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I am looking at this document, which tries to establish the Lagrangian of the Lorentz force. Everything is fine, but I don't see why:

$$\frac{dA_i}{dt}=\frac{\partial A_i}{\partial t}+\frac{\partial r_j}{\partial t}.\frac{\partial A_i}{\partial r_j}$$

have $j$ components in the $i$th component of the total derivative? For example, if $A$ has $x$ and $y$ components, then the $x$ component should be $$\frac{\partial A_x}{\partial t}+\frac{\partial x}{\partial t}.\frac{\partial A_x}{\partial x}$$ but the expression above implies otherwise. What am I missing?

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    $\begingroup$ What do you know about the chain rule? $\endgroup$ – Kyle Kanos May 11 '15 at 3:59
  • $\begingroup$ Chain rule does not mix components. $\endgroup$ – student1 May 11 '15 at 4:16
  • $\begingroup$ Imagine you compute the potential at two times, firstly you will evaluate it at different places if the particle moved, secondly the value of the vector potential itself can change over time. Since the particle could move in the y direction $\partial \vec A / \partial y$ matters. Similarly for x and z. $\endgroup$ – Timaeus May 11 '15 at 4:47
  • $\begingroup$ @student1: What do you mean that the chain rule "does not mix components"? $\endgroup$ – Kyle Kanos May 11 '15 at 11:21
  • $\begingroup$ Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic Jul 27 '15 at 18:01
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$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}$ $\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$As indicated in the comments, the reason why you get a term of the kind $\pdv{A_x}{x}\pdv{x}{t}$ is the chain rule. Notice that the magnetic potential can be written in components as $A_i(x,y,z,t)$. However the components $x,y \text{ and } z$ also time dependent so it actually better to write $A_i\Big(x(t),y(t),z(t),t \Big)$.

Nobody does that because the time dependence is understood (at least the author convinced himself that the time dependence is understood), it takes a while to write it out and it takes print while publishing a book.

Returning to the derivative of the potential. Note that

$$\dv{A_i}{t}=\pdv{A_i}{t}+\sum_j\pdv{A_i}{r_j}\pdv{r_j}{t} \tag{1}$$

,where $j=1,2,3$. I implicitly indicated the summation, because I didn't know whether there was a confusion with the Einstein summation convention. Einstein convention is basically telling you to sum over all indices, which are repeated once upstairs and once downstairs. In your special case the term with upstairs index is $\pdv{A_i}{r_j}$ and one with the downstairs index is $\pdv{r_j}{t}$. Hence you sum over this index as I did in the expression (1).

You said in the comments that the chain rule doesn't mix coordinate. It does on the contrary almost always mix coordinates. Take this example:

$$\pdv{\sin(r)}{x}=\pdv{\sin(r)}{r}\pdv{r}{x}=\text{whatever the result}$$

,where $r=\sqrt{x^2+y^2+z^2}$ as usual.

I hope this resolved your confusion.

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  • $\begingroup$ So taking 2D for simplicity, $$\frac{dA}{dt}|_x=\frac{\partial A_x}{\partial t}+\frac{\partial y}{\partial t}.\frac{\partial A_x}{\partial y}$$ is correct? $\endgroup$ – student1 May 11 '15 at 15:24
  • $\begingroup$ It kind of make sense because at the end you need to get Lorentz force back, which has a curl, and you expect mixing like that... $\endgroup$ – student1 May 11 '15 at 15:26
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Each of the components of $A^i$ depends on $(x,y,z,t)$ thus you must use the chain rule and evaluate $(dA^i/dt) + (dA^i/dj * dj/dt)$ where the j's are summed over x, y, and z.

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