Acceleration without a force in special relativity

Question

Let us consider a relativistic particle of mass $m$ and charge $q$ in a constant electric field $\mathbf{E}=E\mathbf{\hat{j}}$ moving in two spatial dimensions, a relativistic extension of the well-known problem of projectile motion. For the sake of making computations easier, I will use the Hamiltonian to find the equations of motion. It is $H=\sqrt{m^2c^4+(p_x^2+p_y^2)c^2}-qEy$, and Hamilton's equations¹ give the following equations of motion for the positions and momenta: $$\dot{p}_x=0, \dot{p}_y=qE$$ $$\dot{x}=\frac{p_xc^2}{\sqrt{m^2 c^4+(p_x^2+p_y^2)c^2}}, \dot{y}=\frac{p_yc^2}{\sqrt{m^2 c^4+(p_x^2+p_y^2)c^2}},$$ where a dot over a variable represents a derivative WRT the coordinate time $t$ and while in non-relativistic mechanics the trajectory is a parabola, in relativistic mechanics it is a hyperbola. The formulas for the components of the momentum as a function of coordinate time in the $x$ and $y$ directions are trivially found to be $p_x(t)=p_{x,0}$ and $p_y(t)=p_{y,0}+qEt$. We may substitute these results in the equations for $x$ and $y$ to find that in the $x$ direction, and factorizing to eliminate $c$'s we have $$\dot{x}=\frac{p_{x,0}c}{\sqrt{m^2 c^2+p_{x,0}^2+(p_{y,0}+qEt)^2}},$$ however it is easily seen that the $x$ component of the velocity is not constant, it is decreasing. There is a nonzero acceleration even though there is no force acting on it. How is that possible? What is the physical reason for this relativistic phenomenon?

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¹:in their standard form $\dot{p}_i=-\frac{\partial H}{\partial x_i}$ and $\dot{x}_i=\frac{\partial H}{\partial p_i}$

Answer 1

The relativistic equivalent of Newton's second law is$$f^\mu=\frac{d}{d\tau}p^\mu=\gamma\frac{d}{dt}\binom{\gamma mc}{\gamma\beta^i mc}.$$Setting $f^i=0$ only requires $\gamma\beta^i=\frac{\beta^i}{\sqrt{1-\beta^j\beta^j}}$ to be conserved, not $\beta^i$. In this case$$\beta^i=\frac{p^ic^2}{\sqrt{m^2c^4+p^2c^2}}\implies\frac{\beta^i}{\sqrt{1-\beta^j\beta^j}}=\frac{p^i}{m}.$$Since $p^x$ is conserved, so is $\gamma\beta^x$, as you expected.

Answer 2

As stated in the other answer, force is the change in momentum, not mass times acceleration. The physical reason for the phenomenon is that momentum is not simply $mv$ in relativity. This leads to the situation in relativity where the acceleration is not generally parallel to the force.

For there to be acceleration there still needs to be a force, but the force and the acceleration are not necessarily co-linear. So breaking a force into components to get components of acceleration doesn’t work.

Using four-vectors is recommended for this reason.