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The problem (From Electromagnetism by Pollack and Stump, exercise 11.34)
A stationary charge of charge e and mass m encounters a electromagnetic wave with vector potential
[tex]\vec{A}=\vec{j}f(x-ct)[/tex]
where
[tex]\vec{j}[/tex]
is the unit vector in the y-direction. The scalar potential is zero.
What are the components of the velocity, as a function of time,
[tex]v_{x}, v_{y},v_{z}[/tex] ?
What I´m asking for is some insight into why the y-velocity is not a more complicated function. The y-velocity is:
[tex]v_{y}=\frac{e}{m}f(x-ct)[/tex]
More specifically, when the charge is stationary, it will begin to move in the y-direction because of the changing vector potential but the motion in the y-direction will result in the magnetic field causing a motion in the x-direction which in turn should (?) effect the y-velocity, but apparently it does not. Why?
I wrote the equations of motion but they don´t seem to give any insight into the y-motion. I determined the equations of motion by starting with the following equations:
[tex]\vec{E}=-\frac{\partial A}{\partial t}[/tex]
(In the equation above, A is a vector)
[tex]\vec{B}=\nabla X \vec{A} [/tex]
[tex]\vec{F}=e(\vec{E}+\vec{v}X\vec{B})[/tex]
The function f(x-ct) isn´t a problem if we use the chain rule
w = x -ct
[tex]\frac{\partial f}{\partial t}=\frac{df}{dw}\frac{\partial w}{\partial t}=-c\frac{df}{dw} [/tex]
likewise,
[tex]\frac{\partial f}{\partial x}=\frac{df}{dw}\frac{\partial w}{\partial x}=\frac{df}{dw} [/tex]
Note, I´ve been a little sloppy about notation regarding partials and full derivatives.
The resulting equations of motion are:
[tex]\frac{dv_{x}}{dt}=\frac{e}{m}v_{y}\frac{df}{dw}[/tex]
[tex]\frac{dv_{y}}{dt}=\frac{e}{m}(c\frac{df}{dw}-v_{x}\frac{df}{dw})[/tex]
[tex]\frac{dv_{z}}{dt}=0[/tex]
Thus the z-component of the velocity is zero since it started at rest.
The equations do give the right answer for the x-velocity if I assume the value of the y-velocity. But they aren´t helpful in giving me the y-velocity. Note that the two equations involving vy and vx have an additional unknown df/dw. I´ve tried everything, even eliminating df/dw and getting an equation that appears to have just x-dependence on one side and y-dependence on the other which one could solve by setting each to a constant but that doesn´t seem to work either.
A stationary charge of charge e and mass m encounters a electromagnetic wave with vector potential
[tex]\vec{A}=\vec{j}f(x-ct)[/tex]
where
[tex]\vec{j}[/tex]
is the unit vector in the y-direction. The scalar potential is zero.
What are the components of the velocity, as a function of time,
[tex]v_{x}, v_{y},v_{z}[/tex] ?
What I´m asking for is some insight into why the y-velocity is not a more complicated function. The y-velocity is:
[tex]v_{y}=\frac{e}{m}f(x-ct)[/tex]
More specifically, when the charge is stationary, it will begin to move in the y-direction because of the changing vector potential but the motion in the y-direction will result in the magnetic field causing a motion in the x-direction which in turn should (?) effect the y-velocity, but apparently it does not. Why?
I wrote the equations of motion but they don´t seem to give any insight into the y-motion. I determined the equations of motion by starting with the following equations:
[tex]\vec{E}=-\frac{\partial A}{\partial t}[/tex]
(In the equation above, A is a vector)
[tex]\vec{B}=\nabla X \vec{A} [/tex]
[tex]\vec{F}=e(\vec{E}+\vec{v}X\vec{B})[/tex]
The function f(x-ct) isn´t a problem if we use the chain rule
w = x -ct
[tex]\frac{\partial f}{\partial t}=\frac{df}{dw}\frac{\partial w}{\partial t}=-c\frac{df}{dw} [/tex]
likewise,
[tex]\frac{\partial f}{\partial x}=\frac{df}{dw}\frac{\partial w}{\partial x}=\frac{df}{dw} [/tex]
Note, I´ve been a little sloppy about notation regarding partials and full derivatives.
The resulting equations of motion are:
[tex]\frac{dv_{x}}{dt}=\frac{e}{m}v_{y}\frac{df}{dw}[/tex]
[tex]\frac{dv_{y}}{dt}=\frac{e}{m}(c\frac{df}{dw}-v_{x}\frac{df}{dw})[/tex]
[tex]\frac{dv_{z}}{dt}=0[/tex]
Thus the z-component of the velocity is zero since it started at rest.
The equations do give the right answer for the x-velocity if I assume the value of the y-velocity. But they aren´t helpful in giving me the y-velocity. Note that the two equations involving vy and vx have an additional unknown df/dw. I´ve tried everything, even eliminating df/dw and getting an equation that appears to have just x-dependence on one side and y-dependence on the other which one could solve by setting each to a constant but that doesn´t seem to work either.
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