# Minkowski force

1. Dec 12, 2012

### 71GA

Our professor derived a Minkowski force like this:

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

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

$F^\mu = \begin{bmatrix} \gamma(e\vec{E} + e(\vec{v_x}\times \vec{B}))\\ \gamma(e\vec{E} + e(\vec{v_y}\times \vec{B}))\\ \gamma(e\vec{E} + e(\vec{v_z}\times \vec{B}))\\ \gamma \frac{e \vec{E} \vec{v}}{c} \end{bmatrix}$

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?

2. Dec 12, 2012

### Mentz114

$$f^\mu = F^{\mu\nu}J_\nu$$
then you can boost the field tensor F and J ( current) with $\Lambda$ to get the transformed force. The Wiki article might be relevant

http://en.wikipedia.org/wiki/Electromagnetic_tensor

3. Dec 12, 2012

### 71GA

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.

4. Dec 12, 2012

### Mentz114

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

$f'=F'\cdot J'$, $F' = \Lambda(v)\cdot F\cdot\Lambda(v)$ and $J' = \Lambda(-v)\cdot J$.

Last edited: Dec 12, 2012
5. Dec 12, 2012

### Meir Achuz

You have some misprints.
It should be
$F^\mu = \begin{bmatrix} \gamma(e\vec{E} + e(\vec{v}\times \vec{B}))_x\\ \gamma(e\vec{E} + e(\vec{v}\times \vec{B}))_y\\ \gamma(e\vec{E} + e(\vec{v}\times \vec{B}))_z\\ \gamma \frac{e \vec{v}\cdot\vec{E} }{c} \end{bmatrix}$.
Then F^\mu transforms like any other four-vector.

Last edited: Dec 12, 2012