2
$\begingroup$

One of the definitions of magnetic field (in free space): Force on magnetic north pole per unit pole.

Magnetic field is an axial vector. So it changes its sign when we use left handed coordinate system.

So does this imply that force on magnetic north pole becomes opposite just by introducing a left handed coordinate system? Isn't this an absurd consequence?

Improve this question
$\endgroup$
6
  • $\begingroup$ I think this is a great question. My guess is that if we take E-field to be a true vector and electric charge density to be a scalar field, as B-field is a tensor density (or pseudovector), a "magnetic monopole density" would need to be a scalar density for forces on magnetic monopoles to come out as a vector. I'm sure someone can confirm whether or not this is true. $\endgroup$
    – Puk
    Commented Jul 25, 2019 at 9:48
  • 2
    $\begingroup$ Maybe I am mistaken here. But under a reflection, your test north-pole turns into a south-pole (there is no monopole) and thus the force reverses unless the magnetic field changes sign too. $\endgroup$
    – Nephente
    Commented Jul 25, 2019 at 10:10
  • $\begingroup$ @Nephente: If I am not mistaken, it seems to me that you have made a very good point. $\endgroup$
    – N.G.Tyson
    Commented Jul 25, 2019 at 10:14
  • $\begingroup$ @Nephente: For the record I'm not insisting magnetic monopoles exist, I just don't know how to quantitatively describe B-field as a "force per pole" without them. But of course north poles turning south (and vice versa) makes sense and is rather obvious once you point it out. $\endgroup$
    – Puk
    Commented Jul 25, 2019 at 10:21
  • $\begingroup$ @Puk The best way to realise a "monopole" to define the magnetic field as force per polestrength is to take a very thin solenoid that extends into infinity at one end. $\endgroup$
    – Nephente
    Commented Jul 25, 2019 at 10:33

1 Answer 1

Reset to default
1
$\begingroup$

It does not matter which coordinate system you use. The vectors, axial or not, do not change. Only their coefficients in the new coordinate system are different.

Improve this answer
$\endgroup$

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