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- Thread starter alfredblase
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Meir Achuz

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The inhomegenity can be made large enough so that there is significant deflection before much rotation takes place. You can see what is needed from the equations. You would have to assume a classical moment of inertia of the neutral particle.

Since QM is needed, this classical explanation is really irrelevant.

The QM explanation: The particle (if spin 1/2) can only be either up or down in the direction of the B field. The energy difference between the two states is \Delta E=2B\cdot\mu. The spin can only be flipped by an oscillating magnetic field of frequency \hbar\omega=\Delta E.

The field in the SG experiment are static, so no spin flip takes place

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Meir Achuz said:The energy difference between the two states is [tex]\Delta E=2B\cdot\mu[/tex]. The spin can only be flipped by an oscillating magnetic field of frequency [tex]\hbar\omega=\Delta E[/tex]

Thanks for your reply. Just thought i'd quote with the equations showing in tex. Still not sure I understand exactly how (in the classical view) increasing inhomogeneity decreases the ammount of rotation. Would I be right in saying that because the dipole is moving and passing through an effecctively random changing field it never has time to align itself with any field lines? But in all diagrams I have seen the shape of the magnets is the same all allong the path of the beam and therefore it would be as if the dipole had been in a particular B field all through its path and so it would be rotated regardless of the inhomogeneity along the axis perpendicular to its path... I guess I dont understand.

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Actually I think I see now. What is meant by inhomogeneity is not necessarily about the diretion of the field lines but more about the increasing strength of the field in the z direction. If this gradient is made large enough the dipole will be deflected significantly by this magnetic gradient before it has time to rotate much, and the deflection vector depends upon the dipole magnetic moment vector. =)

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