0
$\begingroup$

I've never quite well understood the definition of vector used by physicists based on transformation laws.

From a mathematical point of view a vector in space is:

  1. A point derivation on the germs of smooth functions at the point
  2. One equivalence class of curves with 1st order contact at the point

While a vector field is just the association of one vector to each point.

Now I'm really trying to understand the physicists definition, so I thought on considering some simple examples: the vector potential for the magnetic field, and the electric and magnetic fields.

The point here is: I really can't understand what it means to show that those objects behave as vectors. I mean, by the very definition of $\mathbf{A}$ being $\mathbf{B} = \nabla \times \mathbf{A}$ it is already implied, for me that $\mathbf{A}$ is a vector.

Futhermore, since $\mathbf{e}_i$ forms a basis of $\mathbb{R}^3$, for any three numbers $a^i$ we have that $a^i\mathbf{e}_i$ is an element of $\mathbb{R}^3$ and hence a vector.

My question here is: how can one actually show that $\mathbf{A}$ is a vector from the physicists point of view? How can we show that $\mathbf{A},\mathbf{B}$ and $\mathbf{E}$ are all vectors from the definition based on transformation laws?

$\endgroup$
  • $\begingroup$ Related: math.stackexchange.com/q/789956 $\endgroup$ – David Z Aug 27 '16 at 18:42
  • $\begingroup$ This really depends on what assumptions you want to start from. Usually, we start from the fact that $\mathbf{A}$ is a vector (or that $A^\mu$ is a four-vector). Introductory textbooks start in the middle (e.g. with Coulomb's law) and then work backwards, showing that $\mathbf{A}$ being a vector is self-consistent. $\endgroup$ – knzhou Aug 27 '16 at 18:44
  • $\begingroup$ Also, $\mathbf{B}$ is not a vector, it's a pseudovector. $\endgroup$ – knzhou Aug 27 '16 at 18:45
  • $\begingroup$ @DavidZ, that's indeed related. To tell you the truth I never understood the problem of the fruits and what they have to do with coordinates anyway. The whole point is that as $\mathbb{R}^3$ is by definition the set of all three tuples of reals, $(x,y,z)$ , I never could see why some three touples are not qualified just because they represent number of fruits while others are. $\endgroup$ – user1620696 Aug 27 '16 at 18:51
  • 1
    $\begingroup$ If you do a reflection $x\rightarrow -x, y\rightarrow -y, z\rightarrow -z $, then what will happen to $A$ and what will happen to $B$. In fact what will happen to any cross product under such a transformation? $\endgroup$ – Kolandiolaka Aug 27 '16 at 19:17

Your Answer

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

Browse other questions tagged or ask your own question.