Deriving a Minkowski Force Matrix: Exploring 4-force

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71GA
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Our professor derived a Minkowski force like this:

[itex] F^\mu = \left[ \gamma(e\vec{E} + e(\vec{v}\times \vec{B})) , \gamma \frac{e \vec{E} \vec{v}}{c} \right][/itex]

Does this mean that i can write 4-force like this?

[itex] F^\mu = <br /> \begin{bmatrix}<br /> \gamma(e\vec{E} + e(\vec{v_x}\times \vec{B}))\\<br /> \gamma(e\vec{E} + e(\vec{v_y}\times \vec{B}))\\<br /> \gamma(e\vec{E} + e(\vec{v_z}\times \vec{B}))\\<br /> \gamma \frac{e \vec{E} \vec{v}}{c}<br /> \end{bmatrix}[/itex]

Short anwser would be ok. How do i put this into a matrix form from which i can get Lorentz matrix ##\Lambda## for boost in $x$ direction?
 
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It's easier if you start with the Faraday tensor (F) and write the force (f) as,
[tex] f^\mu = F^{\mu\nu}J_\nu[/tex]
then you can boost the field tensor F and J ( current) with [itex]\Lambda[/itex] to get the transformed force. The Wiki article might be relevant

http://en.wikipedia.org/wiki/Electromagnetic_tensor
 
Mentz114 said:
It's easier if you start with the Faraday tensor (F) and write the force (f) as,
[tex] f^\mu = F^{\mu\nu}J_\nu[/tex]
then you can boost the field tensor F and j ( current) with [itex]\Lambda[/itex] to get the transformed force.

I am new to this and have never encountered tensors before. This is why i have to use basic math. Some day when things are clear to me in simple math i can learn advanced math and swich to tensors.
 
71GA said:
I am new to this and have never encountered tensors before. This is why i have to use basic math. Some day when things are clear to me in simple math i can learn advanced math and swich to tensors.

OK.

I think what your professor has written is the electrostatic force after the field has been boosted by v = √(vx2+vy2+vz2). A 'boost' is a Lorentz, transformation, in case you're not familiar with the term. In this case, the tensor expressions are like matrix and vector operations so that

[itex]f'=F'\cdot J'[/itex], [itex]F' = \Lambda(v)\cdot F\cdot\Lambda(v)[/itex] and [itex]J' = \Lambda(-v)\cdot J[/itex].
 
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71GA said:
Our professor derived a Minkowski force like this:

[itex] F^\mu = \left[ \gamma(e\vec{E} + e(\vec{v}\times \vec{B})) , \gamma \frac{e \vec{E} \vec{v}}{c} \right][/itex]

Does this mean that i can write 4-force like this?

[itex] F^\mu = <br /> \begin{bmatrix}<br /> \gamma(e\vec{E} + e(\vec{v_x}\times \vec{B}))\\<br /> \gamma(e\vec{E} + e(\vec{v_y}\times \vec{B}))\\<br /> \gamma(e\vec{E} + e(\vec{v_z}\times \vec{B}))\\<br /> \gamma \frac{e \vec{E} \vec{v}}{c}<br /> \end{bmatrix}[/itex]

Short anwser would be ok. How do i put this into a matrix form from which i can get Lorentz matrix ##\Lambda## for boost in $x$ direction?
You have some misprints.
It should be
[itex] F^\mu = <br /> \begin{bmatrix}<br /> \gamma(e\vec{E} + e(\vec{v}\times \vec{B}))_x\\<br /> \gamma(e\vec{E} + e(\vec{v}\times \vec{B}))_y\\<br /> \gamma(e\vec{E} + e(\vec{v}\times \vec{B}))_z\\<br /> \gamma \frac{e \vec{v}\cdot\vec{E} }{c}<br /> \end{bmatrix}[/itex].
Then F^\mu transforms like any other four-vector.
 
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