Tag Info

New answers tagged

0

I think I've figured it out with the help provided in the comments by Rahul. I'm pretty sure it's all just a matter of notation. The vector identity $$\vec\nabla\times(\vec A \times \vec B) = \vec A(\vec\nabla.\vec B) - \vec B(\vec\nabla.\vec A)$$ Can be used here IF you keep track on what $\vec\nabla$ will work in on: On the left side it worked in on ...


1

Your professor is correct assuming $\vec{A}$ is a constant vector field. To answer your particular problem, since $\vec{m}$ is constant, \begin{align}\vec{B} &= \frac{\mu_0}{4\pi}\left\{\vec{m}\left(\vec{\nabla} \cdot \frac{\vec{r}}{r^3}\right) - (\vec{m} \cdot \vec{\nabla})\frac{\vec{r}}{r^3}\right\}\\ &= ...


0

Here's how to prove it with index notation: $$[(a \times b)\times c]_p = \epsilon_{pqr}(a\times b)_qc_r = \epsilon_{pqr}\epsilon_{qst}a_sb_tc_r = \epsilon_{qrp}\epsilon_{qst}a_sb_tc_r = (\delta_{rs}\delta_{pt}-\delta_{rt}\delta_{ps})a_sb_tc_r = \delta_{rs}\delta_{pt}a_sb_tc_r-\delta_{rt}\delta_{ps}a_sb_tc_r = a_rb_pc_r-a_pb_rc_r = a_rc_rb_p-b_rc_ra_p = ...


1

The normal vector $\vec n$ has to be not just perpendicular, but in the direction specified by the right-hand rule. So it is $-\vec k$, not $\vec k$


1

I've just realized by skipping to the text ahead that the cross product with respect to the bilinear form b is not the same as the standard cross product on R3. How would, in general, one find a cross product with respect to the given bilinear form? It's not too hard to answer this, provided one doesn't mind a little abstract linear algebra. Given a ...


0

Found it on page 137. For given vectors ${u,v}$ and a bilinear form $b$ there exists a unique vector $w$ such that for all $z$ we have $b\left( {w,z} \right) = \sqrt {\det B} \det \left( {z,u,v} \right)$, where $B$ is a matrix representation of $b$ in any base. Then we say that $w = u{ \times _b}v$. For the given bilinear form from the question, we get ...


1

Hint: Recall definition of cross product of $2$ vectors in $\mathbb{R}^n$ is: $|\overrightarrow{a}\times \overrightarrow{b}| = |\overrightarrow{a}|\cdot |\overrightarrow{b}|\cdot \sin (\overrightarrow{a},\overrightarrow{b})$. Can you take it from here.



Top 50 recent answers are included