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This was used in the answer here, in the derivation of the Lorentz force law from the Lagrangian. $u$ and $A$ are vectors, the velocity of the particle and the spacetime dependent Magnetic field

As part of the Euler Lagrange equation, we had to calculate $$\frac{\partial L}{\partial x}=\frac{\partial {(u\cdot A) }}{\partial x}$$

Since the Lagrangian treats $x$ and $u$, i.e. position and velocity, as independent variables, I think this partial derivative should treat $u$ as a constant to give:

$$\frac{\partial L}{\partial x}=\frac{\partial {(u\cdot A) }}{\partial x}$$

$$=\frac{\partial {(u_x A_x + u_yA_y +u_z A_z) }}{\partial x}$$

$$=u\cdot \frac{\partial A}{\partial x}$$

The vector form of this would be :

$$\frac{\partial L}{\partial r}=u\cdot \nabla A$$

I only got the first term. I don't understand where $u\times ({\nabla \times A})$ came from. Please help

Ryder Rude
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    This question should be in the Physics website I think. I'm not sure (because I don't know much about physics), but my guess would be that you cannot say that $u$ is independent from the position. – Falcon Nov 18 '22 at 04:07
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    @Falcon I'd say it's fair game for Math SE--it's about derivatives after all. – Nick F Nov 18 '22 at 04:47
  • Yes maybe, but doesn't the second term come from the fact that the acceleration $a$ of a particule in a magnetic field is given by $a = u \times B = u \times (\nabla \times A)$ ? Looks like it – Falcon Nov 18 '22 at 05:05
  • Which is definitely a physical argument – Falcon Nov 18 '22 at 05:08
  • @Falcon No, that's what we're trying to derive. The derivation only involves taking derivatives according to the Euler Lagrange equation. Also see this for the treatment of velocity as independent. – Ryder Rude Nov 18 '22 at 06:05
  • @Falcon see my answer – Ryder Rude Nov 18 '22 at 08:05

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Recall that for any vectors $a, b, c$ $$ a\times(b\times c) = (a\cdot c)b - (a\cdot b)c. $$ It is valid to manipulate $\nabla$ as a vector so long as you are aware of what you are differentiating. In the expression $$ u\times(\nabla\times A), $$ we are differentiating $A$. Let us keep track of this by putting a dot over $\nabla$ and $A$. We then see $$ u\times(\dot\nabla\times\dot A) = (u\cdot\dot A)\dot\nabla - (u\cdot\dot\nabla)\dot A = \dot\nabla(u\cdot\dot A) - (u\cdot\dot\nabla)\dot A. $$ The last equality is just from moving around the scalar quantity $u\cdot\dot A$. Since $u$ is to be treated as constant, we can now drop the dots and write in standard notation $$ u\times(\nabla\times A) = \nabla(u\cdot A) - (u\cdot\nabla) A. $$ Rearranging, we finally get $$ \nabla(u\cdot A) = (u\cdot\nabla) A + u\times(\nabla\times A). $$

Nicholas Todoroff
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  • However: $\bf{A}\times(\nabla \times A)= \frac{1}{2}\nabla(A \cdot A)-(A\cdot \nabla)A$. – Kurt G. Nov 18 '22 at 18:18
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    @KurtG. Yes; this is not a contradiction. You have to keep track of what you differentiate. On the RHS, each $A$ in $A\cdot A$ is being differentiated. This is very different from what I wrote, where $u$ is not being differentiated. The following is true:$$A\times(\dot\nabla\times\dot A)= \dot\nabla(A\cdot\dot A)-(A\cdot\dot\nabla)\dot A.$$Then using a variant of the product rule $$\dot\nabla(\dot A\cdot\dot A)=\dot\nabla(\dot A\cdot A)+\dot\nabla(A\cdot\dot A)=2\dot\nabla(A\cdot\dot A)$$ so now $$A\times(\dot\nabla\times\dot A)=\frac12\dot\nabla(\dot A\cdot\dot A)- (A\cdot\dot\nabla)\dot A.$$ – Nicholas Todoroff Nov 19 '22 at 04:16
  • Nice ! Compared to md2perpe's brute force solution to this your method is crystal clear and shows why some of those expressions resemble the Grassmann identity and some don't. – Kurt G. Nov 19 '22 at 07:13
  • For example, $\nabla\times(\nabla\times A)=\nabla(\nabla\cdot A)-\Delta A$ is Grassmann like. (I know this does not surprise you. Just think examples help.) – Kurt G. Nov 19 '22 at 07:42
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OMG this is such ambiguous notation.

The thing is:

$$\vec{a} \cdot (\nabla \vec{b}) \neq (\vec{a}\cdot \nabla) \vec{b}$$

The answer that I linked derived a formula involving $ (\vec{a}\cdot \nabla) \vec{b}$ . This is why they got an extra cross product term.

I derived a formula involving $ \vec{a} \cdot (\nabla \vec{b}) $, which is why I didn't get the cross product term.

$(\vec{a}\cdot \nabla) \vec{b}$ can unambiguously be interpreted as $$a^i \partial ^i b^j$$

$\vec{a}\cdot (\nabla\vec{b})$ has to be interpreted carefully because the $\nabla \vec{b}$ is a matrix. When we matrix multiply this with $\vec{a}$, we have to contract the right indices. It is to be interpreted as the matrix multiplication :

$$a^j \partial ^i b^j$$

To see this, let's look at the derivation of this formula:

$$\partial ^i (v^jw^j) =w^j\partial ^i v^j +v^j \partial ^i w^j$$

In my post $v^j$ is a constant, so the first term vanishes.

Ryder Rude
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  • The ambiguity in the notation comes from a poor choice of the components of the vector gradient. It is much, much better to use $(\nabla v)_{ij}=\partial_jv_i$ than to use $\partial_iv_j$. It follows the formalism of the covariant derivative and transforms as a tensor. – K.defaoite Nov 18 '22 at 13:09
  • Using the correct $(\nabla v)^i{}_j=\nabla_jv^i$ one finds that $(u\cdot \nabla)v$ and $u\cdot(\nabla v)$ are the same, since the first is $(u^j\nabla_j)v^i$ and the second is $u^j(\nabla_jv^i)$ which are the same. – K.defaoite Nov 18 '22 at 13:10
  • @K.defaoit See the edit. You could say that $u\cdot (\nabla v) $ is unambiguously defined to be what you wrote. But in the context of this post, it's not what you wrote, because the formula has been derived here from the Euler Lagrange equation – Ryder Rude Nov 18 '22 at 13:39
  • @K.defaoite In the post, I'm not careful about the distinction between vectors and dual vectors because the metric is Euclidean. But yeah, I should write $a^jb^j$ as $a^jb_j$. Now you can see how the index contraction can be ambiguous after introducing the derivative, because $a^j$ can now be contracted with either of $\partial$ and $b$, and it will transform tensorially either way. The dot product symbol does not make it clear which index you're supposed to contract with. – Ryder Rude Nov 18 '22 at 13:49