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mmcf
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Hey I have a question that was almost completely answered in this thread:
https://www.physicsforums.com/showthread.php?t=116098
Specifically someone interpreted his question to be how exactly do you get that
the term in the Lagrangian corresponding to the magnetic potential is
[tex]-q\vec{v}\cdot\vec{A}[/tex]
Sadly, I had already worked up to the second to last line prior to his post, and I'm not clear
on what happened at that point. I would assume he added in [tex]\partial \phi/\partial x[/tex] because the electric potential is zero and that expression will look more like the typical [tex]E_x[/tex] that you see. The part that loses me though is how did he turn the term
[tex]\dot{x}\frac{\partial A_x}{\partial x}+\dot{y}\frac{\partial A_y}{\partial x}+\dot{z}\frac{\partial A_z}{\partial x}[/tex] into [tex]\dot{y}(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})+\dot{z}(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z})[/tex]
Just from the looks of things it might be that:
[tex]\dot{x}\frac{\partial A_x}{\partial x} + \dot{y}\frac{\partial A_x}{\partial y} + \dot{z}\frac{\partial A_x}{\partial z} = 0[/tex]
I don't really see why that should be true, so is there another way he could have done that, or what is the reasoning behind that equation? Also, sorry in advance. I did take a look at Jackson but he jumps into some relativistic stuff that I'm not ready for right now and we don't have a copy of Franklin in our library.
https://www.physicsforums.com/showthread.php?t=116098
Specifically someone interpreted his question to be how exactly do you get that
the term in the Lagrangian corresponding to the magnetic potential is
[tex]-q\vec{v}\cdot\vec{A}[/tex]
da_willem said:The only proofs I have seen rely on the Lagrangian principle. E.g.
The Lagrangian
[tex]L=T-V=\frac{1}{2}mu^2 + q\vec{u} \cdot \vec{A} [/tex]
Subsituted in the Euler Lagrange equations
[tex]\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial x}[/tex]
Using generalised coordinates x, y, z with generalized velocities [tex]\dot{x}, \dot{y}, \dot{z}[/tex] and [tex]u^2=\dot{x}^2+\dot{y}^2+\dot{z}^2[/tex]
yields
[tex]\frac{d}{dt}(m\dot{x}+qA_x)=q\frac{\partial}{\partial x}(\dot{x}A_x+\dot{y}A_y+\dot{z}A_z)[/tex]
[tex]m \ddot{x}+q \frac{\partial A_x}{\partial t} =q(\dot{x}\frac{\partial A_x}{\partial x}+\dot{y}\frac{\partial A_y}{\partial x}+\dot{z}\frac{\partial A_z}{\partial x})[/tex]
or after rearranging
[tex]m \ddot{x}=-q(\frac{\partial \phi}{\partial x}+\frac{\partial A_x}{\partial t})+q[\dot{y}(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})+\dot{z}(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z})]=qE_x+q(\dot{y}B_z-\dot{z}B_y)[/tex]
Which is exactly the x-component of the usual Lorentz force. So the potential term [tex]V=-q\vec{u} \cdot \vec{A} [/tex] can be motivated by the fact that it yields the right equation of motion.
Sadly, I had already worked up to the second to last line prior to his post, and I'm not clear
on what happened at that point. I would assume he added in [tex]\partial \phi/\partial x[/tex] because the electric potential is zero and that expression will look more like the typical [tex]E_x[/tex] that you see. The part that loses me though is how did he turn the term
[tex]\dot{x}\frac{\partial A_x}{\partial x}+\dot{y}\frac{\partial A_y}{\partial x}+\dot{z}\frac{\partial A_z}{\partial x}[/tex] into [tex]\dot{y}(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})+\dot{z}(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z})[/tex]
Just from the looks of things it might be that:
[tex]\dot{x}\frac{\partial A_x}{\partial x} + \dot{y}\frac{\partial A_x}{\partial y} + \dot{z}\frac{\partial A_x}{\partial z} = 0[/tex]
I don't really see why that should be true, so is there another way he could have done that, or what is the reasoning behind that equation? Also, sorry in advance. I did take a look at Jackson but he jumps into some relativistic stuff that I'm not ready for right now and we don't have a copy of Franklin in our library.