BeauGeste
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I'm not sure where to put this but I though DEQ would be a good start.
I have been dealing with the following differential equations:
<br /> <br /> \frac{d \mathbf{m_1}}{dt} = \gamma_1 \mathbf{m_1} \times \mathbf{B} - \frac{n_{2}}{\gamma_{cr}} \mathbf{m_1} +<br /> \frac{n_1}{\gamma_{cr}} \mathbf{m_2}<br /> <br /> <br />
<br /> <br /> \frac{d \mathbf{m_2}}{dt} = \gamma_2 \mathbf{m_2} \times \mathbf{B} - \frac{n_{1}}{\gamma_{cr}} \mathbf{m_2} +<br /> \frac{n_2}{\gamma_{cr}} \mathbf{m_1}<br /> <br />
The weird things is that \frac{d (\mathbf{m_1} + \mathbf{m_2})}{dt} is not independent of \gamma_{cr} even though when you add bottom and top deq's the \gamma_{cr} terms cancel out. All I know is that it has something to do with \gamma_{1} \neq \gamma_{2} .
Also when you convert to polar coordinates (i assume the m-vectors are only in the x-y direction and B is in the z-direction), the 2 diff. eq.'s take on a form where it's obvious that the \gamma_{cr}'s stick around.
I'm not too fluent in differential equations so I'm not sure if there's an obvious answer here.
Thanks.
I have been dealing with the following differential equations:
<br /> <br /> \frac{d \mathbf{m_1}}{dt} = \gamma_1 \mathbf{m_1} \times \mathbf{B} - \frac{n_{2}}{\gamma_{cr}} \mathbf{m_1} +<br /> \frac{n_1}{\gamma_{cr}} \mathbf{m_2}<br /> <br /> <br />
<br /> <br /> \frac{d \mathbf{m_2}}{dt} = \gamma_2 \mathbf{m_2} \times \mathbf{B} - \frac{n_{1}}{\gamma_{cr}} \mathbf{m_2} +<br /> \frac{n_2}{\gamma_{cr}} \mathbf{m_1}<br /> <br />
The weird things is that \frac{d (\mathbf{m_1} + \mathbf{m_2})}{dt} is not independent of \gamma_{cr} even though when you add bottom and top deq's the \gamma_{cr} terms cancel out. All I know is that it has something to do with \gamma_{1} \neq \gamma_{2} .
Also when you convert to polar coordinates (i assume the m-vectors are only in the x-y direction and B is in the z-direction), the 2 diff. eq.'s take on a form where it's obvious that the \gamma_{cr}'s stick around.
I'm not too fluent in differential equations so I'm not sure if there's an obvious answer here.
Thanks.