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| bio | website | |
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| location | Kolkata,India[91-33-25514464] | |
| age | ||
| visits | member for | 1 year, 6 months |
| seen | Mar 27 at 12:06 | |
| stats | profile views | 225 |
Author/Teacher from India. Interested in General Relativity and other areas of physics
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Aug 14 |
comment |
How does one prove:$\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}=0$? Spatial variation of $\vec{B}$ can produce an electric field:$\nabla \times \vec{B}=\mu_0 \vec{j}+\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$.If $\vec{E}\cdot\vec{dr}=\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}$ to produce non-zero work then,$\vec{E}=\nabla(\vec{\mu_m}\cdot\vec{B})$. How do you prove that by direct calculation?[here you have a view of the problem through the differential equations which are supposed to represent the physical conditions accurately.] |
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Aug 13 |
comment |
How does one prove:$\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}=0$? Let the magnetic field be in the x direction.If the observer moves along the y direction with a uniform speed v,we have $\vec{E}=\gamma(\vec{v}\times\vec{B})$. The electric field accelerates the charged changing its speed and doing work. Initially there was a acceleration normal to the direction of motion. But now we have acceleration due to change of speed also . If the net acceleration changes the radiating power should be different from the two frames.But energy is not an invariant wrt frame transformation though it is a conserved quantity. It might have different values in differentframes |
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Aug 13 |
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How does one prove:$\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}=0$? You can think of redistribution of charges and currents elsewhere(gadgets producing the field) to produce an electric field due to variations of magnetic field at some fixed point. But a pure magnetic field which does not vary with time cannot perform any work.What would be your mechanism for accounting for such variations of $\vec{B}$ in evaluating:$ \nabla (\vec{\mu_m}.\vec{B}).\vec{dr}$. What would be the mechanism of including electric field in the said evaluation of elementary work? |
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Aug 13 |
asked | How does one prove:$\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}=0$? |
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Aug 12 |
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The Stern Gerlach Experiment Revisited In so far as the last equation in the answer is concerned, the value of x is the same for all particles passing though the region of interaction of the magnetic field. The value of y is confined to the dimemsions of the region of interaction. Its becomes negligible due to the large vale of $\frac{C_1}{k}$. We have in effect:$\frac{C_1}{k}z+\frac{c_2}{k}=(V\pm v_m \frac{C_1}{k})t$ |
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Aug 12 |
answered | The Stern Gerlach Experiment Revisited |
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Aug 11 |
revised |
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Aug 11 |
revised |
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Aug 11 |
revised |
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Aug 11 |
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Aug 11 |
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Aug 11 |
awarded | Critic |
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Aug 11 |
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The Stern Gerlach Experiment Revisited We may take $\langle v_x \rangle=\langle v_y \rangle$. They correspond to internal motion.The averages masy be taken over time or space and they should be equal in consideration of the Ergodicity principle |
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Aug 11 |
revised |
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Aug 11 |
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The Stern Gerlach Experiment Revisited Do you find anything counter-intuitive with the above considerations(ie, the answer provided)?Classical prejudice seems to be working out excellently ,at least with the problem at hand---the Stern-Gerlach Experiment. |
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Aug 11 |
revised |
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Aug 11 |
answered | The Stern Gerlach Experiment Revisited |
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Aug 10 |
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The Stern Gerlach Experiment Revisited $\vec{F}=\nabla(\vec{\mu}.\vec{B})\approx\vec{e}_{z}\mu_{z}\frac{\partial B_{z}}{\partial z}=\vec{e}_{z}F_{z}$ Force corresponding to the above formula acts along the direction of $\vec{B}=\vec{e_z}B_z$, if $B_y=B_z=0$ The above force is not identical with the force given by $F=q(\vec{v}\times \vec{B})$. Magnetic force cannot act along the magnetic line of force |
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Aug 10 |
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The Stern Gerlach Experiment Revisited Classical theory fails to predict the "excluded middle". This goes in conjunction with the "classic" exclusion or unaccountability of how spin generates retardation in the x-y plane if $B_x=B_y=0$. Work done by magnetic field for any infinitesimal path is zero. This idea is there in the classical theories and needs to be respected in any type of classical reasoning. |
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Aug 10 |
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The Stern Gerlach Experiment Revisited How is the retarding force (in the above comment) generated by spin?if $B_x=B_y\approx 0$ then acceleration in the x and the y-directions due to spin alone should be zero. |
