Deriving an identity using Einstein's summation notation

[Arman777]

Homework Statement

Derive the Identitiy using Einstein Summation Notation

Relevant Equations

$$\vec{\nabla} \times (\frac{\vec{m} \times \hat{r}}{r^2}) = ?$$

I have an identity

$$\vec{\nabla} \times (\frac{\vec{m} \times \hat{r}}{r^2})$$

which should give us

$$3(\vec{m} \cdot \hat{r}) \hat{r} - \vec{m}$$

But I have to derive it using the Einstein summation notation.

How can I approach this problem to simplify things ?

Should I do something like $\vec{k}=\vec{m} \times \hat{r}$ ? and then

$$\vec{\nabla} \times (\frac{\vec{k}}{r^2}) = \frac{r^2 \nabla \times \vec{k} - \nabla(r^2) \times \vec{k}}{ r^4} $$ ? But it seems like things getting more complicated this way.

 

[fresh_42]

You have three vectors and a scalar, which also depends on the same coordinates than one of the vectors. Einstein notation is an abbreviation for sums. The summands are written in coordinates. So write down your expression in coordinates and build the cross product.

 

[Arman777]

You have three vectors and a scalar, which also depends on the same coordinates than one of the vectors. Einstein notation is an abbreviation for sums. The summands are written in coordinates. So write down your expression in coordinates and build the cross product.

I guess its like this

$$(∂_w(m_i)r_j/r_l^2 + ∂_w(r_j)m_j/r_l^2 + ∂_w(1/r_l^2)m_ir_j) \epsilon_{ijk}\epsilon_{wkh}\hat{e}_h$$ what will happen to

 

[fresh_42]

I don't know how the cross product is written with Einstein. I would write the entire expression in all three coordinates and then repeat it in the shorter notation to see how it is used and what for. It's important in physics to become fit in it. I am not. That's why I would go the long path.

$r^2=r_1^2+r_2^2+r_3^2=r^ir_i$ is a scalar, so you can pull it out of the second, inner cross product, but the outer one are differential operators which apply to $r_i$.

 

[Arman777]

I don't know how the cross product is written with Einstein. I would write the entire expression in all three coordinates and then repeat it in the shorter notation to see how it is used and what for. It's important in physics to become fit in it. I am not. That's why I would go the long path.

$r^2=r_1^2+r_2^2+r_3^2=r^ir_i$ is a scalar, so you can pull it out of the second, inner cross product, but the outer one are differential operators which apply to $r_i$.

Hmm I see. I am sharing a pdf. I am using that kind of notations. I don't know upper index notation ..

I find something like $$(∂_w(m_i)r_j/r_l^2 + ∂_w(r_j)m_j/r_l^2 + ∂_w(1/r_l^2)m_ir_j) \epsilon_{ijk}\epsilon_{wkh}\hat{e}_h$$

 

[Orodruin]

Hmm I see. I am sharing a pdf. I am using that kind of notations. I don't know upper index notation ..

I find something like $$(∂_w(m_i)r_j/r_l^2 + ∂_w(r_j)m_j/r_l^2 + ∂_w(1/r_l^2)m_ir_j) \epsilon_{ijk}\epsilon_{wkh}\hat{e}_h$$

In Cartesian coordinates, you can forget about the difference between upper and lower index, it is only in curvilinear coordinate systems (and on manifolds) that it makes a difference.

Note that $\vec m$ is a constant vector and that $\partial x^i/\partial x^j = \delta_{ij}$. You can also apply the $\epsilon$-$\delta$ relation to your expression.

I have an identity

$$\vec{\nabla} \times (\frac{\vec{m} \times \hat{r}}{r^2})$$

which should give us

$$3(\vec{m} \cdot \hat{r}) \hat{r} - \vec{m}$$

Note that this cannot possibly be true just based on dimensional analysis. The derivative has dimensions 1/L and what you differentiate in the first expression has dimensions of $[m]/\mathsf L^2$, making the full first expression have dimension $[m]/\mathsf L^3$, whereas your second expression has dimension $[m]$.