- #1
TheDestroyer
- 402
- 1
How do we derive the lagrangnian of a charged particle in 3 dimentions?
Actually this is not the real question but it's a step to it, In three dimentions we have:
[tex]{\cal L} = - m_ \circ c^2 \sqrt {1 - \frac{{v^2 }}{{c^2 }}} + q\overrightarrow v \overrightarrow A - q\phi [/tex]
And As we Know in four dimensional system:
We Have:
[tex]
\eqalign{
Speed:U_\mu = \gamma \left( {\overrightarrow v ,ic} \right) \Rightarrow U_\mu ^2 = \gamma ^2 \left( {v^2 - c^2 } \right) = - c^2 \cr
,,,{\rm{Vector Potential: }}A_\mu = \left( {\overrightarrow A ,i{\phi \over c}}\right)\Rightarrow \cr { \rm{A}}_\mu U_\mu = \overrightarrow A \cdot \overrightarrow v - \phi \cr}
[/tex]
Substituting This to the Lagrangian we get, and note that gamma is taken in hamiltons principle ds=L dt in 4-d becomes ds= L dt/Gamma:
[tex]
{\cal L} = m_ \circ U_\mu ^2 + qA_\mu U_\mu
[/tex]
My Professor says that we should write it like this without doing anything else:
[tex]
{\cal L} = {1 \over 2}m_ \circ U_\mu ^2 + qA_\mu U_\mu
[/tex]
And that is to make it like the 3d system, 1/2 mv^2, IS THIS CORRECT? THIS IS MY FIRST QUESTION
And my second question, If we used the 4d motion equation on this lagrangian:
[tex]
{d \over {dt}}{{\partial {\cal L}} \over {\partial U_\mu }} - {{\partial {\cal L}} \over {\partial x_\mu }} = 0
[/tex]
do we get the lorentz force like the 3d system? i didn't get it when i tried, Any one can help?
Can you explain the solution of this motion equation? my professor didn't know it even !
Thanks, Please hurry with the answer
Actually this is not the real question but it's a step to it, In three dimentions we have:
[tex]{\cal L} = - m_ \circ c^2 \sqrt {1 - \frac{{v^2 }}{{c^2 }}} + q\overrightarrow v \overrightarrow A - q\phi [/tex]
And As we Know in four dimensional system:
We Have:
[tex]
\eqalign{
Speed:U_\mu = \gamma \left( {\overrightarrow v ,ic} \right) \Rightarrow U_\mu ^2 = \gamma ^2 \left( {v^2 - c^2 } \right) = - c^2 \cr
,,,{\rm{Vector Potential: }}A_\mu = \left( {\overrightarrow A ,i{\phi \over c}}\right)\Rightarrow \cr { \rm{A}}_\mu U_\mu = \overrightarrow A \cdot \overrightarrow v - \phi \cr}
[/tex]
Substituting This to the Lagrangian we get, and note that gamma is taken in hamiltons principle ds=L dt in 4-d becomes ds= L dt/Gamma:
[tex]
{\cal L} = m_ \circ U_\mu ^2 + qA_\mu U_\mu
[/tex]
My Professor says that we should write it like this without doing anything else:
[tex]
{\cal L} = {1 \over 2}m_ \circ U_\mu ^2 + qA_\mu U_\mu
[/tex]
And that is to make it like the 3d system, 1/2 mv^2, IS THIS CORRECT? THIS IS MY FIRST QUESTION
And my second question, If we used the 4d motion equation on this lagrangian:
[tex]
{d \over {dt}}{{\partial {\cal L}} \over {\partial U_\mu }} - {{\partial {\cal L}} \over {\partial x_\mu }} = 0
[/tex]
do we get the lorentz force like the 3d system? i didn't get it when i tried, Any one can help?
Can you explain the solution of this motion equation? my professor didn't know it even !
Thanks, Please hurry with the answer
Last edited: