# Why is Lorentz force frame independent?

Recently I started studying Magnetics and I came across a statement in my book which confused me.It read:

Magnetic force is frame dependent, Electric force is frame dependent but lorentz force is frame independent

I can understand magnetic force being frame dependent as it is a velocity dependent force and velocity depends on the frame from which it is measured.However, I can't understand why electric force is frame dependent.Does frame dependent here mean that in a non inertial frame of reference or a moving inertial frame, the electric force becomes magnetic force as charge which is at rest moves with respect to this frame?

Also, Lorentz force being frame independent makes me think that some components that arise due to shifting frame of reference, cancel each other out making the resultant lorentz force same.However, I am not able to get the complete picture.

Can anyone give me the reason for this without using complex mathematics/theory of relativity? What's the intuition behind this?

• The correct way to approach the question, "why is the electric field frame dependent", is to ask yourself, "is it frame independent?". If you cannot prove the latter (or give a counter-example where it is frame dependent) then in general the first question is answered. Jul 7, 2020 at 10:20
• As to the reason behind the invariance of the Lorentz force, you would need to explicitly show it using Lorentz transformation for a force and other tensors/pseudotensors. I don't think there is an intuitive way to get to this. Jul 7, 2020 at 10:22
• Which book ? The Lorentz force 3-vector is frame dependent while the Lorentz force 4-vector is also frame dependent. In what sense your book states that the Lorentz force 3-vector is frame independent ??? This is incorrect. Jul 7, 2020 at 12:52
• Vectors like force are certainly not Lorentz invariants. Jul 7, 2020 at 18:30

The Lorentz force 3-vector is only frame independent under a Galilean transformation. For instance, suppose you translate into a new frame moving at $$\mathbf{V}_{o}$$ relative to the original frame, then the transformed vectors would be: \begin{align} \mathbf{E}' & = \mathbf{E} + \mathbf{V}_{o} \times \mathbf{B} \tag{0a} \\ \mathbf{B}' & = \mathbf{B} \tag{0b} \\ \mathbf{v}' & = \mathbf{v} - \mathbf{V}_{o} \tag{0c} \end{align} If we then put these into the Lorentz force we find: \begin{align} \mathbf{F}' & = q \left[ \mathbf{E}' + \mathbf{v}' \times \mathbf{B}' \right] \tag{1a} \\ & = q \left[ \left( \mathbf{E} + \mathbf{V}_{o} \times \mathbf{B} \right) + \left( \mathbf{v} - \mathbf{V}_{o} \right) \times \mathbf{B} \right] \tag{1b} \\ & = q \left[ \mathbf{E} + \mathbf{v} \times \mathbf{B} \right] \tag{1c} \end{align}
Under a proper Lorentz transformation, the Lorentz force 3-vector and 4-vector is not frame independent. The electric and magnetic field (in cgs units now) 3-vectors transform as: \begin{align} \mathbf{E}' & = \gamma \left( \mathbf{E} + \frac{ \mathbf{V}_{o} }{ c } \times \mathbf{B} \right) - \frac{ \gamma^{2} }{ \gamma + 1 } \frac{ \mathbf{V}_{o} }{ c } \left( \frac{ \mathbf{V}_{o} }{ c } \cdot \mathbf{E} \right) \tag{2a} \\ \mathbf{B}' & = \gamma \left( \mathbf{B} - \frac{ \mathbf{V}_{o} }{ c } \times \mathbf{E} \right) - \frac{ \gamma^{2} }{ \gamma + 1 } \frac{ \mathbf{V}_{o} }{ c } \left( \frac{ \mathbf{V}_{o} }{ c } \cdot \mathbf{B} \right) \tag{2b} \end{align} while the 3-vector velocity transforms according to the addition of velocities and $$\gamma$$ is the Lorentz factor. One can see that Equations 2a and 2b reduce to 0a and 0b in the limit $$V_{o} \ll c$$ because $$\gamma \rightarrow 1$$ and the 2nd term in Equations 2a and 2b are 2nd order in $$\tfrac{ V_{o} }{ c }$$. The $$\tfrac{ \mathbf{V}_{o} }{ c } \times \mathbf{E}$$ drops out when you convert back to SI units because $$B \rightarrow c \ B$$ thus there is a factor of $$\tfrac{ V_{o} }{ c^{2} }$$ in that term.