Conserved quantity for a particle in Magnetic Field

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SUMMARY

The discussion centers on the conservation of the quantity Q for a particle with mass m and charge e in a uniform magnetic field B = Bẑ. The Lagrangian is defined as L = (m/2)(x'^2 + y'^2 + z'^2) + (Be/2)(xy' - yx'). The participant successfully derived the Hamiltonian, which is purely kinetic energy, and confirmed that Q = (xy' - yx')B + (eB^2/2)(x^2 + y^2) is conserved, as its time derivative equals zero. The absence of a potential term is attributed to the magnetic field doing no work on the particle.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with Hamiltonian dynamics
  • Knowledge of vector calculus, particularly cross products
  • Basic concepts of magnetic fields and their effects on charged particles
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Physicists, students of classical mechanics, and anyone studying the dynamics of charged particles in magnetic fields will benefit from this discussion.

Siberion
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Homework Statement



Consider a particle of mass m and electric charge e moving in a uniform magnetic field given by B = Bẑ. Then the Lagrangian is given by:

L = \frac{m}{2}(x'^2 + y'^2 + z'^2) + \frac{Be}{2}(xy' - yx')

Prove that Q={L} \cdot{B} + \frac{e}{2}((r \times B)(r \times B)) is a constant of motion, where L = r x p is the angular momentum.

The Attempt at a Solution



I proceeded to calculate the conjugated momenta,

Px = mx' - ey/2
Py = my' + ex/2
Pz = mz'

Then I calculated the Hamiltonian. A lot of terms vanished and I ended up with just kinetic energy.

H = \frac{m}{2} (x'^2 + y'^2 + z'^2) which is a conserved quantity.

I was expecting to also end up with a term for the potential. Is this caused by the fact that the magnetic field does no work on the particle so it doesn't affect the total energy?

I expressed the hamiltonian in terms of the generalized momenta:

H= \frac{m}{2}(Px^2 + Py^2 + Pz^2 + \frac{(Be^2)(x^2+y^2)}{4}+eB(yPx-xPy))I tried re-arranging Q in terms of the vector position r = xî + yĵ + zk , while p would correspond to m(x' + y' + z')

After doing dot and cross product operations, I ended up with the following expression for Q:

Q = (xy' - yx')B + (e{B^2}/2) (x^2 + y^2)

which doesn't depend on z nor z', which gives me trouble when relating terms to the hamiltonian.

Is there any flaw in my procedure? I've tried rearranging terms, playing with algebra, but I don't come up with anything satisfactory.

Also, is the quantity Q a well known quantity? Is there something too obvious I'm missing here?

Thanks for your help. It is really, really appreciated.
 
Last edited:
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Siberion said:
Then I calculated the Hamiltonian. A lot of terms vanished and I ended up with just kinetic energy.

H = \frac{m}{2} (x'^2 + y'^2 + z'^2) which is a conserved quantity.

I was expecting to also end up with a term for the potential. Is this caused by the fact that the magnetic field does no work on the particle so it doesn't affect the total energy?

I expressed the hamiltonian in terms of the generalized momenta:

H= \frac{m}{2}(Px^2 + Py^2 + Pz^2 + \frac{(Be^2)(x^2+y^2)}{4}+eB(yPx-xPy))

In a uniform magnetic field, the scalar potential ##\phi## can be chosen to be zero. The vector potential ##\textbf{A}## can be chosen as ##\textbf{A} = \textbf{B} \times \textbf{r}/2 ##, so you could express the Hamiltonian in terms of the vector potential if you wished.

I tried re-arranging Q in terms of the vector position r = xî + yĵ + zk , while p would correspond to m(x' + y' + z')

After doing dot and cross product operations, I ended up with the following expression for Q:

Q = (xy' - yx')B + (e{B^2}/2) (x^2 + y^2)

which doesn't depend on z nor z', which gives me trouble when relating terms to the hamiltonian.

Is there any flaw in my procedure? I've tried rearranging terms, playing with algebra, but I don't come up with anything satisfactory.

It all looks ok to me. Try taking the time derivative of ##Q## and see if it equals zero.
 
Thanks a lot TSny, according to what I calculated, using E-L equations, its derivative indeed equals to zero. I got lost trying to rearrange the Hamiltonian, I should have checked the derivative of Q at first.
 

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