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I have a question about the Lorentz force on a charge $q$:

$$\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B}).$$

I understand that if one performs a Lorentz boost then the electric field $\mathbf{E}$, magnetic field $\mathbf{B}$ and the velocity of the particle $\mathbf{v}$ will all change.

Does the Lorentz force $\mathbf{F}$ change under a Lorentz boost as well?

I would have thought that it shouldn't as the force is something that has an observable effect: it can be measured directly using an accelerometer.

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2 Answers

In special relativity, the frame-independent quantity is known as the 4-force, and it is defined as the time derivative of the 4-momentum with respect to proper time,

$$ \mathbf{F} = \frac{d\mathbf{P}}{d\tau} $$

The more familiar force vector is the spatial component of the 4-force, hence it is not invariant and depends on the lorentz frame chosen; in particular, it changes when we perform a boost to transform between lorentz frames. Here's a link to a derivation of the exact form of the transformation law.

To answer your question about the accelerometers, acceleration is changed by a boost as well!

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To properly deal with E&M in special relativity, you start with the vector potential $A_{a} = (c\phi, A_{x},A_{y},A_{z})$ (remember that ${\vec B} = {\vec \nabla} \times {\vec A}$). Then, the maxwell tensor is $F_{ab} = \nabla_{a}A_{b} - \nabla_{b}A_{a}$ ${}^{1}$. Then, for a four-velocity $u^{a}$, we have $u^{a}\nabla_{a}u^{b} = F_{ca}u^{a}\eta^{bc}$. This will obviously transform under a boost like a vector on both sides. So, the 3-force will go $F_{x} \rightarrow v\gamma F_{x}$ under an x-boost, with the other two spatial vectors unchanged.

${}^{1}$Note that $\nabla_{a}$ is now the four-dimensional one

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This works, but since the question can be answered based on generic facts about the 4-force and 3-force, as in zodiac's answer, that approach seems preferable to me. –  Ben Crowell Sep 7 '13 at 16:51
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