In an infinite cylindrical field, which component of momentum is conserved
Answers
Hola mate...
Think about conservation laws as direct consequences of the symmetries of a system. In general, when you have a symmetry, you have a conserved quantity.
Let us be clearer with an example: I'll take the (a) from your list and leave you with the rest. In that case, you have an infinite, homogeneous plane that we'll assume from now on to be coinciding with the (x,y) plane.
You can measure the field that such a distribution generates by using a test particle. Symmetries of the system are ways in which you can move your test particle (but NOT the distribution of charges) and end up in the same situation.
In you particular case, you can rotate the particle along an axis that is orthogonal to the plane: the distance of the particle from the plane won't change, and the particle will feel the same field. So we can affirm that the z part of the angular momentum, Lz, is conserved. As you can also translate the test particle along the (x,y) directions without changing the physical situation, you can state that px and py, the linear momenta along x and y, are conserved. The linear momentum pz is not conserved, as a force is acting on the z axis (and the component vz of the velocity will change).
To be more concrete, when you have a charge distribution you have a potential V(r⃗ ,aI), where r is the position of the test particle and aI are parameters specifying the charge distribution. In the case of the homogeneous plane, the aI parameters are the height of the plane and its total charge. They totally determine the potential for all r⃗ .
A transformation of the system is a law r⃗ ′(r⃗ ), changing your coordinates. In particular, a translation by a vector v⃗ is defined as r⃗ ′=r⃗ +v⃗ , and a rotation is r⃗ ′=Mr⃗ , with M an SO(3) matrix. Transofrmation of coordinates must be invertible, so a law r⃗ (r⃗ ′) always exists for a good coordinate change.
You have a symmetry of the system when the potential is unchanged under the transformation. As an example, the potential in your case is V(r⃗ )=Frz: if you plug r⃗ ′ instead of r⃗ you can verify if you have a symmetry. If you translate along the z axis, your potential is transformed to F(rz+vz), so the translation along z is not a symmetry. Translations along the other axes are symmetries. And, as rotations around z do not change the z coordinate, those rotations are a symmetry.
With this in mind, try to approach the other points of the problem, by solving them in a geometrical way (by thinking about transformations). This will develop your intuition about symmetries. After that, you can try to write potentials and verify if your intuitions are correct through transformations
Its to long i know but if u read it u will clearly understand :)
Hope this helps u...
Please make it BRAINLIEST answer..