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I am investigating a lagrangian with the end game of determining a Hamiltonian to describe a SUSY charged particle in an electric field however I am having a fundamental difficulty with my maths.

If we have

$\overrightarrow{A}=A_x\hat{x}+A_y\hat{y}+A_z\hat{z}$

and I am trying to determine

$\frac{d\overrightarrow{A}}{dx}$

How do I do this as I thought this would be as simple as

$\frac{d\overrightarrow{A}}{dx}=\frac{dA_x}{dx}$

However this appears to be wrong, can someone explain what I do understand here ?

I then need to do the same for a velocity

$\overrightarrow{v}=v_x\hat{x}+v_y\hat{y}+v_z\hat{z}$

Once I can solve my first problem is the method the same for $\overrightarrow{v}$?

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    $\begingroup$ The other components potentially depend on $x$ too! $\endgroup$ – Nephente Mar 5 '17 at 15:12
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    $\begingroup$ Exactly, $$\frac{\partial \vec{A}}{\partial x_i}=\sum_j\frac{\partial A_j}{\partial x_i}\hat{x}_j$$ in case of cartesian coordinates. $\endgroup$ – Soba noodles Mar 5 '17 at 15:54
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It seems if you're struggling with these basic concepts, I would recommend revising these before trying to study super-symmetric theories. In this question, it's as simple as,

$$\frac{\partial}{\partial x} \vec A = \frac{\partial A_x}{\partial x}\hat e_x + \frac{\partial A_y}{\partial x}\hat e_y + \frac{\partial A_z}{\partial x} \hat e_z.$$

As a heads up, you may also find,

$$\frac{\partial A^\mu}{\partial x^\mu} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}$$

which is $\nabla \cdot \vec A$ and clearly not the same as $\frac{\partial}{\partial x} \vec A$.

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  • $\begingroup$ Shouldn't $\nabla\cdot \vec{A}$ be a scalar, and not a vector? $\endgroup$ – Soba noodles Mar 5 '17 at 17:22
  • $\begingroup$ @Sobanoodles $\partial_\mu A^\mu = \nabla \cdot \vec A$ is a scalar; I have the arrow on the $A$ which is a vector - I did not put an arrow over the whole thing :) $\endgroup$ – JamalS Mar 5 '17 at 17:56
  • $\begingroup$ That's true, but I was referring to the r.h.s. of your second equation, which is a vectorial quantity. $\endgroup$ – Soba noodles Mar 5 '17 at 17:57
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    $\begingroup$ @Sobanoodles Oh, yes, sorry - thanks for spotting that. $\endgroup$ – JamalS Mar 5 '17 at 18:55

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