathbf r) = \frac{\partial }{\partial x}(bz - cy) + \frac{\partial }{\partial y}(cx - az) + \frac{\partial }{\partial z}(ay - bx) $$ That is, we should have no $\mathbf {i,j,k}$ in our answer. Also, all of these partial derivatives are zero since, as you said in the comment, $a,b,c$ are constant.


Part c) is pretty tedious. You can save some effort if you avoid writing out cross products. In particular, we can write $$ \nabla \times (\mathbf m \times \mathbf r) = \mathbf i \times \frac{\partial }{\partial x}(\mathbf m \times \mathbf r) + \mathbf j \times \frac{\partial }{\partial y}(\mathbf m \times \mathbf r) + \mathbf k \times \frac{\partial }{\partial z}(\mathbf m \times \mathbf r) \\ = \mathbf i \times \left(\mathbf m \times \frac{\partial \mathbf r}{\partial x}\right) + \mathbf j \times \left(\mathbf m \times \frac{\partial \mathbf r}{\partial y}\right) + \mathbf k \times \left(\mathbf m \times \frac{\partial \mathbf r}{\partial z}\right) $$ and then use the BAC-CAB formula.

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  • $\begingroup$ Thank you this is also an excellent response. Can I ask you what the difference is between divergence and gradient. From my knowledge, divergence is calculating the partial derivatives wrt, to x, y and z. And I thought gradient was the same thing. More over, the questions had the notation: \nabla \cdot and \nabla is there any difference for this. Thank you again for your help. $\endgroup$ – Alexis Sanchez Jul 3 '17 at 13:08
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    $\begingroup$ $\nabla$ alone encodes the gradient, and $\nabla \cdot$ encodes the divergence. The gradient is something usually done to a scalar field which outputs a vector. We compute it by putting all 3 partial derivatives into a vector. That is, $$ \nabla f = (f_x,f_y,f_z) $$ The divergence is something usually done to a vector field which outputs a scalar. We compute it by adding the three "matching" partial derivatives. That is, $$ \nabla \cdot \mathbf F = \nabla \cdot (f^{(1)},f^{(2)},f^{(3)}) = f^{(1)}_x + f^{(2)}_y + f^{(3)}_z $$ $\endgroup$ – Omnomnomnom Jul 3 '17 at 13:14
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    $\begingroup$ Note that writing $\nabla \cdot f$ doesn't make sense, since we don't have the separate components to work with. Similarly, computing $\nabla \mathbf F$ would require extending the usual definition of a gradient. $\endgroup$ – Omnomnomnom Jul 3 '17 at 13:18
  • $\begingroup$ Thank you very much for your help, I have since found that part c is equivalent to 2a+2b+2c = 2m. Thanks for your help and the explanation. $\endgroup$ – Alexis Sanchez Jul 3 '17 at 13:23
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    $\begingroup$ @AlexisSanchez $2a + 2b + 2c \neq 2\mathbf m$. The quantity on the left is a scalar, the quantity on the right is a vector. $\endgroup$ – Omnomnomnom Jul 3 '17 at 13:27

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