A particle moves rectilinearly under a rest mass preserving force in some inertial frame. Show that the product of its rest mass and its instantaneous proper acceleration equals the magnitude of the relativistic three-force acting on the particle in that frame. [Hint: (14.1) and (35.14).] Show also that this is not necessarily true when the motion is not rectilinear.

We need to use equations (14.1) and (35.14). We have: (14.1): \(\textbf{F} = \frac{dp}{d\tau}\), where \(\textbf{F}\) is the relativistic four-force, \(p\) is momentum, and \(\tau\) is the proper time. (35.14): \(\pm |F_{\sigma}| =\frac{dP_{\sigma}}{d\tau} = M\frac{ma^{\sigma}}{(1-v^2/c^2)^{3/2}}\), where \(F_{\sigma}\) is the relativistic three-force, \(P_{\sigma}\) is the component of relativistic momentum, \(M\) is the rest mass, \(a^{\sigma}\) is the proper acceleration, \(v\) is the velocity, and \(c\) is the speed of light. From these equations, we are asked to show that \(M \times a^{σ} = |F_{\sigma}|\) under rectilinear motion.

We need to find the relationship between proper acceleration and relativistic three-force for rectilinear motion. Under rectilinear motion, we can say that \(v=v^{\sigma}\) (velocity is aligned with the \(\sigma\) direction) and \(a^\sigma = a = \frac{d^2x}{d\tau^2}\). Therefore, the equation (35.14) can be written as: \(\pm |F_{\sigma}| = M\frac{a}{(1-v^2/c^2)^{3/2}}\) Now, we need to show that \(M \times a = |F_{\sigma}|\).

Multiply both sides of the equation by \((1-v^2/c^2)^{3/2}\): \(M \times a \times (1-v^2/c^2)^{3/2} = \pm |F_{\sigma}| \times (1-v^2/c^2)^{3/2}\) With \(\beta = v/c\): \(M \times a \times (1-\beta^2)^{3/2} = \pm |F_{\sigma}| \times (1-\beta^2)^{3/2}\) Notice that both sides have the same factor of \((1-\beta^2)^{3/2}\), so we can cancel them out: \(M \times a = \pm |F_{\sigma}|\)

For non-rectilinear motion, consider the simplest case of the particle moving in a circle. In this case, the proper acceleration is directed towards the center of the circle and is given by \(a_r = \frac{mv^2}{r}\), where \(m\) is the mass of the particle, \(v\) is the instantaneous velocity, and \(r\) is the radius of the circle. The force acting on this particle has a centrifugal component, and the relativistic three-force has a tangential component. The proper acceleration and relativistic three-force are therefore not aligned in the non-rectilinear case. Since \(M \times a_r\) and \(F_{\sigma}\) have different directions, they cannot be equal. In conclusion, we have shown that under rectilinear motion, the product of the rest mass and the instantaneous proper acceleration is equal to the magnitude of the relativistic three-force acting on the particle. However, when the motion is non-rectilinear, this relationship does not necessarily hold true.