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Electron boosted by an electric field

 
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Aug24-12, 06:00 PM   #1
 

Electron boosted by an electric field


I have an electric field,

[itex]\vec{E}=E_0 e^{-(x_1-ct)^2}\vec{e_2}[/itex]

that boosts an electron that initially is at rest at (0,0,0). I have to calculate the motion of the electron (supposing v<<c and Newton Law is valid). Then, calculate total flux of radiation.

As v<<c I suppose I can assume that the induced magnetic field due to the time-dependent electric field is negligible. Using Newton equation:

[itex]ma=q(E+E_{rad})[/itex]

In the text they give me the expression of the radiation electric field,

[itex]E_{rad}(\vec{x})=-∫d^3x' \frac{\dot{[\vec{j}]}}{c^2 |\vec{x}-\vec{x'}|}[/itex]

But I don't know how to calculate this.

My other doubt is about the calculation of the total flux, what expression shoud I use?

Thank you!
 
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Aug26-12, 11:28 PM   #2
 
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Are you trying to calculate the total power radiated ([itex]\oint \mathbf{S}_{\text{rad}} \cdot d\mathbf{a}[/itex]), or just the flux of [itex]\mathbf{E}_{\text{rad}}[/itex]? If it is the power you want, you can probably just use the Larmor formula.
 
Aug27-12, 03:23 AM   #3
 
Quote by gabbagabbahey View Post
Are you trying to calculate the total power radiated ([itex]\oint \mathbf{S}_{\text{rad}} \cdot d\mathbf{a}[/itex]), or just the flux of [itex]\mathbf{E}_{\text{rad}}[/itex]? If it is the power you want, you can probably just use the Larmor formula.
It is said 'total flux', but maybe it's only the contribution of the radiation field... I don't know.

Any idea for the first part?

thanks for your help
 
Aug27-12, 03:20 PM   #4
 
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Electron boosted by an electric field


Quote by lailola View Post
It is said 'total flux', but maybe it's only the contribution of the radiation field... I don't know.
I interpret that as the total power (flux of the Poyting vector at infiinity) radiated. So I think you should just be able to use the Larmor formula.

Any idea for the first part?
You mean calculating the motion of the electron? If so, there are only two forces affecting it. One is the electric force from the external field [itex]\mathbf{E}=E_0 e^{-(x_x1 - ct)^2)} \mathbf{e}_2[/itex]. Can you think of the other one?
 
Aug28-12, 10:45 AM   #5
 
Quote by gabbagabbahey View Post
I interpret that as the total power (flux of the Poyting vector at infiinity) radiated. So I think you should just be able to use the Larmor formula.



You mean calculating the motion of the electron? If so, there are only two forces affecting it. One is the electric force from the external field [itex]\mathbf{E}=E_0 e^{-(x_x1 - ct)^2)} \mathbf{e}_2[/itex]. Can you think of the other one?
The electric force from the radiation electric field?
 
Aug28-12, 03:02 PM   #6
 
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Quote by lailola View Post
The electric force from the radiation electric field?
If you mean the so-called radiation reaction force, then yes. Have you come across the Abraham-Lorentz formula yet? If so, use it and you will get a differential equation you can solve for the acceleration of the electron.
 
Aug28-12, 04:27 PM   #7
 
Quote by gabbagabbahey View Post
If you mean the so-called radiation reaction force, then yes. Have you come across the Abraham-Lorentz formula yet? If so, use it and you will get a differential equation you can solve for the acceleration of the electron.
But, then, I won't use the expression for the radiation electric field that is given in the text.
 
Aug29-12, 06:15 PM   #8
 
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Quote by lailola View Post
But, then, I won't use the expression for the radiation electric field that is given in the text.
The radiation electric field of the electron doesn't act on the electron. There is no force [itex]q\mathbf{E}_{rad}[/itex] on the electron. Rather, the radiation part of the electric and magnetic fields carry away momentum from the particle off to infinity. The rate at which that momentum is carried off (i.e. the reaction force on the electron) is given by the Abraham Lorentz formula.

Is your problem taken from a textbook (if so, please give the name of the text and the problem number)? Does the problem explicitly tell you to use the formula for [itex]\mathbf{E}_{rad}[/itex]?
 
Aug29-12, 06:19 PM   #9
 
No, it's just given.
 
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