This question is about my curiosity about the relativistic Kepler equation of which I am reading in a recent paper. Actually, I am only interested in an introductory concept stated in paper.
Let $$ m\ddot{x} = -\alpha\frac{x}{|x|^3}\quad x\in\mathbb{R}^n\setminus\{0\},$$ be the the Kepler problem of classical mechanics, where $\alpha = m_1 m_1 G$ with $m_1, m_2$ denote the mass of the two objects and $G$ the gravitational constant.
As a well-known fact, it is stated that including relativistic effects in the above problem can be done in two ways:
- by using the relativistic kinetic energy (instead of the usual one), which makes the previous equation turns into
$$ \frac{d}{dt}\left(\frac{m \dot{x}}{\sqrt{1- |\dot{x}|^2/c^2}}\right) =-\alpha \frac{x}{|x|^3} \quad x\in\mathbb{R}^n\setminus\{0\},$$
where $c>0$ is the speed of light.
- A second relativistic variant comes imposing a correction for the gravitational potential, precisely $$V(x) = -\frac{k(c)}{|x|} -\frac{2\lambda(c)}{|x|^2},$$
where $k(c)$ and $\lambda(c)$ are suitable positive constants, so that the starting problem turns into $$m\ddot{x} = -\frac{k(c)}{|x|} -\frac{2\lambda(c)}{|x|^2} \quad x\in\mathbb{R}^n\setminus\{0\}.$$
The thing that I found relevant, that both the equations are said to be relativistic.
My question is: is that a prerogative only of this Kepler problem or maybe this "double" relativistic formulation can be stated also for the Lorentz force equation?
I know the relativistic Lorentz force equation only in the version
$$ \frac{d}{dt}\left(\frac{m \dot{x}}{\sqrt{1- |\dot{x}|^2/c^2}}\right) =q(E(t, x) +\dot{x}\times B(t, q)) \quad x\in\mathbb{R}^n\setminus\{0\},$$ where $E$ and $B$ denote the electric and magnetic fields respectively.
I hope someone could give any reference or help in understanding if it holds also for Lorentz force equation (also "roghly speaking" answers will be accepted).
Thank you in advance.
Trying to make sense of this question:
The non-relativistic version of the Lorentz force equation is $$\tag{1} m\,\ddot{\boldsymbol x}=q\,(\boldsymbol{E}+\dot{\boldsymbol x}\times \boldsymbol{B})\,. $$ According to §3.1 of Gravitation by Misner, Thorne & Wheeler the relativistic Lorentz force equation is $$\tag{2} \frac{d\boldsymbol{P}}{d\tau}=\gamma \,q\,(\boldsymbol{E}+\boldsymbol{v}\times \boldsymbol{B}) $$ where $\tau$ is proper time and $\boldsymbol{P}$ is the spacial component of four-momentum.
Since both sides of (1) need to be corrected to arrive at a relativistic equation (2) you are free to call this "double" relativistic. Most people call (2) frame-independent, that is, regardless of what inertial frame you are in you will measure the same change of momentum w.r.t. proper time $\tau\,$. The heart of SR is not that we come up with relativistic corrections to classic equations. It is rather the insight that we are looking for quantities that are the same for observers in different inertial frames. Proper time, proper length, four-velocity, four-acceleration, four-momentum and proper force are examples of this.