0
$\begingroup$

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 $\overrightarrow{v}$?

$\endgroup$
2
  • 2
    $\begingroup$ The other components potentially depend on $x$ too! $\endgroup$
    – Nephente
    Mar 5, 2017 at 15:12
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 5, 2017 at 15:54

1 Answer 1

1
$\begingroup$

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 $\frac{\partial}{\partial x} \vec A$.

$\endgroup$
4
  • $\begingroup$ Shouldn't $\nabla\cdot \vec{A}$ be a scalar, and not a vector? $\endgroup$ Mar 5, 2017 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, 2017 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. $\endgroup$ Mar 5, 2017 at 17:57
  • 1
    $\begingroup$ @Sobanoodles Oh, yes, sorry - thanks for spotting that. $\endgroup$
    – JamalS
    Mar 5, 2017 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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