0

I am reading Mathematical Methods of Classical Mechanics by Arnold. I got stucked on a problem about central fields. Usually in literature central fields are given by $\mathbf{F} = \Phi(r) \mathbf{e}_r$, where $\mathbf{e}_r$ is the radial unit vector.

However, Arnold asks to derive this form by imposing the invariance of the field with respect to the group of motions of the plane which fix $0$. Can somebody give a hint?

gmirsan
  • 13
  • Use rotations with a common rotation axis to see that the field along that axis has to point in the direction of the axis, can have no components orthogonal to it. Then rotate that line with its vector field to every other possible direction. /// Also note that in $x=r{\bf e}_r$, the first $r=|{\bf x}|$ is a number, while the second is a label for the radial vector field ${\bf e}_r=\hat {\bf x}=\frac{{\bf x}}{|{\bf x}|}$. – Lutz Lehmann Jan 17 '22 at 11:00
  • Thank you for your comment. First, I'd like to prove this in $\mathbb{R}^2$, since the generalization to three dimensions should be straightforward. In order to use your argument, should I write the field in polar coordinates or is easier than that? – gmirsan Jan 18 '22 at 08:56
  • I believe this only works if you include reflections in the motions of the 2D plane. If you only take the rotations, then there will be more solutions that are invariant. // You can take the line $(x,0)$ as the reference line, then there is not much difference between the polar and Cartesian approach. – Lutz Lehmann Jan 18 '22 at 09:00
  • Reflections are included in the exercise, it says motions fixing $0$, so a reflection with respect to an axis containing $0$ is also allowed. Sorry for not being clear. – gmirsan Jan 18 '22 at 09:03
  • So if you have at $(x,0)$ the vector $(v_x,v_y)$, then it has, under the invariance, to be equal to $(v_x,-v_y)$, thus $v_y=0$... – Lutz Lehmann Jan 18 '22 at 11:39
  • Yes, exactly, now I understand it, thank you. So if we have a vector field $F(x_1, x_2))$ then taking a reflection with respect to the axis passing through $(x_1, x_2)$ forces our vector field to be radial (otherwise $RF(x_1, x_2) = F(R(x_1, x_2)) = F(x_1, x_2)$ does not hold, where $R$ is the reflection). The fact that $\Phi$ only depends on $r$ can be seen by taking an arbitrary rotation for example. – gmirsan Jan 18 '22 at 11:46
  • Yes. Or rotate everything to the x-axis, so that the overall vector field has to follow the pattern there. – Lutz Lehmann Jan 18 '22 at 11:48
  • Right, perfect, thank you for your help. – gmirsan Jan 18 '22 at 11:52

0 Answers0