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Natural Science Forum / Physics / Research / October 2004Angular momentum and magnetic monopoles
I was thinking about magnetic monopoles, and I thought there might be some trouble with angular momentum. If you calculate the angular momentum density rXEXB for a point charge (e) at (0,0,d/2) and a magnetic monopole (m) at (0,0,-d/2), you get for the z-component something like: Lz = e*m*d*R^2 /(r_e^3 r_m^3) where R = (x^2+y^2)^(1/2) and r_e is the distance from the e_pole, and r_m is the distance from the m_pole. If the 2 poles were fired at each other along the z-axis, suppose they would miss each other and continue along the z-axis, then the total field angular momentum would undergo changes as this happens. A striking feature would be a sign flip as (d) goes from positive to negative. Is this a problem? The poles e and m could be spinless particles, so where does the angular momentum go/ come from? Gerard Hello - > I was thinking about magnetic monopoles, and I thought there > might be some trouble with angular momentum. [snip] > If the 2 poles were fired at each other along the z-axis, suppose > they would miss each other and continue along the z-axis, then [quoted text clipped - 8 lines] > The poles e and m could be spinless particles, so where does > the angular momentum go/ come from? Caveat: I haven't really thought this through. However, is it reasonable kinematically to think that the charge and magnetic monopoles will fly straight past each other along the z-axis? i.e. In a frame co-moving with the charged particle, the magnetic field from the monopole will be transformed into some electric field that will push on the charge, and vice versa. Presumably, when the final post-pass "orbital" angular momentum around the z-axis is calculated, the "orbital" plus "field" total will be conserved.... Too lazy to do the real calculation, Doug [..] > Caveat: I haven't really thought this through. However, > is it reasonable kinematically to think that the charge and [quoted text clipped - 6 lines] > calculated, the "orbital" plus "field" total will be > conserved.... Hmm, maybe they could orbit each other... interesting. But then, why should there be a force between the 2 poles? A moving electric charge feels a force from the magnetic field proportional to its speed. But the speed (v) could be slow, and because the forces are then small, remain slow. hmm, OK, slow speed means a long time for the force to act, and Force*time = change in momentum ~ e v B s/v ~ e B s ~ independent of speed (v) OK, yeh. Thinking about it, the 2 poles will not orbit each other. This is clear if you think about the fields, the velocities and the forces using the "right hand rule" ( or was it the left hand?). What will happen: Suppose the particles move toward each other along the z-axis, but very slightly offset along the y-axis so that they narrowly miss each other. The the Lorentz force on the e-pole, and its magnetic analog on the m-pole will be in the direction. This will produce a momentum change in the z-direction. I agree that to do the actual calculation would be a bit too laborious. But I am starting to believe that the solution is that the 2 poles will do something like: ^ ^ / | ee X ee -> eeeeeeeee mmmmmm <- Z -> mm mm / V They need a very slight offset to initiate this, but the smaller the offset, the bigger the force, because of the 1/r^2 Coulomb law. Gerard |
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