I Calculating Force & Acceleration in Electrostatic Field (B=0)

[MichPod]

For a case of electrostatic field (B is equal zero), how should the force acting on a moving charge be calculated if we want to take into account all the relativistic effects? Also would it be correct to calculate the acceleration of the charge as a=F/m, or should some other formula be used? For simplicity, as a special case, let's consider the velocity of the charge to be orthogonal to the vector E of the electric field.

 

[ergospherical]

In an electromagnetic field the Lagrangian of the charge is $L = -m \sqrt{1-v^2} + q \mathbf{A} \cdot \mathbf{v} - q \phi$ therefore by Lagrange's equation $$\dfrac{d}{dt} \left( \dfrac{m\mathbf{v}}{\sqrt{1-v^2}} + q \mathbf{A} \right) = q\left[ (\mathbf{v} \cdot \nabla) \mathbf{A} + \mathbf{v} \times (\nabla \times \mathbf{A}) - \nabla \phi \right]$$Since the electric and magnetic field vectors are respectively defined $\mathbf{E} = - \dfrac{\partial \mathbf{A}}{\partial t} - \nabla \phi$ and $\mathbf{B} = \nabla \times \mathbf{A}$ this becomes $\dfrac{d}{dt} \left(\dfrac{m\mathbf{v}}{\sqrt{1-v^2}} \right) = q( \mathbf{E} + \mathbf{v} \times \mathbf{B})$

 

[pervect]

For a case of electrostatic field (B is equal zero), how should the force acting on a moving charge be calculated if we want to take into account all the relativistic effects? Also would it be correct to calculate the acceleration of the charge as a=F/m, or should some other formula be used? For simplicity, as a special case, let's consider the velocity of the charge to be orthogonal to the vector E of the electric field.

You might want to look at wiki, https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity, or a textbook, if you have access and the appropriate background to read a textbook.

With B=0, you will see that F = qE, where E is the electric field.

You will also see the relativisitic transformation laws for the electric fields E and B, which do not transform as vectors. This only matters if you need to transform frames of reference. If you pick a frame of reference and stick with it, it may not be necessary to know how to correctly transform between frames.

The relationship between force and acceleration in SR is moderately complex, acceleration is not necessarily in the direction of the force (though in some special circumstances it can be).

One simple treatment that gives the correct results is the concept of longitudinal and transverse mass. See for instance https://en.wikipedia.org/wiki/Mass_in_special_relativity#Transverse_and_longitudinal_mass. As wiki mentions, these concepts were used by Einstein early on, but later abandoned by him, and others. So they're not a modern treatment, but may be useful as an introduction to someone who doesn't want the full treatment.

 

[vanhees71]

A simple treatment can also be found in Sects. 2.1 and 2.2 of

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

The upshot of the heuristics is that you can assume that Newtonian mechanics is (at least approximately) correct in the momentaneous rest frame of the particle. Then via Lorentz invariance you can extend the dynamics to the general case, which leads you to the conclusion that for massive particles you can just use the proper time of the particle instead of coordinate time, leading to the definition of the four-velocity and the four-momentum
$$p^{\mu}=m \mathrm{d}_{\tau} x^{\mu}.$$
since
$$(\mathrm{d}_{\tau} x^{\mu}) (\mathrm{d}_{\tau} x_{\mu})=c^2=\text{const}$$
the "on-shell condition"
$$p_{\mu} p^{\mu}=m^2 c^2$$
holds and thus the relativistic equation of motion
$$\mathrm{d}_{\tau} p^{\mu}=K^{\mu},$$
where $K^{\mu}$ is in general a function of $x^{\mu}$ and $p^{\mu}$, implies the constraint
$$p_{\mu} \mathrm{d}_{\tau} p^{\mu}=0=p_{\mu} K^{\mu}.$$
That's why only three equations of motion are indpendent, and the fourth follows, and you can as well write the equation of motion for the spatial part of the above manifestly covariant version in terms of coordinate-time derivative, using $\mathrm{d}_{\tau}=\gamma \mathrm{d}_t$. From this you get
$$\vec{p}=m \mathrm{d}_{\tau} \vec{x}=m \gamma \mathrm{d}_t \vec{x}.$$
and
$$\mathrm{d}_{\tau} \vec{p}=\gamma \mathrm{d}_t \vec{p}=\vec{K}$$
or
$$m \mathrm{d}_t (\gamma \mathrm{d}_t \vec{x})=\frac{1}{\gamma} \vec{K}=\vec{F}.$$
Now of course all the beauty of the covariant formulation is gone, but it's in a similar form as in Newtonian physics and still fully relativistic.