Relation between Force and Potential in a relativistic context

Question

In a relativistic context we define the four-force as, $$\frac{\mbox{d}}{\mbox{d}s}p=F$$ In the particle frame the four-force must have the form $$F=\left(0,\ \mathbf{f}\right)$$ As it must be orthogonal to the four-momentum. Thus, we should have in the laboratory frame a force given by the Lorentz transform, $$F=\left(\gamma\frac{\mathbf{f}\cdot\mathbf{v}}{c^2},\ \mathbf{f}+\frac{\gamma^2}{c^2\left(\gamma+1\right)}\left(\mathbf{f}\cdot\mathbf{v}\right)\mathbf{v}\right)$$ My question is, what should one use the four force equal to? As an example, if one would like to study a harmonic oscillator it must set what equal to $-m\omega^2\mathbf{x}$? We should set the force in the particle frame, or in the lab frame? My intuition is to set the force in the particle frame, as this is the one to which all observers agree on. But, what it seems is that this does not agree with the lagrangian formulation of relativistic mechanics, for example for a relativistic lagrangian would be like, $$L=-\frac{mc^2}{\gamma}-\frac{m\omega^2}{2}{\mathbf{x}}^2$$ Which gives a equation of motion, $$\frac{\mbox{d}}{\mbox{d}t}\mathbf{p}=-m\omega^2\mathbf{x}$$ $$\frac{\mbox{d}}{\mbox{d}s}\mathbf{p}=-m\omega^2\gamma\mathbf{x}$$ This surely isn't equivalent to the equation of motion given by defining the force in the particle frame, which is given as, $$\frac{\mbox{d}}{\mbox{d}s}\mathbf{p}=-m\omega^2\left[\mathbf{x}+\left(\sqrt{1+\frac{{\mathbf{p}}^2}{m^2c^2}}-1\right)\left(\mathbf{x}\cdot\mathbf{p}\right)\mathbf{p}\right]$$ So why exists this difference? And what is the correct way of defining the relativistic force?

Answer 1

There are many possible relativistic versions for the harmonic oscillator. Following your approach, I will only consider the case of a particle in a background field. Physically, this corresponds to the motion of a charged particle in an appropriate electromagnetic field.

The interaction Lagrangian of a particle in a background EM field is: $$ L_I=q(\vec v\cdot \vec A-V) $$ with $A$ the potential vector and $V$ the electric potential.

To obtain the harmonic oscillator, you consider the EM field in a certain inertial frame: $$ V=\frac{1}{2}\vec x^2\\ \vec A=0 $$

Using the kinetic term of the Lagrangian (setting $m=1$): $$ L_0=\sqrt{1-\vec v^2} $$ I get the total Lagrangian (setting $q=\omega=1$): $$ L= \sqrt{1-\vec v^2}-\frac{1}{2}\vec x^2 $$ This gives you the Euler-Lagrange equation: $$ \frac{d\vec p}{dt}+\vec x=0\\ \vec p = \gamma \vec v $$ And in the non relativistic limit, $\gamma=1$ so you recover the harmonic oscillator.

When you change reference frame, $(V,\vec A)$ changes like a 4-vector which is why the force changes. In particular, the velocity dependent force can be interpreted as a magnetic Lorentz force. Indeed, even if in the original frame there is no magnetic field, this is not the case when you change frames.

From this perspective, the relativistic force is not necessarily the most convenient quantity to look at. This is because it is easier to stay in the original inertial frame rather than staying in the non inertial particle’s frame.

Btw, another approach could be to take inspiration from general relativity. The harmonic potential could come from the spatial variations of $g_{tt}$, the time component of the metric tensor. This would lead to a different Lagrangian.

Hope this helps.

Answer to comment

Geometrically, it makes sense to talk about proper acceleration. However, the external field has its most simple form when staying in the lab frame. This is why the latter approach is more convenient. Once again, the different expressions of the forces are consistent and can be interpreted physically as an electric force and a magnetic force.

Another issue is that when you are thinking in terms of a harmonic oscillator, you need to define a distance with respect to the spatial equilibrium point. Intuitively, the lab frame would correspond to the proper frame of this equilibrium point, which is natural to consider. Due to the relativity of simultaneity, switching frames would change the plane of simultaneity which would lead to corrections to the distance and the force.