1
$\begingroup$

Say we have scalar potential in a form $$ U = A \ln (\vec{a} \times \vec{r})^2 e^{-\vec{b} \cdot \vec{r}}. $$ How would one calculate gradient $\vec{E}=-\nabla U$ of such potential?

A is a constant, $\vec{a}$ is a constant vector and $\vec{r}=(x, y, z)$.

$\endgroup$
  • 1
    $\begingroup$ what exactly is the natural log of a cross product? This is a typo? $\endgroup$ – James S. Cook Apr 9 '13 at 22:41
  • $\begingroup$ Ivana what are the variables of its potential function? $\endgroup$ – MathOverview Apr 9 '13 at 22:45
  • 1
    $\begingroup$ By the square of a vector, do you mean the square of its length? $\endgroup$ – Javier Apr 9 '13 at 22:52
  • 1
    $\begingroup$ Yes, I was trying to solve it under assumption that square was a square of length (though it wasn't sprecified). $\endgroup$ – Ivana Apr 9 '13 at 22:55
  • 1
    $\begingroup$ @Ivana Ivana expression $ \ln(\vec{a} \times \vec{r})^2e^{-\vec{b} \cdot \vec{r}}$ should be understood as the logarithm of only $(\vec{a}\times\vec{r})^2 $ or of the expression ($ \vec{a} \times \vec{r})^2 e^{- \vec{b} \cdot \vec{r}} $? Please be clear in your question. $\endgroup$ – MathOverview Apr 9 '13 at 23:00
1
$\begingroup$

$(\vec{a}\times \vec{r})_k = \epsilon_{ijk}a_ix_j$ hence $(\vec{a}\times \vec{r})^2 = \epsilon_{ijk}a_ix_j\epsilon_{lmk}a_lx_m$ consequently, \begin{align} \partial_n (\vec{a}\times \vec{r})^2 = \epsilon_{ijk}\epsilon_{lmk}a_ia_l\partial_n(x_mx_j)&= \epsilon_{ijk}\epsilon_{lmk}a_ia_l[\delta_{nm}x_j+x_m\delta_{nj}] \\ &= \epsilon_{ijk}\epsilon_{lnk}a_ia_lx_j+\epsilon_{ink}\epsilon_{lmk}a_ia_lx_m\\ &= 2\epsilon_{ijk}\epsilon_{lnk}a_ia_lx_j \\ &= 2(\delta_{il}\delta_{jn}-\delta_{in}\delta_{jl})a_ia_lx_j \\ &= 2(\vec{a} \cdot \vec{a} \, x_n - a_n \vec{a} \cdot \vec{r}) \end{align} On the other hand, $\partial_n \vec{b} \cdot \vec{r} = \partial_n b_ix_i = b_i \delta_{in} = b_n$ assuming that $\vec{b}$ is constant. So, \begin{align} \partial_n U &= \partial_n [A\ln(\vec{a}\times \vec{r})^2e^{-\vec{b}\cdot \vec{r}}] \\ &= A\left(\partial_n [\ln(\vec{a}\times \vec{r})^2]e^{-\vec{b}\cdot \vec{r}}+\ln(\vec{a}\times \vec{r})^2\partial_n[e^{-\vec{b}\cdot \vec{r}}]\right) \\ &= A\left(\frac{2(\vec{a} \cdot \vec{a} \, x_n - a_n \vec{a} \cdot \vec{r})}{(\vec{a}\times \vec{r})^2}-b_n\ln(\vec{a}\times \vec{r})^2\right)e^{-\vec{b}\cdot \vec{r}} \\ \end{align} Thus, provided I haven't made some silly mistake, $$ \vec{E} = -\nabla U = A\left(\frac{2[(\vec{a} \cdot \vec{a}) \, \vec{r} - (\vec{a} \cdot \vec{r})\vec{a}]}{(\vec{a}\times \vec{r})^2}-\ln(\vec{a}\times \vec{r})^2\vec{b}\right)e^{-\vec{b}\cdot \vec{r}} $$

$\endgroup$
1
$\begingroup$

Cross-product by a constant is linear, taking the square of a vector is quadratic so by differencing composed functions, $$ d(ln((a\times r)^2)) = \frac{1}{(a\times r)^{2}} d((a\times r).(a\times r)) = \frac{2}{(a\times r)^{2}} d(a \times r).(a \times r) = \frac{2}{(a\times r)^{2}} (a \times dr).(a\times r) $$ for the exp part : $$ d(e^{-b.r}) = e^{-b.r} d(-b.r) = e^{-b.r} (-b.dr) $$ and by combining the two with product differentiation rule : $$ dU = \frac{2}{(a\times r)^{2}} e^{-b.r} (a\times r).(a \times dr) + ln((a\times r)^2) e^{-b.r} (-b.dr) = e^{-b.r}\left(\frac{2(a\times r).(a \times dr)}{||a\times r||^2} - ln(||a\times r||^2) b.dr\right) $$ We want to rewrite it as an inner product of dr, so we use the rules of scalar triple product then vector triple product $$ dU = e^{-b.r}\left(\frac{2 dr.((a\times r \times a)}{||a\times r||^2} - ln(||a\times r||^2) b.dr\right) = e^{-b.r}\left(\frac{2 (||a||^2 r - (r.a)a).dr}{||a\times r||^2} - ln(||a\times r||^2) b.dr\right) = e^{-b.r}\left(2\frac{||a||^2 r - (r.a)a}{||a\times r||^2} - ln(||a\times r||^2) b\right).dr $$ And you read gradU by removing .dr .

$\endgroup$
  • $\begingroup$ Thank you, this is helpful. $\endgroup$ – Ivana Apr 12 '13 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.